Question

I get this error when I m trying to malloc some memory for my struct of ints.

typedef struct _values {
random ints;
} values;

我尝试了下文的内容,但我的汇编者却不喜欢。我如何确定错误?

values v;
v = malloc(sizeof(values));

Answer 1

希望在<代码>数值<>/代码>后和在<代码>v之前添加星号(*),将其标记为点: Values *v;

The way you set it now, the v (without the asterisk) is defined as a stack variable and would be allocated on the stack and discarded once the function ends. It s type will be simply values. malloc is used to allocate memory on the heap and returns a pointer to the memory. Sine the function has no way of knowing the type it returns it as a void * type - Which gives you your error - you re attempting to assign a void * type into a struct type, which the compiler can t do, nor can the compiler find a legitimate cast that could resolve the problem.

Answer 2

这里不需要<代码>小孔记忆;<代码>值s v; 已经分配记忆(栏目”)。如果你离开目前的范围,就会自动取消。

如果你真的希望把记忆“一刀切”,以保持目前的范围,那么你需要<条码>v<>/代码>作为点;即:

values *v;
v = malloc(sizeof(values));

改为free。

Answer 3

values v;

这是在 st上分配的结构的例子。

v = malloc(sizeof(values));

小型公司在一份<条码>上填满了,点子和编汇人打了笔记,使你能够指派一名证人出庭。

你们需要申报点,然后指定小点返回点。

......

values *v = NULL;

v = malloc(sizeof(values));

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