更改基本类型和类类型的返回值

Question

有这样的代码:

#include <iostream>
#include <string>

int returnnumber() { return 2; }
std::string returntext() { return "siema"; }

int main() {

    std::cout << (returntext() += "cze") << std::endl; // siemacze
    //std::cout << (returnnumber() += 2) << std::endl; error: lvalue required as left operand of assignment

    return 0;
} 

为什么可以更改std :: string的返回值,而不是int？

Answer 1

因为std::string是具有已定义+=运算符作为成员函数的类类型.

并且标准允许您在rvalues上调用成员函数.

这是一个愚蠢的结果

struct S { int x; };
S foo() { return S(); }

int main()
{
    foo() = S();    // OK, uses member assignment operator.
    foo().x = 666;  // !Nah, can't assign to rvalue of built-in type.
}

编译结果:

Comeau C/C++ 4.3.10.1 (Oct  6 2008 11:28:09) for ONLINE_EVALUATION_BETA2
Copyright 1988-2008 Comeau Computing.  All rights reserved.
MODE:strict errors C++ C++0x_extensions

"ComeauTest.c", line 7: error: expression must be a modifiable lvalue
      foo().x = 666;  // !Nah, can't assign to rvalue of built-in type.
      ^

1 error detected in the compilation of "ComeauTest.c".

然而,编制者对于他们如何严格地应用这一微妙的规则,或者根本不同,有所不同(或曾经不同).

干杯和hth.,