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Detecting if an expression is constant in C

sleirsgoevy

The Linux kernel has even a way to determine whether the expression is compile-time, WITHOUT aborting compilation in either case.

The trick is this (copied vebratim from Linux):

#define __is_constexpr(x) (sizeof(int) == sizeof(*(8 ? ((void *)((long)(x) * 0l)) : (int *)8)))

Explanation: if x is a constant expression, then multiplying it by zero yields a constant 0, and casting a constant 0 to void* makes a null pointer constant. And the ternary expression, if one of its sides is a null pointer constant, collapses to the type of the other side (thus the type of the returned pointer will be int*, and the sizeof will match). And if x was not constant, then the lefthand side would not be considered a null pointer constant by type inference, the type of the ternary expression will be void*, and the sizeof check will not match.

With a few more clever tricks, it's even possible to implement a compile-time "type ternary expression", like this: TYPE_IF(2 * 2 == 4, int, long). This is left as an exercise for the reader.

cperciva

With a few more clever tricks...

I did this with my PARSENUM macro (https://github.com/Tarsnap/libcperciva/blob/master/util/pars...) to parse strings into floating-point, unsigned integer, or signed integer types (and check bounds) using a single interface.

amelius

This reminds me of the days when Boost was a thing. It was full of tricks like this.

usrnm

It still is a thing, though.

wahern

> This works. But both gcc and clang warn about the enum being anonymous... even though that's exactly what I wanted to do. And this cannot be silenced with #pragma since it's a macro, so the warning occurs at the location where the macro is invoked.

You can use _Pragma instead of #pragma. E.g.

  #define C(x) ( \
    _Pragma("clang diagnostic push") \
    _Pragma("clang diagnostic ignored \"-Wvisibility\"") \
    (x) + 0*sizeof(void (*)(enum { tmp = (int)(x) })) \
    _Pragma("clang diagnostic pop") \
  )
EDIT: Alas, GCC is a little pickier about where _Pragma is allowed so you may need to use a statement expression. Also, it seems GCC 14 doesn't have a -W switch that will disable the anonymous enum warning.

bobbyi

I thought this would work:

#define C(x) (sizeof(char[x]), x)

sizeof is a compile-time operation so x need to be known at compile time.

It didn't work as expected. It turns out there is an exception and the standard says that sizeof is actually calculated at runtime specifically for variable length arrays:

> The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.

pjc50

It's remarkable that people will say that doing this kind of thing is better than learning a language which actually lets you enforce this with the type system.

(or even just insist that users use the version of the language which supports "constexpr"!)

uecker

I do not think anybody said this. The point is that these macros work for early versions of C. If you need to support early versions of C, learning another language is not a solution. If you don't have to, you can use C23's constexpr.

o11c

The problem is that no such language exists.

There are many languages that provide one particular feature that C doesn't provide, but they do this at the cost of excluding numerous other features that C widely relies on.

trealira

C used to seem like a beautiful and simple language to me, but as I used it and learned more about it, it seemed more complex under the surface, and kind of janky as well. It's just utilitarian.

oguz-ismail

What language is that? Is it available everywhere (everywhere) C is?

mitthrowaway2

Indeed, usually if I'm using C these days it's because I only have access to a c compiler for my target platform, or because I'm modifying an existing C codebase.

wat10000

Learning such a language doesn’t mean I can use it.

kjs3

"I have no idea what problem you're trying to solve, what the constraints are, what the use cases might be, what tools are available on the platform, what the job or regulations require, what the skillsets of the people involved are, what the timeline is...but I'm absolutely, unshakably certain that I have a magic bullet that will make all your problems go away."

FTFY.