c++ - Different casting operators used by different compilers -

the following c++ program compiles without warnings in compilers have tried (gcc 4.6.3, llvm 3.0, icc 13.1.1, solarisstudio 12.1/12.3):

struct cclass {   template<class t>   operator t() const { return 1; }    operator int() const { return 2; } };  int main(void) {   cclass x;   return static_cast<char>(x); } 

however, solarisstudio compilers return 2, solarisstudio (either version) returns 1, consider logical result.

using return x.operator char(); results in compilers returning 1.

obviously, since figuring out, have been using latter notation. however, know of compilers correct , why. (one think majority rules, still doesn't explain why.)

this question seems related questions here, here, , here, these "only" give solutions problems, no explanations (that able apply particular problem anyway).

note adding additional overloaded casting operator, operator float() const { return 3; } results in compilers except solarisstudio complaining ambiguity.

the first (template) overload should picked.

paragraph 13.3.3/1 of c++11 standard specifies:

[...] viable function f1 defined better function viable function f2 if arguments i, icsi(f1) not worse conversion sequence icsi(f2), , then

— argument j, icsj(f1) better conversion sequence icsj(f2), or, if not that,

— the context initialization user-defined conversion (see 8.5, 13.3.1.5, , 13.3.1.6) and standard conversion sequence return type of f1 destination type (i.e., type of entity being initialized) is better conversion sequence standard conversion sequence return type of f2 destination type. [ example:

struct {     a();     operator int();     operator double(); } a; int = a; // a.operator int() followed no conversion            // better a.operator double() followed            // conversion int float x = a; // ambiguous: both possibilities require conversions,              // , neither better other 

—end example ] or, if not that,

— f1 non-template function , f2 function template specialization, or, if not that,

[...]

as can see the, fact first conversion operator template becomes relevant when standard conversion sequence return type (char, in case) destination type (char, in case) not better standard conversion sequence return type of non-template overload (int, in case) destination type (char, in case).

however, standard conversion char char exact match, while standard conversion int char not. therefore, third item of § 13.3.3/1 not apply, , second item does.

this means the first (template) overload should picked.