我可以删除[]'一个指针指向一个分配的数组，但不是开始吗？

问题描述

我特别想知道下面的情况（我发现在一些代码中我必须使用）：

  c   $  这是标准的报价。
 
 
 
 §3.7.4.2-3  
 
 
 
 如果解除分配函数通过抛出异常终止，未定义。提供给解除分配函数的第一自变量的值可以是空指针值;如果是，并且如果解除分配函数是在标准库中提供的，则调用没有效果。否则，在标准库中向操作符 delete（void *）提供
的值应为以前调用操作符 new（std :: size_t）或运算符new（std :: size_t，const std :: nothrow_-t&）标准库中，并且提供给标准库中的操作符 delete []（void *）的值应为之前调用 operator new []（std :: size_t）或
  c $ c>在标准库中。 
 
I'm wondering specifically about the following situation (which I discovered in some code I have to work with):
SomeClass *ar = new SomeClass[2];
ar++;
delete[] ar;
This code seems to be working fine - i.e. not crashing (win32, built with VS2005).
Is this "legal"? It certainly doesn't feel right.
 解决方案 
No, it is undefined to pass any address to delete which was not returned by new.
Here is the quote from the Standard.
§ 3.7.4.2-3
If a deallocation function terminates by throwing an exception, the behavior is undefined. The value of the first argument supplied to a deallocation function may be a null pointer value; if so, and if the deallocation function is one supplied in the standard library, the call has no effect. Otherwise, the value suppliedto operator delete(void*) in the standard library shall be one of the values returned by a previous invocation of either operator new(std::size_t) or operator new(std::size_t, const std::nothrow_-t&) in the standard library, and the value supplied to operator delete[](void*) in the standard library shall be one of the values returned by a previous invocation of either operator new[](std::size_t) oroperator new[](std::size_t, const std::nothrow_t&) in the standard library.
                        
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