Pointer argument to function

We can also pass the pointer variable to function.

In other words, we can pass the address of a variable to the function instead of variable value.




How to declare a function which accepts a pointer as an argument?

If a function wants to accept an address of an integer variable then the function declaration will be,

return_type function_name(int*);

* indicates that the argument is an address of a variable not the value of a variable.


Similarly, If a function wants to accept an address of two integer variable then the function declaration will be,

return_type function_name(int*,int*);



What will happen to the argument?

Here, we are passing the address of a variable.

So, if we change the argument value in the function, it will modify the actual value of a variable.


Example

#include<stdio.h>

void setToZero(int *a)
{
  *a = 0;
}

int main()
{
  int a = 10;
  printf("Before calling the function a = %d\n",a);

  setToZero(&a);
  printf("After calling the function a = %d\n",a);

  return 0;
}




Output

Before calling the function a = 10

After calling the function a = 0





Visual Representation

Pointer Argument to Function

step 1. Address of a variable a 1024 is passed to function setToZero().

step 2. *a = 0 will modify the original value of a. So, a value in main function will be 0 after the function execution.




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