Using printf with a pointer to float gives an error

c, floating-point, pointers, printf

Solution

void main()
{
    float x;
    x=6.5;
    printf("Value of x is %f, address of x %ld\n", x, &x);
}

The immediate problem is that you're missing the required `#include <stdio.h>`, but that's not the only problem with your code. Some of these things are errors that you can probably get away with (compilers may not complain, and may generate code that does what you expect), but there's no reason not to do it right.

#include <stdio.h>
int main(void)
{
    float x;
    x = 6.5;
    printf("Value of x is %f, address of x %p\n", x, (void*)&x);
}

To explain the changes I made:

- `#include <stdio.h>` is required for any program that calls `printf`. More precisely, a declaration of `printf` is required, and `<stdio.h>` provides it. (In principle you could write your own declaration instead, but there's no good reason to do so.)

- The correct definition of `main` is `int main(void)`. `void main()` may be accepted by some compilers, but it's useful mostly as a way to detect bad books. If you're using a book that tells you to use `void main()`, its author does not know the language very well, and may have given you other misinformation. Find a better book. (Caveat: `void main()`, or more likely `void main(void)` might actually be the preferred implementation-defined form for some embedded systems. but you're probably not using such a system.)

- The `"%ld"` format requires an argument of type `long int`. The only correct format for printing a pointer value is `"%p"`. Since `"%p"` requires an argument of type `void*`, you should explicitly cast your pointer value to `void*`. Omitting the cast is likely to "work", but `float*` and `void*` are distinct types, and are not guaranteed to have the same representation or to be passed to functions in the same way.

Problem

When I try to compile this code: ``` void main() { float x; x=6.5; printf("Value of x is %f, address of x %ld\n", x, &x); } ``` it gives me this error: pruebaso.c: In function ‘main’: pruebaso.c:5:9: warning: incompatible implicit declaration of built-in function ‘printf’ [enabled by default] printf("Value of x is %f, address of x %ld\n", x, &x); ^ pruebaso.c:5:9: warning: format ‘%ld’ expects argument of type ‘long int’, but argument 3 has type ‘float *’ [-Wformat=] I've seen in another forum the solution is to make a cast to a void pointer first: http://www.linuxquestions.org/questions/programming-9/beginning-c-programming-how-to-print-memory-locations-printf-conversion-number-927305/ But making this change, ``` printf("Value of x is %f, address of x %ld\n", (double)x, (void *)&x); ``` now gives me a warning: pruebaso.c: In function ‘main’: pruebaso.c:5:9: warning: incompatible implicit declaration of built-in function ‘printf’ [enabled by default] printf("Value of x is %f, address of x %ld\n", (double)x, (void *)&x); ^ pruebaso.c:5:9: warning: format ‘%ld’ expects argument of type ‘long int’, but argument 3 has type ‘void *’ [-Wformat=] Could someone explain me how could I solve it without getting a warning? Thank you

Original source