明确布尔操作符 - 不能返回，测试，初始化布尔

Question

我刚刚尝试使用explicit operator bool()第一次，它的行为对我来说是非常意外的。有人可以请说明为什么以下部分标有// does not work。

该convertible类将例如，是一个智能指针类，能够检查包含数据的有效性。

struct convertible 
{ 
    explicit operator bool() const 
    { 
    return ptr; 
    } 

    void* ptr = nullptr; 
}; 

bool testReturn() 
{ 
    convertible test; 
    // does not work 
    return test; 
} 

bool testReturn2() 
{ 
    convertible test; 
    // does not work 
    return test == true; 
} 

bool testReturn3() 
{ 
    convertible test; 
    // works ?! 
    return !test; 
} 

int main() 
{ 
    convertible test; 
    // works 
    if (test) { } 
    // works 
    bool init(test); 
    bool tested = test; 
    bool tested2 = testReturn(); 
    bool tested3 = testReturn2(); 
    bool tested4 = testReturn3(); 
    return 0; 
} 

GCC的，我得到：

milian@minime:/tmp$ g++ --version 
g++ (GCC) 4.8.2 20131219 (prerelease) 
Copyright (C) 2013 Free Software Foundation, Inc. 
This is free software; see the source for copying conditions. There is NO 
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. 

milian@minime:/tmp$ g++ -std=c++11 test.cpp 
test.cpp: In function ‘bool testReturn()’: 
test.cpp:15:10: error: cannot convert ‘convertible’ to ‘bool’ in return 
    return test; 
     ^
test.cpp: In function ‘bool testReturn2()’: 
test.cpp:22:15: error: no match for ‘operator==’ (operand types are ‘convertible’ and ‘bool’) 
    return test == true; 
      ^
test.cpp:22:15: note: candidate is: 
test.cpp:22:15: note: operator==(int, int) <built-in> 
test.cpp:22:15: note: no known conversion for argument 1 from ‘convertible’ to ‘int’ 
test.cpp: In function ‘int main()’: 
test.cpp:39:17: error: cannot convert ‘convertible’ to ‘bool’ in initialization 
    bool tested = test; 

铿锵它类似于：

milian@minime:/tmp$ clang++ --version 
clang version 3.4 (tags/RELEASE_34/final) 
Target: x86_64-unknown-linux-gnu 
Thread model: posix 
milian@minime:/tmp$ clang++ -std=c++11 test.cpp 
test.cpp:15:10: error: no viable conversion from 'convertible' to 'bool' 
    return test; 
     ^~~~ 
test.cpp:22:15: error: invalid operands to binary expression ('convertible' and 'int') 
    return test == true; 
     ~~~~^~~~~ 
test.cpp:39:8: error: no viable conversion from 'convertible' to 'bool' 
    bool tested = test; 
    ^  ~~~~ 
3 errors generated. 

Answer 1

如果你的目标是禁用某些转换，只需通过delete符禁用它们：

struct convertible 
{ 
    operator bool() const //(Implicit) Conversion to bool allowed 
    { 
     return true; 
    } 

    operator int() const = delete; //Implicit/Explicit conversions to int disallowed 
            //(Results in compilation error). 
}; 


int main() 
{ 
    convertible c; 

    if(c) 
     ... 

    int a = c; //ERROR 
    int b = (int)c; //ERROR (The same for static_cast<int>(c)) 
} 

你也可以删除所有对象，除了其正在使用的转换操作符的模板版本明确重载：

struct foo 
{ 
    operator bool() const //Explicit overload for bool 
    { 
     return true; 
    } 

    template<typename T> 
    operator T() const = delete; //Everithing which is not a bool (Everithing which 
           //does not fit in the explicit overload) would 
           //resolve to this operator and will fail. 
}; 


int main() 
{ 
    foo f; 

    bool b = f; //OK 
    int i = f; //ERROR 
    char c = f; //ERROR 
    etc... 
} 

Answer 2

如果你的转换运算符是explicit，你需要转换它，像：

return static_cast<bool>(test);