Seg故障后将项目压入STL容器 [英] Seg fault after is item pushed onto STL container

问题描述

  typedef struct temp 
 {
 int a，b; 
 char * c; 
 temp（）{c =（char *）malloc（10）;}; 
〜temp（）{free（c）;}; 
} temp; 
 
 int main（）
 {
 temp a; 
 list< temp> l1; 
 l1.push_back（a）; 
 l1.clear（）; 
 return 0; 
 
} 
  

解决方案

您没有复制构造函数。

你将'a'推入列表，它被复制。
因为你没有一个拷贝构造函数（为c分配内存并从旧c分配到新c）c是a中的同一个指针，而且是列表中的a的副本。

两个a的析构函数被调用，第一个将成功，第二个将失败，因为内存c指向已经被释放。

要查看发生了什么，在构造函数和析构函数中加入一些cout，然后逐步执行代码。

typedef struct temp  
{  
        int a,b;  
        char  *c;  
        temp(){	c = (char*)malloc(10);};  
        ~temp(){free(c);};  
}temp;  

int main()  
{  
   temp a;  
   list<temp>   l1;  
   l1.push_back(a);  
   l1.clear();  
   return 0;  

}

giving segmentation fault.

解决方案

You don't have a copy constructor.

When you push 'a' into the list, it gets copied. Because you don't have a copy constructor (to allocate memory for c and copy from old c to new c) c is the same pointer in a and the copy of a in the list.

The destructor for both a's gets called, the first will succeed, the second will fail because the memory c points to has already been freed.

You need a copy constructor.

To see whats happening, put some couts in the constructors and destructors and step through the code.

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