Increment char arrays int values by 1

Question

Hey everyone I've got a char array which holds an ID field. The id field looks as follows i0001. I want to increment the ID by one when i create a new link in my linked list.

So I sort through the linked list to the last value, which is for example; i0005, now I want to create my new field as i0006. Obviously I can't just increment the value of the array because it's a char, how should I go about this?

char id[5];

I tried:

id[4] = id[4]+1;

The issue with this is it doesn't increment id[3] when it hits 10, instead it starts replacing the numerical value with symbols (i assume these are the ascii values after number).

Anyone got any idea of how they would approach this? I'm a bit stumped!

If the format is constant, i.e. `iXXX` then you can simply get rid of `i` (move pointer to the string by one), convert string to number (use `strtol()` function), increase the number by 1 and finally write it back to string (use `sprtinf()` function). — Crozin, Oct 13 '15 at 12:02
I would store it as a number and generate an id string as needed. — molbdnilo, Oct 13 '15 at 12:07
I agree with @molbdnilo. However, if you are going to do this - `id[4] = id[4]+1;` - don't: instead, use `++id[4];`. A sane compiler would optimise them to be the same, but why type more when you don't have to? — underscore_d, Oct 13 '15 at 12:11
@molbdnilo: with the (possibly important) side note that these are not valid strings to begin with. — Jongware, Oct 13 '15 at 12:12
It's not a good idea to increase on string values. As when you have a value `0009`, would you turn this to `0010`. Using an `int` value maybe a better solution. And it's not hard to turn the int to a string you want. — luoluo, Oct 13 '15 at 12:14
Definitely, the problem you cited is why this thread exists! My suggestion was more general, that we'd want to avoid referencing any variable multiple times when simpler operators exist (increment, etc.). It makes typing/reading easier (& might slow a _really_ bad _theoretical compiler) — underscore_d, Oct 13 '15 at 12:23

score 2 · Accepted Answer · answered Oct 13 '15 at 12:08

2

Implementing carry by yourself should be good.

void inclementArray(char *id) {
    int i;
    for (i = 4; i >= 1; i--) {
        id[i]++;
        if (id[i] > '9') {
            id[i] = '0';
            /* do carry and proceed to next digit */
        } else {
            /* no carry, the calculation is completed */
            break;
        }
    }
}

Note: char id[5]; is sufficient for storing {'i', '0', '0', '0', '5'}, but it is insufficient to store null-terminated string "i0005".

answered Oct 13 '15 at 12:08

MikeCAT

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3

`inclementArray`? is it rainy? :-) – underscore_d Oct 13 '15 at 12:14
Thanks man! This seems to be the most simple way to solve my issue, appreciate it friend! – Ryan Oct 13 '15 at 12:30
Seems like there's some off-by-one bug here when the last digit is handled. What will end up in `id[0]`? – Lundin Oct 13 '15 at 12:59
The first digit is always to be i so we don't change that, thats why id[0] isn't changed – Ryan Oct 13 '15 at 13:19

ameyCU · Answer 2 · 2015-10-13T12:20:33.540

0

Here is a simple program which will give you an idea how to do that -

#include <stdio.h>
#include <stdlib.h>

int main(void){
   char id[6]="i0001";            // your id 
   char a[5];                     // array to hold first part of id
   int i;                         // integer variable to hold number (non-zero)
   sscanf(id,"%[^1-9]%d",a,&i);   // assign non zero number to i
   i=i+1;                         // add 1 to i 
   sprintf(id,"%s%d",a,i);        // store both a and i again into id
   printf("%s",id);
}

Check output here.

edited Oct 13 '15 at 12:20

answered Oct 13 '15 at 12:10

ameyCU

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try `char id[6]="i0009";` – BLUEPIXY Oct 13 '15 at 12:24

score 0 · Answer 3 · answered Oct 13 '15 at 12:11

0

Something like this:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define PREFIX "i"

int main(void)
{
    char id[] = "i0001";
    printf("before: %s\n", id);
    int digit = atoi(strtok(id, PREFIX));
    ++digit;
    snprintf(id, sizeof(id), "%s%04d", PREFIX, digit);
    printf("after: %s\n", id);
    exit(0);
}

answered Oct 13 '15 at 12:11

Arkadiusz Drabczyk

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using `strtok` for this simple case is just overkill – phuclv Oct 13 '15 at 12:43
I agree, I think [this answer](http://stackoverflow.com/a/33102322/3691891) is the best. – Arkadiusz Drabczyk Oct 13 '15 at 12:52
see also [Why shouldn't I use atoi()?](https://stackoverflow.com/q/17710018/995714) – phuclv Aug 02 '19 at 04:35

phuclv · Answer 4 · 2019-08-02T04:38:20.133

Doing simply like this

id[4] = id[4]+1;

will not carry to the next digit. You must carry it yourself, but that of course need many more lines of code.

A simple way is to convert the number to value, increase and print back

char id[] = "i0005";
int idval;

sscanf(id, "i%d", &idval);
snprintf(id, sizeof(id), "i%d", idval+1);

Alternatively scan the number ignoring the i prefix like this

scanf(id+1, "%d", &idval);
snprintf(id+1, sizeof(id)-1, "%d", idval+1);

Increment char arrays int values by 1

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