Type deduction with auto at runtime-CodePudding

all. Very quick question. Why statement like below can be compiled with recent gcc or MSVC, but same thing with custom types is not possible?

const auto& a = isParamFP16 ? 0 : "aaa"; - possible const auto& a = isParamFP16 ? paramFP32 : paramFP16; - impossible (error C2446: ':': no co nversion from 'const T' to 'const T' for MSVC and operands to ?: have different types ‘const param<FP32>’ and ‘const param<FP16>’ for gcc)

I know that all types should be resolved at compile time, but why first variant works?

Thank you.

CodePudding user response：

With

const auto& a = isParamFP16 ? 0 : "aaa";

"aaa"'s type will decay to a const char *. 0 is an int, but it is also a valid value for a const char * since it is a null pointer constant. That means the compiler will deduce the entire conditional operation as yielding a const char * no matter which value is returned.

With

const auto& a = isParamFP16 ? paramFP32 : paramFP16;

You have a param<FP32> for the first result and a param<FP16> for the second. There is no way to convert one to the other, so a compiler error is issued as the conditional operator can only yield a single type.

CodePudding user response：

The first variant works, because an integer literal 0 is a null pointer constant, which means it can be implicitly converted to any pointer type (resulting in a null pointer value).

So, 0 can be converted to const char* and the string literal "aaa" decays to a pointer const char*. The types match, and so the type of a will be char const* const&.

Try it with any other value, e.g. 1 or some math e.g. 1-1, and it won't compile.

CodePudding user response：

The last two operands of the ? expression must have identical types (to an implicit conversion) so that the type of the whole expression can be deduced at compile-time. In the first case they do (pointer). Not in the second.