RangeError: precision is out of range

Message

RangeError: toExponential() argument must be between 0 and 100 (V8-based & Safari)
RangeError: toFixed() digits argument must be between 0 and 100 (V8-based & Safari)
RangeError: toPrecision() argument must be between 1 and 100 (V8-based & Safari)
RangeError: precision -1 out of range (Firefox)

Error type

What went wrong?

There was an out of range precision argument in one of these methods:

• Number.prototype.toExponential(), which requires the arguments to be between 0 and 100, inclusive.
• Number.prototype.toFixed(), which requires the arguments to be between 0 and 100, inclusive.
• Number.prototype.toPrecision(), which requires the arguments to be between 1 and 100, inclusive.

Examples

Invalid cases

(77.1234).toExponential(-1); // RangeError
(77.1234).toExponential(101); // RangeError

(2.34).toFixed(-100); // RangeError
(2.34).toFixed(1001); // RangeError

(1234.5).toPrecision(-1); // RangeError
(1234.5).toPrecision(101); // RangeError

Valid cases

(77.1234).toExponential(4); // 7.7123e+1
(77.1234).toExponential(2); // 7.71e+1

(2.34).toFixed(1); // 2.3
(2.35).toFixed(1); // 2.4 (note that it rounds up in this case)

(5.123456).toPrecision(5); // 5.1235
(5.123456).toPrecision(2); // 5.1
(5.123456).toPrecision(1); // 5

See also

• Number.prototype.toExponential()
• Number.prototype.toFixed()
• Number.prototype.toPrecision()