Пояснення вказівників на С - вони не такі складні, як ви думаєте

Вказівники - це, мабуть, найскладніша особливість C для розуміння. Але це одна з особливостей, яка робить C чудовою мовою.

У цій статті ми перейдемо від самих основ покажчиків до їх використання з масивами, функціями та структурою.

Тож розслабтеся, візьміть кави і готуйтеся дізнатися все про вказівки.

Теми

А. Основи
  1. Що саме являють собою покажчики?
  2. Визначення та позначення
  3. Деякі спеціальні вказівки
  4. Арифметика покажчика
B. Масиви та рядки
  1. Чому вказівники та масиви?
  2. 1-D масиви
  3. 2-D масиви
  4. Струни
  5. Масив покажчиків
  6. Вказівник на масив
C. Функції
  1. Дзвінок за значенням v / s Дзвінок за посиланням
  2. Покажчики як аргументи функції
  3. Покажчики як повернення функції
  4. Покажчик на функцію
  5. Масив покажчиків на функції
  6. Покажчик на функцію як аргумент
D. Структура
  1. Вказівник на структуру
  2. Масив структури
  3. Вказівник на структуру як аргумент
E. Вказівник на вказівник
F. Висновок

А. Визначення, позначення, типи та арифметика

1. Що саме являють собою покажчики?

Перш ніж ми дійдемо до визначення покажчиків, давайте зрозуміємо, що відбувається, коли ми пишемо такий код:

int digit = 42; 

Блок пам'яті зарезервований компілятором для зберігання intзначення. Ім'я цього блоку - digitі значення, яке зберігається в цьому блоці, - 42.

Тепер, щоб запам'ятати блок, йому присвоюється адреса або номер місцезнаходження (скажімо, 24650).

Значення номера місцезнаходження для нас не важливо, оскільки це випадкове значення. Але ми можемо отримати доступ до цієї адреси, використовуючи &(амперсанд) або адресу оператора.

printf("The address of digit = %d.",&digit); /* prints "The address of digit = 24650. */ 

Ми можемо отримати значення змінної digitз її адреси, використовуючи інший оператор *(зірочку), який називається непрямим або невідповідним або значенням за адресовим оператором.

printf("The value of digit = %d.", *(&digit); /* prints "The value of digit = 42. */ 

2. Визначення та позначення

Адреса змінної може зберігатися в іншій змінній, відомій як змінна покажчика. Синтаксис для збереження адреси змінної до покажчика:

dataType *pointerVariableName = &variableName; 

Для нашої digitзмінної це можна записати так:

int *addressOfDigit = &digit; 

або ось так:

int *addressOfDigit; addressOfDigit= &digit; 

Декларація та визначення

Це можна прочитати як - Вказівник на int(ціле число) addressOfDigitзберігає address of(&)digitзмінну.

Кілька моментів, щоб зрозуміти:

dataType- Нам потрібно сказати комп’ютеру, який тип даних є змінною, адресу якої ми збираємось зберігати. Тут intбув тип даних digit.

Це не означає, що addressOfDigitбуде зберігати значення типу int. Ціле число покажчик (наприклад addressOfDigit) може тільки зберігати адреси змінних цілочисельного типу.

int variable1; int variable2; char variable3; int *addressOfVariables; 

*- Змінна покажчика - це спеціальна змінна в тому сенсі, що вона використовується для зберігання адреси іншої змінної. Щоб відрізнити його від інших змінних, які не зберігають адресу, ми використовуємо *як символ у декларації.

Тут ми можемо призначити адресу variable1і variable2цілочисельному покажчику, addressOfVariablesале не, variable3оскільки він має тип char. Нам знадобиться змінна вказівника символу, щоб зберегти її адресу.

Ми можемо використовувати нашу addressOfDigitзмінну покажчика для друку адреси та значення, digitяк показано нижче:

printf("The address of digit = %d.", addressOfDigit); /* prints "The address of digit = 24650." */ printf("The value of digit = %d.", *addressOfDigit); /*prints "The value of digit = 42. */ 

Тут *addressOfDigitможна прочитати як значення за адресою, що зберігається в addressOfDigit.

Зверніть увагу, що ми використовувались %dяк ідентифікатор формату для addressOfDigit. Ну, це не зовсім правильно. Правильним буде ідентифікатор %p.

Використовуючи %p, адреса відображається у шістнадцятковому значенні. Але адреса пам'яті може відображатися як у цілих числах, так і в восьмеричних значеннях. Проте, оскільки це не зовсім правильний спосіб, відображається попередження.

int num = 5; int *p = # printf("Address using %%p = %p",p); printf("Address using %%d = %d",p); printf("Address using %%o = %o",p); 

Вихідні дані відповідно до компілятора, який я використовую, такі:

Address using %p = 000000000061FE00 Address using %d = 6422016 Address using %o = 30377000 
Це попередження відображається, коли ви використовуєте% d - "попередження: формат '% d' очікує аргумент типу 'int', але аргумент 2 має тип 'int *'".

