相同的浮点值作为比较不相等 [英] Identical float values comparing as inequal

问题描述

我在做一些'确定'的输出在C问题我遇到了这个问题，它看着它的脸很简单，但在运行code后，给我留下了疑惑。

我的预期产量为真。但是一旦运行，这是假。而当我检查的值f用printf（），它表现出0.1。是否有人可以解释为什么f为0.1分配不为IF语句返回true？

有是从那里我拿起这个问题了答案没有解释，我是不是能够找到一个答案我自己。

 的#include＆LT;＆stdio.h中GT;诠释的main（）
{
    浮动F = 0.1;
    如果（F = = 0.1）
        的printf（真）;
    其他
        的printf（假）;
}
 

解决方案

的问题是，变量˚F定义为具有类型浮动而浮动恒 0.1 的类型为双击。键入双击有更多的precision类型不是浮动。它有更多的二进制数字来重新present部分。因此，在这个声明

 浮动F = 0.1;
 

有一个截断。

要得到你应该写预期的结果至少

 如果（F == 0.1F）
 

此外，要确保没有四舍五入code看起来应该

 的#include＆LT;＆stdio.h中GT;INT主要（无效）
{
    浮动F = 0.1F;    如果（F == 0.1F）
        的printf（真\\ n）;
    其他
        的printf（假\\ n）;
}
 

I was working on some 'determine the output' questions in C. I came across this question which looked simple on the face of it, but after running the code left me puzzled.

The output I expected was "True". However upon running, it was "False". And when I checked the value of f using printf(), it showed 0.1. Can someone please explain why f being assigned 0.1 doesn't return true for the IF statement?

There was no explanation for the answer from where I picked the question up, and I wasn't able to find an answer myself.

#include <stdio.h>

int main()
{
    float f = 0.1;
    if (f == 0.1)
        printf("True");
    else
        printf("False");
}

解决方案

The problem is that variable f defined as having type float while float constant 0.1 has type double. Type double has more precision than type float. it has more binary digits to represent fraction. So in this statement

        float f = 0.1;

there is a truncation.

To get the expected result you should write at least

        if (f == 0.1f)

Also that to be sure that there is no rounding the code should look as

#include <stdio.h>

int main( void )
{
    float f = 0.1f;

    if ( f == 0.1f )
        printf("True\n");
    else
        printf("False\n");
}

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