How to check a string literal in static_assert?

Question

I want to check, that a string literal doesn't have some specific value.

#include <assert.h>
static_assert("aaa"[0] != 'a');

However, this results in this error:

error: expression in static assertion is not constant

How can this be achieved? Why doesn't it work, given that all data is known at compile time?

Seems to be a difference between C and C++ in what's considered a compile-time constant. This appears to be a valid failing assertion for C++17 onwards (and in C++11, if we add a "message" argument). — Toby Speight
– Toby Speight, Commented Mar 21 at 16:50
Historically C string literals are char[N], as const was not available until later. — BoP
– BoP, Commented Mar 21 at 18:23
@BoP Being const or not does not make the object constant as in the error message. — chux
– chux, Commented Mar 21 at 19:15
@Caulder, "Why doesn't it work" --> I have worked with some compilers that did allow something close to it (embedded processor ones). Per the C language - I suspect it is not possible. Yet maybe C23 offers some way or a compiler extension? — chux
– chux, Commented Mar 21 at 19:20
Before C23, code needs a 2nd argument as in static_assert("aaa"[0] != 'a', "some string here");. — chux
– chux, Commented Mar 21 at 19:22

Lundin · Accepted Answer · 2025-03-31 07:57:01Z

1

static_assert requires an integer constant expression (not to be confused with const) as parameter. So the issue is what C regards as a an integer constant expression. It's rather picky with this (C23 6.6):

An integer constant expression shall have integer type and shall only have operands that are integer constants, named and compound literal constants of integer type, character constants, sizeof expressions whose results are integer constants, alignof expressions, and floating, named, or compound literal constants of arithmetic type that are the immediate operands of casts.

String literals isn't among any of these, hence the error.

However, the standard allows implementation-defined (meaning compiler-specific) forms on constant expressions, why for example some compilers like clang might accept the code. But code relying on that is not portable.

Similarly, C++ is more lenient to various forms of constant expressions than C, so if you happen to compile the code as C++ by accident, that would work too.

answered Mar 31 at 7:57

Lundin

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1 Comment

Caulder Mar 31 at 21:39

Does that mean, that it is not possible to check that value in the source code at compiletime at all, or is there some hack how to still do it?

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