C ++函数式转换的目的究竟是什么或是什么？ [英] What exactly is or was the purpose of C++ function-style casts?

问题描述

我在说type（value） - 风格的cast。我读过的书很快地传递过来，只是说它们在语义上等同于C风格的cast，（type）value，并且应该避免它们。如果他们的意思是一个老式的演员做同样的事情，为什么他们被添加到语言？另外，因为声明可以包含多余的括号，这段代码：T x（T（y））;不做有意使用函数式cast的人会期望;它声明一个名为x的函数接受T并返回一个T，而不是通过将y转换为T来构造一个名为x的T变量。

I am talking about "type(value)"-style casts. The books I have read pass over them quickly, saying only that they are semantically equivalent to C-style casts, "(type) value", and that they should be avoided. If they mean the same thing an old-style cast does, why were they ever added to the language? Also, because declarations can contain superfluous parentheses, this code: "T x(T(y));" doesn't do what someone intending to use the function-style casts would expect; it declares a function named x accepting a T and returning a T rather than constructing a T variable named x by casting y to a T.

如果他们在设计中有错误？

Were they a mistake in the design of the language?

推荐答案

函数式转换为原始类型和用户定义类型提供一致性。这在定义模板时非常有用。例如，以这个非常愚蠢的例子：

Function style casts bring consistency to primitive and user defined types. This is very useful when defining templates. For example, take this very silly example:

template<typename T, typename U>
T silly_cast(U const &u) {
  return T(u);
}

我 silly_cast 将适用于原始类型，因为它是一个函数式的转换。它也适用于用户定义的类型，只要T类有一个参数构造函数，它接受一个U或U const&。

My silly_cast function will work for primitive types, because it's a function-style cast. It will also work for user defined types, so long as class T has a single argument constructor that takes a U or U const &.

template<typename T, typename U>
T silly_cast(U const &u) {
    return T(u);
}

class Foo {};
class Bar {
public:
    Bar(Foo const&) {};
};

int main() {
    long lg = 1L;
    Foo f;
    int v = silly_cast<int>(lg);
    Bar b = silly_cast<Bar>(f);
}

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