Question

enum ev_type { ARRIVAL, DEP_Q0, DEP_Q1, DEP_Q2, DEP_Q3 };
struct node {
    double time;
    double duration;
    enum ev_type event;
    struct node * next;
}*evlist;
typedef struct node *nptr;
double min = 0;
void generate_new_customer(double clock, double duration)
{
    nptr newev = (nptr)malloc(sizeof(nptr));
    newev->duration = duration;
    newev->time = min;
    newev->event = ARRIVAL;

    if (evlist == NULL)
    {
        evlist = newev;
        evlist->next = NULL;
    }
    else
    {
        newev->next = evlist;
        evlist = newev;
    }
    printf("%lf\n", evlist->time);
}
int main() {
    start();
    while (1) {
        generate_new_customer(1, 6);
        if (min > 480)
            break;
        min++;
    }
    return 0;
}

我正在尝试使用功能进行内存分配。该函数按最小值工作。如果最小值增加480，它将停止并中断。我检查了但是仍然会发生调试断言失败的错误。我该怎么办？

Answer 1

您的malloc不正确。只是在需要 pointer 的大小时才将其分配给您的结构。

更改：

nptr newev = (nptr) malloc(sizeof(nptr));

进入：

nptr newev = malloc(sizeof(struct node));

更新：

我注意到您的功能正在将您的列表推到最前面。这不是错误，但是您可以简化一下：

您的原始代码：

if (evlist == NULL) {
    evlist = newev;
    evlist->next = NULL;
}
else {
    newev->next = evlist;
    evlist = newev;
}

轻微的重新排序：

if (evlist == NULL) {
    evlist = newev;
    newev->next = NULL;
}
else {
    newev->next = evlist;
    evlist = newev;
}

一些重新排序：

if (evlist == NULL) {
    newev->next = NULL;
    evlist = newev;
}
else {
    newev->next = evlist;
    evlist = newev;
}

由于newev->next = NULL在if被保证为evlist的{{1}}子句的NULL内，我们可以将其更改为：

if (evlist == NULL) {
    newev->next = evlist;
    evlist = newev;
}
else {
    newev->next = evlist;
    evlist = newev;
}

现在，if和else都是相同的。因此，我们可以消除冗余，并且有：

newev->next = evlist;
evlist = newev;

while中的动态内存分配

1 个答案: