69,368
社区成员
发帖
与我相关
我的任务
分享怎么样才能正确运行？
#include <stdio.h>
#include <math.h>
main()
{double a=2.00,b=4.00,c;
c=pow(a,b);
c=100.00%c;
printf("%lf",c);
}
#include <math.h>
main()
{double a=2.00,b=4.00,c;
c=pow(a,b);
c=100.00%c;
printf("%lf",c);
}
...全文
请发表友善的回复…
发表回复
neptunez 2002-04-10
- 打赏
- 举报
% 整数
prototype 2002-04-10
- 打赏
- 举报
the above posts are correct. namely, operator '%' is for
integral types (int, char, short, long, maybe bool).
for your problem, the correct thing you should use is 'fmod'
example:
#include <math.h>
...
c=fmod(100.00 , c); // 'c' gets the remainder of 100.0
// divided by 'c'.
integral types (int, char, short, long, maybe bool).
for your problem, the correct thing you should use is 'fmod'
example:
#include <math.h>
...
c=fmod(100.00 , c); // 'c' gets the remainder of 100.0
// divided by 'c'.
ddszhan 2002-04-10
- 打赏
- 举报
%的两个操作数一定是定点型的，如整型，字符型等。
