deque推回错误

Question

我正在编写一个编译器并使用deque来存储类的方法标签,这里是示例代码:

#include <deque>
#include <iostream>
#include <string>

using std::cout;
using std::deque;
using std::endl;
using std::string;

int main()
{
  deque<const char *> names;

  string prefix = "___";
  const char *classname = "Point";

  const char *methodname[] = {"Init", "PrintBoth", "PrintSelf", "equals"};

  for (int i = 0; i < 4; i++)
    {
      string label = prefix + classname + "." + methodname[i];
      names.push_back(label.c_str());
    }

  for (int i = 0; i < 4; i++)
    cout << names[i] << endl;

  return 0;
}

但是,结果并不是我所期望的:

___Point
___Point.PrintSelf
___Point.PrintSelf
___Point.equals

另外,我注意到我只是推回了methodname

names.push_back(methodname[i])

我按顺序获得所有方法名称.

我在这做错了什么？

Answer 1

for (int i = 0; i < 4; i++)
{
  string label = prefix + classname + "." + methodname[i];
  names.push_back(label.c_str()); //what you're pushing? a temporary!

} //<--- `label` is destroyed here and it's memory is freed.

这label是一个变量,它在闭合括号中被破坏,并在每次迭代中再次创建.

这意味着,你所追求的names是一个暂时的价值.这导致了问题.

我建议你用这个:

std::deque<std::string> names;

然后这样做:

names.push_back(label); //a copy is pushed to the deque!