###彩蛋一#####
今天debug一个程序,发现一个很有趣的问题:
float 类型不能表示 16777217这个数。
比如:
float i=0; while(true){ i++; if( i >= 16777217 ){ break; } }
如果这样写的话,就会死循环,因为 float 型浮点数不能表示16777217这个数,所以当 i 等于 16777216的时候,往上加 1 ,还是16777216,所以永远跳不出那个while循环。
为什么会这样子呢?这就牵扯到浮点数的表示问题了。引用stackoverflow上一个人的回答如下(链接):
Short roundup of IEEE-754 floating point numbers (32-bit) off the top of my head:
- 1 bit sign (0 means positive number, 1 means negative number)
- 8 bit exponent (with -127 bias, not important here)
- 23 bits "mantissa"
- With exceptions for the exponent values 0 and 255, you can calculate the value as:
(sign ? -1 : +1) * 2^exponent * (1.0 + mantissa)
- The mantissa bits represent binary digits after the decimal separator, e.g.
1001 0000 0000 0000 0000 000 = 2^-1 + 2^-4 = .5 + .0625 = .5625
and the value in front of the decimal separator is not stored but implicitly assumed as 1 (if exponent is 255, 0 is assumed but that's not important here), so for an exponent of 30, for instance, this mantissa example represents the value1.5625
Now to your example:
16777216 is exactly 224, and would be represented as 32-bit float like so:
- sign = 0 (positive number)
- exponent = 24 (stored as 24+127=151=
10010111
)- mantissa = .0
- As 32 bits floating-point representation:
0 10010111 00000000000000000000000
- Therefore: Value =
(+1) * 2^24 * (1.0 + .0) = 2^24 = 16777216
Now let's look at the number 16777217, or exactly 224+1:
- sign and exponent are the same
- mantissa would have to be exactly 2-24 so that
(+1) * 2^24 * (1.0 + 2^-24) = 2^24 + 1 = 16777217
- And here's the problem. The mantissa cannot have the value 2-24 because it only has 23 bits, so the number 16777217 just cannot be represented with the accuracy of 32-bit floating points numbers!
假设你的机器的浮点数表示是符合IEEE-754标准的(一般都是),那么,表示16777216这个数的时候,情况是这样的:
- sign = 0(表示正数)
- exponent=24
- mantissa(尾数) = .0
- 结合起来就是 (+1) * 2^24 * (1.0 + .0) = 2^24 = 16777216
现在,要给这个数 + 1 的时候会发生什么?
显然不能是 2^25,那就太大了。
按照浮点数运算的特点,“方向” 应该是这样: (+1) * 2^24 * ( 1.0 + 2^(-24) ) = 2^24 + 1 = 16777217 。然而,mantissa(尾数)部分只有 23 位,凑不成一个 1 ,所以,float类型根本就不能表示 16777217这个数。
BONUS:如果要表示16777218这个数呢?这是可以的,只需要:(+1) * 2^24 * ( 1.0 + 2^(-23) ) = 2^24 + 2 = 16777218
同样的,double型的数也是,不能表示 199,254,740,993 (2^53+1)这个数(符合IEEE-754标准的double型的尾数(mantissa)是53位的)
参考:stackoverflow_1, stackoverflow_2
####彩蛋二####
用gcc编译C程序,按标准,假如是C89(-std=c89 / -ansi)那么,是不能使用Variable Length Array(VLA)的,也就是,不能这样:
void foo(int n){ int arr[n]; ... }
只有C99之后才能这样。
然而,gcc是不遵守这个规范的【1】【2】,即使你使用了-std=c89.....(FUCK)......如果要按照C89规范来,要加上 -pedantic / -pedantic-errors。
在C++中,无论哪个版本,都不支持VLA,然而g++还是默认开启了 (FUCK....)
Clang也是,FUCK ......
:)