删除指向自动变量的指针
请看一下这段代码，
int i = 10; //line 1
int *p = &i; //line 2
delete p; //line 3
cout << "*p = " << *p << ", i = " << i << endl; //line 4
i = 20; //line 5
cout << "*p = " << *p << ", i = " << i << endl; //line 6
*p = 30; //line 7
cout << "*p = " << *p << ", i = " << i << endl; //line 8
这段代码的结果是什么？尤其是3、5、7号线？他们会调用未定义的行为吗？输出会是什么？
编辑：我尝试使用 g++ 运行它，它编译并运行良好！我在 Windows 7 上使用 MinGW。
标准在这种情况下说了什么？
Please look at this code
int i = 10; //line 1
int *p = &i; //line 2
delete p; //line 3
cout << "*p = " << *p << ", i = " << i << endl; //line 4
i = 20; //line 5
cout << "*p = " << *p << ", i = " << i << endl; //line 6
*p = 30; //line 7
cout << "*p = " << *p << ", i = " << i << endl; //line 8
What is the result of this code? Especially of line 3, 5 and 7? Do they invoke undefined behavior? What would be the output?
EDIT : I tried running it using g++, and it's compiling and running fine! I'm using MinGW on Windows 7.
What does Standard say in this context?
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如果您曾经使用 new 动态分配过指针，则只能删除该指针。在这种情况下，您没有使用 new 分配指针，而是简单地将其定义并初始化为指向 int 类型的局部变量。
在未使用 new 动态分配的指针上调用删除称为未定义行为。简而言之，这意味着当执行这样的代码时，地球上任何事情都可能发生，并且您不能向这个星球上的任何人抱怨一点。
You can delete only a pointer if you have ever allocated it dynamically using new. In this case you have not allocated the pointer using new but simply defined and initialized it to point to a local variable of type int.
Invoking delete on a pointer not allocated dynamically using new is something called Undefined Behavior. In short, it means that anything on the earth can happen when such a code is executed and you can't complaint a bit to anyone on this planet.
delete p;是 UB，因此无法预测或依赖任何进一步的行为。您的程序可能会立即崩溃，或者花掉所有的钱，或者只是退出main()并假装什么也没发生。delete p;is UB and so any further behavior can't be predicted or relied upon. You program might crash immediately or spend all your money or just exit frommain()and pretend nothing happened.第 3 行绝对是未定义的行为，因为您试图删除不在堆上的地址处的内存。
Line 3 is definitely undefined behaviour, since you're trying to deleting memory at an address that is not on the heap.