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 mad

Answer 1

#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;
}

Answer 2

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;
}

Answer 3

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 *)

Answer 4

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"..