3. Деякі спеціальні вказівки

Дикий вказівник

char *alphabetAddress; /* uninitialised or wild pointer */ char alphabet = "a"; alphabetAddress = &alphabet; /* now, not a wild pointer */ 

When we defined our character pointer alphabetAddress, we did not initialize it.

Such pointers are known as wild pointers. They store a garbage value (that is, memory address) of a byte that we don't know is reserved or not (remember int digit = 42;, we reserved a memory address when we declared it).

Suppose we dereference a wild pointer and assign a value to the memory address it is pointing at. This will lead to unexpected behaviour since we will write data at a  memory block that may be free or reserved.

Null Pointer

To make sure that we do not have a wild pointer, we can initialize a pointer with a NULL value, making it a null pointer.

char *alphabetAddress = NULL /* Null pointer */ 

A null pointer points at nothing, or at a memory address that users can not access.

Void Pointer

A void pointer can be used to point at a variable of any data type. It can be reused to point at any data type we want to. It is declared like this:

void *pointerVariableName = NULL; 

Since they are very general in nature, they are also known as generic pointers.

With their flexibility, void pointers also bring some constraints. Void pointers cannot be dereferenced as any other pointer. Appropriate typecasting is necessary.

void *pointer = NULL; int number = 54; char alphabet = "z"; pointer = &number; printf("The value of number = ", *pointer); /* Compilation Error */ /* Correct Method */ printf("The value of number = ", *(int *)pointer); /* prints "The value at number = 54" */ pointer = &alphabet; printf("The value of alphabet = ", *pointer); /* Compilation Error */ printf("The value of alphabet = ", *(char *)pointer); /* prints "The value at alphabet = z */ 

Similarly, void pointers need to be typecasted for performing arithmetic operations.

Void pointers are of great use in C. Library functions malloc() and calloc() which dynamically allocate memory return void pointers. qsort(), an inbuilt sorting function in C, has a function as its argument which itself takes void pointers as its argument.

Dangling Pointer

A dangling pointer points to a memory address which used to hold a variable. Since the address it points at is no longer reserved, using it will lead to unexpected results.

main(){ int *ptr; ptr = (int *)malloc(sizeof(int)); *ptr = 1; printf("%d",*ptr); /* prints 1 */ free(ptr); /* deallocation */ *ptr = 5; printf("%d",*ptr); /* may or may not print 5 */ } 

Though the memory has been deallocated by free(ptr), the pointer to integer ptr still points to that unreserved memory address.

4. Pointer Arithmetic

We know by now that pointers are not like any other variable. They do not store any value but the address of memory blocks.

So it should be quite clear that not all arithmetic operations would be valid with them. Would multiplying or dividing two pointers (having addresses) make sense?

Pointers have few but immensely useful valid operations:

  1. You can assign the value of one pointer to another only if they are of the same type (unless they're typecasted or one of them is void *).
int ManU = 1; int *addressOfManU = &ManU; int *anotherAddressOfManU = NULL; anotherAddressOfManU = addressOfManU; /* Valid */ double *wrongAddressOfManU = addressOfManU; /* Invalid */ 

2.   You can only add or subtract integers to pointers.

int myArray = {3,6,9,12,15}; int *pointerToMyArray = &myArray[0]; pointerToMyArray += 3; /* Valid */ pointerToMyArray *= 3; /* Invalid */ 

When you add (or subtract) an integer (say n) to a pointer, you are not actually adding (or subtracting) n bytes to the pointer value. You are actually adding (or subtracting) n-times the size of the data type of the variable being pointed bytes.

int number = 5; /* Suppose the address of number is 100 */ int *ptr = &number; int newAddress = ptr + 3; /* Same as ptr + 3 * sizeof(int) */ 

The value stored in newAddress will not be 103, rather 112.

3.  Subtraction and comparison of pointers is valid only if both are members of the same array. The subtraction of pointers gives the number of elements separating them.

int myArray = {3,6,9,12,15}; int sixthMultiple = 18; int *pointer1 = &myArray[0]; int *pointer2 = &myArray[1]; int *pointer6 = &sixthMuliple; /* Valid Expressions */ if(pointer1 == pointer2) pointer2 - pointer1; /* Invalid Expressions if(pointer1 == pointer6) pointer2 - pointer6 

4.  You can assign or compare a pointer with NULL.

The only exception to the above rules is that the address of the first memory block after the last element of an array follows pointer arithmetic.

