Conversion of string to integer array?

**jigamuffin** · 03-19-2012

Hi.
I'm trying to make a program that reads an input of a barcode number, then validates the "check digit" (the last digit) with an algorithm. I am stuck at the start. In order for the algorithm to work, the barcode must be in separate digits, not a 'number' so to speak.

I need the input to be a char string because it may start with '0', which wouldn't count if it were an integer.

Code:

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


#define SIZE 12


int main() {
    char str[SIZE];
    int digit[SIZE];

    int i;


    
    fgets(str, 14, stdin);


    
    for(i = 0; i < 12; i++){
          str[i] = digit[i];
          }
          
    system("pause");
    return 0;


}

I know this is completely wrong, but it is all I have so far. I'm not worried about the algorithm or output, just the conversion of the string of numbers into an integer array.

**laserlight** · 03-19-2012

The idea here is that you already have the individual digits, except that they encoded as characters of a character set rather than having the desired values in the range [0, 9]. The solution is to subtract '0' from each character that represents a digit, which works because it is guaranteed that the characters '0' to '9' will be contiguous and in order in the character set.

**memcpy** · 03-19-2012

Remove "system("pause")". No idea why this is so popular, but read this > SourceForge.net: Pause console - cpwiki

Also, if anything, you have the order wrong:

Code:

str[i] = digit[i]

This will be filling the string with whatever's in digit's memory. Most likely it won't be useful data.

And even if you are trying to make your own "strtoul" function (hint hint), you need to actually convert them into digits:

Code:

str[i] - '0'

Put this in the right place, and it will subtract the ascii number 0 from the character, turning it into the actual numeral you want.

Also note that you should probably do some kind of error checking to make sure you're actually converting a digit.

**jigamuffin** · 03-19-2012

Thank you both. From what I gathered from you two, something like this should convert the digits.

Code:

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


#define SIZE 12


int main() {
    char str[SIZE];
    int digit[SIZE];
    int i;


    
    fgets(str, 14, stdin);


    for(i=0; i<SIZE; i++){
             str[i] - '0';
             }
    
    for(i = 0; i < 12; i++){
          digit[i] = str[i];
          }


    
    
    printf("%d, %d, %d", digit[1],digit[2],digit[3]);
  
          
    getch();
    return 0;


}

the final printf outputs something like 50, 51, 52 every time?

I think I have a misunderstanding of strings, or atleast unknown strings which are obtained from input. I thought say,

Code:

char str[81];    
        printf("Enter a string: "); /*user enters "hello"*/
        fgets(str, 81, stdin);
        
        printf("%s", str[2]); /*output would be "l"*/

this would output "l" but it does not. what have I missed here?

**laserlight** · 03-19-2012

This computes the digit's value, but you don't assign the result to anything:

Code:

str[i] - '0';

**whiteflags** · 03-19-2012

@jigamuffin: A string can be further divided into characters, so the type of string foo and foo[i] are different. foo[i] is a character.

**jigamuffin** · 03-19-2012

Originally Posted by laserlight

This computes the digit's value, but you don't assign the result to anything:

Code:

str[i] - '0';

Haha I see now. I feel silly.

You have helped me with several of my problems laserlight. I greatly appreciate your help. Thank you.

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