英文:
Time and Space complexity of String Concatenation in C++
问题
x = x + x
x += x
x.append(x)
上面的三种方法是否相同（考虑时间和空间复杂性）？
我认为使用 "+" 会创建一个新字符串，而 "append" 只是将内容添加到现有字符串，因此 "append" 更快且占用更少的空间。
这是一个更偏向理论和知识的问题...
英文:
x = x+x
x+=x
x.append(x)
Are the three ways above the same (thinking of time and space complexity)
I’m thinking using “+” creates a new string while append just adds to the existing string, so append is faster and less space usage.
This is a questions that is more theory/knowledge based...
答案1
得分: 2
你正在调用三个（实际上是四个）不同的函数：operator=(s)、operator+(s)、operator+=(s) 和 append(s)，时间和空间要求（以及相关问题）将取决于这些函数的确切代码。对于你的问题，没有通用的答案。即使对于像 std::string 这样的给定类，它也取决于具体的实现。
请注意，在第一种情况中（x=x+x），你执行了两个操作（也就是调用了两个函数）。还要注意，operator=(s)、operator+=(s) 和 append(s) 直接作用于主对象的引用（我指的是 this，而不是额外的参数）。operator+=(s) 和 append(s) 的参考手册中说明：“复杂性：没有标准的复杂性保证，典型的实现行为类似于 std::vector::insert。”
如果“追加”()之所以更快，是因为它“只是添加到现有字符串”，那么如果“现有字符串”有保留的空间（这不是情况）可能是正确的。
英文:
You're calling three (actually four) different functions: operator=(s), operator+(s), operator+=(s) and append(s), and the time and space requirements (and concerns) will depend on the exact code of these functions. There's no generic answer to your question. Even for a given class, like std::string, it would be implementation dependant.
Please note that in the first case (x=x+x), you're doing two operations (that is, calling two functions). And also note that operator=(s), operator+=(s) and append(s) act directly on a reference to the main object (I'm talking about the this, and not about the additional parameter). The reference manual for operator+=(s) and append(s) states "Complexity: There are no standard complexity guarantees, typical implementations behave similar to std::vector::insert."
The idea of append() being faster because it "just adds to the existing string" could be right if the "existing string" had reserved space (which is not the case).
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