Print value and address of pointer defined in function?

Question

I think this is a really easy thing to code, but I\'m having trouble with the syntax in C, I\'ve just programmed in C++.

#include 
#include 

        

Answer 1

Read the comments

#include 
#include 
    
void pointerFuncA(int* iptr){
  /*Print the value pointed to by iptr*/
  printf("Value:  %d\n", *iptr );
    
  /*Print the address pointed to by iptr*/
  printf("Value:  %p\n", iptr );

  /*Print the address of iptr itself*/
  printf("Value:  %p\n", &iptr );
}
    
int main(){
  int i = 1234; //Create a variable to get the address of
  int* foo = &i; //Get the address of the variable named i and pass it to the integer pointer named foo
  pointerFuncA(foo); //Pass foo to the function. See I removed void here because we are not declaring a function, but calling it.
   
  return 0;
}

Output:

Value:  1234
Value:  0xffe2ac6c
Value:  0xffe2ac44