why sizeof ran on struct must have parenthesis

Question

So I was trying to get the size of a struct on my C program and write the following code:

typedef struct Msg_Header Msg_Header; // the struct Msg_Header was defined earlier

int size = sizeof Msg_Header;

This produced a compilation error I cant understand. In order for this to work I found that I must user parenthesis like that:

int size = sizeof(Msg_Header);

This is weird since sizeof works perfectly on simple variable types like int without parenthesis, same weird behaviour accures when I use struct Msg_Header instead if the alias.

Can someone explain what is going on here?

EDIT: The compilation error says: "expected expression before Msg_Header"

What is the compilation error exactly? – phantom Nov 02 '14 at 16:31 — phantom, Nov 02 '14 at 16:31

Grzegorz Szpetkowski · Answer 1 · 2014-11-02T16:41:39.717

The sizeof operator can be used without parantheses only in case of expressions, not types (nor type synonyms created by typedef keyword) itself. See following example for possible valid usage:

#include <stdio.h>

struct Msg_Header { // struct tag
    int n;
};

typedef struct Msg_Header Msg_Header; // typedef synonym

int main(void) {
    Msg_Header msg_header;

    printf("%zu\n", sizeof(struct Msg_Header));
    printf("%zu\n", sizeof(Msg_Header));
    printf("%zu\n", sizeof msg_header);

    return 0;
}

Life sample here: http://ideone.com/tMB5cY

score 1 · Answer 2 · answered Nov 02 '14 at 16:37

Although sizeof is an operator and not a function, when used with a type you need the parentheses. Basically you hand it a type-cast of the type, which requires parentheses.

From the standard 6.5.3 Unary operators:

unary-expression:
    postfix-expression
    ++ unary-expression
    -- unary-expression
    unary-operator cast-expression
    sizeof unary-expression
    sizeof ( type-name )

why sizeof ran on struct must have parenthesis

2 Answers2