Re: increment a hex number

If you have the numbers 0x000000, 0x010000, 0x020000 and 0x030000, then you can increment them using addition ($array[-1] += 1;) or using the pre- or post-increment operator (++$array[-1]; or $array[-1]++;).

my @nums = (
    0x000000,
    0x010000,
    0x020000,
    0x030000,
);

++$nums[-1];

printf("0x%05X\n", $nums[-1]);
[download]

If you have the strings 0x000000, 0x010000, 0x020000 and 0x030000, then you need to convert them to numbers first using hex.

my @formatted_nums = (
    '0x000000',
    '0x010000',
    '0x020000',
    '0x030000',
);

$formatted_nums[-1] = sprintf("0x%05X\n", hex($formatted_nums[-1]) + 1
+);

print("$formatted_nums[-1]\n");
[download]

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Re^2: increment a hex number by Anonymous Monk on May 10, 2010 at 22:30 UTC
thanks for the input, but I cannot use 0x030001. I meant to say I need it to be increment from 0x030000 to 0x040000. ideas?	[reply]
Re^3: increment a hex number by graff (Chancellor) on May 10, 2010 at 22:47 UTC
Instead of using "+1" (or "++") for the "increment", you would use "+ 0x10000". That still involves treating the values as numbers rather than as strings: `$_ = "0x030000\n"; s/0x([\da-f]+)/sprintf("%#.*x", length($1), hex($1) + 0x10000)/ei; print;` [download] That snippet will print "0x040000".	[reply] [d/l]
Re^4: increment a hex number by Anonymous Monk on May 10, 2010 at 23:00 UTC
thats dude....that will work! The code I do isn't much. I have a hex # converted to binary then using base64 math, I did some division built a key=>value such as 0x010000 => 65536 and thought this way seems silly and what if I miss a hex number. say my array only has 0x010000 to 0x050000....	[reply]
Re^3: increment a hex number by toolic (Bishop) on May 10, 2010 at 22:47 UTC
ideas? Yes. I think it's time you showed us the code you are running, the output you are getting and the output you expect. Try to look at this from our perspective: you are asking us how to add two numbers together. Why doesn't this work for you? `use strict; use warnings; my $n = 0x30000; $n += 0x10000;` [download] See also: Scalar value constructors How do I convert between numeric representations/bases/radixes?	[reply] [d/l]
Re^3: increment a hex number by almut (Canon) on May 10, 2010 at 23:07 UTC
Like the others said. BTW, the same arithmetic principle applies to decimal numbers, too: `30000 + 10000 ------- 40000` [download] Or any other radix.	[reply] [d/l]