浮点比较 [英] Floating point comparison

问题描述

  INT的main（）
{
    浮动= 0.7;
    浮动B = 0.5;
    如果（一个或下; 0.7）
    {
       如果（B小于0.5）的printf（2是右）;
       其他的printf（1是正确的）;
    }
    其他的printf（0是对的）;
}
 

我本来期望这code的输出为 0是正确。
但令我失望的是输出为 1是正确为什么？

解决方案

  INT的main（）
{
    浮动= 0.7，B = 0.5; //这些花车
    如果（A＆LT; .7）//这是一个双
    {
      如果（B＆LT; .5）//这是一个双
        的printf（2是正确的）;
      其他
        的printf（1是正确的）;
    }
    其他
      的printf（0是对的）;
}
 

花车获得比较期间晋升为双打，并且由于浮子precise增加了一倍少，0.7浮法是的不的相同的0.7为双。在这种情况下，当它被提升0.7浮法变差至0.7为双。正如Christian表示，0.5是2的权力始终用来重新presented准确，所以测试按预期工作： 0.5 LT; 0.5 是假的。

因此​​，无论

• 更改浮动到双击或


• 更改 0.7 和 .5，到 .7f 和 .5f ，

，你会得到预期的行为。

int main()
{
    float a = 0.7;
    float b = 0.5;
    if (a < 0.7)
    {
       if (b < 0.5) printf("2 are right");
       else         printf("1 is right");
    }
    else printf("0 are right");
}

I would have expected the output of this code to be 0 are right. But to my dismay the output is 1 is right why?

解决方案

int main()
{
    float a = 0.7, b = 0.5; // These are FLOATS
    if(a < .7)              // This is a DOUBLE
    {
      if(b < .5)            // This is a DOUBLE
        printf("2 are right");
      else
        printf("1 is right");
    }
    else
      printf("0 are right");
}

Floats get promoted to doubles during comparison, and since floats are less precise than doubles, 0.7 as float is not the same as 0.7 as double. In this case, 0.7 as float becomes inferior to 0.7 as double when it gets promoted. And as Christian said, 0.5 being a power of 2 is always represented exactly, so the test works as expected: 0.5 < 0.5 is false.

So either:

• Change float to double, or:
• Change .7 and .5 to .7f and .5f,

and you will get the expected behavior.

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