Code::Blocks does not print pointer value

Question

I developed a C code to print the value and address of the variable x using a p pointer, but Code::Blocks does not print the values.

    #include <stdio.h>
    #include <conio.h>

    int main()
    {
    int x=2;
    int *p;
    p = &x;

    printf("O valor de X e: %p\n", *p);

    printf("O valor de X e: %d\n", p);

    return(0);

    }

Answer 1

Do not use conio.h, even if you don’t have to. And if you have a bad compiler IDE (Dev-C++), stop doing this. Alias Code::Blocks does nothing in executing the same code, understand who’s who in programming.

From what I understand the first you want to print the address of the memory where the X since you used %p, and the second wants to print the value of X. So you have to tidy up the text to give correct information.

There’s a problem that’s reversed. p is already a memory address, so just use it in the first. The second wants the value and catch it indirectly by the variable p which is an address, so now you have to do the reverse operation you had done to get the address of X, then you have to do what’s called derreferenciar, which is done with the operator *. The *P means "take the amount that is in the address of p".

I put a (void *) Because good compilers with the right configuration prevents direct compilation, they force you to be explicit in what you are using to indicate that you are not confusing what you are sending. But it is possible not to use in certain circumstances (I did not use in Coding Ground, look there), under risk of doing something wrong in more complex real code.

#include <stdio.h>

int main() {
    int x = 2;
    int *p = &x;
    printf("O endereço de X e: %p\n", (void *)p);
    printf("O valor de X e: %d\n", *p);
}

Behold working in the ideone. And in the repl.itd. Also put on the Github for future reference.

See more in What is the meaning of the operator "&" (and commercial) in the C language? and Operator & and * in functions.

Answer 2

int x=5;
int *p;
p = &x;
printf("O valor de X e: %d\n", *p);
printf("O valor de X e: %d\n", p);

The first printf() prints the value of x, the second prints the address of the pointer.

All your code is fine, that 'c' is 12 in hexadecimal.