C++ std::transform() 和 toupper() ..为什么会失败?

2022-01-07 00:00:00 string transform c++ stl

我有 2 个 std::string.我只想，给定输入字符串:

I have 2 std::string. I just want to, given the input string:

  1. 大写每个字母
  2. 将大写字母分配给输出字符串.

这是如何工作的:

  std::string s="hello";
  std::string out;
  std::transform(s.begin(), s.end(), std::back_inserter(out), std::toupper);

但这不会(导致程序崩溃)?

but this doesn't (results in a program crash)?

  std::string s="hello";
  std::string out;
  std::transform(s.begin(), s.end(), out.begin(), std::toupper);

因为这有效(至少在同一个字符串上:

because this works (at least on the same string:

  std::string s="hello";
  std::string out;
  std::transform(s.begin(), s.end(), s.begin(), std::toupper);

推荐答案

out 中没有空格.C++ 算法不会自动增长其目标容器.您必须自己腾出空间，或者使用插入式适配器.

There is no space in out. C++ algorithms do not grow their target containers automatically. You must either make the space yourself, or use a inserter adaptor.

要在 out 中腾出空间，请执行以下操作:

To make space in out, do this:

out.resize(s.length());

[edit] 另一种选择是使用此构造函数创建具有正确大小的输出字符串.

[edit] Another option is to create the output string with correct size with this constructor.

std::string out(s.length(), 'X');

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