How to disable a warning within a C++ macro

Question

In Visual C++ you can temporarily disable a warning by using pragma:

#pragma warning(suppress: 4307)

How can I disable a warning within a macro, e.g., when I cause an "integral constant overflow" warning like this:

#define TIMES_A_MILLION(x) x * 1000000
int value = TIMES_A_MILLION(4711);

I don't want to repeat the warning at every place where the macro is used, but want the suppression to be part of the macro it self.

It is obviously not possible to do like this:

#define TIMES_A_MILLION(x) \
#pragma warning(suppress: 4307) \
  x * 1000000

Why would you want to suppress this warning instead of fixing the cause? — user7860670, Jan 18 '19 at 09:23
Possible duplicate of [What is \_\_pragma and what are the differences between \_\_pragma and #pragma](https://stackoverflow.com/questions/23790112/what-is-pragma-and-what-are-the-differences-between-pragma-and-pragma) — Max Vollmer, Jan 18 '19 at 09:26
The question I voted yours to be duplicate of approaches this from exactly the other direction, but answers your question perfectly. There is also [this article](https://learn.microsoft.com/en-us/cpp/preprocessor/pragma-directives-and-the-pragma-keyword?view=vs-2017) by Microsoft. — Max Vollmer, Jan 18 '19 at 09:26
"I don't want to repeat the warning at every place where the macro is used" - I would! — Bathsheba, Jan 18 '19 at 09:42
The macro is just an example, in my case it is a hash calculation which uses a multiplication that intentionally overflows. — Jordfräs, Jan 18 '19 at 09:47
@Jordfräs That begs two questions: 1. Why not use (possibly `constexpr` and/or templated) functions instead? 2. Storing something that overflows into an `int` is extremely suspicious because signed integer overflow is undefined behavior. — Max Langhof, Jan 18 '19 at 10:20
Because I'm an old fart that haven't discovered constexpr, but I have already realized I probably should try that. My actual hash calculation is unsigned, if that makes more sense. — Jordfräs, Jan 18 '19 at 13:53

score 9 · Accepted Answer · answered Jan 18 '19 at 09:22

9

In your case you have to use the extension __pragma

__pragma

instead of

#pragma

answered Jan 18 '19 at 09:22

darune

1 Answers1