std::thread 对象如何复制参数？

Question

问题已更新！

我有一个 std::string 对象，std::thread 应该将它复制到它的内部存储中，为什么下面的代码仍然不起作用？（已修复）

如果我分离 t(f, 5, std::string(buf))，则根本没有输出！ （已修复）

#include <thread>
#include <iostream>
#include <unistd.h>

using namespace std;

void f(int i, const std::string &str) {
  printf("i = %d, str = %s\n", i, str.c_str());
}

void oops(int ival) {
  char *buf = new char[32]{0};
  snprintf(buf, 32, "%i", ival);
  // maybe danling pointer, if it's a danling pointer, the content of buf is undefined
  // in this program it happens to be 1999.
  std::thread t(f, 3, buf); 
  t.detach();

  delete [] buf; 
}

void not_oops(int ival) {
  char *buf = new char[32]{0};
  snprintf(buf, 32, "%i", ival);
  std::thread t(f, 5, std::string(buf));
  t.detach();

  delete [] buf;
}

int main() {
  oops(1992); // This is buggy
  not_oops(1999);
}

预期输出：

i = 3, str = 1992
i = 5, str = 1999

实际输出：

i = 3, str = 1999
i = 5, str = 1999

Answer 1

你的程序在它完成之前完成，你的函数在线程完成之前退出作用域，杀死它。

void oops(int ival) {
  char *buf = new char[32]{0};
  snprintf(buf, 32, "%i", ival);
  std::thread t(f, 3, buf); // maybe danling pointer
  t.detach();     // no more waiting for it

  delete [] buf;  // thread is not started yet (probably)    
} // thread is destroyed ,  and std::string dies.

在不对代码进行大量更改的情况下使其工作的唯一方法是允许在外部管理线程。此外，您必须确保创建了一个新的 std::string，或者线程似乎对两者使用相同的存储。

void f(int i, const std::string &str) {
  printf("i = %d, str = %s\n", i, str.c_str());
}

std::thread *t;
void oops(int ival) {
  char *buf = new char[32]{0};
  snprintf(buf, 32, "%i", ival);
  t = new std::thread(f, 3, std::string(buf)); // maybe danling pointer
  //t.detach();

  delete [] buf; 
}

void not_oops(int ival) {
  char *buf = new char[32]{0};
  snprintf(buf, 32, "%i", ival);
  std::thread t(f, 5, std::string(buf)); // FIXME: why it won't work ?
  t.join();
  delete [] buf;
}

int main() {
  oops(1992); // This is buggy
  not_oops(1999);
  t->join();
  delete t;
}