Quick help needed on simple example

This is a discussion on Quick help needed on simple example within the C Programming forums, part of the General Programming Boards category; Code: char c=0; int letter [3]; letter[0]=0; letter[1]=0; letter[2]=0; printf("please enter a letter"); scanf("%c", &c); switch (c) { case 'a': ...

  1. #1
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    Quick help needed on simple example

    Code:
    char c=0;
    int letter [3];
    
    letter[0]=0; letter[1]=0; letter[2]=0;
    
    printf("please enter a letter");
    scanf("%c", &c);
    
    switch (c) {
    case 'a': letter[0]++;
    case 'b': letter[1]++;
    default: letter[2]++;
    }
    printf("%d %d %d", letter[0], letter[1], letter[2]);
    Can someone please explain what line2 is actually doing. Is it there to define how many "letter" variations there are (i.e letter[0]...), or does it do something else?

    Also because letter[0]=0 for example isnt given a value type, but it follows on from an int being defined, is it reasonable to assume letter[0] is being defined as an int as well?

  2. #2
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    Hello.

    Line 2 says: "letter is an array of 3 integers.". There is nothing more to it.

  3. #3
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    Is this even being used in the code though. I can see letter[0], letter[1] and letter[2], but no mention of just letter.

    Sorry if im being dumb.

  4. #4
    Kernel hacker
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    letter[0] is the first element of the array declared on line 2. letter[1] is the second element of the array, etc.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  5. #5
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    Ah i understand now, thank you.

  6. #6
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    Quote Originally Posted by hiven View Post
    Code:
    char c=0;
    int letter [3];
    
    letter[0]=0; letter[1]=0; letter[2]=0;
    
    printf("please enter a letter");
    scanf("%c", &c);
    
    switch (c) {
    case 'a': letter[0]++;
    case 'b': letter[1]++;
    default: letter[2]++;
    }
    printf("%d %d %d", letter[0], letter[1], letter[2]);
    Can someone please explain what line2 is actually doing. Is it there to define how many "letter" variations there are (i.e letter[0]...), or does it do something else?

    Also because letter[0]=0 for example isnt given a value type, but it follows on from an int being defined, is it reasonable to assume letter[0] is being defined as an int as well?
    Your code is really ate up. I'm not really sure what your trying to acheive.
    Code:
    #include <stdio.h>
    
    int main(void)
    {
       char c;
       int letter [3] = {0};
    
    
       printf("please enter \"a\" or \"b\": ");
       scanf("%c", &c);
    
       switch (c) 
       {
          case 'a': 
             ++letter[0];
             break;
          case 'b': 
             ++letter[1];
             break;
          default: 
             ++letter[2];
       }
       printf("\nletter[0] contains: %d\nletter[1] contains:"
                " %d\nletter[2] contains: %d", letter[0], letter[1], letter[2]);
       
       fflush(stdin);
       getchar();
       return 0;
    }
    You really need a break in those cases or else all the cases will be performed.
    edit: unless its your intention to add one to letter[1] and letter[2] when you enter a to add
    one to letter[0]
    Last edited by strickyc; 05-12-2009 at 03:58 PM.

  7. #7
    CSharpener vart's Avatar
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    do not fflush stdin - read FAQ
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

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