The C standards committee has been broken since C99 introduced long long, thus silently breaking conforming code. How do you print out a size_t, portably, with

The C standards committee has been broken since C99 introduced long long, thus silently breaking conforming code. How do you print out a size_t, portably, without losing information? The standard says size_t is an unsigned integer type, but doesn't say way size. However, the C89 spec explicitly stated that there are no integers sizes longer than long- so the conforming way to do this is to explicitly cast the size_t to unsigned long (which, while it may add bits, is guaranteed not to lose them), and print the unsigned long. In an attempt to save all the broken, non-conforming code that assumed that sizeof(int) == sizeof(long), C99 introduced the long long type. And, among other things, allowed size_t to now be unsigned long long. Which meant the conforming C89 code that wants to print out a size_t is now wrong. Worse yet, it's silently broken- because the type cast is explicit, the compiler has to assume it's correct. They added a new way to print size_t's, granted- but this is no help for the legacy code (or code that needs to continue to support C89-only compilers). Of course, the punchline here is that I have yet to see code that assumed sizeof(int) == sizeof(long) that didn't also assume sizeof(void *) == sizeof(int). So all that broken non-conforming code they were trying to save? It still needed to get fixed for 64-bit.