在下面的代码片段中,我试图取消分配用于使用New创建Set的动态内存,但是在擦除SET的节点后,如果我尝试删除.
#include <iostream>
#include <set>
#include <string>
using namespace std;
int main()
{
std::set<char *> myset;
std::set<char *>::iterator it;
char *l_ptr1, *l_ptr2;
std::cout <<"Before Insertion::Address of myset " << &myset<<endl;
//l_ptr1 = (char*)malloc(256*sizeof(char*));
//l_ptr2 = (char*)malloc(256*sizeof(char*));
l_ptr1 = new char(256);
l_ptr2 = new char(256);
printf("Before Insertion ::Address of l_ptr1 %x\n", l_ptr1);
printf("Before Insertion ::Address of l_ptr2 %x\n", l_ptr2);
std::cin>>l_ptr1;
myset.insert(l_ptr1);
std::cin>>l_ptr2;
myset.insert(l_ptr2);
std::cout <<"After Insertion::Address of myset " << &myset<<endl;
std::cout << "Myset Contains :: Value and It's Address "<<endl;
it = myset.begin();
std::cout<<*it<<"\t"<<&(*it)<<endl; ++it;
std::cout<<*it<<"\t"<<&(*it)<<endl;
std::cout << endl;
it = myset.begin();
myset.erase(it++);
std::cout <<"After First Node Deletion ::Address of myset " << &myset<<endl;
std::cout <<"Before free(l_ptr1) :: l_ptr1:: " <<l_ptr1 <<endl;
//free(l_ptr1);
delete[] l_ptr1;
std::cout <<"After free(l_ptr1) :: l_ptr1:: " <<l_ptr1 <<endl;
myset.erase(it);
std::cout <<"After Second Node Deletion ::Address of myset " << &myset<<endl;
std::cout <<"Before free(l_ptr2) :: l_ptr2:: " <<l_ptr2 <<endl;
//free(l_ptr2);
delete[] l_ptr2;
std::cout <<"After free(l_ptr2) :: l_ptr2:: " <<l_ptr2 <<endl;
printf("After All Node Deletion ::Address of l_ptr1 %x\n", l_ptr1);
printf("After All Node Deletion ::Address of l_ptr2 %x\n", l_ptr2);
return 0;
}
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我收到了以下错误.
HEAP CORRUPTION DETECTED:在正常块(#150)之后的0x005B5380处.CRT检测到应用程序在堆缓冲区结束后写入内存.
首先让我引用一位程序员,我从中学到了很多东西:
您的C++代码包含一个错误:您没有使用std :: string.
也就是说,如果你坚持让生活变得更加困难并使用char数组,你应该检查你的代码是否真的符合你的要求:
l_ptr1 = new char(256);
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这一行分配一个char并分配数字256.
l_ptr1 = new char[256];
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这一行分配256个字符.