在Objective-C中的浮动问题 [英] Trouble with floats in Objective-C

问题描述

我有一个小问题，我找不到解决方案！

我的代码是（这只是一个示例代码，但我的原始代码做某事像这样）：

  float x = [@2.45floatValue]; 
 
 
 for（int i = 0; i <100; i ++）
 x + = 0.22; 
 
 NSLog（@％f，x）; 
  

输出为52.450001而不是52.450000！

我不知道，因为这发生！

感谢任何帮助！

〜SOLVED〜 b
$ b

感谢大家！是的，我已经解决了double类型！

解决方案

浮点数是一个精度的数字表示。不是每个值都可以用这种格式表示。请参见此处。

你可以很容易想到为什么会是这种情况：在intervall（1..1）中有一个无限数量的数字，有限数量的位表示（-MAXFLOAT..MAXFLOAT）中的所有数字。

更合适的是：在32位整数表示中，有一个可计数的整数表示，但有无限的无数的实数值不能完全表示在32或64位的有限表示。因此，不仅对最高和最低可表示的实际价值有限制，而且对准确性也有限制。

因此，为什么在浮点数后面有一个数字受影响？因为表示是基于二进制系统而不是十进制，所以其他数字很容易表示，而十进制。

I've a small problem and I can't find a solution!

My code is (this is only a sample code, but my original code do something like this):

float x = [@"2.45" floatValue];


for(int i=0; i<100; i++)
    x += 0.22;

NSLog(@"%f", x);

the output is 52.450001 and not 52.450000 !

I don't know because this happens!

Thanks for any help!

~SOLVED~

Thanks to everybody! Yes, I've solved with the double type!

解决方案

Floats are a number representation with a certain precision. Not every value can be represented in this format. See here as well.

You can easily think of why this would be the case: there is an unlimited number of number just in the intervall (1..1), but a float only has a limited number of bits to represent all numbers in (-MAXFLOAT..MAXFLOAT).

More aptly put: in a 32bit integer representation there is a countable number of integers to be represented, But there is an infinite innumerable number of real values that cannot be fully represented in a limited representation of 32 or 64bit. Therefore there not only is a limit to the highest and lowest representable real value, but also to the accuracy.

So why is a number that has little digits after the floating point affected? Because the representation is based on a binary system instead of a decimal, making other numbers easily represented then the decimal ones.

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