String.length woes [英] String.length woes

问题描述

编辑：解决方案必须编译Microsoft Visual Studio 2012.

Solutions must compile against Microsoft Visual Studio 2012.

我想使用已知的字符串长度来声明另一个相同长度的字符串。

I want to use a known string length to declare another string of the same length.

推理是第二个字符串将作为对第一个字符串进行操作的容器，对于第一个字符串必须是非易失性的。

The reasoning is the second string will act as a container for operation done to the first string which must be non volatile with regards to it.

例

const string messy "a bunch of letters";

string dostuff(string sentence) {
    string organised NNN?????  // Idk, just needs the same size.
    for ( x = 0; x < NNN?; x++) {
        organised[x] = sentence[x]++; // Doesn't matter what this does.
    }
}

在上述两种情况下，声明和退出条件，NNN？

In both cases above, the declaration and the exit condition, the NNN? stands for the length of 'messy'.

如何在编译时发现长度？

How do I discover the length at compile time?

推荐答案

std :: string 有两个构造函数。

第一个是复制构造函数：

The first, a copy constructor:

string organised(sentence);

第二个是一个构造函数，它接受一个字符和一个计数。您可以使用临时字符初始化字符串。

The second, a constructor which takes a character and a count. You could initialize a string with a temporary character.

string organised(sentence.length(), '_'); 

或者，您可以：

• 使用空字符串，并在其后添加（ + = ）文本，或


• 使用 std :: stringstream 同样的目的。

• Use an empty string and append (+=) text to it as you go along, or
• Use a std::stringstream for the same purpose.

。

>总的来说，如果长度已知，我宁愿选择复制构造函数。

Overall, I would prefer the copy constructor if the length is known.

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