What is the meaning of int(expr) in C++?

Question

When looking over some code from a friend's project, I recently saw syntax that looked like this.

#include <iostream>

int main(){
    std::cout<< int(32.5/5)  << std::endl;
}

When you run the above code, you get 6, which is the expected value if the use of int functions like a cast.

However, I have never seen this syntax before and I could not find documentation for it on the web. I also did an experiment and noticed that this syntax is not valid in C.

Can someone explain the meaning of this syntax with documentation references?

Questions asking us to recommend or find a book, tool, software library, tutorial **or other off-site resource** are off-topic for Stack Overflow — nhgrif, Feb 14 '15 at 19:34
This is just another syntax for a cast. See http://stackoverflow.com/questions/32168/c-cast-syntax-styles — fvannee, Feb 14 '15 at 19:35
Is it actually a different syntax for cast or a constructor call? — Codor, Feb 14 '15 at 19:37
This question is asking for link only answers which also aren't acceptable to post. How can a question which can only be answered with unacceptable answers be on topic? — nhgrif, Feb 14 '15 at 19:37
@nhgrif, If you copy the proper documentation into an answer, it is no longer a link-only request. — merlin2011, Feb 14 '15 at 19:42
I'm very surprised to see a question like this from a 22.9k user. Surely you know how SO works, by now? — Lightness Races in Orbit, Feb 14 '15 at 20:02
@wilx: Such _what_? He's right. He's literally quoting from the site rules. — Lightness Races in Orbit, Feb 14 '15 at 20:02
@gnidmoo: Please do not write answers as comments. Write answers as _answers_. — Lightness Races in Orbit, Feb 14 '15 at 20:02
@Codor that's the advantage of this notation : same syntax for built-in types and object constructors. Extra handy in template code. — Quentin, Feb 14 '15 at 20:03

score 9 · Accepted Answer · answered Feb 14 '15 at 19:59

9

This is not a constructor call or a "function". There is no "int function".

This is functional cast notation; it's just a cast.

It's the same as (int)(32.5/5) (in this particular case).

And, no, C does not have it.

answered Feb 14 '15 at 19:59

Lightness Races in Orbit

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I think the fact that you could write a reasonable answer to this question implies that it is not exactly a terrible fit for our format. I agree that I could have phrased it better but you understood the intent fairly clearly. – merlin2011 Feb 14 '15 at 20:12
@merlin2011: Actually I was a bit naughty, because I did not answer your question (which _is_ a bad fit) but some imaginary modified version of your question! But, yes, I gave you the benefit of the doubt. Now perhaps you can edit the question to make it fit the site's model, retaining your intent..? – Lightness Races in Orbit Feb 14 '15 at 22:03
I have edited the question. Let me know if the edit is up to scratch. :) – merlin2011 Feb 14 '15 at 22:10
Adding to this a bit, the same syntax can involve a constructor call depending on the type. For example if `T` is a template parameter then `T(32.5/5);` is valid, and instantiating with a class type will call a constructor, and instantiating with `int` won't. – M.M Feb 14 '15 at 22:43
@MattMcNabb: Right, but that's a factor of how explicit conversions work, not of what this syntax specifically does. If you catch my drift. – Lightness Races in Orbit Feb 14 '15 at 23:35

What is the meaning of int(expr) in C++?

1 Answers1