Parsing two character input from a string in C

Question

Okay so I am parsing a 18 character string consisting of '?'s and '0' - '9'. What I am trying to do is use the atoi function to convert two character chunks of the string into integers. The characters that I need to parse are in an array. I am lost on how to implement such a solution.

  char *str = "01";
  int n = atoi(str);
  printf("The string %s as an integer is = %d\n",str,n);

Which gives you "The string 01 as an integer is = 1"

Could you give some examples? For instance, do you have to handle the string "?123?456?789?12345" and if so, what two integers do you intend to extract from it? Oh, and do you mean "two (character chunks)", or two-character chunks? If the latter then I wouldn't necessarily bother with atoi, you can convert a pair of digits to an integer with str[0] * 10 + str[1] - '0' * 11. Furthermore, if "?4" is a possible two-character chunk then atoi can't convert it anyway, so you aren't necessarily saving any special-casing by using it. — Steve Jessop
– Steve Jessop, Commented Feb 2, 2011 at 2:50
The question marks come in pairs so ??23??130105??1223 so what I would want is the first int is unknown, the second int is 23, the third int is unknown, the fourth int is 13 and so on — foo
– foo, Commented Feb 2, 2011 at 2:56
I don't understand this: str[0] * 10 + str[1] - '0' * 11 why is it * 10 then * 11 thanks for your help by the way :) — foo
– foo, Commented Feb 2, 2011 at 3:00
Question: Do you expect 010 to parse as decimal (10) or octal (8)? — Chris Lutz
– Chris Lutz, Commented Feb 2, 2011 at 3:22

mpenkov · Accepted Answer · 2011-02-02 04:13:10Z

1

Here's something that may work.

Advance pointer to next decimal or end of string
If reached end of string, you're done
Use strtol, keep the result. This will advance the pointer to the next non-decimal or the end of the string.
Go back to step 1.

Here's some source.

#include <stdlib.h>
#include <stdio.h>
#define ARR_LEN 18    
int
main(void)
{
    char *str = "12?456?8??12????78";
    char *ptr = str;
    int result[ARR_LEN];
    int i = 0;
    int j = 0;

    i = 0;
    for (; ;)
    {
        while (*ptr == '?' && *ptr != '\0')
            ++ptr;
        if (*ptr == '\0')
            break;
        result[i++] = (int)strtol(ptr, &ptr, 10);
    }    
    for (j = 0; j < i; ++j)
        printf("%d ", result[j]);
    printf("\n");
    return 0;
}

edited Feb 2, 2011 at 4:13

answered Feb 2, 2011 at 3:07

mpenkov

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4 Comments

Chris Lutz Over a year ago

Combine 3 and 4 into "Use strtol." result = strtol(ptr, &ptr, 10); will do the same thing as result = atoi(ptr); while(isdigit(*ptr) && ptr != '\0') ++ptr; and probably be more efficient.

Chris Lutz Over a year ago

Now I'm worried about that cast.

mpenkov Over a year ago

@Chris Yeah, so am I. strtol takes in a char ** whereas &ptr is a char const **. Not having the cast there causes a compiler warning. I've removed all the consts cause they cause more trouble than they're worth in this case -- the OP is probably not going to be dealing with constant string input.

Chris Lutz Over a year ago

Probably not, but this may get abstracted into a function, and someone later will ask, "Why is this function's parameter not const correct?" and they'll try to do this all again. It seems like there should be a way to do this, but I can't think of it.

kaoD · Accepted Answer · 2011-02-02 03:18:47Z

#define UNKNOWN_CONTROL -1           // integer to recognize unknown pairs
#define CONSTANT_STRING_LENGTH 18    // The string length

int i;
char string[CONSTANT_STRING_LENGTH]; // This is your string
int pairs[CONSTANT_STRING_LENGTH/2]; // Array to store results

for (i=0; i<CONSTANT_STRING_LENGTH/2; i++) {       // For each pair in the string
  if (string[i*2] == '?' || string[i*2]+1 == '?')  // Is it a '??' pair?
    pairs[i] = UNKNOWN_CONTROL;                    // Store some constant (> 99)
  else
    pairs[i] = string[i*2] * 10 + string[i*2+1] - '0' * 11; // Compute the number and store
}

And then your array pairs would be filled with your desired results if I got your question right.

EDIT: To understand the computing bit, you've got to understand that ASCII characters (printable, the ones you store on strings) don't correspond to their integer counterparts. ASCII '0' is integer 48, ASCII '1' is integer 49, ASCII '2' is integer 50, and forth...

By multiplying the first character by ten and adding the first characters, you're summing the ASCII values, not the integer ones, so you've got to subtract the bias. Substracting '0' (the base number for ASCII) would work for one character (e.g. ASCII '2' - ASCII '0' == 2), but you've got to multiply it by 11 for two characters.

Keep in mind '0' * 11 is the same as '0' * 10 + '0'. Doing some redistribution of the math you can see exactly what's being done there:

pairs[i] = (string[i*2] - '0') * 10 + string[i*2+1] - '0';

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