算术运算后删除指针 [英] Delete a pointer after the arithmetics

问题描述

int main() {
  int* i = new int(1);
  i++;
  *i=1;
  delete i;
}

这是我的逻辑：

I将I递增1，然后为其分配一个值。然后删除I，以便在泄漏原始内存的同时释放内存位置。我的问题在哪里？

I increment I by 1, and then assign a value to it. Then I delete the I, so I free the memory location while leaking the original memory. Where is my problem?

我还尝试了其他版本。每次，只要我做算术并删除指针，我的程序就会崩溃。

I also tried different versions. Every time, as long as I do the arithmetics and delete the pointer, my program crashes.

推荐答案

您的程序显示的内容有几个行为未定义的情况：

What your program shows is several cases of undefined behaviour:

1. 您写入未分配的内存（ * i = 1 ）


2. 释放未分配的内容，实际上是删除i + 1 。

1. You write to memory that hasn't been allocated (*i = 1)
2. You free something that you didn't allocate, effectively delete i + 1.

您必须调用 delete ，其指针值与从<$返回的指针值完全相同c $ c> new -没有别的。假设其余代码有效，则在 int * i = new int（1）之后执行 int * j = i; ; ，然后删除j; 。 [例如， int * i = new int [2]; 将使您的 i ++; * i = 1; 有效代码]

You MUST call delete on exactly the same pointer-value that you got back from new - nothing else. Assuming the rest of your code was valid, it would be fine to do int *j = i; after int *i = new int(1);, and then delete j;. [For example int *i = new int[2]; would then make your i++; *i=1; valid code]

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