c++ - 将结构复制(使用赋值)到 union 内的结构导致段错误

Question

我写了下面的代码:

#include <iostream>
#include <string>
#include <cstring>

struct bar
{
  std::string s3;
  std::string s4;
}Bar;

union foo
{
  char * s1;
  char * s2;
  bar    b1;

  foo(){};
  ~foo(){};
}Foo;


int main ()
{
  foo f1;
  bar b2;

  std::string temp("s3");
  b2.s3 = temp;
  b2.s4 = temp;

  //f1.b1 = b2;                           //-- This Fails (Seg faults)

  /*
    #0  0x00002b9fede74d25 in std::string::_Rep::_M_dispose(std::allocator<char> const&) [clone .part.12] ()
        from /usr/local/lib64/libstdc++.so.6
    #1  0x00002b9fede75f09 in std::string::assign(std::string const&) () from /usr/local/lib64/libstdc++.so.6
    #2  0x0000000000400ed1 in bar::operator= (this=0x7fff3f20ece0) at un.cpp:5
    #3  0x0000000000400cdb in main () at un.cpp:31
  */

  memcpy( &f1.b1, &b2, sizeof(b2) );  //-- This Works 

  std::cout << f1.b1.s3 << " " << f1.b1.s4 << std::endl;
  return 0;
} 

你能解释一下为什么会出现段错误吗？我无法破译回溯中的数据所暗示的内容。

Answer 1

union foo 无法初始化 bar 对象(它如何知道要调用哪个成员的初始化程序？)因此无法初始化 std::string .如果你想使用foo里面的bar，那么你需要手动初始化它，像这样......

new (&f1.b1) bar; // Placement new
f1.b1 = b2;
// And later in code you'll have to manually destruct the bar, because
//   foo doesn't know to destruct the bar either...
f1.b1.~bar();

或者，您可以尝试自己将此功能添加到 union 的构造函数和析构函数中。

foo() : b1() {}
// Or you construct like this, which you might need to for a non-trivial union...
// foo() { new (&b1) bar; }  // Placement new.
~foo() { b1.~bar(); }

注意复制也需要特殊处理。