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;); 

or

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

or

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);   

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