Basic beginner loop (with arrays) does NOT work but why?

This is a discussion on Basic beginner loop (with arrays) does NOT work but why? within the C Programming forums, part of the General Programming Boards category; Hi. I am frustrated with a piece of code I wrote because it malfunctions when the user inputs (command argument ...

  1. #1
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    Basic beginner loop (with arrays) does NOT work but why?

    Hi. I am frustrated with a piece of code I wrote because it malfunctions when the user inputs (command argument style) any string (keyword) with the letter a or A, and I have no idea why and I hate not knowing what's going on. Can you please tell me what I am doing wrong and why the code doesn't work when the user inputs a string with the letter a or A?

    Code:
    #include <cs50.h>
    #include <stdio.h>
    #include <string.h>
    
    int main(int argc, string argv[])
    {
         string keyword = argv[1];
         for (int j = 0; j < strlen(keyword); j++)
        {
            if (keyword[j] >= 'a' && keyword[j] <= 'z')
            {
                keyword[j] = keyword[j] - 'a';
            }
            else if (keyword[j] >= 'A' && keyword[j] <= 'Z')
            {
                keyword[j] = keyword[j] - 'A';
            }
            printf("%i\n", keyword[j]);    
        } 
    }
    P.S. The code works fine with every other letter. Just not A or a.

    Thank you!
    Last edited by Rubik; 03-07-2013 at 03:50 PM.

  2. #2
    Registered User TheBigH's Avatar
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    It's a subtle little bug. If the string contains an 'A' or an 'a', then keyword[j] gets set to zero. Once that happens, keyword[j] = 0 = '\0' = end of string. So strlen(keyword) thinks the string ends there, and halts the loop. You want this:

    Code:
    length_of_word = strlen(keyword);
    for (int j = 0; j < length_of_word; j++) {
    /*and so on */
    Code:
    while(!asleep) {
       sheep++;
    }

  3. #3
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    How does strlen() work?
    What is the integer value of the character '\0'?
    What is the integer value of 'a' - 'a' ?

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    Hi TheBigH! Thank you for your answer!

    I tried putting a while loop at the bottom inside a for-loop, but it doesn't work; here's what I did:
    Code:
    printf("%i", keyword[j]);
    while (keyword[j] == 0)
        j++;
    Then, I tried writing an if-condition within the if statement:
    Code:
    for (int j = 0; j < strlen(keyword); j++)
        if (keyword[j] >= 'a' && keyword[j] <= 'z')
        {
           if (keyword[j] == 'a')
           {     
                keyword[j] = keyword[j] - 'a';
                j++; 
            }
            else 
                 keyword[j] = keyword[j] - 'a';
         }
    but that gives me completely different values when I enter a (it messes up the j count).

    What am I doing wrong?

  5. #5
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    Quote Originally Posted by TheBigH View Post
    Code:
    length_of_word = strlen(keyword);
    for (int j = 0; j < length_of_word; j++) {
    /*and so on */
    Ummm.

  6. #6
    young grasshopper jwroblewski44's Avatar
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    Quote Originally Posted by Tclausex View Post
    Ummm.
    It's not exactly the answer to the question posted, but a good idea nonetheless.

  7. #7
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    Quote Originally Posted by Tclausex View Post
    How does strlen() work?
    It counts characters until it finds one with a value zero.

    Quote Originally Posted by Tclausex View Post
    What is the integer value of the character '\0'?
    Zero.
    Quote Originally Posted by Tclausex View Post
    What is the integer value of 'a' - 'a' ?
    Seriously? You're asking this?
    Right 98% of the time, and don't care about the other 3%.

  8. #8
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    @Grumpy: I'm pretty sure Tclausex was asking those questions of the OP to get them thinking about the problems they were having with the code.

  9. #9
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    I'm a fan of the Socratic method /shrug.

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