使用define将一个结构成员分配给同一结构的另一个成员

Question

typedef struct{
               [..]
               type_t member1;
               type_t member2;
               [...]       

}structType_t

I want to assign member1 to member2. As this is a repeated operation I thought of putting the assignmet in a #define:

#define op (structType_t).member1=(structType_t).member2

However, this seems to be wrong how would the compiler know that it is the members of the same struct and I can't see a way of using it. Any ideas? I know I have other options such as macro or function but my question is this way possible?

Answer 1

Writing such macros is never a good idea. It is probably not possible to write something better and more readable than mystruct.member1 = mystruct.member2; .

If you want to encapsulate this for whatever reason, use a function:

void structTypeMemberCopy (structType_t* obj)
{
  obj->member1 = obj->member2;
}

This is likely going to get inlined and replaced with the equivalent of mystruct.member1 = mystruct.member2; in the machine code.

And finally there is the bad idea macro, which can be made type safe:

#define structTypeCopy(obj) _Generic((obj), structType_t: (obj).member1 = (obj).member2)

...

structType_t mystruct = { ... };
structTypeCopy(mystruct);

Answer 2

You need to define the macro like this:

#define a(T1) T1.n1 = T1.n2

the macro a receive one parameter T1 and does the asignement.

Something like this:

#include <iostream>

#define a(T1) T1.n1 = T1.n2

typedef struct {
    int n1;
    int n2;
} estructure;

int _tmain(int argc, _TCHAR* argv[])
{
    estructure test;

    test.n1 = 1;
    test.n2 = 2;

    a(test);

    std::cout << test.n1 << " " << test.n2 << std::endl;

    return 0;
}