plz explain a small function

This is a discussion on plz explain a small function within the C Programming forums, part of the General Programming Boards category; //atoi: converts s to integer int atoi(char s[]) { int i,n; n=0; for(i=0; s[i]>='0' && s[i]<='9'; i++) n=10*n + (s[i] ...

  1. #1
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    plz explain a small function

    //atoi: converts s to integer

    int atoi(char s[])
    {
    int i,n;

    n=0;

    for(i=0; s[i]>='0' && s[i]<='9'; i++)

    n=10*n + (s[i] - '0');

    return n;
    }

    how the above func. works!

    thanx

  2. #2
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    You've been told before... use code tags.
    This has to be your homework. Make an effort.
    DavT
    -----------------------------------------------

  3. #3
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    Why not use the atoi() function included in stdlib?? anyways..
    i'm not even sure if the above function works.

    but you could always play knowing that the ASCII code 48 is 0. substract 48 to every ASCII code that represents a number and knowing how many spaces to the left you could multiply by ten.

    EDIT: lapsus mentis.. sorry lol i just explained what it did without really looking at the function, i just thought it didn't work because of the parameter, i would write

    Code:
    atoi(char *s)
    Last edited by ROCKaMIC; 11-11-2004 at 09:25 AM. Reason: ...

  4. #4
    Just Lurking Dave_Sinkula's Avatar
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    Debugging with printf:
    Code:
    #include <stdio.h>
    
    int atoi(char s[])
    {
       int i, n = 0;
       for ( i=0; s[i]>='0' && s[i]<='9'; i++ )
       {
          printf("10 * n = 10 * %4d = %5d, ", n, 10 * n);
          printf("s[%d] - '0' = '%c' - '0' = %d - %d = %d, ",
                     i,         s[i],       s[i], '0', s[i] - '0');
          n = 10 * n + (s[i] - '0');
          printf("n = %d\n", n);
       }
       return n;
    }
    
    int main ( void )
    {
       int result = atoi("12345");
       printf("result = %d\n", result);
       return 0;
    }
    
    /* my output
    10 * n = 10 *    0 =     0, s[0] - '0' = '1' - '0' = 49 - 48 = 1, n = 1
    10 * n = 10 *    1 =    10, s[1] - '0' = '2' - '0' = 50 - 48 = 2, n = 12
    10 * n = 10 *   12 =   120, s[2] - '0' = '3' - '0' = 51 - 48 = 3, n = 123
    10 * n = 10 *  123 =  1230, s[3] - '0' = '4' - '0' = 52 - 48 = 4, n = 1234
    10 * n = 10 * 1234 = 12340, s[4] - '0' = '5' - '0' = 53 - 48 = 5, n = 12345
    result = 12345
    */
    Last edited by Dave_Sinkula; 11-11-2004 at 09:25 AM. Reason: Added another printf after the calculation of n.
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*

  5. #5
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    thanx a lot mate!
    code tags means complete code which can be run

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