c++ convert a fractional part of a number into integer

问题

I needed to convert a fractional part of a number into integer without a comma, for example I have 3.35 I want to get just 35 part without zero or a comma,

Because I used the modf() function to extract the the fractional part but it gives me a 0.35 if there is any way to do that or to filter the '0.' part I will be very grateful if you show me how with the smaller code possible,

回答1:

A bit more efficient than converting to a string and back again:

int fractional_part_as_int(double number, int number_of_decimal_places) {
    double dummy;
    double frac = modf(number,&dummy);
    return round(frac*pow(10,number_of_decimal_places));
}

回答2:

#include <iostream>
#include <cmath>

double round(double r) {
    return (r > 0.0) ? std::floor(r + 0.5) : std::ceil(r - 0.5);
}

double floor_to_zero(double f) {
    return (f > 0.0) ? std::floor(f) : std::ceil(f);
}

double sign(double s) {
    return (s < 0.0) ? -1.0 : 1.0;
}

int frac(double f, int prec) {
    return round((f - floor_to_zero(f)) * prec) * sign(f);
}

int main() {
    double a = 1.2345;
    double b = -34.567;
    std::cout << frac(a, 100) << " " << frac(b, 100) << std::endl; // 23 57
}

回答3:

another solution

int precision= 100;
double number = 3.35;
int f = floor(xx);
double temp = ( f - number ) * -1;
int fractional_part = temp * precision;

回答4:

IF you need it as a string, a quite easy C style solution would be (should work for variable number of decimal places):

    double yourNumber = 0.35f;
    char buffer[32];
    snprintf(buffer, 32, "%g", yourNumber);
    strtok(buffer, "."); // Here we would get the part before . , should still check
    char* fraction = strtok(NULL, ".");
    int fractionAsInt = atoi(fraction);

This example lacks error handling in case of a bad string and is not feasible if you just need a fixed number of decimal places, since the arithmetic approaches work better there.

回答5:

Something like this should work:

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

static int get_frac(double value, unsigned short precision)
{
    return (int)((value - (long)value) * pow(10, precision));
}

static int get_frac_no_trailing_zeros(double value, unsigned short precision)
{
    int v = get_frac(value, precision);
    while (v % 10 == 0)
        v /= 10;
    return v;
}

int main(int argc, char *argv[])
{
    double v;

    v = 123.4564;
    printf("%.4f = %d\n", v, get_frac(v, 2));
    printf("%.4f = %d\n", v, get_frac(v, 4));
    printf("%.4f = %d\n", v, get_frac(v, 6));
    printf("%.4f = %d\n", v, get_frac_no_trailing_zeros(v, 6));

    return EXIT_SUCCESS;
}

You may also want to either avoid calling pow by having a user supply a number in a power of 10 in a first place, or use a lookup table.

回答6:

Using some stl magic, here is the sample code:

typedef std::pair<int, int> SplitFloat;

SplitFloat Split(float value, int precision)
{
    // Get integer part.
    float left = std::floor(value);
    // Get decimal part.
    float right = (value - left) * float(std::pow(10, precision));

    return SplitFloat(left, right);
}

It can be improved, but is pretty straightforward.

回答7:

I just did something close to what you are trying to do, though I'm still pretty new. None the less, maybe this will help someone in the future as I landed here looking for results for my problem.

The first step is making sure that the variable that contains 3.35 is a double, but that's probably obvious.

Next, create a variable that is only an integer and set it's value equal to the value of the double. It will then only contain the whole number.

Then subtract the whole number (int) from the double. You will be left with the fraction/decimal value. From there, just multiply by 100.

Beyond the 100ths decimal value, you would have to do a little more configuring obviously, but it should be fairly simple to do with an if statement. If the decimal value is greater than .99, multiply 1000 instead etc..

回答8:

Here's how I would do it.

#include <sstream>
#include <string>

int main()
{
    double d = yourDesiredNumber; //this is your number
    std::ostringstream out;
    out << setprecision(yourDesiredPrecision) << std::fixed 
        << std::showpoint << d;
    std::istringstream in(out.str());
    std::string wholePart; //you won't need this. 
    int fractionalPart;  
    std::getline(in, wholePart, '.');
    in >> fractionalPart;
    //now fractionalPart contains your desired value.
}

I'm pretty sure that instead of two different istringstream and ostringstream objects you could have gotten away with just one stringstream object, but I am not sure about the details (never used that class) so I didn't use it in the example.

来源：https://stackoverflow.com/questions/12801062/c-convert-a-fractional-part-of-a-number-into-integer

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c++

math

integer