I don't think so. ES5 specified the algorithm of toExponential. “The number is the same” is not the only requirement.
See the algorithm below: ( please pay attention to 8.a ) 1. Let *x* be this Number value <http://es5.github.com/#this-Number-value>. 2. Let *f* be ToInteger <http://es5.github.com/#x9.4>(*fractionDigits*). 3. If *x* is *NaN*, return the String *"NaN"*. 4. Let *s* be the empty String. 5. If *x* *<* 0, then 1. Let *s* be *"-"*. 2. Let *x* = �C*x*. 6. If *x* = +∞, then 1. Return the concatenation of the Strings *s* and *"Infinity"*. 7. If *fractionDigits* is not *undefined* and (*f* < 0 or *f* > 20), throw a *RangeError <http://es5.github.com/#x15.11.6.2>* exception. 8. If *x* = 0, then 1. *Let f = 0.* 2. Let *m* be the String consisting of *f*+1 occurrences of the character ‘0’. 3. Let *e* = 0. 9. Else, *x* ≠ 0 1. If fractionDigits is not *undefined*, then 1. Let *e* and *n* be integers such that 10*f* ≤ *n* < 10*f*+1 and for which the exact mathematical value of *n* × 10*e*�C*f* �C *x* is as close to zero as possible. If there are two such sets of *e* and *n*, pick the *e* and *n* for which *n* × 10*e*�C*f* is larger. 2. Else, *fractionDigits* is *undefined* 1. Let *e*, *n*, and *f* be integers such that *f* ≥ 0, 10*f* ≤ *n* < 10*f*+1, the number value for n× 10*e*�C*f* is *x*, and *f* is as small as possible. Note that the decimal representation of*n* has * f*+1 digits, *n* is not divisible by 10, and the least significant digit of *n* is not necessarily uniquely determined by these criteria. 3. Let *m* be the String consisting of the digits of the decimal representation of *n* (in order, with no leading zeroes). 10. If *f* ≠ 0, then 1. Let *a* be the first character of *m*, and let *b* be the remaining *f * characters of *m*. 2. Let *m* be the concatenation of the three Strings *a*, *"."*, and *b*. 11. If *e* = 0, then 1. Let *c* = *"+".* 2. Let *d* = *"0".* 12. Else 1. If *e* > 0, then let *c* = *"+".* 2. Else, *e* ≤ 0 1. Let *c* = *"-"*. 2. Let *e* = �C*e*. 3. Let *d* be the String consisting of the digits of the decimal representation of *e* (in order, with no leading zeroes). 13. Let *m* be the concatenation of the four Strings *m*, *"e"*, *c*, and *d* . 14. Return the concatenation of the Strings *s* and *m*. According to this algorithm, (0).toExponential(1) should return "0e+0".(f is set to 0) But it looks none of V8, spidermonkey and JScript did that thing. Their result is "0.0e+0". So is there anyone could give some message about why does ES5 define this behavior like this? Or is it simply a spec bug? 2012/3/8 Jens Nockert <j...@nockert.se> > > On H.24/03/08, at 12:39, DX Jin <dox....@gmail.com> wrote: > > > (0).toExponential(1) should, according to the algorithm in 15.7.4.6, > return "0e+0". > > > > But all browser implementations return "0.0e+0" instead. > > > > Which is wrong? The specification or the implementations? > > Neither, the result is correct. The number is the same, just different > representations as strings. > _______________________________________________ > es-discuss mailing list > es-discuss@mozilla.org > https://mail.mozilla.org/listinfo/es-discuss >
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