command line argument help

Code:

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//detab - replace tabs with appropriate number of blank spaces to reach next
//tabstop. Assume a fixed number of tabstops.

#include <stdio.h>

#define MAXLINE 81

int gotline(char s[],int lim);

int main(int argc,char argv[]){
    //int len;                    /* length of current line */
    //int getmore = 0;
    int tabstops = (argv[1] - '0');    //here is the line in question
   
    //char line[MAXLINE];            /* current line being read */
   
   
    printf("%d\n",tabstops);
    getchar();
    return 0;
}
/*int gotline(char s[],int lim){
    int c,i;
    for(i=0;i < lim-1 && (c = getchar()) != EOF && c != '\n';++i)
        s[i] = c;
    if(c == '\n'){
        s[i] = c;
        ++i;
    }
    else if(c == EOF && i > 0){
        s[i] = '\n';
        ++i;
    }
    s[i] = '\0';
    return i;
}*/

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commented code is not the problem... I'm trying to learn how to convert an interger entered on the command line as an argument from a char pointer to an integer data type of the same respective value, (ie argv[1] is a char pointer to an integer, say 14, and I would like to place the value "14" into "tabstops"

You got the arguments to main listed wrong, should be:

Code:

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

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argv should be an array of pointers (one pointer to each command line argument) to char. You currently have an array of characters which is not the same thing. Then, assuming you enter a single character digit you'd do:

Code:

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int tabstops = argv[1][0] - '0';

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...this would pick up the first (and hopefully only) character of the second command line arg and convert from a char to it's integer representation.

another way would be

Code:

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count = sscanf(argv[1],"%d",&tabstops);
if (count != 1) {
    invalid input
}

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this would convert any number, not just a single digit.

thank you for your quick responses. i would need to be able to convert all characters in the second argument, not just the first one. But I also cannot use sscanf() because I am not far enough in the book to use this function.

Code:

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if(argc != 2)
        printf("Usage: %s n [tabstop every n columns]\n",argv[0]);
    else{
        //tabstops = argv[1][0] - '0';    //converts only first char
        int c,i;
       
        for(i = 0;((c = argv[1][i]) != '\0');++i)
            tabstops += (c - '0');
        printf("%d\n",tabstops);
    }

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with this, im assuming that the char array will end with a null character(wrong assumption im guessing?). So far im getting garbage.

Quote:

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Originally Posted by jwroblewski44

thank you for your quick responses. i would need to be able to convert all characters in the second argument, not just the first one. But I also cannot use sscanf() because I am not far enough in the book to use this function.

Code:

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if(argc != 2)
        printf("Usage: %s n [tabstop every n columns]\n",argv[0]);
    else{
        //tabstops = argv[1][0] - '0';    //converts only first char
        int c,i;
       
        for(i = 0;((c = argv[1][i]) != '\0');++i)
            tabstops += (c - '0');
        printf("%d\n",tabstops);
    }

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with this, im assuming that the char array will end with a null character(wrong assumption im guessing?). So far im getting garbage.

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You've got the idea. First, converting a strign to an integer is a natural unit of programming. So why not wrap it up in a function.

Code:

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int stringtointeger(char *str)

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Secondly, the logic is that you don't add the digits together. You take the first digit, subtract '0' for it to get the integer, as you have done. Then you look at the next character. If it is a digit, multiply the number you have by ten, subtract the '0', and add. Then repeat until there are no new digits waiitng. You can get the next character simply by indexing [i+1].

Quote:

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Originally Posted by jwroblewski44

thank you for your quick responses. i would need to be able to convert all characters in the second argument, not just the first one. But I also cannot use sscanf() because I am not far enough in the book to use this function.

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There will be a lot of other functions your books don't touch on because the point of the book is to teach you the language not necessarily the entire C library. So why not start a, let's say cheeky adventure, and start using cool functions like sscanf which mind you ALSO help you learn something new in the process :).

Quote:

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Originally Posted by claudiu

There will be a lot of other functions your books don't touch on because the point of the book is to teach you the language not necessarily the entire C library. So why not start a, let's say cheeky adventure, and start using cool functions like sscanf which mind you ALSO help you learn something new in the process :).

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*le gasp* Such scandalous audacity!

Quote:

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Originally Posted by Malcolm McLean

You've got the idea. First, converting a strign to an integer is a natural unit of programming. So why not wrap it up in a function.

Code:

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int stringtointeger(char *str)

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Secondly, the logic is that you don't add the digits together. You take the first digit, subtract '0' for it to get the integer, as you have done. Then you look at the next character. If it is a digit, multiply the number you have by ten, subtract the '0', and add. Then repeat until there are no new digits waiitng. You can get the next character simply by indexing [i+1].

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thank you for the advice on multiplying by ten and adding the digit. Is there a specific char I can look for to notify the end of the string? Obviously '\0' is not the choice.

Quote:

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Originally Posted by jwroblewski44

Obviously '\0' is not the choice.

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Why not? That's how the end of a string is denoted in C normally

Quote:

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Originally Posted by jwroblewski44

thank you for the advice on multiplying by ten and adding the digit. Is there a specific char I can look for to notify the end of the string? Obviously '\0' is not the choice.

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Use the isdigit() function. If you don't know how this works, it's simply.

Code:

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int isdigit(int ch)
{
  if(ch >= '0' && ch <= '9')
    return 1;
  else
    return 0;

}

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At some stage you'll need to decide what you do if the first non-digit character is not nul. But you can leave that for now. nul or '/0' or zero is a char value. It's only end of string by convention, and you can read and manipulate it like other char values.

Thank you sincerely to Malcolm McLean to your fast and helpful reponses. Here's my attempt at the following assignment..

Code:

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/* detab- write a program 'detab' that replaces tabs in the input with
            the proper number of blanks to space to the next tab stop.
          Assume a fixed set of tab stops, say every n columns. Should
          n be a variable of a symbolic parameter? */
#include <stdio.h>

int stringtointeger(char **str);

int main(int argc,char *argv[]){
    if(argc != 2)
        printf("Usage: %s n [tabstop every n columns]\n",argv[0]);
    else{
        int tabstop,k,c;
        if((tabstop = stringtointeger(argv)))
            if(tabstop > 10)
                printf("Tab stop must be less than 10\n");
            else{
                k = tabstop;      //holds the current tabstop locator
                while((c = getchar()) != EOF){
                    if(c != '\t'){
                        printf("%c",c);
                        if(c == '\n')
                            k = tabstop;
                    }
                    else{
                        int i;for(i = k;i > 0;--i)
                                printf(" ");
                    }
                    if(k > 0)
                        --k;
                    else if(k == 0)
                        k = tabstop;
                }
            }
        else
            printf("Input Error\n");
    }
    return 0;
}
int stringtointeger(char **str){
    int i= 0,c,t = 0;
   
    if((c = str[1][i]) < '0' || c > '9')
        return 0;
    else{
        for(i = 0;((c = str[1][i]) != '\0' && c >= '0' && c <= '9');++i)
            t = (t * 10) + (c - '0');
        if(c != '\0')
            return 0;
        return t;
    }
}

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How does that look?