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How to convert a number from unsigned to signed?

signed: -32768 to 32767 unsigned: 0 to 65535

I am solving the problem in JavaScript. The situation is that I have a number that goes e.g. from 0 to 65535 and I want to convert it to a reasonable signed value.

e.g.: 65535 should bee -1.

Please do not use any bit related operations but something arithmetical.

I guess this should be language independent assuming that we use a data type that is big enough.

Update: Implementation according to the answer further down:

function convertWordToShort(ival) {
    if (isNaN(ival) === false) {
        if (ival > 32767) {
            ival = ival - 65536;
        }
    }
    return ival;
}
function convertShortToWord(ival) {
    if (isNaN(ival) === false) {
        if (ival < 0) {
            ival = ival + 65536;
        }
    }
    return ival;
}
function convertIntToDWord(ival) {
    if (isNaN(ival) === false) {
        if (ival < 0) {
            ival = ival + 4294967296;
        }
    }
    return ival;
}
function convertDWordToInt(ival) {
    if (isNaN(ival) === false) {
        if (ival > 2147483647) {
            ival = ival - 4294967296;
        }
    }
    return ival;
}

How to convert a number from unsigned to signed?

signed: -32768 to 32767 unsigned: 0 to 65535

I am solving the problem in JavaScript. The situation is that I have a number that goes e.g. from 0 to 65535 and I want to convert it to a reasonable signed value.

e.g.: 65535 should bee -1.

Please do not use any bit related operations but something arithmetical.

I guess this should be language independent assuming that we use a data type that is big enough.

Update: Implementation according to the answer further down:

function convertWordToShort(ival) {
    if (isNaN(ival) === false) {
        if (ival > 32767) {
            ival = ival - 65536;
        }
    }
    return ival;
}
function convertShortToWord(ival) {
    if (isNaN(ival) === false) {
        if (ival < 0) {
            ival = ival + 65536;
        }
    }
    return ival;
}
function convertIntToDWord(ival) {
    if (isNaN(ival) === false) {
        if (ival < 0) {
            ival = ival + 4294967296;
        }
    }
    return ival;
}
function convertDWordToInt(ival) {
    if (isNaN(ival) === false) {
        if (ival > 2147483647) {
            ival = ival - 4294967296;
        }
    }
    return ival;
}

• javascript
• jquery
• c

Share Improve this question Follow edited Mar 14, 2013 at 9:33 Matthias asked Mar 13, 2013 at 15:31 MatthiasMatthias 1,43644 gold badges2727 silver badges5959 bronze badges 11

• I don't quite get your rules, why is 65535 = -1? – Agi Sferro Commented Mar 13, 2013 at 15:34
• That is because if you do that conversion in C programming language this is exactly what es out. UInt16 x = 65535; var y = (Int16)x; // y = -1 – Matthias Commented Mar 13, 2013 at 15:36
• 2 Different languages have their own rules about what to do in these situations. – teppic Commented Mar 13, 2013 at 15:39
• 1 @Matthias Your "C" syntax is pretty exotic. :) – unwind Commented Mar 13, 2013 at 15:42
• Its C but with a # at the end I guess. unsigned short x = 65535; short y = (short)x//y=-1 – Matthias Commented Mar 13, 2013 at 15:54

 |  Show 6 more ments

3 Answers 3

Sorted by: Reset to default 10

Just test if the number is over halfway, then subtract the modulus.

if(x > 32767) {x = x - 65536;}

In the case that you would be looking for bitwise operation related answers, you could try the following:

• to convert something akin to a 16 bit unsigned integer (i.e 0 <= n <= 65535) to a 16 bit signed one:

(number << 16) >> 16

• Or:

new Int16Array([number])[0]

Where number in both cases is your 16 bit number.

As a side note the reason the first solution works is because if you bit shift to the right 16 times, the most significant bit of your 16 bit number will actually bee the most significant bit of the 32 bit JavaScript integer (so if the most significant bit was a 1, it'd make the number negative), and so when you shift it to the left 16 times it'd shift while keeping the standard 2s plement form and retain the value/sign it gained from being shifted to the right previously, see this Wikipedia article for more:
https://en.m.wikipedia/wiki/Arithmetic_shift

function signed(bits, value) { return value & (1 << (bits - 1)) ? value - (1 << bits) : value; }

signed(8, 0xFF); // returns -1

signed(16, 0xFF); // returns 255

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