Pointer and arrays exist together. These valid manipulations of pointers are immensely useful with arrays, which will be discussed in the next section.

B. Arrays and Strings

1. Why pointers and arrays?

In C, pointers and arrays have quite a strong relationship.

The reason they should be discussed together is because what you can achieve with array notation (arrayName[index]) can also be achieved with pointers, but generally faster.

2. 1-D Arrays

Let us look at what happens when we write int myArray[5];.

Five consecutive blocks of memory starting from myArray[0] to myArray[4] are created with garbage values in them. Each of the blocks is of size 4 bytes.

Thus, if the address of myArray[0] is 100 (say), the address of the rest of the blocks would be 104, 108, 112, and 116.

Have a look at the following code:

int prime[5] = {2,3,5,7,11}; printf("Result using &prime = %d\n",&prime); printf("Result using prime = %d\n",prime); printf("Result using &prime[0] = %d\n",&prime[0]); /* Output */ Result using &prime = 6422016 Result using prime = 6422016 Result using &prime[0] = 6422016 

So, &prime, prime, and &prime[0] all give the same address, right? Well, wait and read because you are in for a surprise (and maybe some confusion).

Let's try to increment each of &prime, prime, and &prime[0] by 1.

printf("Result using &prime = %d\n",&prime + 1); printf("Result using prime = %d\n",prime + 1); printf("Result using &prime[0] = %d\n",&prime[0] + 1); /* Output */ Result using &prime = 6422036 Result using prime = 6422020 Result using &prime[0] = 6422020 

Wait! How come &prime + 1 results in something different than the other two? And why are prime + 1 and &prime[0] + 1 still equal? Let's answer these questions.

prime and &prime[0] both point to the 0th element of the array prime. Thus, the name of an array is itself a pointer to the 0th element of the array.

Here, both point to the first element of size 4 bytes. When you add 1 to them, they now point to the 1st element in the array. Therefore this results in an increase in the address by 4.

&prime, on the other hand, is a pointer to an int array of size 5. It stores the base address of the array prime[5], which is equal to the address of the first element. However, an increase by 1 to it results in an address with an increase of 5 x 4 = 20 bytes.

In short, arrayName and &arrayName[0] point to the 0th element whereas &arrayName points to the whole array.

1-D масив

We can access the array elements using subscripted variables like this:

int prime[5] = {2,3,5,7,11}; for( int i = 0; i < 5; i++) { printf("index = %d, address = %d, value = %d\n", i, &prime[i], prime[i]); } 

We can do the same using pointers which are always faster than using subscripts.

int prime[5] = {2,3,5,7,11}; for( int i = 0; i < 5; i++) { printf("index = %d, address = %d, value = %d\n", i, prime + i, *(prime + i)); } 

Both methods give the output:

index = 0, address = 6422016, value = 2 index = 1, address = 6422020, value = 3 index = 2, address = 6422024, value = 5 index = 3, address = 6422028, value = 7 index = 4, address = 6422032, value = 11 

Thus, &arrayName[i] and arrayName[i] are the same as arrayName + i and  *(arrayName + i), respectively.

3. 2-D Arrays

Two-dimensional arrays are an array of arrays.

int marks[5][3] = { { 98, 76, 89}, { 81, 96, 79}, { 88, 86, 89}, { 97, 94, 99}, { 92, 81, 59} }; 

Here, marks can be thought of as an array of 5 elements, each of which is a one-dimensional array containing 3 integers. Let us work through a series of programs to understand different subscripted expressions.

printf("Address of whole 2-D array = %d\n", &marks); printf("Addition of 1 results in %d\n", &marks +1); /* Output */ Address of whole 2-D array = 6421984 Addition of 1 results in 6422044 

Like 1-D arrays, &marks points to the whole 2-D array, marks[5][3]. Thus, incrementing to it by 1 ( = 5 arrays X 3 integers each X 4 bytes = 60) results in an increment by 60 bytes.

printf("Address of 0th array = %d\n", marks); printf("Addition of 1 results in %d\n", marks +1); printf("Address of 0th array =%d\n", &marks[0]); printf("Addition of 1 results in %d\n", &marks[0] + 1); /* Output */ Address of 0th array = 6421984 Addition of 1 results in 6421996 Address of 0th array = 6421984 Addition of 1 results in 6421996 

If marks was a 1-D array, marks and &marks[0] would have pointed to the 0th element. For a 2-D array, elements are now 1-D arrays. Hence, marks and &marks[0] point to the 0th array (element), and the addition of 1 point to the 1st array.

printf("Address of 0th element of 0th array = %d\n", marks[0]); printf("Addition of 1 results in %d\n", marks[0] + 1); printf("Address of 0th element of 1st array = %d\n", marks[1]); printf("Addition of 1 results in %d\n", marks[1] + 1); /* Output */ Address of 0th element of 0th array = 6421984 Addition of 1 results in 6421988 Address of 0th element of 1st array = 6421996 Addition of 1 results in 6422000 

And now comes the difference. For a 1-D array, marks[0] would give the value of the 0th element. An increment by 1 would increase the value by 1.

