use of strsep in C while parsing argv Variable doesen't work

Question

I have a problem in C while parsing the argv Variable. This is the task:

I got some Parameters form the Command line into my C programm. One of them looks like "-r=20-5". There I have to find the "r". this works with:

if (argv[i][0] != 0 && argv[i][1] != 0 && argv[i][2] != 0 &&
    argv[i][0] == '-' && argv[i][2] == '=') { /* Check the Variables */
    switch (argv[i][1]) {
        case 'r':
            /* what have to be here? */
            break;
}

Now I have the "r". But I need the 20 and the 5 in two different variables too. Here is what i thougt about, but it did't work. I tried something with

strsep(argv[i]+3, "-");

But my Compiler (Xcode) throws a warning (warning: passing argument 1 of 'strsep' from incompatible pointer type)

Does anyone has any idea (or link) how I can solve the problem?

After it's solved there was still a warning, so I was answered to post my entire new code:

int parse(int argc, const char *argv[]) {
    int i, x, y;
    int intTmp;
    long longTmp;
    char *strTmp;
    char *end;

    for (i = 1; i < argc; i++) {
        /* check for (-?=) */
        if (argv[i][0] != 0 && argv[i][1] != 0 && argv[i][2] != 0 &&
            argv[i][0] == '-' && argv[i][2] == '=') {

            /* IGNORE THIS
             * longTmp = strtol(argv[i]+3, &end, 10);
             * intTmp = analyseEndptr(longTmp, argv[i]+3, end);
             */

            switch (argv[i][1]) {
                case 'r':
                    strTmp = argv[i]+3;                 /* <= Here I got the warning */
                    x = atoi(strsep(&strTmp, "-"));
                    y = atoi(strsep(&strTmp, "-"));
                    printf("x=%d, y=%d\n", x, y);
                    break;
            }
        }
    }
}

The warning is: warning: passing argument 1 of 'strsep' from incompatible pointer type

My compiler is: gcc on MacOS using XCode as IDE

`strsep` takes a `char **` as its first argument, not a `char *` — Paul R, May 02 '12 at 20:22
Or you can write your own parser. It will take some time, but it will be working as you wish. All you need to do is just iterate over the string. As a separator you can use '-' symbol. — besworland, May 02 '12 at 20:27
@Paul R How do I get a char** from a char*? (I quickly google that, but with no good results) — waXve, May 02 '12 at 20:29

score 1 · Accepted Answer · answered May 02 '12 at 21:12

1

Do you know sscanf? Have a look at:

  unsigned a,b;
  if( 2==sscanf(argv[i],"-r=%u-%u",&a,&b) )
    printf("%u-%u",a,b);

answered May 02 '12 at 21:12

user411313

3,930
19
16

That one is much simpler as the [other one](http://stackoverflow.com/a/10421687/1319755). Althow, the other one works too. – waXve May 02 '12 at 21:29

dpp · Answer 2 · 2012-05-02T21:15:05.813

0

Here is how you pass a string to strsep()

    switch (argv[i][1]) {
        char *s;
        int x, y;
        case 'r':
            s = argv[i] + 3;
            x = atoi(strsep(&s, "-"));
            y = atoi(strsep(&s, "-"));
            printf("x=%d, y=%d\n", x, y);
            break;
    }

Will produce:

x=3, y=4

Do not use it as is. You will have to do additional NULLchecks on x and y appropriately for your case.

edited May 02 '12 at 21:15

answered May 02 '12 at 21:02

dpp

1,748
11
7

Cool. that works, but I still got the warning. Why? (Warnings aren't good) – waXve May 02 '12 at 21:07

use of strsep in C while parsing argv Variable doesen't work

2 Answers2