c++ - const 引用默认值

Question

这个问题在这里已经有了答案:

关闭 9年前。

Possible Duplicate:
how to initialize function arguments that are classes with default value

#include <string>

void foo1(const std::string& s = std::string());

void foo2(std::string& s = std::string());

void foo3(const std::string s = std::string());

void foo4(std::string s = std::string());

error at foo2(): default argument for ‘std::string& s’ has type ‘std::string {aka std::basic_string<char>}’

我理解编译器的意思，但我不明白这不适用于 foo1()也是。

Answer 1

您不能像 foo2 那样对临时对象进行非常量引用。

请注意，这不是专门的默认参数。对于函数变量，您会得到相同的错误:http://ideone.com/g7Tf7L

#include <string>
using std::string;

#include <iostream>
using std::cout; using std::endl;

int main()
{
    string s1        = string("s1"); // OK, copy it
    const string& s2 = string("s2"); // OK, const reference to it
    string& s3       = string("s3"); // ERROR! non-const reference not allowed!

    cout
            << s1 << ", "
            << s2 << ", "
            << s3 << endl;
    return 0;
}

当您对临时对象进行 const 引用时，临时对象的生命周期会延长到引用的生命周期(第 12.2 节，引 self 的 C++11 草案 n3337 拷贝):

There are two contexts in which temporaries are destroyed at a different point than the end of the fullexpression.

...

The second context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference except:

• A temporary bound to a reference member in a constructor’s ctor-initializer (12.6.2) persists until the constructor exits.
• A temporary bound to a reference parameter in a function call (5.2.2) persists until the completion of the full-expression containing the call.
• The lifetime of a temporary bound to the returned value in a function return statement (6.6.3) is not extended; the temporary is destroyed at the end of the full-expression in the return statement.
• A temporary bound to a reference in a new-initializer (5.3.4) persists until the completion of the full-expression containing the new-initializer.