But, in a 2-D array, marks[0] points to the 0th element of the 0th array. Similarly, marks[1] points to the 0th element of the 1st array. An increment by 1 would point to the 1st element in the 1st array.

printf("Value of 0th element of 0th array = %d\n", marks[0][0]); printf("Addition of 1 results in %d", marks[0][0] + 1); /* Output */ Value of 0th element of 0th array = 98 Addition of 1 results in 99 

This is the new part. marks[i][j] gives the value of the jth element of the ith array. An increment to it changes the value stored at marks[i][j]. Now, let us try to write marks[i][j] in terms of pointers.

We know marks[i] + j would point to the ith element of the jth array from our previous discussion. Dereferencing it would mean the value at that address. Thus, marks[i][j] is the same as  *(marks[i] + j).

From our discussion on 1-D arrays, marks[i] is the same as *(marks + i). Thus, marks[i][j] can be written as *(*(marks + i) + j) in terms of pointers.

Here is a summary of notations comparing 1-D and 2-D arrays.

Expression1-D Array2-D Array
&arrayNamepoints to the address of whole array

adding 1 increases the address by 1 x sizeof(arrayName)

points to the address of whole array

adding 1 increases the address by 1 x sizeof(arrayName)

arrayNamepoints to the 0th element

adding 1 increases the address to 1st element

points to the 0th element (array)

adding 1 increases the address to 1st element (array)

&arrayName[i]points to the the ith element

adding 1 increases the address to (i+1)th element

points to the ith element (array)

adding 1 increases the address to the (i+1)th element (array)

arrayName[i]gives the value of the ith element

adding 1 increases the value of the ith element

points to the 0th element of the ith array

adding 1 increases the address to 1st element of the ith array

arrayName[i][j]Nothinggives the value of the jth element of the ith array

adding 1 increases the value of the jth element of the ith array

Pointer Expression To Access The Elements*( arrayName + i)*( *( arrayName + i) + j)

4. Strings

A string is a one-dimensional array of characters terminated by a null(\0). When we write char name[] = "Srijan";, each character occupies one byte of memory with the last one always being \0.

Similar to the arrays we have seen, name and &name[0] points to the 0th character in the string, while &name points to the whole string. Also, name[i] can be written as *(name + i).

/* String */ char champions[] = "Liverpool"; printf("Pointer to whole string = %d\n", &champions); printf("Addition of 1 results in %d\n", &champions + 1); /* Output */ Address of whole string = 6421974 Addition of 1 results in 6421984 printf("Pointer to 0th character = %d\n", &champions[0]); printf("Addition of 1 results in %d\n", &champions[0] + 1); /* Output */ Address of 0th character = 6421974 Addition of 1 results in a pointer to 1st character 6421975 printf("Pointer to 0th character = %d\n", champions); printf("Addition of 1 results in a pointer to 1st character %d\n", champions + 1); /* Output */ Address of 0th character = 6421974 Addition of 1 results in 6421975 printf("Value of 4th character = %c\n", champions[4]); printf("Value of 4th character using pointers = %c\n", *(champions + 4)); /* Output */ Value of 4th character = r Value of 4th character using pointers = r 

A two-dimensional array of characters or an array of strings can also be accessed and manipulated as discussed before.

/* Array of Strings */ char top[6][15] = { "Liverpool", "Man City", "Man United", "Chelsea", "Leicester", "Tottenham" }; printf("Pointer to 2-D array = %d\n", &top); printf("Addition of 1 results in %d\n", &top + 1); /* Output */ Pointer to 2-D array = 6421952 Addition of 1 results in 6422042 printf("Pointer to 0th string = %d\n", &top[0]); printf("Addition of 1 results in %d\n", &top[0] + 1); /* Output */ Pointer to 0th string = 6421952 Addition of 1 results in 6421967 printf("Pointer to 0th string = %d\n", top); printf("Addition of 1 results in %d\n", top + 1); /* Output */ Pointer to 0th string = 6421952 Addition of 1 results in 6421967 printf("Pointer to 0th element of 4th string = %d\n", top[4]); printf("Pointer to 1st element of 4th string = %c\n", top[4] + 1); /* Output */ Pointer to 0th element of 4th string = 6422012 Pointer to 1st element of 4th string = 6422013 printf("Value of 1st character in 3rd string = %c\n", top[3][1]); printf("Same using pointers = %c\n", *(*(top + 3) + 1)); /* Output */ Value of 1st character in 3rd string = h Same using pointers = h 

5. Array of Pointers

Like an array of ints and an array of chars, there is an array of pointers as well. Such an array would simply be a collection of addresses. Those addresses could point to individual variables or another array as well.

