c++ - 函数式转换与构造函数

Question

我在 C++ 中学到了这一点，

typedef foo* mytype;

(mytype) a        // C-style cast

和

mytype(a)         // function-style cast

做同样的事情。

但我注意到函数式转换与构造函数共享相同的语法。 是否存在模棱两可的情况，我们不知道它是强制转换还是构造函数？

char s [] = "Hello";
std::string s2 = std::string(s);     // here it's a constructor but why wouldn't it be ...
std::string s3 = (std::string) s;    // ... interpreted as a function-style cast?

Answer 1

在句法上，它始终是一个转换。该转换可能恰好调用了构造函数:

char s [] = "Hello";
// Function-style cast; internally calls std::basic_string<char>::basic_string(char const*, Allocator)
std::string s2 = std::string(s);
// C-style cast; internally calls std::basic_string<char>::basic_string(char const*, Allocator)
std::string s3 = (std::string) s;