帮忙写一个基础函数

[wtadminxjeri]

要求写一个函数， amt是传入值的指针，n是保留的小数位数。类似四舍五入函数，但进位有所不同
要求用纯C写


double roundm(amt,n)
double *amt;
int n;
{
////////代码
////////代码
................
...............
}

//调用
double x,y,z;
x=6.0068003
y=6.0068008
z=6.0068000
roundm(&x,6) = 6.006801
roundm(&y,6) = 6.006801
roundm(&z,6) = 6.006800



求中间代码

[wtadminxjeri]

调用说的有点问题，应该是
调用完roundm(&x,6)
roundm(&y,6)
roundm(&z,6) 后
x=6.006801
y=6.006801
z=6.006800

[liangbch]

你这那是4舍五入呀，x=6.0068003，roundm(&x,6)应该是6.006800而不是6.006801

[DataChat.Club]

意思是：输入一个x和n,先判断x的第n+1位小数是不是0,如果是,那么x的第n位小数不变,否则x的第n位小数加1....

这样？？

[liangbch]

下贴出运行结果
roundm(5.234567893400,1)=5.200000000000
roundm(5.234567893400,2)=5.230000000000
roundm(5.234567893400,3)=5.235000000000
roundm(5.234567893400,4)=5.234600000000
roundm(5.234567893400,5)=5.234570000000
roundm(5.234567893400,6)=5.234568000000
roundm(5.234567893400,7)=5.234567900000

再给出代码

#include <stdlib.h>
#include <stdio.h>
#include <math.h>

double roundm(double f,int n)
{
	f+= 0.5* pow(10.0,-1.0 * n);
	f=  floor( f * pow(10.0,n));
	f=  f / pow(10.0, n);
	return f;
}


int main()
{
	double f=5.2345678934;

	printf("roundm(%.12f,%d)=%.12f\n",f,1,roundm(f,1));
	printf("roundm(%.12f,%d)=%.12f\n",f,2,roundm(f,2));
	printf("roundm(%.12f,%d)=%.12f\n",f,3,roundm(f,3));
	printf("roundm(%.12f,%d)=%.12f\n",f,4,roundm(f,4));
	printf("roundm(%.12f,%d)=%.12f\n",f,5,roundm(f,5));
	printf("roundm(%.12f,%d)=%.12f\n",f,6,roundm(f,6));
	printf("roundm(%.12f,%d)=%.12f\n",f,7,roundm(f,7));
}

[DataChat.Club]

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

double roundm(double *amt, int n)
{
       

        if ((int)(*amt * pow(10.0, n + 1)) % 10 == 0)
        {
                *amt = (int)(*amt * pow(10.0, n)) / pow(10.0, n);
        }
        else
        {
                *amt = (((int )*amt * pow(10.0, n)) + 1) / pow(10.0, n);
        }
        return *amt;

}

int main()
{
        double amt = 6.0068003;
        roundm(&amt, 6);
        printf("%f\n", amt);
        return 0;
}

[wtadminxjeri]

我先测测哦，不好意思，我想表达的意思如四楼所说。

[wtadminxjeri]

6楼打出的值是多少？ 是不是6.006801 ?

[赵4老师]

//Round(1.234,2) = 1.23
//Round(1.234,0) = 1.0
//Round(123.4,-1) = 120.0
double Round(double dVal, short iPlaces) {
    double dRetval;
    double dMod = 0.0000001;
    if (dVal<0.0) dMod=-0.0000001;
    dRetval=dVal;
    dRetval+=(5.0/pow(10.0,iPlaces+1.0));
    dRetval*=pow(10.0,iPlaces);
    dRetval=floor(dRetval+dMod);
    dRetval/=pow(10.0,iPlaces);
    return(dRetval);
}

double round(double dVal, short iPlaces) //iPlaces>=0
{
    unsigned char s[20];
    double dRetval;

    sprintf(s,"%.*lf",iPlaces,dVal);
    sscanf(s,"%lf",&dRetval);
    return (dRetval);
}

