Using sizeof on a typedef instead of a local variable

Question

Like in this example (in C):

typedef int type;

int main()
{
    char type;
    printf("sizeof(type) == %zu\n", sizeof(type)); // Outputs 1
}

The output is always the size of the local variable type.

When C++ removed the need to write struct before each use of a structure it still preserved the struct {type} syntax and introduced an alias (class {type}) to explicitly refer to a structure or class.

Example (in C++):

struct type {
    int m;
};

int main()
{
    char type;
    printf("sizeof(type) == %u\n", sizeof(type)); // Outputs 1
    printf("sizeof(struct type) == %u\n", sizeof(struct type)); // Outputs 4
    printf("sizeof(class type) == %u\n", sizeof(class type)); // Outputs 4
}

My question is if there is a way to explicitly refer to a typedef in C or C++. Something like sizeof(typedef type) perhaps (but that does not work).

I know that it is common practice to use different naming conventions for variables and types to avoid these kinds of situations but I would still like to know if there is a way within the langauge to do this or if there is not. :)

Answer 1

There is no way to resolve this one but if your structure is defined globally you can use this,

Scope resolution operator ::.

printf("sizeof(type) == %zu\n", sizeof(::type));

Answer 2

In C++, use :: operator to get the answer as 4.

printf("sizeof(::type) == %u\n", sizeof(::type));

The :: is used for accessing global variables in C++. In C, there is no direct way i think. You can do it using functions.

The :: operator works even if it is not a class or struct.

typedef int type1;

int main() {
 int type1;
 cout<<sizeof(::type1);
 return 0;
}

This will also give the answer as 4.

Answer 3

In C this is not possible. You are hiding the type type. You cannot use it as a type after you declare the char:

typedef int type;

int main(void) {
    char type;
    type t;      // error: expected ‘;’ before ‘t'
    printf( "%d %d\n", sizeof type, sizeof t );
    return 0;
}

However, if you create an alias for type or declare a type before you declare the char you can use that:

int main(void) {
    type t;
    char type;
    printf( "%d %d\n", sizeof type, sizeof t );
    return 0;
}


int main(void) {
    typedef type type_t;
    char type;
    printf( "%d %d\n", sizeof type, sizeof( type_t ) );
    return 0;
}

C++ has the scope resolution operator :: which you can use to reference the type using the qualified name, i.e. ::type or my_namespace::type.