C++ 指针“错误：双重释放或损坏（输出）”
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/9901339/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
C++ pointer "error: double free or corruption (out)"
提问by Kam
This is my code:
这是我的代码：
uint16_t * ptemparr = new uint16_t[20];
for (int x=0;x<2;x++)
{
function(ptemparr);
ptemparr ++;
}
delete[] ptemparr;
When I do that I get this error:
当我这样做时，我收到此错误：
double free or corruption (out)
EDITED: Thank you I understood why I get this error, now do you think this is a better idea?
编辑：谢谢我明白为什么我会收到这个错误，现在你认为这是一个更好的主意吗？
uint16_t temparr[20];
uint16_t * ptemparr = temparr;
for (int x=0;x<2;x++)
{
function(ptemparr);
ptemparr ++;
}
This way I make the pointer on the stack and there's no memory leak issue. Also, this code above has to run every 1 sec, so please bare this in mind before letting me know what's the best coding practice for this situation
通过这种方式，我将指针放在堆栈上，并且没有内存泄漏问题。此外，上面的代码必须每 1 秒运行一次，所以在让我知道这种情况的最佳编码实践是什么之前请记住这一点
回答by Alok Save
You need to pass the same address to delete []which was returned by new [].
Also, make sure function()does notdeallocate the memory by callingdelete`on the passed pointer.
您需要传递由delete []返回的相同地址new []。
另外，请确保function()不要deallocate the memory by calling删除传递的指针。
回答by tartar
you need to reset ptemparr to its home address since you incremented it in your for loop. so, I suggest decrementing it by 2 before deleting it.
您需要将 ptemparr 重置为其家庭地址，因为您在 for 循环中增加了它。所以，我建议在删除它之前将其减少 2。
ptemparr-=2;
ptemparr-=2;
回答by Jonathon Reinhart
You need to deletethe same address returned by new.
您需要delete返回相同的地址new。
I always make my loops line this work on a copy of the original pointer, and never modify the original pointer returned by mallocor new.
我总是让我的循环在原始指针的副本上工作，并且从不修改由mallocor返回的原始指针new。
