What is a common C/C++ macro to determine the size of a structure member?

问题

In C/C++, how do I determine the size of the member variable to a structure without needing to define a dummy variable of that structure type? Here's an example of how to do it wrong, but shows the intent:

typedef struct myStruct {
  int x[10];
  int y;
} myStruct_t;

const size_t sizeof_MyStruct_x = sizeof(myStruct_t.x);  // error

For reference, this should be how to find the size of 'x' if you first define a dummy variable:

myStruct_t dummyStructVar;

const size_t sizeof_MyStruct_x = sizeof(dummyStructVar.x);

However, I'm hoping to avoid having to create a dummy variable just to get the size of 'x'. I think there's a clever way to recast 0 as a myStruct_t to help find the size of member variable 'x', but it's been long enough that I've forgotten the details, and can't seem to get a good Google search on this. Do you know?

Thanks!

回答1:

In C++ (which is what the tags say), your "dummy variable" code can be replaced with:

sizeof myStruct_t().x;

No myStruct_t object will be created: the compiler only works out the static type of sizeof's operand, it doesn't execute the expression.

This works in C, and in C++ is better because it also works for classes without an accessible no-args constructor:

sizeof ((myStruct_t *)0)->x

回答2:

I'm using following macro:

#include <iostream>
#define DIM_FIELD(struct_type, field) (sizeof( ((struct_type*)0)->field ))
int main()
{
    struct ABC
    {
        int a;
        char b;
        double c;
    };
    std::cout << "ABC::a=" << DIM_FIELD(ABC, a) 
              << " ABC::c=" << DIM_FIELD(ABC, c) << std::endl;

    return 0;
}

Trick is treating 0 as pointer to your struct. This is resolved at compile time so it safe.

回答3:

You can easily do

sizeof(myStruct().x)

As sizeof parameter is never executed, you'll not really create that object.

回答4:

Any of these should work:

sizeof(myStruct_t().x;);

myStruct_t *tempPtr = NULL;
sizeof(tempPtr->x)

sizeof(((myStruct_t *)NULL)->x);

Because sizeof is evaluated at compile-time, not run-time, you won't have a problem dereferencing a NULL pointer.

回答5:

In C++11, this can be done with sizeof(myStruct_t::x). C++11 also adds std::declval, which can be used for this (among other things):

#include <utility>
typedef struct myStruct {
  int x[10];
  int y;
} myStruct_t;

const std::size_t sizeof_MyStruct_x_normal = sizeof(myStruct_t::x);
const std::size_t sizeof_MyStruct_x_declval = sizeof(std::declval<myStruct_t>().x);

回答6:

From my utility macros header:

#define FIELD_SIZE(type, field) (sizeof(((type *)0)->field))

invoked like so:

FIELD_SIZE(myStruct_t, x);

来源：https://stackoverflow.com/questions/1400660/what-is-a-common-c-c-macro-to-determine-the-size-of-a-structure-member

标签

c++

sizeof