CODEWITHSUNDEEP
c++signalsvolatile
Tootsie
Tootsie

Reputation: 867

Why is volatile not part of sig_atomic_t

On my platform (X86, Fedora, gcc 9.1.1) sig_atomic_t is typedef'd to a plain int.

In the C++ standard, sig_atomic_t is always used with the volatile qualifier.
I understand why volatile is needed, but why is it not part of the type then ?

Something like:

using sig_atomic_t = volatile int;

Upvotes: 6

Views: 1107

Answers (2)

Petr Skocik
Petr Skocik

Reputation: 60067

C89 says that it's

the integral type of an object that can be accessed as an atomic entity, even in the presence of asynchronous interrupts.

volatile not specified, probably because qualifiers were a new thing when the first standard was made.

C99 adds "possibly volatile-qualified.

I suppose it's backwards compatibility from then on combined with "nobody cares enough", since signal-handling is a relatively minor part of most project.

Also somebody could presumably use it in a context where volatile is not required (e.g., to store a copy of a flag used for communication with signal handlers) and in non-GNU C (again, backwards compatibility) it's basically impossible to map a type to a less qualified version of that type, which makes an implementation that chooses to omit the qualifier more flexible.

Upvotes: 3

1201ProgramAlarm
1201ProgramAlarm

Reputation: 32727

This is inherited from C. The C definition, while allowing for sig_atomic_t to be volatile qualified, does not require it. All the example uses in the standard doc I've looked (N1570) at are given as volatile sig_atomic_t.

These days it may be better to use std:atomic and the other capabilities specified in the <atomic> header when feasible. (Also see sig_atomic_t on cppreference.)

Upvotes: 3

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