Hello I am a beginner in C programming language. Recently I read about call by value and call by address. I have learned that in call by address changes in the called functions reflects the callee. However the following code does not work like that.
int x = 10,y = 20;
void change_by_add(int *ptr) {
ptr = &y;
printf("\n Inside change_by_add\t %d",*ptr);
// here *ptr is printing 20
}
void main(){
int *p;
p = &x;
change_by_add(p);
printf("\nInside main\t %d", *p);
// here *p is still pointing to address of x and printing 10
}
When I am passing address then why the changes made by called function does not reflect caller?
The function is assigning a new address to the pointer but the pointer itself is being passed by value, as all arguments are in C. To change the value of a pointer variable the address of the pointer itself must be passed:
void change_by_add(int **ptr)
{
*ptr = &y;
}
change_by_add(&p);
See C FAQ Question 4.8.
Passing by reference does not exist in C but can be achieved by passing the address of the variable who's value is to be changed to a function. For example:
void add_to_int(int* a_value, int a_increment)
{
*a_value += a_increment;
}
You are simply setting the value of the pointer in the function, not the value of the pointed to variable. The function should use the following code:
*ptr = y;
This derefences the pointer (exposing the value pointed to), and therefore when you use the equals operator, the memory pointed at is modified, not the pointer itself. I hope this helps to clarify things.
There is no such thing as call by address in C. There is only call by value. What one does when a function needs to modify an argument in a way that is visible to the caller is to have the caller pass a pointer to something, and have the called function write the update though that pointer. Note that the pointer itself is still sent as call-by-value - that is: the called function gets its own copy of the pointer and could change it to point to anything else if it wants to.