Functions are building blocks of C programming language. In this chapter we shall see different ways to pass pointers to a function.
We shall concentrate on 2 aspects.
1. Sending pointers to a function
2. Returning pointers from a function.
Passing variable by pointer
This type of passing is also called as pass by value.
The use of passing by pointer is to change the data hold by the variable.
Simple example is a swapping example:
Example 1:
In the below example, we have a swap function that accepts 2 int pointer. And we get the data by deferring during the swap operation.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void swap(int *a, int *b)
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}
int main()
{
int a = 10;
int b = 20;
printf("before swap a = %d, b = %d\n", a, b);
swap(&a, &b);
printf("after swap a = %d, b = %d\n", a, b);
return 0;
}
Output:
before swap a = 10, b = 20 after swap a = 20, b = 10
Passing a variable by value:
If you pass by value, then the copy of the arguments are stored in called function.
Hence even if we modify the values in the swap function, the same will not be reflected in the calling function.
Example:
In the same swap example, we are sending the variables in call by value.
Hence a copy of 2 variables will be there in swap function.
Hence any modification made on the arguments in the called function will not be reflected back to calling function.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void swap(int a, int b)
{
int temp;
temp = a;
a = b;
b = temp;
}
int main()
{
int a = 10;
int b = 20;
printf("before swap a = %d, b = %d\n", a, b);
swap(a, b);
printf("after swap a = %d, b = %d\n", a, b);
return 0;
}
Output:
before swap a = 10, b = 20 after swap a = 10, b = 20
How to pass strings to a function in C?
We can pass strings in the similar way that we pass normal integer variables by pointer as shown below:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void show(char *str)
{
printf("The string is = %s \n", str);
}
int main()
{
char *str = "Hello World";
char str_1[] = "www.ProDeveloperTutorial.com";
show(str); // note that we dont need to send "&str"
show(str_1);
return 0;
}
Output:
The string is = Hello World The string is = www.ProDeveloperTutorial.com
Returning a pointer from a function in C
To return a pointer, we need to declare the return type as the pointer to the appropriate data type.
In the below example, we allocate memory for an array and then return the address of that allocated memory.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int * allocate_memory(int len)
{
int *arr = (int*) malloc(len * sizeof(int));
for (int i = 0; i < len; ++i)
{
arr[i] = i;
}
return arr;
}
int main()
{
int *arr = allocate_memory(5);
for (int i = 0; i < 5; ++i)
{
printf("%d\n",arr[i]);
}
return 0;
}
Output:
0 1 2 3 4
Returning a pointer to a local data from a function in C
Generally programmers tend to do this easy mistake, if they dont know how stack works.
If suppose you allocate memory statically and then try to return that address. The compiler will not give any warning.
But the address of the array returned is no longer valid once the function returns because the function’s stack frame is popped off the stack.
Even-though you will be able to print the value, it may be overwritten if another function is called.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int * allocate_memory(int len)
{
int arr[len] ;
for (int i = 0; i < len; ++i)
{
arr[i] = i;
}
return arr;
}
int main()
{
int *arr = allocate_memory(5);
for (int i = 0; i < 5; ++i)
{
printf("%d\n",arr[i]);
}
return 0;
}
Output:
0 1 2 3 4
Passing pointer to a pointer (double pointer) to a function in C
From the above examples WKT,
1.If a variable is pass by value, then a copy of arguments will be passed to the function. Hence any modification for those variables inside the function will not be reflected in the called function.
2. If a variable is pass by reference, then any modification made in called function will be reflected in the calling function.
3. Now, if you want to modify any data pointed by the pointer then we need to send pointer to a pointer.
Example:
In the below example, we have a function called as allocate_memory, that will take pointer to pointer.
This is because we are allocating a memory (modifying) pointed by the pointer, then you need to pass a pointer to pointer to pointer.
if you want to modify a pointer in C, you need to pass a pointer to a pointer. If you need to modify a pointer to a pointer, you need to pass a pointer to a pointer to a pointer.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void allocate_memory(int **arr, int size)
{
*arr = (int*)malloc(size * sizeof(int));
if(*arr != NULL)
{
for(int i=0; i<size; i++)
{
*(*arr+i) = i;
}
}
}
int main()
{
int *arr = NULL;
allocate_memory(&arr, 5);
for (int i = 0; i < 5; ++i)
{
printf("%d\n",arr[i]);
}
return 0;
}
Output:
0 1 2 3 4
