"half free" 一块内存来缩小它安全吗？
Is it safe to "half free" a chunk of memory to shrink it?
有人告诉我，即使它是安全的，也绝对是不好的做法。我的问题是，这样做真的安全吗，还是完全未定义的行为？假设我已经分配了一个 100 字节的内存块，但我想将它缩小到 50。将 50 添加到指针并释放它是否安全？这是代码示例：
char *ptr = malloc(100);
//...
free(ptr+50);
//Presumably, since I freed the second half only, would that be equivalent to ptr = realloc(ptr, 50);?
这是cross-platform/undefined行为吗？这种缩内存的方法安全吗？
您只能将从 malloc、realloc、calloc 返回的指针传递给 free。传递其他任何内容是 undefined behavior.
关于 free 函数的 C standard 第 7.22.3.3p2 节指出：
The free function causes the space pointed to by ptr to be
deallocated, that is, made available for further allocation. If
ptr is a null pointer, no action occurs. Otherwise, if the
argument does not match a pointer earlier returned by a
memory management function, or if the space has been
deallocated by a call to free or realloc, the behavior is
undefined.
这里正确的做法是分配一个新的内存块，使用 memcpy 复制字节，free 旧块。
有人告诉我，即使它是安全的，也绝对是不好的做法。我的问题是，这样做真的安全吗，还是完全未定义的行为？假设我已经分配了一个 100 字节的内存块，但我想将它缩小到 50。将 50 添加到指针并释放它是否安全？这是代码示例：
char *ptr = malloc(100);
//...
free(ptr+50);
//Presumably, since I freed the second half only, would that be equivalent to ptr = realloc(ptr, 50);?
这是cross-platform/undefined行为吗？这种缩内存的方法安全吗？
您只能将从 malloc、realloc、calloc 返回的指针传递给 free。传递其他任何内容是 undefined behavior.
关于 free 函数的 C standard 第 7.22.3.3p2 节指出：
The
freefunction causes the space pointed to byptrto be deallocated, that is, made available for further allocation. Ifptris a null pointer, no action occurs. Otherwise, if the argument does not match a pointer earlier returned by a memory management function, or if the space has been deallocated by a call tofreeorrealloc, the behavior is undefined.
这里正确的做法是分配一个新的内存块，使用 memcpy 复制字节，free 旧块。