Here's an utility that rounds (instead of truncating) a double to specified number of decimal places.

For example:

Use with AIround(200.3456, 2); // returns 200.35

Original version; watch out with this

Use with AIpublic static double round(double value, int places) {
    if (places < 0) throw new IllegalArgumentException();

    long factor = (long) Math.pow(10, places);
    value = value * factor;
    long tmp = Math.round(value);
    return (double) tmp / factor;
}

This breaks down badly in corner cases with either a very high number of decimal places (e.g. round(1000.0d, 17)) or large integer part (e.g. round(90080070060.1d, 9)). Thanks to Sloin for pointing this out.

I've been using the above to round "not-too-big" doubles to 2 or 3 decimal places happily for years (for example to clean up time in seconds for logging purposes: 27.987654321987 -> 27.99). But I guess it's best to avoid it, since more reliable ways are readily available, with cleaner code too.

So, use this instead

(Adapted from this answer by Louis Wasserman and this one by Sean Owen.)

Use with AIpublic static double round(double value, int places) {
    if (places < 0) throw new IllegalArgumentException();

    BigDecimal bd = BigDecimal.valueOf(value);
    bd = bd.setScale(places, RoundingMode.HALF_UP);
    return bd.doubleValue();
}

Note that HALF_UP is the rounding mode "commonly taught at school". Peruse the RoundingMode documentation, if you suspect you need something else such as Bankers’ Rounding.

Of course, if you prefer, you can inline the above into a one-liner:
new BigDecimal(value).setScale(places, RoundingMode.HALF_UP).doubleValue()

And in every case

Always remember that floating point representations using float and double are inexact. For example, consider these expressions:

Use with AI999199.1231231235 == 999199.1231231236 // true
1.03 - 0.41 // 0.6200000000000001

For exactness, you want to use BigDecimal. And while at it, use the constructor that takes a String, never the one taking double. For instance, try executing this:

Use with AISystem.out.println(new BigDecimal(1.03).subtract(new BigDecimal(0.41)));
System.out.println(new BigDecimal("1.03").subtract(new BigDecimal("0.41")));

Some excellent further reading on the topic:

• Item 48: "Avoid float and double if exact answers are required" in Effective Java (2nd ed) by Joshua Bloch
• What Every Programmer Should Know About Floating-Point Arithmetic

If you wanted String formatting instead of (or in addition to) strictly rounding numbers, see the other answers.

Specifically, note that round(200, 0) returns 200.0. If you want to output "200.00", you should first round and then format the result for output (which is perfectly explained in Jesper's answer).

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.

double someValue = 2.346;    
String.Format("{0:0.00}", someValue);

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.

Just use: (easy as pie)

double number = 651.5176515121351;

number = Math.round(number * 100);
number = number/100;

The output will be 651.52

This works:

inputValue = Math.Round(inputValue, 2);

Here's some examples:

decimal a = 1.994444M;

Math.Round(a, 2); //returns 1.99

decimal b = 1.995555M;

Math.Round(b, 2); //returns 2.00

You might also want to look at bankers rounding / round-to-even with the following overload:

Math.Round(a, 2, MidpointRounding.ToEven);

There's more information on it here.