如果新大小更小，realloc可以移动指针吗？
我想知道 C 或 C++ 标准是否保证当以较小（非零）大小调用 realloc 时指针不会更改：
size_t n=1000;
T*ptr=(T*)malloc(n*sizeof(T));
//<--do something useful (that won't touch/reallocate ptr of course)
size_t n2=100;//or any value in [1,n-1]
T*ptr2=(T*)realloc(ptr,n2*sizeof(T));
//<-- are we guaranteed that ptr2==ptr ?
基本上，操作系统是否可以自行决定，既然我们释放了一个大内存块，他想要占用利用所有重新分配来对内存进行碎片整理，并以某种方式移动 ptr2 ？
I am wondering whether the C or C++ standard guarantees that a pointer is not changed when realloc is called with a smaller (nonzero) size:
size_t n=1000;
T*ptr=(T*)malloc(n*sizeof(T));
//<--do something useful (that won't touch/reallocate ptr of course)
size_t n2=100;//or any value in [1,n-1]
T*ptr2=(T*)realloc(ptr,n2*sizeof(T));
//<-- are we guaranteed that ptr2==ptr ?
Basically, can the OS decide on its own that since we freed a large memory block, he wants to take advantage of all reallocs to defragment the memory, and somehow move ptr2 ?
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http://opengroup.org/onlinepubs/007908775/xsh/realloc.html
不，没有保证
http://opengroup.org/onlinepubs/007908775/xsh/realloc.html
Nope, no guarantee
无法保证
realloc会返回相同的位置、周期。There's no guarantee
reallocwill return the same location, period.使用
realloc，您绝对无法保证内存之后的位置。我相信 libc 的默认 malloc 只会勉强复制内存，所以实际上来说你可能没问题。但不要指望它。With
realloc, you get absolutely no guarantees about where the memory will live afterwords. I believe that libc's default malloc will only begrudgingly copy memory around, so practically speaking you may be OK. But don't count on it.realloc不需要将块保留在适当的位置，即使它适合，事实上，最简单的存根实现是一个可能不需要的示例：malloc：调用sbrk。realloc：调用malloc和memcpy。免费：无操作。这听起来可能很荒谬，但有时对于嵌入式系统来说，像我刚刚描述的那样的实现实际上是最佳的实现。
reallocis not required to leave the block in place even if it would fit, and in fact the simplest stub implementation is an example where it might not:malloc: callsbrk.realloc: callmallocandmemcpy.free: no-op.This may sound ridiculous, but sometimes for embedded systems an implementation like I've just described is actually the optimal one.
在我看来，所有当前答案（在回答这个问题时）都没有引用任何标准文档。
对于 C++，我将参考 工作草案，编程语言 C++ 标准，文档编号：N3337，日期：2012-01-16，修订：N3291，根据 https://isocpp.org/std/the-standard，是最接近非自由官方C++11标准文档的免费文档；我们在 20.6.13 C 库 中找到：
所以现在我们必须参考C标准。
根据https://stackoverflow.com/a/83763/15485最接近非自由官方C11的免费文档标准文档是编程语言 - C，N1570 委员会草案 — 2011 年 4 月 12 日 ISO/IEC 9899:201x；在这里，我们在 7.22.3.5 处找到了 realloc 函数：
我的母语不是英语，因此由您来解释“可能有”的含义。
It seems to me that all the current answers (at the time of this answer) do not refer to any standard document.
For C++ I will refer to Working Draft, Standard for Programming Language C++, Document Number: N3337, Date: 2012-01-16, Revises: N3291 that, according to https://isocpp.org/std/the-standard, is the closest free document to the non-free official C++11 standard document; here we find at 20.6.13 C library:
So now we have to refer to the C standard.
According to https://stackoverflow.com/a/83763/15485 the closest free document to the non-free official C11 standard document is Programming languages — C, N1570 Committee Draft — April 12, 2011 ISO/IEC 9899:201x; here we find at 7.22.3.5 The realloc function:
I am not a native English speaker and so is up to you to interpret the meaning of "may have".
在 Windows 上，C 运行时获取堆，然后从该堆分配内存。因此操作系统不会知道各个内存分配，因此不会移动东西。
On Windows, the C-Runtime grabs a heap, and then allocates memory from that heap. So the OS won't know about individual memory allocations, and thus won't move things around.