仅使用特定类型调用强制函数

Question

我正在考虑在C++ 11中将char*转换为bool时强制执行类型安全性,如果你这样做,则建议你这样做

template<typename T>
void foo(T) = delete;

void foo(int f) {}

foo只有在给出明确的int论证时,这才有效.我做了一个测试用例:

template<typename T>
void foo(T) = delete;

void foo(int f) {}

int main()
{
    foo(1);
    foo(3.0);
    foo(short(5));
    foo(float(7.0));
    foo(long(9));
}

我用coliru用g++ -std=c++14 -O2 -Wall -pedantic -pthread main.cpp && ./a.out(实例)编译代码,我得到了以下错误:

main.cpp: In function 'int main()':
main.cpp:9:12: error: use of deleted function 'void foo(T) [with T = double]'
     foo(3.0);
            ^
main.cpp:2:6: note: declared here
 void foo(T) = delete;
      ^
main.cpp:10:17: error: use of deleted function 'void foo(T) [with T = short int]'
     foo(short(5));
                 ^
main.cpp:2:6: note: declared here
 void foo(T) = delete;
      ^
main.cpp:11:19: error: use of deleted function 'void foo(T) [with T = float]'
     foo(float(7.0));
                   ^
main.cpp:2:6: note: declared here
 void foo(T) = delete;
      ^
main.cpp:12:16: error: use of deleted function 'void foo(T) [with T = long int]'
     foo(long(9));
                ^
main.cpp:2:6: note: declared here
 void foo(T) = delete;
      ^

^{用clang编译也产生了类似的错误}

现在,当我读到= delete关于cppreference它说

如果函数过载,则首先发生重载解析,如果选择了删除的函数,则程序只会格式错误.

因此,如果cppreference是正确的并且我的程序是格式错误的,这只是意味着它不会编译,还是未指定或未定义的行为？

Answer 1

你的程序格式不正确.首先,对于每次调用foo,我们执行重载决策.那会叫:

foo(1);            // foo(int )
foo(3.0);          // foo<T>, T=double
foo(short(5));     // foo<T>, T=short
foo(float(7.0));   // foo<T>, T=float
foo(long(9));      // foo<T>, T=long

其中四个函数是明确的deleted,来自[dcl.fct.def.delete]:

除了声明它之外,隐式或显式引用已删除函数的程序是不正确的.

这不是未定义或未指定的行为.它应该根本不编译.