如下代码。
我想将向量中的元素向后移动。
例如:
[(1),2,3,4]-> [2,3,4,(1)]
但是,这会导致双重释放问题。
此代码中的逻辑很简单。
我想我滥用了擦除功能。
是真的吗谁能告诉我细节吗?
感谢您的阅读。
输出:*** Error in '/home/ubuntu/workspace/hello-cpp-world.cc.o': double free or corruption (out): 0x00000000016ffca0 ***
这是代码片段:
#include <iostream>
#include <vector>
int main() {
std::vector<int*> targets;
int* a = new int;
*a = 1;
int* b = new int;
*b = 2;
targets.push_back(a);
targets.push_back(b);
int i =0;
for (std::vector<int*>::iterator obj1 = targets.begin(); obj1 != targets.end(); i++)
{
if(i==0)
{
int* tmp = *obj1;
targets.push_back(tmp);
targets.erase(obj1);
}
else obj1++;
}
}
最佳答案
对push_back
调用erase
或std::vector
可能会使迭代器无效。使用索引更容易。
#include <iostream>
#include <vector>
int main() {
std::vector<int*> targets;
int* a = new int;
*a = 1;
int* b = new int;
*b = 2;
targets.push_back(a);
targets.push_back(b);
int i =0;
for(size_t obj1 = 0; obj1 < targets.size(); i++)
{
if(i==0)
{
int* tmp = targets[obj1];
targets.push_back(tmp);
targets.erase(targets.begin() + obj1);
}
else obj1++;
}
}
由于除了
obj1
时,除了递增以外不使用i==0
,因此您可以编写得更简单#include <iostream>
#include <vector>
int main() {
std::vector<int*> targets;
int* a = new int;
*a = 1;
int* b = new int;
*b = 2;
targets.push_back(a);
targets.push_back(b);
int* tmp = targets[0];
targets.push_back(tmp);
targets.erase(targets.begin());
}