The syntax for declaring a pointer array is the following:

dataType *variableName[size]; /* Examples */ int *example1[5]; char *example2[8]; 

Following the operators precedence, the first example can be read as -  example1 is an array([]) of 5 pointers to int. Similarly, example2 is an array of 8 pointers to char.

We can store the two-dimensional array to string top using a pointer array and save memory as well.

char *top[] = { "Liverpool", "Man City", "Man United", "Chelsea", "Leicester", "Tottenham" }; 

top will contain the base addresses of all the respective names. The base address of "Liverpool" will be stored in top[0], "Man City" in top[1], and so on.

In the earlier declaration, we required 90 bytes to store the names. Here, we only require ( 58 (sum of bytes of names) + 12 ( bytes required to store the address in the array) ) 70 bytes.

The manipulation of strings or integers becomes a lot easier when using an array of pointers.

If we try to put "Leicester" ahead of "Chelsea", we just need to switch the values of top[3] and top[4] like below:

char *temporary; temporary = top[3]; top[3] = top[4]; top[4] = temporary; 

Without pointers, we would have to exchange every character of the strings, which would have taken more time. That's why strings are generally declared using pointers.

6. Pointer to Array

Like "pointer to int" or "pointer to char", we have pointer to array as well. This pointer points to whole array rather than its elements.

Remember we discussed how &arrayName points to the whole array? Well, it is a pointer to array.

A pointer to array can be declared like this:

dataType (*variableName)[size]; /* Examples */ int (*ptr1)[5]; char (*ptr2)[15]; 

Notice the parentheses. Without them, these would be an array of pointers. The first example can be read as - ptr1 is a pointer to an array of 5 int(integers).

int goals[] = { 85,102,66,69,67}; int (*pointerToGoals)[5] = &goals; printf("Address stored in pointerToGoals %d\n", pointerToGoals); printf("Dereferncing it, we get %d\n",*pointerToGoals); /* Output */ Address stored in pointerToGoals 6422016 Dereferencing it, we get 6422016 

When we dereference a pointer, it gives the value at that address. Similarly, by dereferencing a pointer to array, we get the array and the name of the array points to the base address. We can confirm that *pointerToGoals gives the array goals if we find its size.

printf("Size of goals[5] = %d, *pointerToGoals); /* Output */ Size of goals[5] = 20 

If we dereference it again, we will get the value stored in that address. We can print all the elements using pointerToGoals.

for(int i = 0; i < 5; i++) printf("%d ", *(*pointerToGoals + i)); /* Output */ 85 102 66 69 67 

Pointers and pointer to arrays are quite useful when paired up with functions. Coming up in the next section!

C. Functions

1. Call by Value vs Call by Reference

Have a look at the program below:

#include  int multiply(int x, int y){ int z; z = x * y; return z; } main(){ int x = 3, y = 5; int product = multiply(x,y); printf("Product = %d\n", product); /* prints "Product = 15" */ } 

The function multiply() takes two int arguments and returns their product as int.

In the function call multiply(x,y), we passed the value of x and y ( of main()), which are actual arguments, to multiply().

The values of the actual arguments are passed or copied to the formal argumentsx and y ( of multiply()). The x and y of multiply() are different from those of main(). This can be verified by printing their addresses.

#include  int multiply(int x, int y){ printf("Address of x in multiply() = %d\n", &x); printf("Address of y in multiply() = %d\n", &y); int z; z = x * y; return z; } main(){ int x = 3, y = 5; printf("Address of x in main() = %d\n", &x); printf("Address of y in main() = %d\n", &y); int product = multiply(x,y); printf("Product = %d\n", product); } /* Output */ Address of x in main() = 6422040 Address of y in main() = 6422036 Address of x in multiply() = 6422000 Address of y in multiply() = 6422008 Product = 15 

Since we created stored values in a new location, it costs us memory. Wouldn't it be better if we could perform the same task without wasting space?

Call by reference helps us achieve this. We pass the address or reference of the variables to the function which does not create a copy. Using the dereferencing operator *, we can access the value stored at those addresses.

We can rewrite the above program using call by reference as well.

