帮忙写一个基础函数解决方法

帮忙写一个基础函数
 
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


[解决办法]

C/C++ code

//Round(1.234,2) = 1.23//Round(1.234,0) = 1.0//Round(123.4,-1) = 120.0double 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);}
[解决办法]
C/C++ code
#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;}~
[解决办法]
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>#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));}
[解决办法]
// 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。
*/