js中的toFixed神坑

四舍六入五看缘分。

知乎上看到一个靠谱的分析

学会查标准
摘自ECMA-262 5.1 Edition

15.7.4.5 Number.prototype.toFixed (fractionDigits)

Return a String containing this Number value represented in decimal fixed-point notation with fractionDigits 
digits after the decimal point. If fractionDigits is undefined, 0 is assumed. Specifically, perform the following 
steps: 
1. Let f be ToInteger(fractionDigits). (If fractionDigits is undefined, this step produces the value 0). 
2. If f < 0 or f > 20, throw a RangeError exception. 
3. Let x be this Number value. 
4. If x is NaN, return the String "NaN". 
5. Let s be the empty String. 
6. If x < 0, then 
	a. Let s be "-". 
	b. Let x = –x. 
7. If x >= 10^21, then 
	a. Let m = ToString(x). 
8. Else, x < 10^21
	a. Let n be an integer for which the exact mathematical value of n / 10^f – x is as close to zero as possible. If there are two such n, pick the larger n. 
	b. If n = 0, let m be the String "0". Otherwise, let m be the String consisting of the digits of the decimal representation of n (in order, with no leading zeroes). 
	c. If f != 0, then 
		i. Let k be the number of characters in m. 
		ii. If k ≤ f, then 
			1. Let z be the String consisting of f+1–k occurrences of the character '0'. 
			2. Let m be the concatenation of Strings z and m. 
			3. Let k = f + 1. 
		iii. Let a be the first k–f characters of m, and let b be the remaining f characters of m. 
		iv. Let m be the concatenation of the three Strings a, ".", and b. 
9. Return the concatenation of the Strings s and m. 

根据上述步骤

9.955.toFixed(2)的过程为

1. f = 2
3. x = 9.955
5. s = ''
8.

	a. n = 996
		原因
		当 n = 995 时 n / 10^f – x = -0.005000000000000782
		当 n = 996 时 n / 10^f – x = 0.005000000000000782
		同样靠近0, 选择大的那个
	b. m = '996'
	c.
		i. k = 3
		iii. a = '9', b = '96'
		iv. m = a + '.' + b = '9.96'
9. Return s + m = '9.96'

9.655.toFixed(2)的过程为

1. f = 2
3. x = 9.655
5. s = ''
8.

	a. n = 965
		原因
		当 n = 965 时 n / 10^f – x = -0.004999999999999005
		当 n = 966 时 n / 10^f – x = 0.005000000000000782
		965更靠近0
	b. m = '965'
	c.
		i. k = 3
		iii. a = '9', b = '65'
		iv. m = a + '.' + b = '9.65'
9. Return s + m = '9.65'

简而言之其实还是浮点误差的锅