C ++获得整数除法和余数的最佳方法

问题描述

我只是想知道，如果我想要除以b，并且对结果c和余数感兴趣（例如，我有秒数，并想分裂成分和秒），什么是最好的方法吗？

是

  int c =（int）a / b; 
 int d = a％b; 
  

或

  int c =（int）a / b; 
 int d = a  -  b * c; 
  

或

  double tmp = a / b; 
 int c =（int）tmp; 
 int d =（int）（0.5+（tmp-c）* b）; 
  

或

也许有魔法

$ p

解决方案

在x86上，剩余部分是除法本身的副产品，体面的编译器应该能够使用它（而不是再次执行 div ）。

  int c =（int）a / b; 
 int d = a％b; / *可能使用除法的结果。 / / b $ b  

I am just wondering, if I want to divide a by b, and am interested both in the result c and the remainder (e.g. say I have number of seconds and want to split that into minutes and seconds), what is the best way to go about it?

Would it be

int c = (int)a / b;
int d = a % b;

or

int c = (int)a / b;
int d = a - b * c;

or

double tmp = a / b;
int c = (int)tmp;
int d = (int)(0.5+(tmp-c)*b);

or

maybe there is a magical function that gives one both at once?

解决方案

On x86 the remainder is a by-product of the division itself so any half-decent compiler should be able to just use it (and not perform a div again). This is probably done on other architectures too.

int c = (int)a / b;
int d = a % b; /* Likely uses the result of the division. */

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