If you just want to round the number for output purposes, then the "%.2f" format string is indeed the correct answer. However, if you actually want to round the floating point value for further computation, something like the following works:

#include <math.h>

float val = 37.777779;

float rounded_down = floorf(val * 100) / 100;   /* Result: 37.77 */
float nearest = roundf(val * 100) / 100;  /* Result: 37.78 */
float rounded_up = ceilf(val * 100) / 100;      /* Result: 37.78 */

Notice that there are three different rounding rules you might want to choose: round down (ie, truncate after two decimal places), rounded to nearest, and round up. Usually, you want round to nearest.

As several others have pointed out, due to the quirks of floating point representation, these rounded values may not be exactly the "obvious" decimal values, but they will be very very close.

For much (much!) more information on rounding, and especially on tie-breaking rules for rounding to nearest, see the Wikipedia article on Rounding.

scanf("%.2f",&x);

And i think that will solve a problem

Well, depending on your needs you can choose any of the following. Out put is written against each method

You can choose the one you need

This will round

decimal d = 2.5789m;
Console.WriteLine(d.ToString("#.##")); // 2.58

This will ensure that 2 decimal places are written.

d = 2.5m;
Console.WriteLine(d.ToString("F")); //2.50

if you want to write commas you can use this

d=23545789.5432m;
Console.WriteLine(d.ToString("n2")); //23,545,789.54

if you want to return the rounded of decimal value you can do this

d = 2.578m;
d = decimal.Round(d, 2, MidpointRounding.AwayFromZero); //2.58

Well, depending on your needs you can choose any of the following. Out put is written against each method

You can choose the one you need

This will round

decimal d = 2.5789m;
Console.WriteLine(d.ToString("#.##")); // 2.58

This will ensure that 2 decimal places are written.

d = 2.5m;
Console.WriteLine(d.ToString("F")); //2.50

if you want to write commas you can use this

d=23545789.5432m;
Console.WriteLine(d.ToString("n2")); //23,545,789.54

if you want to return the rounded of decimal value you can do this

d = 2.578m;
d = decimal.Round(d, 2, MidpointRounding.AwayFromZero); //2.58

I use the following:

double x = Math.Truncate(myDoubleValue * 100) / 100;

For instance:

If the number is 50.947563 and you use the following, the following will happen:

- Math.Truncate(50.947563 * 100) / 100;
- Math.Truncate(5094.7563) / 100;
- 5094 / 100
- 50.94

And there's your answer truncated, now to format the string simply do the following:

string s = string.Format("{0:N2}%", x); // No fear of rounding and takes the default number format

This works:

inputValue = Math.Round(inputValue, 2);

To round a double up to 2 decimal places, you can use:

Use with AI#include <iostream>
#include <cmath>

int main() {
    double value = 0.123;
    value = std::ceil(value * 100.0) / 100.0;
    std::cout << value << std::endl; // prints 0.13
    return 0;
}

To round up to n decimal places, you can use:

Use with AIdouble round_up(double value, int decimal_places) {
    const double multiplier = std::pow(10.0, decimal_places);
    return std::ceil(value * multiplier) / multiplier;
}

This method won't be particularly fast, if performance becomes an issue you may need another solution.

What you want is %.2f, not 2%f.

Also, you might want to replace your %d with a %f ;)

#include <cstdio>
int main()
{
printf("When this number: %f is assigned to 2 dp, it will be: %.2f ", 94.9456, 94.9456);
return 0;
}

This will output:

When this number: 94.945600 is assigned to 2 dp, it will be: 94.95

See here for a full description of the printf formatting options: printf

You made simple mistake. This will work:

[NSString stringWithFormat: @"%.2lf", total]