为什么我不能检查 C 中结构的大小？ （未声明的错误）

Question

I am trying to figure out the sizeof(p) where p is the struct defined below; but, when I try and run the following code:

#include <stdio.h>

struct p
{
    char x;
    int y;
};

int main()
{
    printf("%d", sizeof(p));
    return 0;
}

I receive this error:

main.c: In function ‘main’:
main.c:19:25: error: ‘p’ undeclared (first use in this function)
     printf("%d", sizeof(p));
                         ^

I am a beginner in C and I tried to move p 's definition into the main function, changing the definition of p , looking the error up online (none of the posts with the same error answered my question), etc., but I couldn't seem to get it to work. Any suggestions are appreciated.

Answer 1

In C (unlike in C++), the struct p... construct does not define a new type of variable. It only defines p as a particular type of struct . So, in order to get the size of that structure, or to declare a variable of that type, you need to use struct p to refer to it.

Like this:

#include <stdio.h>

struct p {
    char x;
    int y;
};

int main()
{
    printf("%zu\n", sizeof(struct p));
    // Alternatively ...
    struct p q;
    printf("%zu\n", sizeof(q));
    return 0;
}

Also, note that you should use the %zu format specifier for the size_t type.