Question

因此,I ve realised that a += b; is actual a = a + b;。 (将<条码>a改为<条码>>,<条码>b。)

<代码>a = b + a;? I ve Trial a =+ b;, but that don t on work.

Answer 1

插头:铺设,operator+=,是操作者,处理背弃义的电动论点,从而改变论点,并退回参考。由于“密码”+ 用于示例的意思是“申请”,因此不可能有这样一位操作员参加另一轮工作。

This seems to be an annoying limitation at first, but actually there is a pretty good reason for that. std::string dynamically allocate more memory in the same way vector and other dynamic fields do. (-> constant amortized complexity). This operation always allocates memory at the end. So whenever you want to add something to the beginning of a string, you actually have to alter the argument that will be the new prefix or you have to construct an entirely new string.

迫使你改变预先确定由此造成的扼杀的物体。创造全新的目标(以记忆分配和完全复制这两种论点)必然会暗中发生。如果想像<条码>b = + b,则这意味着<条码>b>现在有新添的<>建筑>号指示。这是一项新的任务。 http://www.ohchr.org。

Answer 2

<代码>a = b + a系指处理编号时与a = a + b相同的东西。对于扼杀,没有类似的捷径。

Answer 3

我不认为缩短了工作方式:a = b + a(在处理插图时)。

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