c++基礎知識-double 四舍五入保留N位小數

兩種方法

方法1： 乘10法

去掉整數部分后，剩余小數*10，乘N次，加上0.5后再除回去，最后return 整數部分+小數部分

#include<iostream>
#include<cmath>
using namespace std;

//用long long 取整
/*double round(double number, unsigned int bits)
{
	long long integerpart = number;//取整數
	number -= integerpart;
	for (unsigned int i = 0; i < bits; ++i)//取整后的小數進行乘10處理
	{
		number *= 10;
	}
	number = (long long)(number + 0.5);
	for (unsigned int i=0;i<bits;++i)
	{
		number /= 10;
	}
	return integerpart + number;
}*/

int main()
{

	//方法一：乘10法
	printf("%.2f\n", round(23.54645544545452, 2));
	printf("%.3f\n", round(23.54645544545452, 3));
	printf("%.4f\n", round(23.54645544545452, 4));
	printf("%.5f\n", round(23.54645544545452, 5));
	printf("%.6f\n", round(23.54645544545452, 6));
	return 0;
}

　　

#include<iostream>
#include<cmath>
using namespace std;

//用cmath庫中的floor()取整
double round(double number, unsigned int bits)
{
	double integerpart = floor(number);
	number -= integerpart;
	for (unsigned int i = 0; i < bits; ++i)
	{
		number *= 10;
	}
	number = floor(number + 0.5);
	for (unsigned int i = 0; i < bits; ++i)
	{
		number /= 10;
	}
	return integerpart + number;
}
int main()
{

	//方法一：乘10法
	printf("%.2f\n", round(23.54645544545452, 2));
	printf("%.3f\n", round(23.54645544545452, 3));
	printf("%.4f\n", round(23.54645544545452, 4));
	printf("%.5f\n", round(23.54645544545452, 5));
	printf("%.6f\n", round(23.54645544545452, 6));
	return 0;
}