Is there a way to prevent implicit static_casts when calling a function?

Question

C++11 adds this.

int x;
unsigned int y{ x }; // ERROR

Is it possible to enable something like this.

int x;
void f(unsigned int y);
f(x); //ERROR

Compiler: VC++ 2013

@pranitkothari, I want the compiler to give an error or a warning. — NFRCR, Mar 22 '14 at 15:07
Do not try to change compiler's behavior (because it may be not easily possible). You need to take care of your code, yourself. — Pranit Kothari, Mar 22 '14 at 15:08
Please add which compiler you are using so someone may help. — Pranit Kothari, Mar 22 '14 at 15:11
Your question text currently does not reflect your title question... apparently you only want to forbid narrowing conversions. — Johannes Schaub - litb, Mar 22 '14 at 15:17
In `gcc` you can use `-Werror=conversion`. This will give you an error on conversion during compilation. Not sure about VC++... — lapk, Mar 22 '14 at 15:24
@LightnessRacesinOrbit so you should be able to say what he is asking. — Johannes Schaub - litb, Mar 23 '14 at 15:57
@JohannesSchaub-litb: He wishes to set his compiler such that implicit conversions for function arguments of built-in integral types is disabled. — Lightness Races in Orbit, Mar 23 '14 at 15:58
@LightnessRacesinOrbit Does he want `unsigned short x; f(x);` be an error or not? The corresponding snippet analogous to the first case, `unsigned short x; unsigned int y{ x };` is well-formed. I remarked this yesterday and the question was not yet clarified. — Johannes Schaub - litb, Mar 23 '14 at 15:59

iavr · Answer 1 · 2014-03-22T15:28:36.593

0

Try this (live example):

template <
    typename T,
    typename = typename std::enable_if <std::is_same <T, unsigned int>{}>::type
>
void f(T x) { }

edited Mar 22 '14 at 15:28

answered Mar 22 '14 at 15:14

iavr

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I didn't downvote, but I think OP's author wants more of a compiler switch to forbid implicit casts. – lapk Mar 22 '14 at 15:17
I retracted my downvote because your reply is otherwise correct. Whether it is an answer can't yet be said. – Johannes Schaub - litb Mar 22 '14 at 15:36

score 0 · Answer 2 · edited May 23 '17 at 11:57

No, there's no compiler switch or other general setting to do that. Implicit conversions are part of the language, and cannot be disabled in the general case for built-in types. The closest you can get is a user-defined wrapper class with only explicit constructors, or applying some template meta-hackery to the function you're trying to call (as in iavr's answer).

The premise of your question appears to conflate implicit conversions with the special case of narrowing conversions. The following:

int x = 0;
unsigned int y{ x };

is not an error. You may get a warning about the potential narrowing conversion, and this warning may be turned into an error (with GCC's -Werror, for example), but this is not new in C++11 and it does not inherently prohibit the conversion.

The program is only ill-formed if you actually cause a narrowing conversion. Again, this is specific to the value involved and is not a general rule about implicit conversions.

Is there a way to prevent implicit static_casts when calling a function?

2 Answers2