使用 time_t 解析用户输入
我的想法是，如果用户输入 t = 2.5，那么我将 2 和 0.5 分别提取到 2 个不同的变量中。但我无法做到这一点。
这是代码：
$ export LT_LEAK_START=1.5
$ echo $LT_LEAK_START
1.5
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
double d;
time_t t;
long nsec;
d=strtod(getenv("LT_LEAK_START"), NULL);
t=(time_t)d;
nsec=d-(double)((time_t)d); // Extract fractional part as time_t always whole no.
printf("d = %lf\n",d);
printf("t = %u, nsec = %f\n",d,nsec);
}
输出是：
# ./a.out
d = 1.500000
t = 0, nsec = 0.000000
My idea is, if user enters t = 2.5, then I extract 2 and 0.5 separately in 2 different variables. But I am unable to do that.
Here is the code:
$ export LT_LEAK_START=1.5
$ echo $LT_LEAK_START
1.5
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
double d;
time_t t;
long nsec;
d=strtod(getenv("LT_LEAK_START"), NULL);
t=(time_t)d;
nsec=d-(double)((time_t)d); // Extract fractional part as time_t always whole no.
printf("d = %lf\n",d);
printf("t = %u, nsec = %f\n",d,nsec);
}
Output is:
# ./a.out
d = 1.500000
t = 0, nsec = 0.000000
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评论（3）
你的输出坏了。实际上，您在以下代码中写入了两次
d的值：如果您这样写：
那么您将得到输出：
现在可以更清楚地看出您有舍入错误。在这种情况下，您可以通过将 0.5 分配给
nsec（一个long）来舍弃所有小数位。将nsec改为float。Your output is broken. You're actually writing the value of
dtwice in the following code:If you'd written this:
Then you'd have the output:
It's now clearer that you have a rounding error. In this case, you cast away all the decimal places by assigning 0.5 to
nsec, along. Makensecafloatinstead.您还尝试在 long 中存储小数值。您需要将其乘以 1000 或将
nsec设为双精度。nsec=d-(double)((time_t)d);如果 d 为 1.5，则右侧的结果将为 0.5，当它存储在 nsec 中时，它会隐式向下转换为 0 。
You are also trying to store a fractional value in a long. You either need to multiply this by 1000 or make
nseca double.nsec=d-(double)((time_t)d);If d is 1.5, the result of the right hand side would be 0.5, which will implicitly cast down to 0 when it gets stored in nsec.
您尝试将
.5分配给long，但这不会发生。You're trying to assign
.5to alongwhich isn't going to happen.