这不是数字如何存储的问题，而是如何显示它的问题。当将其转换为字符串时，必须四舍五入到所需的精度，在您的示例中是小数点后两位。

例如:

NSString* formattedNumber = [NSString stringWithFormat:@"%.02f", myFloat];

％．02f告诉格式化程序，您将格式化一个浮点数(%f)和，它应该四舍五入到两个位置，并且应该用0填充。

例如:

%f = 25.000000
%.f = 25
%.02f = 25.00

这里有一些更正

//for 3145.559706

斯威夫特3

let num: CGFloat = 3145.559706
print(String(format: "%f", num)) = 3145.559706
print(String(format: "%.f", num)) = 3145
print(String(format: "%.1f", num)) = 3145.6
print(String(format: "%.2f", num)) = 3145.56
print(String(format: "%.02f", num)) = 3145.56 // which is equal to @"%.2f"
print(String(format: "%.3f", num)) = 3145.560
print(String(format: "%.03f", num)) = 3145.560 // which is equal to @"%.3f"

卷-C

@"%f"    = 3145.559706
@"%.f"   = 3146
@"%.1f"  = 3145.6
@"%.2f"  = 3145.56
@"%.02f" = 3145.56 // which is equal to @"%.2f"
@"%.3f"  = 3145.560
@"%.03f" = 3145.560 // which is equal to @"%.3f"

等等……

使用NSNumberFormatter和maximumFractionDigits，如下所示:

NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];
formatter.maximumFractionDigits = 2;
NSLog(@"%@", [formatter stringFromNumber:[NSNumber numberWithFloat:12.345]]);

你会得到12.35

 lblMeter.text=[NSString stringWithFormat:@"%.02f",[[dic objectForKey:@"distance"] floatValue]];

你也可以尝试使用NSNumberFormatter:

NSNumberFormatter* nf = [[[NSNumberFormatter alloc] init] autorelease];
nf.positiveFormat = @"0.##";
NSString* s = [nf stringFromNumber: [NSNumber numberWithFloat: myFloat]];

你可能还需要设置否定格式，但我认为解决这个问题已经足够聪明了。

我根据上面的回答做了一个快速的扩展

extension Float {
    func round(decimalPlace:Int)->Float{
        let format = NSString(format: "%%.%if", decimalPlace)
        let string = NSString(format: format, self)
        return Float(atof(string.UTF8String))
    }
}

用法:

let floatOne:Float = 3.1415926
let floatTwo:Float = 3.1425934
print(floatOne.round(2) == floatTwo.round(2))
// should be true