Why typedef struct

[deXo-fan]

Hi,

Why would one write something like

typedef struct tagSTRUCT
{
  int n;
} StructName;

... instead of

struct StructName
{
  int n;
};

What's the deal?

[tabstop]

Because the first can be used as

StructName foo;

while the second must be used as

struct StructName foo;

[mingerso]

Other examples while combining both methods:

typedef struct
{
        int a;
} a;

typedef struct b
{
        int b;
} b;

int main(void)
{
        a struct_a;

        b struct_b;
        struct b struct_b2;

        struct_a.a  = 0;
        struct_b.b  = 0;
        struct_b2.b = 0;

        return 0;
}

[deXo-fan]

Hi,

Thanks for your almost immediate replies. But this compiles perfectly:

#include <iostream>

struct s
{
    int a;
};

int main ()
{
    s _s;
    std::cout << _s.a << std::endl;
    return 0;
}

I don't need to type:
struct s _s;

Why is that? I'm using Dev-C++ (the latest version) with the MingW compiler.

[MacGyver]

That's because you're using C++, not C, and you don't need to typedef structs in C++.

[grumpy]

That's because you're using C++, not C, and you don't need to typedef structs in C++.

Using a C++ compiler actually. C++ supports some things that are not supported in C and a lot of ostensible C compilers in use are actually C++ compilers.

You might also try looking for compiler options (eg command line settings) to make your compiler behave as a C compiler rather than as a C++ compiler. To do that, it is necessary to read the documentation for your compiler.

[anon]

I thought iostream, cout and endl were C++.

In C you need the typedef for convenience. In C++ you don't need it all (but the compiler accepts this syntax too).

[Elysia]

Yup, they are. This is C++, not C however much you turn and spin it.