Decrement a hex value stored in a string

Question

Let's assume I have:

char str[]="0x7fffffffe181"

basically I want to decrement the hex to 0x7fffffffe180.

How do I do this?

_{Note: if the final character is letter e.g c after decrement it should be b}

Will the strings represent numbers that can fit in a C `long` or `unsigned long`, or can they be arbitrarily long like `0xf876c6784564645de653654ee65d56310913bbba0009897768b`? — Ray Toal, May 26 '19 at 18:32
@RayToal it could be up to an `x64` architecture memory address or `0xffffffffffffffff` — amiTheregroot, May 26 '19 at 18:34
Convert to a long long int, decrement, convert back to string. How to do the conversions is easily found on the interwebs. — lurker, May 26 '19 at 18:39
Then covert to a long, decrement, and convert back. [Solutions are here](https://stackoverflow.com/q/10156409/831878). — Ray Toal, May 26 '19 at 18:40
See also https://stackoverflow.com/questions/3464194/how-can-i-convert-an-integer-to-a-hexadecimal-string-in-c — TylerH, May 26 '19 at 19:00
You haven't really asked a question. What, exactly, are you stuck on? — lurker, May 26 '19 at 19:07

S.S. Anne · Answer 1 · 2019-05-26T19:06:49.207

1

You can convert it to an integer and back like this:

char *str2;
unsigned long long int len, a = strtoull(str, NULL, 16);
a--;
len = snprintf(NULL, 0, "0x%llx", a);
str2 = malloc(len+1);
sprintf(str2, "0x%llx", a);

edited May 26 '19 at 19:06

answered May 26 '19 at 18:56

S.S. Anne

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What happens if `str` is `"0x0"`? I suppose the result will not fit into `str2` with size 3... – Stephan Lechner May 26 '19 at 18:58
@StephanLechner Please see update. – S.S. Anne May 26 '19 at 19:02
@melpomene Thanks for the edit. Didn't catch that. – S.S. Anne May 26 '19 at 19:03
It's not portable, but `asprintf` would simplify this a bit. – melpomene May 26 '19 at 19:07
I agree, but the operating system wasn't specified in the question. – S.S. Anne May 26 '19 at 19:08
@JL2210 your answer doesn't even decrement – amiTheregroot May 26 '19 at 19:52
@amiTheregroot What does `a--` do then? – S.S. Anne May 26 '19 at 21:49

score 1 · Accepted Answer · edited May 26 '19 at 23:43

1

It only has to work in the range of an unsigned long long.

const char *eptr;
unsigned long long numeric = strtoull(str, &eptr, 16);
if (eptr == str || numeric == 0) {
    /* handle error */
}
sprintf(str, "0x%llx", numeric - 1);

We can't do this for signed hex in place because decrementing 0 or certain negative numbers would make the string longer.

In theory, 0 should work for the third argument to strtoull, but it's buggy in our libc right now so it doesn't.

edited May 26 '19 at 23:43

S.S. Anne

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answered May 26 '19 at 18:57

Joshua

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`strtoull` takes 3 arguments. `base` is undeclared. – melpomene May 26 '19 at 18:59
@melpomene: Fixed. I looked up strtoull and verified that 0 did what I wanted and still forgot to type it. :( – Joshua May 26 '19 at 18:59
@Joshua your usage of `strtoull` is very wrong. You should read up on it before trying to use it. – S.S. Anne May 26 '19 at 19:00
I still don't understand that `eptr == base` check. – melpomene May 26 '19 at 19:02
@melpomene: if strtoull is passed a pointer to something not a number, eptr will be the same as the pointer. If its passed a pointer to something that starts with a number, eptr points to after the end of the number. If it's just a number in the string, eptr is pointing at the null terminator. – Joshua May 26 '19 at 19:03
`base` is not declared anywhere. – melpomene May 26 '19 at 19:05
@melpomene: My mistake. I started with base being the input variable and didn't fix it everywhere. – Joshua May 26 '19 at 19:06
@JL2210: Not according to `man 3 strtoul`. – Joshua May 26 '19 at 19:10
Yep. Just saw that. I guess that needs to be fixed in my implementation. – S.S. Anne May 26 '19 at 19:11
@Joshua, You answer turns `0x7fffffffe171` into `0xffffffffffffe170` – amiTheregroot May 26 '19 at 19:43
@amiTheregroot: Looks like we share the same buggy implementation. Use 16 until it can be fixed. Reporting bug now. – Joshua May 26 '19 at 19:58
@JL2210: You probably don't care, but that wasn't bragging. I was expecting the question to require something too big for an integer type, which is a much harder problem. – Joshua May 26 '19 at 23:43
1

@Joshua OK. I agree, that would be hard :) – S.S. Anne May 26 '19 at 23:52

Decrement a hex value stored in a string

2 Answers2