舍入双精度值到2位小数

Question

我有一个双倍值22.368511我想将它舍入到2位小数.即它应该返回22.37

我怎样才能做到这一点？

Answer 1

与大多数语言一样,格式为

%.2f

你可以在这里看到更多的例子

编辑:如果你担心在25.00的情况下显示点,我也得到了这个

{
    NSNumberFormatter *fmt = [[NSNumberFormatter alloc] init];
    [fmt setPositiveFormat:@"0.##"];
    NSLog(@"%@", [fmt stringFromNumber:[NSNumber numberWithFloat:25.342]]);
    NSLog(@"%@", [fmt stringFromNumber:[NSNumber numberWithFloat:25.3]]);
    NSLog(@"%@", [fmt stringFromNumber:[NSNumber numberWithFloat:25.0]]);
}

2010-08-22 15:04:10.614 a.out[6954:903] 25.34
2010-08-22 15:04:10.616 a.out[6954:903] 25.3
2010-08-22 15:04:10.617 a.out[6954:903] 25

Answer 2

您可以使用以下代码将其格式化为两位小数

NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];

[formatter setNumberStyle:NSNumberFormatterDecimalStyle];
[formatter setMaximumFractionDigits:2];
[formatter setRoundingMode: NSNumberFormatterRoundUp];

NSString *numberString = [formatter stringFromNumber:[NSNumber numberWithFloat:22.368511]];

NSLog(@"Result...%@",numberString);//Result 22.37

斯威夫特4:

let formatter = NumberFormatter()
formatter.numberStyle = .decimal
formatter.maximumFractionDigits = 2
formatter.roundingMode = .up

let str = String(describing: formatter.string(from: 12.2345)!)

print(str)

Answer 3

 value = (round(value*100)) / 100.0;

Answer 4

[label setText:@"Value: %.2f", myNumber];

Answer 5

您可以使用NSDecimalRound函数

Answer 6

在Swift 2.0和Xcode 7.2中:

let pi:Double = 3.14159265358979
String(format:"%.2f", pi)

例:

Answer 7

要从double中删除小数,请查看此输出

对象C.

double hellodouble = 10.025;
NSLog(@"Your value with 2 decimals: %.2f", hellodouble);
NSLog(@"Your value with no decimals: %.0f", hellodouble);

输出将是:

10.02 
10

Swift 2.1和Xcode 7.2.1

let hellodouble:Double = 3.14159265358979
print(String(format:"Your value with 2 decimals: %.2f", hellodouble))
print(String(format:"Your value with no decimals: %.0f", hellodouble))

输出将是:

3.14 
3