How does static casting from long to double in c++ differ from using the double constructor?

Question

I'm in the habit of doing 1. to coerce the code below into a floating point calculation but it just occurred to me: is there any difference in behaviour with 2.? It certainly looks less neater.

long blah = 3;

// 1

double d1 = 1 / static_cast<double>(blah);

// 2 note: not C-style (double) blah

double d2 = 1 / double(blah);

Possible duplicate of [Why use static\_cast(x) instead of (int)x?](https://stackoverflow.com/questions/103512/why-use-static-castintx-instead-of-intx) — Revolver_Ocelot, Sep 16 '17 at 09:31
It is, in everything, but name: http://en.cppreference.com/w/cpp/language/explicit_cast _The functional cast expression [...] is exactly equivalent to the corresponding C-style cast expression._ — Revolver_Ocelot, Sep 16 '17 at 10:02
You can get the same effect without any casts: double d = 1.0 / blah;`. — Pete Becker, Sep 16 '17 at 13:24
Yeah, good idea. This particular bit of code has loads of them so I'm just trying to make it look a little cleaner. — Dean Turbeaux, Sep 16 '17 at 17:39

Potatoswatter · Answer 1 · 2017-09-16T09:33:29.460

4

Using a scalar type like a constructor, and the equivalent "C-style" cast syntax, is shorthand for static_cast, const_cast, and/or reinterpret_cast, whichever is needed. Since it's imprecise, it can mask an error.

In this case, and for all conversions between built-in numeric types, the C-style cast is the same as static_cast. Still, many C++ users (including me) consider it a bad habit even so.

edited Sep 16 '17 at 09:33

answered Sep 16 '17 at 09:30

Potatoswatter

134,909
25
265
421

3

See http://en.cppreference.com/w/cpp/language/explicit_cast for exact order of casts attempted. C-style casts use the first one of these that is successful. – patatahooligan Sep 16 '17 at 09:32

How does static casting from long to double in c++ differ from using the double constructor?

1 Answers1