如何打印数字的浮动部分的最后一位？

Question

I am finding an easy way to print the last floating point digit of a number. For EG : 33.44 ans - 4 .

However, I managed it in JAVA easily, but I am a C/C++ beginner.An SSCCE for JAVA code :

import java.util.Scanner;

 public class Test{
 public static void main(String[] args){
 Scanner scan=new Scanner(System.in);
    Float f;
    f=scan.nextFloat();
    System.out.println(f);
    String g=f.toString(f);
    System.out.println(g.charAt(g.length()-1));
 }
}

I don't think it will be something that easy in C/C++ . Any logics/solutions?

Answer 1

If you want to do it in an equivalent manner, you would do something like (not compiled to check)

int getLast( double value )
{
  int rc;
  char buf[64];
  rc=snprintf(buf,64,"%f",value);
  // TODO:: check rc is in the range 1-64 
  return buf[rc-1]-'0';
}

You should probably check the result of rc and use strtol rather than the ascii hack of -'0' , but you get the picture.

if you want the rightmost non-0 value, the return line should be replace with

char *x=&(buf[rc-1]);
while( *x=='0' ) x--;
return *x-'0';

Answer 2

If the nominal value of a floating-point number is not an integer, unless the fractional part is exactly equal to 0.5, 0.25, or 0.75, the exact value of the fractional part will always end in 125, 375, 625, or 875. In many cases, one isn't actually interested in the exact nominal value of the fractional part, but instead a rounded decimal representation. In that case, the value of the "last" digit will depend upon how much one is willing to have the reported value differ from the number's exact nominal value (eg the 1.0f / 10.0f is a float whose nominal value is exactly 1,342,173/13,421,728, ie 0.100000001490116119384765625. The last digits of the exact nominal value are 625, but the number might also be interpreted as 0.1, or 0.1000000015, 0.10000000149, 0.10000000149012, etc. so one could justify 1, 5, 9, or 2 as being the "last" digit.

Answer 3

Hope this helps you(in C++) :

#include <iostream>
#include<string>
using namespace std;
int main()
{
 string float_data;
 cin>>float_data;
 int len = float_data.length();
 cout<<float_data.at(len-1)<<endl;
  return 0;
}

Answer 4

The below approach can be followed but it has limitation that the float value should not be very big.

void main()
{
int i, m;
float n;
printf("Enter the number n: ");
scanf("%f", &n);
m=(int)n;
i=(m % 10);
printf("last digit = %d", i);
}