[wtadminxjeri]

写代码的最后将调用结果报一下案例就是
//传入金额
x=6.0068003
y=6.0068008
z=6.0068000

保留6位小数，给个结果

[wtadminxjeri]

6楼的值不对
打出的值是6.000001而不是6.006801

[DataChat.Club]

我错了。。。

[Quote=引用 11 楼 的回复:]

6楼的值不对
打出的值是6.000001而不是6.006801
[/Quote]

[DataChat.Club]

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

double roundm(double *amt, int n)
{
//      (int)(*amt * pow(10.0, n + 1)) % 10 == 0 ? *amt = (int)(*amt * pow(10.0, n)) / pow(10.0, n) : *amt = (((int )*amt * pow(10.0, n)) + 1) / pow(10.0, n);

        if ((int)(*amt * pow(10.0, n + 1)) % 10 == 0)
        {
                *amt = (int)(*amt * pow(10.0, n)) / pow(10.0, n);
        }
        else
        {
                *amt = (((int )(*amt * pow(10.0, n))) + 1) / pow(10.0, n);//少加了个括号
        }
        return *amt;

}

int main()
{
        double amt = 6.0068003;
        roundm(&amt, 6);
        printf("%f\n", amt);
        return 0;
}
~     

[liangbch]

照4楼的规则，将我的程序改了一下。

先给出运行结果：
roundm(6.006800300000,6)=6.006801000000
roundm(6.006800800000,6)=6.006801000000
roundm(6.006800000000,6)=6.006800000000
再给出源代码

#include <stdlib.h>
#include <stdio.h>
#include <math.h>


double roundm(double f,int n)
{
	double t1,t2;
	t1= f* pow(10.0,1.0*n);
	t2= t1-floor(t1);
    t1=floor(t1);
	if (t2>=0.1)
		t1+=1.0;
	t1=  t1 / pow(10.0, n);
	return t1;

}


int main()
{
	double x;

	x=6.0068003; printf("roundm(%.12f,%d)=%.12f\n",x,6,roundm(x,6));
	x=6.0068008; printf("roundm(%.12f,%d)=%.12f\n",x,6,roundm(x,6));
	x=6.0068000; printf("roundm(%.12f,%d)=%.12f\n",x,6,roundm(x,6));
}

[lbq199204]

受教了，这张帖，让我了解到floor() pow() ceil() 函数，我想问下，这些函数，c里面有很多很多函数，是不是，要到做项目（何为项目？）的时候才会去查的么？或者说去哪查。就想java的JDK那样么？
[Quote=引用 14 楼 的回复:]
照4楼的规则，将我的程序改了一下。

先给出运行结果：
roundm(6.006800300000,6)=6.006801000000
roundm(6.006800800000,6)=6.006801000000
roundm(6.006800000000,6)=6.006800000000
再给出源代码

C/C++ code
#include <stdlib.h>
#in……
[/Quote]

[Mooker_Lee]

// roundm.cpp
#include <math.h>
#include <stdio.h>
#include <conio.h>
/*
#include <iostream>
#include <iomanip>
*/
double roundm(double* amt,int n)
{
double ABC = pow(10.0,n+1);
int t = (int)(*amt*ABC);
int x = t%10;
if(x)
return (t-x+10)/ABC;
else
return t/ABC;
}

int main()
{
extern double roundm(double *amt,int n);
double x,y,z;
x=6.0068003;
y=6.0068008;
z=6.0068000;
printf("%f\n",roundm(&x,6));
/*
double num;
int m;
std::cin >> m >> num;
std::cout << std::setprecision(16)<< roundm(&num,m) << std::endl;
*/
getch();
return 0;
}

/*
老实说，C语言中的输入输出流是最让我难受的一部分。
*/

[wtadminxjeri]

13楼的我验证了没问题，不好意思其他的没验证