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

To access the value that a pointer points to, you have to use the indirection operator *.

To print the pointer itself, just access the pointer variable with no operator.

And to get the address of the pointer variable, use the & operator.

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

    /*Print the address pointed to by iptr*/
    printf("Address of value: %p\n", (void*)iptr);

    /*Print the address of iptr itself*/
    printf("Address of iptr: %p\n", (void*)&iptr);
}

The %p format operator requires the corresponding argument to be void*, so it's necessary to cast the pointers to this type.

Answer 2

Read the comments

#include <stdio.h>
#include <stdlib.h>
    
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

Answer 3

#include <stdio.h>
#include <stdlib.h>

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

/*Print the address pointed to by iptr*/

/*Print the address of iptr itself*/
}

int main(){
int iptr = 0;
pointerFuncA( &iptr); 

return 0;
}

I think you are looking at something like this, there is no need to re-define the function again in the main....

Answer 4

int* iptr is already a pointer, so you don't need the & in front of it when you write

printf("Address of iptr variable: %x\n", &iptr );

This is how to print a pointer value.

printf("Address of iptr variable: %p\n", (void*)iptr);

Also you have the function prototype for pointerFuncA() in the wrong place, being inside main(). It should be outside of any function, before it is called.

Answer 5

Address are some memory values which are written in hexadecimal notation starting with 0x

/Value pointed to by the pointer iptr/

printf("Value is: %i", *iptr);

Address pointed to by the pointer will be the value of the iptr pointer itself

/print the address pointed to by the iptr/

 printf("Address is: %p", iprt);

/print the address of iptr itself/

 printf("Address of iptr: %p", &iptr )