c++11 use chrono as syntactic sugar

Question

I use function from external library with interface like this: void f(int timeout); . Where timeout is in milliseconds. To make my code more readable, I want to use chrono in such form:

f(std::chrono::milliseconds(10).count());

1. is it possible that std::chrono::milliseconds(10).count() != 10 ?

2. is any "underwater rocks" that prevent modern compilers (clang, gcc, VC++) to convert f(std::chrono::milliseconds(10).count()) to f(10) ?

Answer 1

1. is it possible that std::chrono::milliseconds(10).count() != 10 ?

No. The duration constructor you're using does:

3) Constructs a duration with r ticks.

and count() simply:

Returns the number of ticks for this duration.

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2. is any "underwater rocks" that prevent modern compilers (clang, gcc, VC++) to convert f(std::chrono::milliseconds(10).count()) to f(10) ?

Nope. The duration constructor is constexpr , as is the count() member function you're using - that should be a very simple optimization for a compiler to make.

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To make my code more readable [...]

Personally, I find that questionable. If f() itself took a duration , then f(std::chrono::milliseconds(10)) in of itself is certainly better than just f(10) , with the added benefit that whatever duration you pass in will work correctly. But if it's just taking an int , you're just giving yourself the illusion of security with just lots more typing. So I'm not sure it's any better, personally.

Answer 2

1. No. But beware that the result is not milliseconds, but ticks. So std::chrono::seconds(10).count() is also 10. So you might want to do an assignment to a std::chrono::milliseconds variable first.

2. No, because the method is a constexpr . So any good compiler should respect that.