How to Round Cgfloat

How to round CGFloat

There are already standard functions with behaviors you might need in <math.h> such as: floorf, ceilf,
roundf, rintf and nearbyintf (lasf 'f' means "float" version, versions without it are "double" versions).

It is better to use standard methods not only because they are standard, but because they work better in edge cases.

2013 Update (jessedc)

iOS is no longer only 32 bit. There are a number of other answers to this question that are now more relevant.

Most answers mention importing tgmath.h

• https://stackoverflow.com/a/5352779/184130
• https://stackoverflow.com/a/7695011/184130

Round up a CGFloat in Swift

Update: Apple have now defined some CGFloat-specific versions of common functions like ceil:

func ceil(x: CGFloat) -> CGFloat

...specifically to cope with the 32/64-bit difference. If you simply use ceil with a CGFloat argument it should now work on all architectures.

My original answer:

This is pretty horrible, I think, but can anyone think of a better way? #if doesn't seem to work for CGFLOAT_IS_DOUBLE; I think you're limited to build configurations, from what I can see in the documentation for conditional compilation.

var x = CGFloat(0.5)

#if arch(x86_64) || arch(arm64)
var test = ceil(x)
#else
var test = ceilf(x)
#endif

How to round of CGFloat value

Try this........

NSLog(@"SCALE : %.3f", scale);
NSString *value = [NSString stringWithFormat:@"%.3f", theFloat];

Stolen from Previous SO answer

Swift: Rounding float to nearest custom multiplier

You can use the same approach as rounding to a power of 10 here.

When you want to round to the nearest 0.1, you multiply by 10, round, then divide by 10. When you want to round to the nearest 0.01, you multiply by 100, round, then divide by 100.

See the pattern here? The number you multiply and divide by is always 1 over the granularity!

So for the granularity of 0.25, you multiply and divide by 4:

print(round(0.3 * 4) / 4)
// or
print(round(0.3 / 0.25) * 0.25)

More generally, given a granularity: Double and a number to round x: Double, we can round it like this:

let rounded = round(x / granularity) * granularity

Rounding a double value to x number of decimal places in swift

You can use Swift's round function to accomplish this.

To round a Double with 3 digits precision, first multiply it by 1000, round it and divide the rounded result by 1000:

let x = 1.23556789
let y = Double(round(1000 * x) / 1000)
print(y) /// 1.236

Unlike any kind of printf(...) or String(format: ...) solutions, the result of this operation is still of type Double.

EDIT:

Regarding the comments that it sometimes does not work, please read this: What Every Programmer Should Know About Floating-Point Arithmetic

How to safely floor or ceil a CGFloat to int?

There are a couple misconceptions in your question.

what if my CGFloat is 2.0f but internally it is represented as 1.999999999999f

can't happen; 2.0, like all reasonably small integers, has an exact representation in floating-point. If your CGFloat is 2.0f, then it really is 2.0.

something like 2.0 would never accidentally get ceiled to 2

The ceiling of 2.0 is 2; what else would it possibly be?

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I think the question that you're really asking is "suppose I do a calculation that produces an inexact result, which mathematically should be exactly 2.0, but is actually slightly less; when I apply floor to that value, I get 1.0 instead of 2.0--how do I prevent this?"

That's actually a fairly subtle question that doesn't have a single "right" answer. How have you computed the input value? What are you going to do with the result?

Rounding in Swift with round()

You can do:

round(1000 * x) / 1000

How to round off Float values?

In addition to the other answers:

float theFloat = 1.23456;
int rounded = roundf(theFloat); NSLog(@"%d",rounded);
int roundedUp = ceil(theFloat); NSLog(@"%d",roundedUp);
int roundedDown = floor(theFloat); NSLog(@"%d",roundedDown);
// Note: int can be replaced by float

For rounding to specific decimals, see the question mentioned by Alex Kazaev.

Round up double to 2 decimal places

Use a format string to round up to two decimal places and convert the double to a String:

let currentRatio = Double (rxCurrentTextField.text!)! / Double (txCurrentTextField.text!)!
railRatioLabelField.text! = String(format: "%.2f", currentRatio)

Example:

let myDouble = 3.141
let doubleStr = String(format: "%.2f", myDouble) // "3.14"

If you want to round up your last decimal place, you could do something like this (thanks Phoen1xUK):

let myDouble = 3.141
let doubleStr = String(format: "%.2f", ceil(myDouble*100)/100) // "3.15"

How to round a Double to the nearest Int in swift?

There is a round available in the Foundation library (it's actually in Darwin, but Foundation imports Darwin and most of the time you'll want to use Foundation instead of using Darwin directly).

import Foundation

users = round(users)

Running your code in a playground and then calling:

print(round(users))

Outputs:

15.0

round() always rounds up when the decimal place is >= .5 and down when it's < .5 (standard rounding). You can use floor() to force rounding down, and ceil() to force rounding up.

If you need to round to a specific place, then you multiply by pow(10.0, number of places), round, and then divide by pow(10, number of places):

Round to 2 decimal places:

let numberOfPlaces = 2.0
let multiplier = pow(10.0, numberOfPlaces)
let num = 10.12345
let rounded = round(num * multiplier) / multiplier
print(rounded)

Outputs:

10.12

Note: Due to the way floating point math works, rounded may not always be perfectly accurate. It's best to think of it more of an approximation of rounding. If you're doing this for display purposes, it's better to use string formatting to format the number rather than using math to round it.

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