Is it possible to print out the size of a C++ class at compile-time?

If you really need to to get sizeof(X) in the compiler output, you can use it as a parameter for an incomplete template type:

template<int s> struct Wow;
struct foo {
    int a,b;
};
Wow<sizeof(foo)> wow;

$ g++ -c test.cpp
test.cpp:5: error: aggregate ‘Wow<8> wow’ has incomplete type and cannot be defined

To answer the updated question -- this may be overkill, but it will print out the sizes of your classes at compile time. There is an undocumented command-line switch in the Visual C++ compiler which will display the complete layouts of classes, including their sizes:

That switch is /d1reportSingleClassLayoutXXX, where XXX performs substring matches against the class name.

https://devblogs.microsoft.com/cppblog/diagnosing-hidden-odr-violations-in-visual-c-and-fixing-lnk2022/

EDITED (3jun2020) This trick works IN ALL C COMPILERS. For Visual C++:

struct X {
    int a,b;
    int c[10];
};
int _tmain(int argc, _TCHAR* argv[])
{
    int dummy;

    switch (dummy) {
    case sizeof(X):
    case sizeof(X):
        break;
    }
    return 0;
}

------ Build started: Project: cpptest, Configuration: Debug Win32 ------ cpptest.cpp c:\work\cpptest\cpptest\cpptest.cpp(29): error C2196: case value '48' already used ========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

For other compilers that only print "duplicate case value", see my answer to this question: How can I print the result of sizeof() at compile time in C?

Whats wrong with sizeof? This should work on objects and classes.

void foo( bar* b )
{
  int i = sizeof bar;
  int j = sizeof *b;

  // please remember, that not always i==j !!!
}

Edit:

This is the example I was thinking of, but for some reason it's not working. Can anyone tell me what's wrong?

#include <iostream>
using namespace std;
class bar {
public: int i;
        bar( int ii ) { i = ii; }
        virtual ~bar(){ i = 0; }
        virtual void d() = 0;
};

class bar2: public bar {
public: long long j;
        bar2( int ii, long long jj ):bar(ii){ j=jj; }
        ~bar2() { j = 0; }
        virtual void d() { cout <<  "virtual" << endl; };
};

void foo( bar *b )
{
        int i = sizeof (bar);
        int j = sizeof *b;
        cout << "Size of bar = " << i << endl;
        cout << "Size of *b  = " << j << endl;
        b->d();
}


int main( int arcc, char *argv[] )
{
        bar2 *b = new bar2( 100, 200 );
        foo( b );
        delete b;
        return 0;
}

The application been run on linux (gcc 4.4.2):

[elcuco@pinky ~/tmp] ./sizeof_test
Size of bar = 8
Size of *b  = 8
virtual

sizeof() determines the size at compile time.

It doesn't work until compile time, so you can't use it with the preprocessor.