Question

In my code, I want to ensure that sizeof(a) == sizeof(b).

First approach was to let the preprocessor do the checking:

#if (sizeof(a) != sizeof(b))
#  error sizes don't match
#endif

which doesn’t compile because of fatal error C1017: invalid integer constant expression. Okay. Understand.

Next try:

if(sizeof(a) != sizeof(b)){
  printf("sizes don't match\n");
  return -1;
}

Which results in a warning: warning C4127: conditional expression is constant.

Now, I’m stuck. Is there a warning-and-error-free way to make sure that the two structs a and b have the same size?

---

Edit:
Compiler is Visual Studio 2005, Warning level is set to 4.

Answer 1

Bad array size

// make sure sizeof(a) == sizeof(b):
int size_must_match[sizeof(a) == sizeof(b)];
// will fail if sizeof(a) == sizeof(b) evaluates to 0
// might produce warning: 'size_must_match' is not used 

// to suppress "'size_must_match' is not used", try:
size_must_match[0]; // might produce warning: "statement has no effect"

or

typedef int size_must_match[sizeof(a) == sizeof(b)];

Bad compile-time arithmetic

In C++ is guaranteed that these constant expressions are evaluated by the compiler at compile-time, and I believe the same holds in C:

// make sure sizeof(a) == sizeof(b):
1 / (sizeof(a) == sizeof(b));
// might produce warning: "statement has no effect"

int size_must_match = 1 / (sizeof(a) == sizeof(b));
// might produce warning: 'size_must_match' is unused

assert (1 / (sizeof(a) == sizeof(b)));
// very silly way to use assert!

Switch (just for fun)

switch (0) { // compiler might complain that the 
             // controlling expression is constant
case 0:       
case sizeof(a) == sizeof(b):
    ; // nothing to do
}

You get the idea. Just play around with this until the compiler is 100 % happy.