比较相同的浮点值奇怪的输出？ [英] Strange output when comparing same float values?
问题描述
我读了浮点上面的链接， 。
#include< stdio.h>
int main（）
{
float x = 0.5;
if（x == 0.5）
printf（IF）;
else if（x == 0.5f）
printf（ELSE IF）;
else
printf（ELSE）;
$现在，根据促销规则，不应该 ELSE IF 必须打印？
但是，这里是打印 IF
编辑：是因为二进制中的0.5 = 0.1，而且之后一切都是0，精度的损失因此没有效果，所以比较 IF 返回true。
如果是0.1,0.2,0.3,0.4,0.6,0.7 ....那么 Else如果块返回true。
请原谅我提出同样的问题，因为我从上面的链接浮动比较绝不能做。
但是，这种意外行为的原因是什么？
解决方案
浮点数是不准确的。
这种说法是错误的。一些浮点数是准确的，如1.0，12345.0，12345.5，-2.25。所有这些数字都可以表示为由2的幂作为整数。所有数字不能也不准确。
在您的具体情况下， float x = 0.5 结果为 x ，其值 1.00000000 * 2 ^ -1 。当你把它与 double 0.5 进行比较时，两个操作数都被转换为 double ，所以比较变成 1.000000000000000 * 2 ^ -1 == 1.000000000000000 * 2 ^ -1 ，即可成功
float x = 0.1 ，看起来不一样。该值存储为 1.01010101 * 2 ^ -3 （或类似）。请注意，这已经不准确了。当你把它和 double 0.1 进行比较时，float在最后加零，比较变成 1.010101010000000 * 2 ^ -3 == 1.010101010101010 * 2 ^ -3 ，失败。
Comparing Same Float Values In C
strange output in comparison of float with float literal
Float addition promoted to double?
I read the above links on floating points, but even getting strange output.
#include<stdio.h>
int main()
{
float x = 0.5;
if (x == 0.5)
printf("IF");
else if (x == 0.5f)
printf("ELSE IF");
else
printf("ELSE");
}
Now, according to the promotion rules, Shouldn't "ELSE IF" must be printed ?
But, here it is printing "IF"
EDIT : Is it because 0.5 = 0.1 in binary and everything is 0 after that and loss of precision hence no effects, so comparison IF returns true.
Had it been 0.1, 0.2, 0.3, 0.4, 0.6, 0.7.... , then Else If block returns true.
Pardon me asking same question, because I have read from the above links that floats comparison must never be done.
But, What is the reason of this unexpected behaviour ?
解决方案
Floating point numbers are never accurate.
This statement is wrong. Some floating point numbers are accurate, such as 1.0, 12345.0, 12345.5, -2.25. All these numbers can be represented as integers didived by a power of 2. All numbers that cannot also are not accurate.
In your specific case, float x = 0.5 results in x having the value 1.00000000 * 2^-1. When you compare this against double 0.5, both operands are converted to double, so the comparison becomes 1.000000000000000 * 2^-1 == 1.000000000000000 * 2^-1, which succeeds.
For float x = 0.1, it looks different. The value is stored as 1.01010101 * 2^-3 (or similar). Note that this is already not precise. When you compare this against double 0.1, the float is extended with zeros at the end, the comparison becomes 1.010101010000000 * 2^-3 == 1.010101010101010 * 2^-3, which fails.
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