如何访问的timeval结构的字段 [英] How to access the fields of a timeval structure

问题描述

我想在来打印值timeval结构变量如下：

int main()  
{  

    struct timeval *cur;  
    do_gettimeofday(cur);  
    printf("Here is the time of day: %ld %ld", cur.tv_sec, cur.tv_usec);  

    return 0;  
}  

我不断收到此错误：

I keep getting this error:

request for member 'tv_sec' in something not a structure or union.  
request for member 'tv_usec' in something not a structure or union.

我该如何解决这个问题？

How can I fix this?

推荐答案

由于 CUR 是一个指针。使用

struct timeval cur;
do_gettimeofday(&cur);

在Linux中， do_gettimeofday（）要求用户pre-分配的空间。不只是传递一个没有指向任何一个指针！你可以使用的malloc（），但你最好的选择仅仅是在栈上传递的东西的地址。

In Linux, do_gettimeofday() requires that the user pre-allocate the space. Do NOT just pass a pointer that is not pointing to anything! You could use malloc(), but your best bet is just to pass the address of something on the stack.

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