c - 结构指针释放导致错误

Question

我有以下结构:

typedef struct {
    char* type;
    char* address;
    int area, price;
}Offer;

这两个函数:

Offer* initOffer(char* type, char* address, int area, int price)
{
    Offer* p;
    p = (Offer*)malloc(sizeof(Offer));
    p->type = (char*)malloc(sizeof(type));
    p->address = (char*)malloc(sizeof(address));
    strcpy(p->type, type);
    strcpy(p->address, address);
    p->area = area;
    p->price = price;
    return p;
}

void destroyOffer(Offer* offer)
{
    free(offer->type);
    free(offer->address);
    free(offer);
}

当我调用 destroyOffer 时出现问题，我不知道为什么，但是当我运行代码时，出现错误:检测到堆损坏。如果我删除这两行，它工作正常，但我认为内存没有正确清理:

free(offer->type);
free(offer->address);

Answer 1

问题:

p->type = (char*)malloc(sizeof(type));   // That's just the size of a pointer
p->address = (char*)malloc(sizeof(address)); // Same problem.

之后，行:

strcpy(p->type, type);
strcpy(p->address, address);

最终写入了他们不应该写入的内存。这会导致未定义的行为。

您需要:

p->type = malloc(strlen(type)+1);
p->address = malloc(strlen(address)+1);

参见Do I cast the result of malloc?

如果您的编译器支持，您还可以使用 strdup。如果是这样，您的代码可以简化为:

p->type = strdup(type);
p->address = strdup(address);