绑定到C ++模板中非常量引用的常量参数 [英] Const arguments binding to non-const references in C++ templates

问题描述

请考虑以下内容：

template <typename T>
void f(T& x) 
{
    ....       
}

为什么像 const int 这样的东西绑定到 f（T&）？

Why does something like const int binds to f(T&)?

在我看来，这违反了 const-correctness em>。实际上，如果 f（）采用 non -const T& 引用，则 f（）很可能会修改其参数（否则， f（）会被定义为 void f（const T&））。

This seems to me kind of a violation of const-correctness. In fact, if f() takes a non-const T& reference, then it's very likely that f() will modify its argument (else, f() would have been defined as void f(const T&)).

在这样的代码中：

template <typename T>
inline void f(T& x) 
{
    x = 0;
}

int main() 
{
    int n = 2;
    f(n);

    const int cn = 10;
    f(cn);
}

编译器尝试调用 f（）与 T = const int ，那么由于 x = 0; 在 f（）体内的赋值。

这是来自GCC的错误消息：

the compiler tries to call f() with T = const int, then of course there is an error message because of the x = 0; assignment inside f()'s body.
This is the error message from GCC:

test.cpp: In instantiation of 'void f(T&) [with T = const int]':
test.cpp:13:9:   required from here
test.cpp:4:7: error: assignment of read-only reference 'x'
     x = 0;
       ^

但是为什么编译器尝试绑定 const 参数与带有 non -const参数的函数模板的函数？

But why does the compiler try to bind a const argument with a function template which takes a non-const parameter?

合理价格是多少放在此C ++模板规则后面？

What's the rationale behind this C++ template rule?

推荐答案

您可以使用 std :: enable_if 加上例如 std :: is_const 可以避免 T 绑定到 const

You can use std::enable_if plus e.g. std::is_const to avoid that T binds to a const type.

Re…

Re …

“此C ++模板规则的基本原理是什么？

“What's the rationale behind this C++ template rule?”

它可能在Bjarne的设计中可以找到- -进化书，但最常见的理由是选择规则是为了简单和统一，因此似乎也出现在这里：以特殊方式处理某些类型会引入不必要的复杂性。

it can possibly be found in Bjarne's design-and-evolution book, but about the most common rationale is that the rules have been chosen for simplicity and uniformity, and so it appears to be also here: treating some types in special ways would introduce needless complexity.

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