typedef and variable names

问题:

Ignoring why I would want to do this, just trying to understand what is happening here: This code compiles:

#include <stdio.h>
typedef char byte;

int main (void) 
{
    byte var_byte;
    int byte = 10;

    printf("\n Test program: %d\n", byte);
}  

But, if I change the order in which the variables are declared, it does not compile.

This DOES NOT COMPILE:

#include <stdio.h>
typedef char byte;

int main (void)
{
    int byte = 10;
    byte var_byte;

    printf("\n Test program: %d\n", byte);
}

Compiler error:

b.c:7:8: error: expected ‘;’ before ‘var_byte’
   byte var_byte;
        ^~~~~~~~

Could someone please explain why order matters?

回答1:

In this program

#include <stdio.h>
typedef char byte;

int main (void)
{
    int byte = 10;
    byte var_byte;

    printf("\n Test program: %d\n", byte);
}

the name of the variable byte hides the name of the typedef.

From the C Standard (6.2.1 Scopes of identifiers)

4. ... If an identifier designates two different entities in the same name space, the scopes might overlap. If so, the scope of one entity (the inner scope) will end strictly before the scope of the other entity (the outer scope). Within the inner scope, the identifier designates the entity declared in the inner scope; the entity declared in the outer scope is hidden (and not visible) within the inner scope.

Pay attention to that the name of an identifier and a typedef name belong to the same name space.

The typedef name is declared in the global scope (file scope) while the name of the variable is declared in the inner block scope and the name of the variable hides the name declared in the global scope.

Consider this program.

#include <stdio.h>
typedef char byte;

void f( void );

int main (void)
{
    int byte = 10;

    printf("\n Test program: %d\n", byte);

    f();
}

void f( void )
{
    byte c = 'A';
    printf( "%c\n", c );
}

Within the block scope of the function main (the inner scope relative to the file scope) the name of the typedef is hidden by the declaration of the variable with the same name.

However inside the block scope of the function f the name declared in the typedef is visible because neither other declaration in the block scope of the function hides the name declared in the typedef.

Here is a more interesting program that deals with the point of declaration (it is a C++ term)

#include <stdio.h>

size_t byte = 255;

int main(void) 
{
    typedef int byte[byte];

    {
        byte byte;

        printf( "sizeof( byte ) = %zu\n", sizeof( byte ) );
    }

    return 0;
}

Its output might look like

sizeof( byte ) = 1020

Here is in the file scope there is declared variable with the name byte

size_t byte = 255;

In the outer block scope of the function main there is introduced typedef name byte.

typedef int byte[byte];

It hides the previously declared name byte after declaration of the declarator. That is in this typedef

    typedef int byte[byte];

the name byte in the square braces corresponds to the global name byte.

Then in the inner block scope there is declared an array with the same name byte that hides the typedef name.

byte byte;

Pay attention to that in the expression

sizeof( byte )

there is used the name of the array not the typedef name.

回答2:

Edit here: (misunderstand of the question)

When you declare a variable in C it does not look for typedef structures to see if a structure named the same with the variable. On the first code,

byte var_byte; 

this line of code comes before the declaration of the variable int byte. The computer will look for the most recent reference of the word byte as it was the structure name here.

On the second one the the variable int byte does not return an error because you can still create variables with the same structure type name in C. But after you do so you can't create new structures to that structure as the computer will think that it is referring to the variable name and not the the structure type as the variable was declared most recently