How to change value of arguments in the function

Question

I try it :

#include 
#include 

void foo(int* x)
{
   x = (int *)malloc(sizeof(int));
   *x = 5;
}

int main()
{

   int a;
           


        

Answer 1

You are changing the pointer value in foo by the malloc call - just do this:

void foo(int* x)
{
   *x = 5;
}

In your code - this function:

void foo(int* x)
{
   x = (int *)malloc(sizeof(int));
   *x = 5;
}

Works as follows:

+----+
| a  |
+----+
/
x points here

Then you make x point somewhere else:

+----+                            +----+
| 5  | /* created by malloc*/     | a  |
+----+                            +----+
/ On Heap                         / On Stack
x now points here                a is now in memory somewhere not getting modified.

And you make it 5. Also note - x is a local copy of the &a address of a - in C there is no pass by reference but only pass by value - since the variables value is copied into the function.