C-释放的指针未分配 [英] C - pointer being freed was not allocated

问题描述

我试图从分配给malloc()的向量中释放分配给我的指针，当我尝试删除第一个元素(索引[0])时，当我尝试删除第二个元素(索引[ 1])我收到此错误:

I am trying to free a pointer that I assigned from a vector allocated with malloc(), when I try to remove the first element(index [0]), it works, when I try to remove the second(index [1]) I receive this error:

malloc: *** error for object 0x100200218: pointer being freed was not allocated

代码:

table->t = malloc (sizeof (entry) * tam);
entry * elem = &table->t[1];
free(elem);

推荐答案

您只能在(或需要)free()调用由malloc()和family返回的指针.

You can only call (or need to) free() on the pointer returned by malloc() and family.

引用C11，第7.22.3.3章

Quoting C11, chapter §7.22.3.3

[...]否则，如果 该参数与内存管理器先前返回的指针不匹配 函数，或者如果通过调用free或realloc释放了空间，则 行为是不确定的.

[...] 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.

在您的情况下，table->t(或&table->t[0])是该指针，而不是&table->t[1].

In your case, table->t (or, &table->t[0]) is that pointer, not &table->t[1].

也就是说，free() -ing table->t会释放整个内存块，您不需要(也不能)单独/部分释放.有关更多信息，请参见此答案.

That said, free()-ing table->t frees the whole memory block, you don't need to (you can't, rather) free individually/ partially. See this answer for more info.

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