Unusual static_cast syntax

Question

I managed to track a bug down to the following expression:

foo(static_cast<T>(a, b)); // Executes specialisation 1

The closing bracket was in the wrong place. The correct statement should have been:

foo(static_cast<T>(a), b); // Executes specialisation 2

I've never seen static_cast used with the form (a,b), or seen it described anywhere. What does it mean? The former statement returned b.

It's just a bug, not a special case of a static_cast. It's the comma operator, doing its thing. — Joe, May 29 '15 at 14:08

score 15 · Accepted Answer · edited May 23 '17 at 11:43

15

static_cast is not a function, it's a keyword, so the comma in a, b is not an argument separator; it is the comma operator. It evaluates a but throws away the result. The expression evaluates to b.

edited May 23 '17 at 11:43

Community

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answered May 29 '15 at 14:08

John Kugelman

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Baum mit Augen · Answer 2 · 2015-05-29T14:09:35.010

8

This has nothing to do with static_cast, but "makes use" of the comma operator. Its result is its right hand side, so

foo(static_cast<T>(a, b));

is equivalent to

foo(static_cast<T>(b));

unless a has other effects (which would then be executed and have their result discarded). With the right compiler settings, you will be warned about such things: Live

edited May 29 '15 at 14:09

answered May 29 '15 at 14:07

Baum mit Augen

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Why is ""makes use"" in quotation marks like you're being ironic or hiding a secret meaning? – Kerrek SB May 29 '15 at 14:11
@KerrekSB "Makes use" in the sense "No one actually intended to use this", i.e. no one *actually* makes use of it, they just happened to invoke it on accident. – Baum mit Augen May 29 '15 at 14:12

score 0 · Answer 3 · edited May 29 '15 at 15:42

0

, is comma operator, so a, b is just b

edited May 29 '15 at 15:42

John Kugelman

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answered May 29 '15 at 14:09

Xiaotian Pei

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Unusual static_cast syntax

3 Answers3