#include  int multiply(int *x, int *y){ int z; z = (*x) * (*y); return z; } main(){ int x = 3, y = 5; int product = multiply(&x,&y); printf("Product = %d\n", product); /* prints "Product = 15" */ } 

2. Pointers as Function Arguments

In this section, we will look at various programs where we give int, char, arrays and strings as arguments using pointers.

#include  void add(float *a, float *b){ float c = *a + *b; printf("Addition gives %.2f\n",c); } void subtract(float *a, float *b){ float c = *a - *b; printf("Subtraction gives %.2f\n",c); } void multiply(float *a, float *b){ float c = *a * *b; printf("Multiplication gives %.2f\n",c); } void divide(float *a, float *b){ float c = *a / *b; printf("Division gives %.2f\n",c); } main(){ printf("Enter two numbers :\n"); float a,b; scanf("%f %f",&a,&b); printf("What do you want to do with the numbers?\nAdd : a\nSubtract : s\nMultiply : m\nDivide : d\n"); char operation = '0'; scanf(" %c",&operation); printf("\nOperating...\n\n"); switch (operation) { case 'a': add(&a,&b); break; case 's': subtract(&a,&b); break; case 'm': multiply(&a,&b); break; case 'd': divide(&a,&b); break; default: printf("Invalid input!!!\n"); } } 

We created four functions, add(), subtract(), multiply() and divide() to perform arithmetic operations on the two numbers a and b.

The address of a and b was passed to the functions. Inside the function using * we accessed the values and printed the result.

Similarly, we can give arrays as arguments using a pointer to its first element.

#include  void greatestOfAll( int *p){ int max = *p; for(int i=0; i  max) max = *(p+i); } printf("The largest element is %d\n",max); } main(){ int myNumbers[5] = { 34, 65, -456, 0, 3455}; greatestOfAll(myNumbers); /* Prints :The largest element is 3455" */ } 

Since the name of an array itself is a pointer to the first element, we send that as an argument to the function greatestOfAll(). In the function, we traverse through the array using loop and pointer.

#include  #include  void wish(char *p){ printf("Have a nice day, %s",p); } main(){ printf("Enter your name : \n"); char name[20]; gets(name); wish(name); } 

Here, we pass in the string name to wish() using a pointer and print the message.

3. Pointers as Function Return

#include  int* multiply(int *a, int *b){ int c = *a * *b; return &c; } main(){ int a= 3, b = 5; int *c = multiply (&a,&b); printf("Product = %d",*c); } 

The function multiply() takes two pointers to int. It returns a pointer to int as well which stores the address where the product is stored.

It is very easy to think that the output would be 15. But it is not!

When multiply() is called, the execution of main() pauses and memory is now allocated for the execution of multiply(). After its execution is completed, the memory allocated to multiply() is deallocated.

Therefore, though c ( local to main()) stores the address of the product, the data there is not guaranteed since that memory has been deallocated.

So does that mean pointers cannot be returned by a function? No!

We can do two things. Either store the address in the heap or global section or declare the variable to be static so that their values persist.

Static variables can simply be created by using the keywordstatic before data type while declaring the variable.

To store addresses in heap, we can use library functions malloc() and calloc() which allocate memory dynamically.

The following programs will explain both the methods. Both methods return the output as 15.

#include  #include  /* Using malloc() */ int* multiply(int *a, int *b){ int *c = malloc(sizeof(int)); *c = *a * *b; return c; } main(){ int a= 3, b = 5; int *c = multiply (&a,&b); printf("Product = %d",*c); } /* Using static keyword */ #include  int* multiply(int *a, int *b){ static int c; c = *a * *b; return &c; } main(){ int a= 3, b = 5; int *c = multiply (&a,&b); printf("Product = %d",*c); } 

4. Pointer to Function

Like pointer to different data types, we also have a pointer to function as well.

A pointer to function or function pointer stores the address of the function. Though it doesn't point to any data. It points to the first instruction in the function.

The syntax for declaring a pointer to function is:

 /* Declaring a function */ returnType functionName(parameterType1, pparameterType2, ...); /* Declaring a pointer to function */ returnType (*pointerName)(parameterType1, parameterType2, ...); pointerName = &functionName; /* or pointerName = functionName; */ 

The below example will make it clearer.

int* multiply(int *a, int *b) { int *c = malloc(sizeof(int)); *c = *a * *b; return c; } main() { int a=3,b=5; int* (*p)(int*, int*) = &multiply; /* or int* (*p)(int*, int*) = multiply; */ int *c = (*p)(&a,&b); /* or int *c = p(&a,&b); */ printf("Product = %d",*c); } 

The declaration for the pointer p to function multiply() can be read as ( following operator precedence) - p is a pointer to function with two integer pointers ( or two pointers to int) as parameters and returning a pointer to int.

