将双精度值舍入到小数点后两位 [英] Round double value to 2 decimal places

问题描述

我有一个双精度值 22.368511我想把它四舍五入到小数点后两位.即它应该返回 22.37

我该怎么做?

解决方案

与大多数语言一样，格式是

%.2f

您可以在此处查看更多示例

<小时>

编辑:如果您担心 25.00 时点的显示，我也得到了这个

<代码>{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]]);}

<块引用>

2010-08-22 15:04:10.614 a.out[6954:903] 25.342010-08-22 15:04:10.616 a.out[6954:903] 25.32010-08-22 15:04:10.617 a.out[6954:903] 25

I have a double value as 22.368511 I want to round it to 2 decimal places. i.e. it should return 22.37

How can I do that?

解决方案

As in most languages the format is

%.2f

you can see more examples here

---

Edit: I also got this if your concerned about the display of the point in cases of 25.00

{
    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

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