Problem with count the decimal part by C++?

Question

I'm a newbie in C++. I got trouble when count the decimal part by C++, here is the example and code:

example:

   3.01: count of decimal part is 2
103.939: count of decimal part is 3

code:

int result = 0;
double d   = 4.01;

while(true) {
    cout << "(int)d: " << (int)d << endl;
    if(d == (int)d) break;

    d *= 10;
    cout << "d *= 10: " << d << endl;

    result++;
}
cout << result << endl;

console:

int(d): 4
d *= 10: 40.1
int(d): 40
d *= 401
int(d): 400
d *= 10: 4010
int(d): 4009
...

what happens to 4.01? And the same strange result when the double = 5.01 etc. I know it's the problem of precision when convert DOUBLE to INT, but I'm really curious about how it happens when the test doubles like 4.01, 5.01, etc.

And in addition, how to modify the if state correct to test 4.01?

When you do `double d = 4.01;`, the value of `d` is not 4.01. It's a number very close to 4.01. Important reading: https://stackoverflow.com/questions/588004/is-floating-point-math-broken — aschepler, Feb 22 '21 at 05:49
Hint. The closest `double` to `4.01` is `4.0099999999999997868371792719699442386627197265625`. Nothing particularly strange here. After all, most folk are happy with the fact that the closest `int` to `4.01` is `4`. One way round this is to round your number to 14 significant figures, and output only those digits, writing something clever to not print trailing zeros. — Bathsheba, Feb 22 '21 at 07:09

KL_KISNE_DEKHA_HAI · Accepted Answer · 2021-02-22T12:25:25.870

Since your question has been answered in the comments, I am posting a different approach to count decimal digits ( by taking input as string and treating the entire thing as string operation ).

#include <iostream>

using namespace std;

int main()
{
    string d;
    cin>>d;
    
    int flag=0, count=0, pos;
    
    for(int i = 0 ; i < d.length() ; i++){
        if(d[i]=='.'){
            flag = 1; 
            pos = i;  // for keeping the index of decimal point
            continue;
        }
        
        if(flag==1){
            count++;
        }
        
    }
    

    // for removing trailing zeroes like 0.10000 is actually 0.1
    if(flag==1){
        for(int i=d.length()-1;i>pos;i--){
            
            if(d[i]!='0'){
                break;
            }
            
            if(d[i]=='0'){
                count--;
            }
            
        }
    }
    
    cout<<count<<"\n";

    return 0;
}

UPD 1.0 : Taken input as double as asked by frank.

#include <iostream>

using namespace std;

int main()
{
    
    double num;
    
    string d;
    cin>>num;
    
    d = to_string(num);
    
    int flag=0, count=0, pos;
    
    for(int i = 0 ; i < d.length() ; i++){
        if(d[i]=='.'){
            flag = 1; 
            pos = i;  // for keeping the index of decimal point
            continue;
        }
        
        if(flag==1){
            count++;
        }
        
    }
    

    // for removing trailing zeroes like 0.10000 is actually 0.1
    if(flag==1){
        for(int i=d.length()-1;i>pos;i--){
            
            if(d[i]!='0'){
                break;
            }
            
            if(d[i]=='0'){
                count--;
            }
            
        }
    }
    
    cout<<count<<"\n";

    return 0;
}

Thanks anyway. However, the input of my code must be double. — frank jorsn, Feb 22 '21 at 06:35
the benefit with string is you can also take a value which is outside of double range, and still count the digits without storing the result in a variable ( like is done for int/long) — KL_KISNE_DEKHA_HAI, Feb 22 '21 at 12:30

kdcode · Answer 2 · 2021-02-22T10:17:07.950

0

Check this code out. It worked for me.

int result = 0;
int f = 1;
double d = 4.01;
while((int)(d*f)!=d*f)
{
    f*=10;
    result++;
}
std::cout<<result;

Output:

edited Feb 22 '21 at 10:17

answered Feb 22 '21 at 09:56

kdcode

524
2
7

Problem with count the decimal part by C++?

2 Answers2