Since the name of the function is also a pointer to the function, the use of & is not necessary. Also removing * from the function call doesn't affect the program.

5. Array of Pointers to Functions

We have already seen how to create an array of pointers to int, char, and so on. Similarly, we can create an array of pointers to function.

In this array, every element will store an address of a function, where all the functions are of the same type. That is, they have the same type and number of parameters and return types.

We will modify a program discussed earlier in this section. We will store the addresses of add(), subtract(), multiply() and divide() in an array make a function call through subscript.

#include  void add(float *a, float *b){ float c = *a + *b; printf("Addition gives %.2f\n",c); } void subtract(float *a, float *b){ float c = *a - *b; printf("Subtraction gives %.2f\n",c); } void multiply(float *a, float *b){ float c = *a * *b; printf("Multiplication gives %.2f\n",c); } void divide(float *a, float *b){ float c = *a / *b; printf("Division gives %.2f\n",c); } main(){ printf("Enter two numbers :\n"); float a,b; scanf("%f %f",&a,&b); printf("What do you want to do with the numbers?\nAdd : a\nSubtract : s\nMultiply : m\nDivide : d\n"); char operation = '0'; scanf(" %c",&operation); void (*p[])(float* , float*) = {add,subtract,multiply,divide}; printf("\nOperating...\n\n"); switch (operation) { case 'a': p[0](&a,&b); break; case 's': p[1](&a,&b); break; case 'm': p[2](&a,&b); break; case 'd': p[3](&a,&b); break; default: printf("Invalid input!!!\n"); } } 

The declaration here can be read as - p is an array of pointer to functions with two float pointers as parameters and returning void.

6. Pointer to Function as an Argument

Like any other pointer, function pointers can also be passed to another function, therefore known as a callback function or called function. The function to which it is passed is known as a calling function.

A better way to understand would be to look at qsort(), which is an inbuilt function in C. It is used to sort an array of integers, strings, structures, and so on. The declaration for qsort() is:

void qsort(void *base, size_t nitems, size_t size, int (*compar)(const void *, const void *)); 

qsort() takes four arguments:

  1. a void pointer to the start of an array
  2. number of elements
  3. size of each element
  4. a function pointer that takes in two void pointers as arguments and returns an int

The function pointer points to a comparison function that returns an integer that is greater than, equal to, or less than zero if the first argument is respectively greater than, equal to, or less than the second argument.

The following program showcases its usage:

#include  #include  int compareIntegers(const void *a, const void *b) { const int *x = a; const int *y = b; return *x - *y; } main(){ int myArray[] = {97,59,2,83,19,97}; int numberOfElements = sizeof(myArray) / sizeof(int); printf("Before sorting - \n"); for(int i = 0; i < numberOfElements; i++) printf("%d ", *(myArray + i)); qsort(myArray, numberOfElements, sizeof(int), compareIntegers); printf("\n\nAfter sorting - \n"); for(int i = 0; i < numberOfElements; i++) printf("%d ", *(myArray + i)); } /* Output */ Before sorting - 97 59 2 83 19 97 After sorting - 2 19 59 83 97 97 

Since a function name is itself a pointer, we can write compareIntegers as the fourth argument.

D. Structure

1. Pointer to Structure

Like integer pointers, array pointers, and function pointers, we have pointer to structures or structure pointers as well.

struct records { char name[20]; int roll; int marks[5]; char gender; }; struct records student = {"Alex", 43, {76, 98, 68, 87, 93}, 'M'}; struct records *ptrStudent = &student; 

Here, we have declared a pointer ptrStudent of type struct records. We have assigned the address of student to ptrStudent.

ptrStudent stores the base address of student, which is the base address of the first member of the structure. Incrementing by 1 would increase the address by sizeof(student) bytes.

printf("Address of structure = %d\n", ptrStudent); printf("Adress of member `name` = %d\n", &student.name); printf("Increment by 1 results in %d\n", ptrStudent + 1); /* Output */ Address of structure = 6421984 Adress of member `name` = 6421984 Increment by 1 results in 6422032 

We can access the members of student using ptrStudent in two ways. Using our old friend * or using -> ( infix or arrow operator).

With *, we will continue to use the .( dot operator) whereas with -> we won't need the dot operator.

printf("Name w.o using ptrStudent : %s\n", student.name); printf("Name using ptrStudent and * : %s\n", ( *ptrStudent).name); printf("Name using ptrStudent and -> : %s\n", ptrStudent->name); /* Output */ Name without using ptrStudent: Alex Name using ptrStudent and *: Alex Name using ptrStudent and ->: Alex 

Similarly, we can access and modify other members as well. Note that the brackets are necessary while using * since the dot operator(.) has higher precedence over *.

