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

To make it work rewrite the code as follows -

#include 

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 *)