Objective-C 中的浮点数问题 [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！

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

我的代码是(这只是一个示例代码，但我的原始代码是这样做的):

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);

输出是 52.450001 而不是 52.450000！

the output is 52.450001 and not 52.450000 !

我不知道，因为这种情况发生了！

I don't know because this happens!

感谢您的帮助！

~解决~

谢谢大家！是的，我已经用 double 类型解决了！

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

推荐答案

浮点数是具有一定精度的数字表示.并非每个值都可以用这种格式表示.请参见此处.

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

你可以很容易地想到为什么会这样:在区间 (1..1) 中有无限数量的数字，但浮点数只有有限数量的位来表示 (-MAXFLOAT..MAXFLOAT).

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).

更恰当地说:在 32 位整数表示中，有可数数量的整数要表示，但是在 32 位或 64 位的有限表示中无法完全表示无限数量的实数值.因此，不仅对可表示的最高和最低实值有限制，而且对准确率也有限制.

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.

那么为什么浮点数后面的小数位会受到影响呢?因为表示是基于二进制而不是十进制的，所以其他数字比十进制更容易表示.

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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