2. Array Of Structure

We can create an array of type struct records and use a pointer to access the elements and their members.

struct records students[10]; /* Pointer to the first element ( structure) of the array */ struct records *ptrStudents1 = &students; /* Pointer to an array of 10 struct records */ struct records (*ptrStudents2)[10] = &students; 

Зверніть увагу, що ptrStudent1це вказівник на, student[0]тоді ptrStudent2як це вказівник на весь масив 10 struct records. Додавання 1 до ptrStudent1вказувало б на student[1].

Ми можемо використовувати ptrStudent1цикл для обходу елементів та їх членів.

 for( int i = 0; i name, ( ptrStudents1 + i)->roll); 

3. Вказівник на структуру як аргумент

Ми також можемо передати адресу змінної структури функції.

#include  struct records { char name[20]; int roll; int marks[5]; char gender; }; main(){ struct records students = {"Alex", 43, {76, 98, 68, 87, 93}, 'M'}; printRecords(&students); } void printRecords( struct records *ptr){ printf("Name: %s\n", ptr->name); printf("Roll: %d\n", ptr->roll); printf("Gender: %c\n", ptr->gender); for( int i = 0; i marks[i]); } /* Output */ Name: Alex Roll: 43 Gender: M Marks in 0th subject: 76 Marks in 1th subject: 98 Marks in 2th subject: 68 Marks in 3th subject: 87 Marks in 4th subject: 93 

Зверніть увагу, що структура struct recordsоголошена зовні main(). Це для того, щоб гарантувати, що він доступний у всьому світі та printRecords()може ним користуватися.

Якщо структуру визначено всередині main(), її сфера застосування буде обмежена main(). Також структура також повинна бути оголошена перед оголошенням функції.

Як і структури, ми можемо мати вказівники на об’єднання та мати доступ до членів за допомогою оператора стрілки ( ->).

E. Pointer to Pointer

So far we have looked at pointer to various primitive data types, arrays, strings, functions, structures, and unions.

The automatic question that comes to the mind is – what about pointer to pointer?

Well, good news for you! They too exist.

int var = 6; int *ptr_var = &var; printf("Address of var = %d\n", ptr_var); printf("Address of ptr_var = %d\n", &ptr_var); /* Output */ Address of var = 6422036 Address of ptr_var = 6422024 

To store the address of int variable var, we have the pointer to intptr_var. We would need another pointer to store the address of ptr_var.

Since ptr_var is of type int *, to store its address we would have to create a pointer to int *. The code below shows how this can be done.

int * *ptr_ptrvar = &ptr_var; /* or int* *ppvar or int **ppvar */ 

We can use ptr_ptrvar to access the address of ptr_var and use double dereferencing to access var.

printf("Address of ptr_var = %d\n", ptr_ptrvar); printf("Address of var = %d\n", *ptr_ptrvar); printf("Value at var = %d\n", *(*ptr_ptrvar)); /* Output */ Address of ptr_var = 6422024 Address of var = 6422036 Value at var = 6 

It is not required to use brackets when dereferencing ptr_ptrvar. But it is a good practice to use them. We can create another pointer ptr_ptrptrvar, which will store the address of ptr_ptrvar.

Since ptr_ptrvar is of type int**, the declaration for ptr_ptrptrvar will be

int** *ptr_ptrptrvar = &ptr_ptrvar; 

We can again access ptr_ptrvar, ptr_var and var using ptr_ptrptrvar.

printf("Address of ptr_ptrvar = %d\n", ptr_ptrptrvar); printf("Value at ptr_ptrvar = %d\n",*ptr_ptrptrvar); printf("Address of ptr_var = %d\n", *ptr_ptrptrvar); printf("Value at ptr_var = %d\n", *(*ptr_ptrptrvar)); printf("Address of var = %d\n", *(*ptr_ptrptrvar)); printf("Value at var = %d\n", *(*(*ptr_ptrptrvar))); /* Output */ Address of ptr_ptrvar = 6422016 Value at ptr_ptrvar = 6422024 Address of ptr_var = 6422024 Value at ptr_var = 6422036 Address of var = 6422036 Value at var = 6 

Вказівник на вказівник

If we change the value at any of the pointer(s) using ptr_ptrptrvar or ptr_ptrvar, the pointer(s) will stop pointing to the variable.

Conclusion

Phew! Yeah, we're finished. We started from pointers and ended with pointers (in a way). Don't they say that the curve of learning is a circle!

Try to recap all the sub-topics that you read. If you can recollect them, well done! Read the ones you can't remember again.

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