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Home/ Questions/Q 7601711
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T23:13:52+00:00 2026-05-30T23:13:52+00:00

Why is this behavior difference between parseInt() and parseFloat() ? I have a string

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Why is this behavior difference between parseInt() and parseFloat()?

I have a string that contains 08 in it.

When I write this code:

alert(hfrom[0]);
alert(parseInt(hfrom[0]));
alert(parseFloat(hfrom[0]));

The following output is generated:

08
0
8

Why does parseInt and parseFloat return two different results in this case?

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  1. Editorial Team
    Editorial Team
    2026-05-30T23:13:53+00:00Added an answer on May 30, 2026 at 11:13 pm

    parseInt() assumes the base of your number according to the first characters in the string. If it begins with 0x it assumes base 16 (hexadecimal). Otherwise, if it begins with 0 it assumes base 8 (octal). Otherwise it assumes base 10.

    You can specify the base as a second argument:

    alert(parseInt(hfrom[0], 10)); // 8
    

    From MDN (linked above):

    If radix is undefined or 0, JavaScript assumes the following:

    If the input string begins with “0x” or “0X”, radix is 16
    (hexadecimal). If the input string begins with “0”, radix is eight
    (octal). This feature is non-standard, and some implementations
    deliberately do not support it (instead using the radix 10). For this
    reason always specify a radix when using parseInt. If the input string
    begins with any other value, the radix is 10 (decimal).

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