模量返回时间分量 [英] Returning Time Components with Modulus

问题描述

某人做了20小时42分钟&一秒16秒，总计74536秒. 我如何从该人完成该班次的秒数中得到小时数?

Someone does 20 Hours 42 Minutes & 16 Seconds in one shift totaling 74536 seconds. How do I get the hours from number of seconds the person has done for that shift?

20 * 60 * 60      =    72000
     42 * 60      =     2520
          16      =       16
                  +    -----
Total             =    74536
____________________________
Total % 60        =  Seconds (16)
Total % ?         =  Minutes (42)
Total % ?         =    Hours (20)

已经尝试了84600；事实证明，当数字低于模数时，它确实不是很有用，如果某人仅登录几秒钟，我将必须抓住一些问题...

Tried 84600 already; turns out when a number is lower the modulus, it really is not very helpful, and something I am going to have to catch should someone only sign in for a few seconds ...

推荐答案

您需要同时使用模数和除法:

You need to use both modulus and division:

t = seconds_in_shift;
secs = t % 60;
t /= 60;
mins = t % 60;
t /= 60;
hour = t;

或者:

secs =  ttime % 60;
mins = (ttime / 60) % 60;
hour =  ttime / 3600;

另一个选项使用Standard中的 div() 函数C(<stdlib.h>):

One other option uses the div() function from Standard C (<stdlib.h>):

div_t v1 = div(ttime, 60);
div_t v2 = div(v1.quot, 60);

此后，v1.rem包含秒； v2.rem包含分钟，而v2.quot包含小时，其基于ttime中最初的秒数.

After that, v1.rem contains the seconds; v2.rem contains the minutes, and v2.quot contains the hours based on the number of seconds originally in ttime.

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