为什么在直接分配而不使用条件运算符分配时，此函数指针分配起作用?

问题描述

(本示例中未使用#include，在带有g ++选项-O0 -g3 -Wall -c -fmessage-length = 0的MacOS10.14，Eclipse IDE上编译)

(No #include's were used for this example, compiled on MacOS10.14, Eclipse IDE, with g++, options -O0 -g3 -Wall -c -fmessage-length=0)

假设此变量声明:

int (*fun)(int);

无法通过"std :: toupper和std :: tolower的无效重载"进行编译.

This fails to compile with "invalid overload of std::toupper and std::tolower".

fun = (1 ? std::toupper : std::tolower);   // ERROR, invalid overload

这样编译就可以了:

if (1) {
    fun = std::toupper;   // OK
}
else {
    fun = std::tolower;   // OK
}

推荐答案

std::toupper( 1 和 2 )和std::tolower( 1 和 2 )已超载.在为条件运算符确定它们之间的通用类型时(在分配之前到chr2fun)，无法确定应使用的重载.

std::toupper(1 and 2) and std::tolower(1 and 2) are overloaded. When determining the common type between them for the conditional operator (before the assignment to chr2fun), which overloading should be used can't be determined.

您可以使用 static_cast 来指定应考虑的哪一个. (通常，首先要强制执行过载解析，确定常见类型消失了.)

You can use static_cast to specify which one should be considered. (Presicely, to force the overload resolution happens at first respectively, then the trouble in determining the common type disappears.)

例如

chr2fun = (str2modus == STR2UP ? static_cast<int(*)(int)>(std::toupper) 
                               : static_cast<int(*)(int)>(std::tolower));

对于第二种情况，直接分配chr2fun； chr2fun的类型是明确的，并且可以在过载解析.

For the 2nd case, chr2fun is assigned directly; the type of chr2fun is explicit and the correct overloading would be selected in overload resolution.

(重点是我的)

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