Javascript 为什么toFixed（）只接受0-20个数字

在javascript文档中

toFixed（）只需要0-20秒，所以

123.1234567294239482379483749.toFixed(20)

是

并导致范围误差

123.1234567294239482379483749.toFixed(21)
RangeError: toFixed() digits argument must be between 0 and 20

感谢ECMA文档中的

，toFixed被指定为

注意第2点，这是该范围的原因

15.7.4.5 Number.prototype.toFixed (fractionDigits)
Return a string containing the number represented in 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, go to step 9.
7. Let s be "-".
8. Let x = –x.
9. If x ≥ 1021, let m = ToString(x) and go to step 20.
10. Let n be an integer for which the exact mathematical value of n ÷ 10f – x is as close to zero as
possible. If there are two such n, pick the larger n.
11. 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).
12. If f = 0, go to step 20.
13. Let k be the number of characters in m.
14. If k > f, go to step 18.
15. Let z be the string consisting of f+1–k occurrences of the character ‘0’.
16. Let m be the concatenation of strings z and m.
17. Let k = f + 1.
18. Let a be the first k–f characters of m, and let b be the remaining f characters of m.
19. Let m be the concatenation of the three strings a, ".", and b.
20. Return the concatenation of the strings s and m.

15.7.4.5 Number.prototype.toFixed（分形数字）
返回一个字符串，该字符串包含使用fractionDigits的定点表示法表示的数字
小数点后。如果未定义fractionDigits，则假定为0。具体来说，执行以下操作：
以下步骤：
1.设f为整数（分数位数）。（如果未定义fractionDigits，此步骤将生成值
0).
2.如果f<0或f>20，则抛出RangeError异常。
3.设x为该数值。
4.如果x是NaN，则返回字符串“NaN”。
5.让我们成为空字符串。
6.如果x≥ 0，转至步骤9。
7.让我们成为“-”。
8.设x=-x。
9如果x≥ 1021，让m=ToString（x）并转至步骤20。
10设n为整数，其精确数学值n÷10f–x尽可能接近于零
可能的如果有两个这样的n，选择较大的n。
11如果n=0，则m为字符串“0”。否则，设m是由
n的十进制表示法（按顺序，无前导零）。
12如果f=0，则转至步骤20。
13设k为m中的字符数。
14如果k>f，则转至步骤18。
15设z是由字符“0”的f+1–k个匹配项组成的字符串。
16设m是字符串z和m的串联。
17设k=f+1。
18设a为m的第一个k–f字符，b为m的剩余f字符。
19设m是三个字符串a、“.”和b的串联。
20返回字符串s和m的串联。

在我看来似乎是任意的，可能与JS的浮点精度有关？可能与ieee754的限制有关

15.7.4.5 Number.prototype.toFixed (fractionDigits)
Return a string containing the number represented in 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, go to step 9.
7. Let s be "-".
8. Let x = –x.
9. If x ≥ 1021, let m = ToString(x) and go to step 20.
10. Let n be an integer for which the exact mathematical value of n ÷ 10f – x is as close to zero as
possible. If there are two such n, pick the larger n.
11. 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).
12. If f = 0, go to step 20.
13. Let k be the number of characters in m.
14. If k > f, go to step 18.
15. Let z be the string consisting of f+1–k occurrences of the character ‘0’.
16. Let m be the concatenation of strings z and m.
17. Let k = f + 1.
18. Let a be the first k–f characters of m, and let b be the remaining f characters of m.
19. Let m be the concatenation of the three strings a, ".", and b.
20. Return the concatenation of the strings s and m.