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Forcing single-argument constructors to be explicit in C++?

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c++

By default, in C++, a single-argument constructor can be used as an implicit conversion operator. This can be suppressed by marking the constructor as explicit.

I'd prefer to make "explicit" be the default, so that the compiler cannot silently use these constructors for conversion.

Is there a way to do this in standard C++? Failing that, is there a pragma (or similar) that'll work in Microsoft C++ to do this? What about g++ (we don't use it, but it might be useful information)?

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Roger Lipscombe Avatar asked Oct 06 '08 13:10

Roger Lipscombe


4 Answers

Nope, you have to do it all by hand. It's a pain, but you certainly should get in the habit of making single argument constructors explicit. I can't imagine the pain you would have if you did find a solution and then had to port the code to another platform. You should usually shy away from compiler extensions like this because it will make the code less portable.

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Matt Price Avatar answered Oct 22 '22 16:10

Matt Price


If there was a pragma or command line option that made constructors explicit by default, how would you declare one that is not explicit? There would have to be another compiler-specific token or pragma to make it possible to declare an implicit conversion constructor.

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bk1e Avatar answered Oct 22 '22 16:10

bk1e


It could be rather nasty for any header you have. Like <vector>, or any of the Boost headers. It would also cause quite a few false bugreports. So, no, I don't expect a compiler to add such a #pragma.

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MSalters Avatar answered Oct 22 '22 17:10

MSalters


There is no such option in the compilers, as far as I am aware. But there is a Lint warning for such cases (see http://www.gimpel.com/lintinfo.htm).

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MP24 Avatar answered Oct 22 '22 15:10

MP24