浮点比较的问题 [英] problems in floating point comparison

问题描述

void main()
{
    float f = 0.98;
    if(f <= 0.98)
        printf("hi");
    else
        printf("hello");
    getch();
}

我使用的F I我得到不同的结果，不同的浮点数得到这个问题here.On。
为什么会这样？

I am getting this problem here.On using different floating point values of f i am getting different results. Why this is happening?

推荐答案

˚F使用浮动 precision，但是0.98是双击默认precision，这样的声明 F＆LT; = 0.98 比较使用双击 precision。

f is using float precision, but 0.98 is in double precision by default, so the statement f <= 0.98 is compared using double precision.

的˚F因此在比较转换为双击，但可能使结果比0.98略大

The f is therefore converted to a double in the comparison, but may make the result slightly larger than 0.98.

使用

if(f <= 0.98f)

或使用双击为˚F代替。

在细节......假设浮动 IEEE单precision 和双击是的 IEEE双precision 。

In detail... assuming float is IEEE single-precision and double is IEEE double-precision.

这些类型的浮点数的存储与基地-2再presentation。在基-2-这个数需要一个无限precision重新present，因为它是一个重复的小数

These kinds of floating point numbers are stored with base-2 representation. In base-2 this number needs an infinite precision to represent as it is a repeated decimal:

0.98 = 0.1111101011100001010001111010111000010100011110101110000101000...

A 浮动只能存储显著人物24位，即

A float can only store 24 bits of significant figures, i.e.

       0.111110101110000101000111_101...
                                 ^ round off here
   =   0.111110101110000101001000

   =   16441672 / 2^24

   =   0.98000001907...

A 双击可存储signficant人物53位，因此

A double can store 53 bits of signficant figures, so

       0.11111010111000010100011110101110000101000111101011100_00101000...
                                                              ^ round off here
   =   0.11111010111000010100011110101110000101000111101011100

   =   8827055269646172 / 2^53

   =   0.97999999999999998224...

所以，0.98将成为浮动在稍大，小双。

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