Initialization makes pointer from integer without a cast - C

Question

Sorry if this post comes off as ignorant, but I'm still very new to C, so I don't have a great understanding of it. Right now I'm trying to figure out pointers.

I made this bit of code to test if I can change the value of b in the change function, and have that carry over back into the main function(without returning) by passing in the pointer.

However, I get an error that says.

Initialization makes pointer from integer without a cast
    int *b = 6

From what I understand,

#include <stdio.h>

int change(int * b){
     * b = 4;
     return 0;
}

int main(){
       int * b = 6;
       change(b);
       printf("%d", b);
       return 0;
}

Ill I'm really worried about is fixing this error, but if my understanding of pointers is completely wrong, I wouldn't be opposed to criticism.

Since the change function doesn't return anything useful, you should declare it void. — Barmar
– Barmar, Commented Oct 29, 2015 at 1:52
Also check out "Pointer from integer/integer from pointer without a cast" issues. — Lundin
– Lundin, Commented Sep 4 at 6:16

AStopher · Accepted Answer · 2025-09-02 15:05:45Z

17

To make it work rewrite the code as follows -

#include <stdio.h>

int change(int *b)
{
    *b = 4;
    return 0;
}

int main()
{
    int b = 6; //variable type of b is 'int' not 'int *'

    change(&b);//Instead of b the address of b is passed
    printf("%d", b);

    return 0;
}

The code above will work.

In C, when you wish to change the value of a variable in a function, you "pass the Variable into the function by Reference". You can read more about this here - Pass by Reference

Now the error means that you are trying to store an integer into a variable that is a pointer, without typecasting. You can make this error go away by changing that line as follows (But the program won't work because the logic will still be wrong )

int *b = (int *)6; //This is typecasting int into type (int *)

edited Sep 2 at 15:05

AStopher

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answered Oct 29, 2015 at 1:52

Soren Goyal

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Comments

BobRun · Accepted Answer · 2015-10-29 01:47:50Z

3

Maybe you wanted to do this:

#include <stdio.h>

int change( int *b )
{
  *b = 4;
  return 0;
}

int main( void )
{
  int *b;
  int myint = 6;

  b = &myint;
  change( &b );
  printf( "%d", b );
  return 0;
}

answered Oct 29, 2015 at 1:47

BobRun

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Comments

milevyo · Accepted Answer · 2015-10-29 01:36:09Z

2

#include <stdio.h>

int change(int * b){
     * b = 4;
     return 0;
}

int main(){
       int  b = 6; // <- just int not a pointer to int
       change(&b); // address of the int
       printf("%d", b);
       return 0;
}

answered Oct 29, 2015 at 1:36

milevyo

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Comments

udarH3 · Accepted Answer · 2017-07-29 11:24:32Z

0

Maybe too late, but as a complement to the rest of the answers, just my 2 cents:

void change(int *b, int c)
{
     *b = c;
}

int main()
{
    int a = 25;
    change(&a, 20); --> with an added parameter
    printf("%d", a);
    return 0;
}

In pointer declarations, you should only assign the address of other variables e.g "&a"..

answered Jul 29, 2017 at 11:24

udarH3

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