[SOLVED] C programming isdigit(argv[i]) is getting a seg fault .. why?

BW-userx · 04-13-2017, 10:24 AM

[please take your time and read the text first then look at the code.]

I am just pecking at this one peice at a time. right now I am trying to figure out how to get a line of digits off of the command line - check first to be sure they are digits then store them perhaps in an array. (the last part I have not gotten to yet.)

I have tyied this two seperate ways. which neither are working. the first is off of a example on how to use isdigit(data) which gives me an seg fault.

Code:

/* get char numbers off command line
 * 
 * Apr 13 2017
 * 
 * ***/

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

int main( int argc, char *argv[]) {
  
 int i=1;
 int d = 10;
  
  for ( i = 1 ; i < argc ; i++ ) {
  //  if ( isdigit(atoi(argv[i])) ) {
    if (isdigit(argv[i])) {
    printf("argc is %d, i = %d and is a number%d\n",argc, i, argv[i]);
    }
    else
    {
      printf("argc is %d - i is %d is not a number%c\n", argc, i, argv[i]);
    }
      
  }
  
  return 0;
}

gets seg fault

Code:

userx@slackwhere⚡~/bin/C-files $./a.out 3 4 5                                   
Segmentation fault

then changed it to the upper commented out line so please look again. Because I do not bleive that I have to convert char to int using atoi(data)

Code:

/* get char numbers off command line
 * 
 * Apr 13 2017
 * 
 * ***/

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

int main( int argc, char *argv[]) {
  
 int i=1;
 int d = 10;
  
  for ( i = 1 ; i < argc ; i++ ) {
    if ( isdigit(atoi(argv[i])) ) {
  
      //if (isdigit(argv[i])) {
    printf("argc is %d, i = %d and is a number%d\n",argc, i, argv[i]);
    }
    else
    {
      printf("argc is %d - i is %d is not a number%c\n", argc, i, argv[i]);
    }
      
  }
  
  return 0;
}

but it still is not even getting what is on the command line. Instead I get something completely different.

Code:

userx@slackwhere⚡~/bin/C-files $./a.out 3 4 5                                   
argc is 4 - i is 1 is not a number;
argc is 4 - i is 2 is not a number=
argc is 4 - i is 3 is not a number?

as argc is the total count of whatever is on the command line.
argv[ ] is the array it gets put into.

looping trough it starting at 1 (one) should return the value within the array. yes or no?

example writen as I type

Code:

  for ( int i = 1; i < argc ; i++ ) {

printf("this is argc %d, this is i %d, and this is what is on the command line%c\n",argc, i, argv[i]);

}

Now let me go test that first.
that gets the same thing

Code:

userx@slackwhere⚡~/bin/C-files $./a.out 3 4 5                                   
this is argc 4, this is i 1, and this is what is on the command line;
this is argc 4, this is i 2, and this is what is on the command line=
this is argc 4, this is i 3, and this is what is on the command line?

It should return 3 4 5 not ; = ?

what is wrong with me?

michaelk · 04-13-2017, 11:06 AM

Although you are entering a number argv is still a character string.

Even though, isdigit() takes integer as an argument, character is passed to the function. Internally, the character is converted to its ASCII value for the check. argv is a character string.

Here is a quick snippet.

Code:

  char e;
 
  e=argv[1][0];

  if ( isdigit{e) )
     printf("Is True");
  else
     printf("Is False");

Code:

char e;
  
 e='A';
 printf("e=%c and e=%d\n",e,e);

BW-userx · 04-13-2017, 11:15 AM

Quote:

Originally Posted by michaelk

Although you are entering a number argv is still a character string.

Even though, isdigit() takes integer as an argument, character is passed to the function. Internally, the character is converted to its ASCII value for the check. argv is a character string.

Here is a quick snippet.

Code:

  char e;
 
  e=argv[1][0];

  if ( isdigit{e) )
     printf("Is True");
  else
     printf("Is False");

I just got it (just now)
by changing it argument adding the symbol for a pointer * to the equation it now works.

Code:

/* get char numbers off command line
 * 
 * Apr 13 2017
 * 
 * ***/

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

int main( int argc, char *argv[]) {
  
 int i=1;
 int d = 10;
  
  for ( i = 1 ; i < argc ; i++ ) {
    // here is added the pointer reference to argv[ ]   
    if (isdigit(*argv[i]) )
    {
      printf("argc is %d, i = %d and argv[i] is a number: %d\n",argc, i, atoi(argv[i]));
    }
    else
    {
      printf("argc is %d - i is %d argv[i] is not a number: %s\n", argc, i, argv[i]);
    }
  }
 
 
  return 0;
}

now it returns properly

Code:

userx@slackwhere⚡~/bin/C-files $./a.out 4 goop                                                       
argc is 3, i = 1 and argv[i] is a number: 4
argc is 3 - i is 2 argv[i] is not a number: goop

so isdigit() takes anything and checks to see if it is a digit type char - that is the purpose of it.

Quote:

isdigit()
checks for a digit (0 through 9).

and isalpha()

Quote:

isalpha()
checks for an alphabetic character; in the standard "C" locale, it is equivalent to (isupper(c) ||
islower(c)).

In some locales, there may be additional characters for which isalpha() is true-letters which are
neither upper case nor lower case.

reference:
https://linux.die.net/man/3/isdigit

michaelk · 04-13-2017, 11:21 AM

That works too... I was trying to show a few code examples...

BW-userx · 04-13-2017, 11:25 AM

so this

Code:

 e=argv[1][0];

then the first element is what is on the command line, and the second element is the place keeper it?

Code:

programName 7 8
argv[programName][0]
argv[7][1]
argv[8][2]

that would be how it is stored then. correct?

michaelk · 04-13-2017, 11:34 AM

argv[1][0] = The first character of string argv[1].

BW-userx · 04-13-2017, 11:37 AM

Quote:

Originally Posted by michaelk

argv[1][0] = The first character of string argv[1].

thanks!