getopt - parsing string as argument

I'm trying to write a program to take a string from the UNIX command line, and perform some operations on it depending on command line switches. The format is ./program -(options) (arguments) string, and there are two switches -a, and -b, which takes an integer argument. I'm using getopt in a while loop with a switch statement, but have no idea how to read the string from the command line. So far I have:

Code:

---

int main(int argc,char *argv[]){

char sw; //stores switch
char string[100]; //stores string
int n; //stores integer argument for switch b

opterr=0;

while((sw=getopt(argc,argv,"ab:"))!=-1){
  switch(sw){
    case 'a': //performs a string operation
    return -1;
    case 'b': //performs a different string operation also using the argument for switch -b
    return -1;
    case '?': printf("Invalid switch entered, options are -a or -b");
    return -1;
  }
}

}

---

for me to be able to perform the string operations I need to somehow read the string, and the argument for b into the variables string and n - I'm familiar with sscanf and fgets, but everything I tried ended up in an error.

Can someone tell me how it's done?

Thanks

the command line strings are all in argv. i don't really understand your question.

I can get the code the recognise the different switches, but I don't know how to get the integer argument for b, and the string from the command line, using the format above. either -a or -b may be used, or both, and the string can be up to 100 characters long including spaces.

From my "man 3 getopt":

Quote:

---

The option string optstring may contain the following elements: individ-
ual characters, and characters followed by a colon to indicate an option
argument is to follow. For example, an option string "x" recognizes an
option ``-x'', and an option string "x:" recognizes an option and argu-
ment ``-x argument''. It does not matter to getopt() if a following
argument has leading white space.

---

Presumably yours is similar, but you should check, of course.

you're probably not using getopt the right way. i can tell for sure because i always did that kind of stuff manually.

Code:

---

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

int main(int argc, char **argv)
{
    int i;
    for (i = 1; i < argc; i++)
    {
        if (argv[i][0] == '-')
        {
            if (i+1 == argc || argv[i+1][0] == '-')
            {
                printf("invalid switch entered, no arguments\n");
                return EXIT_FAILURE;
            }
            if (argv[i][1] == 'a')
                printf("argument for -a: %s\n", argv[++i]);
            else if (argv[i][1] == 'b')
                printf("argument for -a: %s\n", argv[++i]);
            else
            {
                printf("invalid switch entered, options are -a or -b\n");
                return EXIT_FAILURE;
            }
        }
    }
    return EXIT_SUCCESS;
}

---

some more relevant info from the getopt() manpage.

Code:

---

The getopt() function shall return the next option character (if one is found) from argv that
matches a character in optstring, if there is one that matches. If the option takes an argument,
getopt() shall set the variable optarg to point to the option-argument as follows:

+ If the option was the last character in the string pointed to by an element of argv, then optarg
  shall contain the next element of argv, and optind shall be incremented by 2. If the resulting
  value of optind is greater than argc, this indicates a missing option-argument, and getopt()
  shall return an error indication.

+ Otherwise, optarg shall point to the string following the option character in that element of
  argv, and optind shall be incremented by 1

---

Code:

---

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

int main(int argc, char *argv[]) {
        char *A;
        int opt, B;
        while ((opt=getopt(argc, argv, "a:b:"))>0)  switch (opt) {
                case '?': puts("Bad argument"); break;
                case 'a': A=malloc(strlen(optarg));
                        strcpy(A,optarg); break;
                case 'b': B=atoi(optarg); break;
                default: break;
        }
        printf("%s %d\n",A,B);
        free(A);
        return 0;
}

---

Your option string in your original post is wrong. "ab:" means that a does not have an argument (it's just a switch). "a:b:" means a and b both have a parameter returned in the char* optarg.

Thanks for the replies. MK27, that's just what I was looking for, except I only want an argument after -b and not -a.

Quote:

---

Originally Posted by bertazoid

Thanks for the replies. MK27, that's just what I was looking for, except I only want an argument after -b and not -a.

---

Sorry, I though by "//performs a string operation" you meant "//performs a different string operation also using the argument for switch". Good luck.

I don't understand why I can't use optarg as an integer. Form one of the string operations I used:

Code:

---

for(i=0;i<optarg;i++){
  printf("%s",string);
}

---

and got a compiler error saying "warning: comparison between pointer and integer". I don't understand this since i is an integer as defined by me and so is optarg isn't it?...When I print optarg using printf("%d",optarg); it prints the integer entered at the command line as the argument for -b. i.e. -b 3, prints optarg as 3. So why is it different when I use it in the for loop? I also tried:

Code:

---

for(i=0;i<*optarg;i++){
  printf("%s",string);
}

---

and instead of printing the string three times, it printed it about 30 times.

Did you read the manpage of getopt() yet? optarg is not an integer but a pointer to char and is declared externally.

I'm reading through it now finding it hard to get my head round.

Quote:

---

Originally Posted by bertazoid

I don't understand why I can't use optarg as an integer. Form one of the string operations I used:

Code:

---

for(i=0;i<optarg;i++){
  printf("%s",string);
}

---

---

That's expected since optarg is of type char*.

Quote:

---

Originally Posted by bertazoid

I also tried:

Code:

---

for(i=0;i<*optarg;i++){
  printf("%s",string);
}

---

and instead of printing the string three times, it printed it about 30 times.

---

That 3 is not an integer but an ASCII character. It should print it about 50 times precisely.

Ah ok now I get it. Thanks