C语言 小数点后N位C四舍五入
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Round in C with N digits after decimal point
提问by testCoder
Possible Duplicate:
Rounding Number to 2 Decimal Places in C
可能的重复：
在 C 中将数字四舍五入到 2 个小数位
I have not found a function with a signature double round(double d, int digits)like herein c.
When i try to build i get a error:
我还没有在 c 中找到具有double round(double d, int digits)像这里这样的签名的函数。当我尝试构建时，出现错误：
error: too many arguments to function 'round'
错误：函数“round”的参数太多
How can I round in C with N digits after the decimal point?
如何在小数点后用 N 位在 C 中舍入？
回答by andand
Using recursion (which is going to be slow for some values of digits)
使用递归（对于某些数字值会很慢）
#include <math.h>
double my_round(double x, unsigned int digits) {
if (digits > 0) {
return my_round(x*10.0, digits-1)/10.0;
}
else {
return round(x);
}
}
A method likely to be somewhat faster, but which relies on a single call to the slow powfunction:
一种可能会更快一些的方法，但它依赖于对慢速pow函数的单个调用：
#include <math.h>
double my_round(double x, unsigned int digits) {
double fac = pow(10, digits);
return round(x*fac)/fac;
}
An even faster method is to precompute a lookup table with the likely powers and use that instead of pow.
一种更快的方法是预先计算具有可能幂的查找表并使用它而不是pow.
#include <math.h>
double fac[]; // population of this is left as an exercise for the reader
double my_round(double x, unsigned int digits) {
return round(x*fac[digits])/fac[digits];
}
回答by Muhammad
here's a (very) simple function,
这是一个（非常）简单的函数，
double round1(double num, int N) {
int temp=(int) num*pow(10,N);
double roundedN= temp/pow(10,N);
return roundedN;
}
回答by md5
In C standard, such function does not exist. Anyway, you can write your own.
在 C 标准中，这样的函数是不存在的。不管怎样，你可以自己写。
#include <math.h>
/* Round `n` with `c` digits after decimal point. */
double nround (double n, unsigned int c)
{
double marge = pow (10, c);
double up = n * marge;
double ret = round (up) / marge;
return ret;
}
See also the comments above about floating points "decimal point".
另请参阅上面关于浮点“小数点”的评论。
回答by Mats Petersson
Whilst "answerd" gives a decent answer, here's one that works for arbitrarily large numbers:
虽然“已回答”给出了一个不错的答案，但这里有一个适用于任意大数的答案：
double round1(double num, int N) {
ASSERT(N > 0);
double p10 = pow(10,N);
return round(num* p10) / p10;
}
Of course, as stated, floating point numbers don't have a set number of decimal digits, and this is NOT guaranteed to PRINT as 3.70000 if you call printf("%8.5f", round1(3.7519, 1));for example.
当然，如前所述，浮点数没有固定的十进制数字，printf("%8.5f", round1(3.7519, 1));例如，如果您打电话，则不能保证打印为 3.70000 。
