C++错误之一 ：invalid initialization of non-const reference of type ‘xxx&’ from an rvalue of type ‘xxx’
1、看代码 2、编译结果 3、分析和解决 就拿f(a + b)来说，a+b的值会存在一个临时变量中，当把这个临时变量传给f时，由于f的声明中，参数是int&，不是常量引用，因为c++编译器的一个关于语义的限制。如果一个参数是以非const引用传入，c++编译器就有理由认为程序员会在函数中修改这个值，并且这个被修改的引用在函数返回后要发挥作用。但如果你把一个临时变量当作非const引
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