my do while loop not working

This is a discussion on my do while loop not working within the C Programming forums, part of the General Programming Boards category; when i run my code even if i enter "yes" for the input it doesnt loop over. any idea? Code: ...

  1. #1
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    my do while loop not working

    when i run my code even if i enter "yes" for the input it doesnt loop over. any idea?

    Code:
    #include <stdio.h>
    #include <string.h>
    int main()
    {
    int x;
    char tin[2];
    printf("\n");
    do{
    printf("enter a ascii value to see its corresponding character - ");
    scanf("%i", &x);
    printf(" \n  the ascii character of %i is %c \n \t Would you like to start over?(yes or no)", x , (char)x);
    
    fgets(tin, 2, stdin);
    getchar();
    }while(tin == "yes");
    }
    Last edited by rodrigorules; 09-06-2005 at 05:41 PM.

  2. #2
    FOX
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    Code:
    while(tin == "yes")
    That's not how you compare strings. Use strcmp or strncmp. Furthermore, tin can never be "yes" since it's only got room for one character plus the nul character. Either make tin 4 bytes, or just store yes or no as a single character (y/n) and then use scanf to read it in.

  3. #3
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    you could try whit this....


    Code:
    #include <stdio.h>
    
    int main()
    {
    char tin[2];
    char z[2];
    int x,flag;
    
    do{
    sprintf(tin,"no\0");
    
    printf("enter a ascii value to see its corresponding character - ");
    scanf("%d",&x);
    printf(" \n  the ascii character of %d is %c \n",x,x);
    printf(" \t Would you like to start over?(yes or no)");
    scanf("%s",&z);
    flag = strcmp(tin,z);
    
    
    }while(flag == 0 );
    }

  4. #4
    FOX
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    Code:
    char tin[2];
    ...
    sprintf(tin,"no\0");
    You're still writing past the array boundary.

  5. #5
    ... kermit's Avatar
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    Code:
    char z[2];
    How many elements will you need if the user enters "yes"? How many if they enter "no"? (Hint, with either "yes" or "no" a 2 element array will not be big enough)

    This is nice:

    Code:
    scanf("%s",&z);
    1) The variable z is not a scalar variable, so you should not be passing the address of z as an argument. Get rid of the '&'.

    2) In a similar fashion to the gets() function, the user would be able to overflow the array z, as there is no limitation on the input . If you wish to use scanf() to get a string into an array, use a width specifier something like this:

    Code:
    char z[5]
    scanf("%4s",&z);  /* Read in at most 4 chars to leave room for '\0' */
    As well, if you are using scanf(), you ought to check its return value, like so:

    Code:
      char z[5]
       if(scanf("%4s",&z) != 1) {
         /* clean up the mess */
       }
    The return value of scanf corresponds to the number of arguments succesfully converted. If possible though, consider using fgets() instead of scanf() to read in a string.

    You forgot to #include <string.h>

    main() ought to return something meaningfulL

    Code:
    int main(void)
    {
       ....
       return 0;
    }
    Last edited by kermit; 09-07-2005 at 04:02 AM.

  6. #6
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    so ur saying i need to always leave room for \o in a string array?
    (so i need [4] if i want to put "yes" in it?

    ACtually - heres the edits i made based on ur comments
    still doesnt work, it doesnt loop, (and i changed the loop type)..

    Code:
    #include <stdio.h>
    #include <string.h>
    
    int main()
    {
    char tin[4];
    char yes[4];
    int x,flag=0;
    
    
    while(flag == 0 )
    {
    strcpy(yes,"yes");
    
    printf("enter a ascii value to see its corresponding character - ");
    scanf("%3d",&x);
    printf(" \n  the ascii character of %d is %c \n",x,x);
    printf(" \t Would you like to start over?(yes or no)");
    fgets(tin, 4 , stdin);
    getchar();
    flag = strcmp(tin,yes);
    }
    }
    Last edited by rodrigorules; 09-07-2005 at 02:04 PM.

  7. #7
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    Yes, you should always leave room for the '\0' character.
    All standard functions expect char arrays to be NULL terminated.

    so i need [4] if i want to put "yes" in it?
    Correct, for "yes" you should reserve room for 4.


    still doesnt work
    a newline character is left in the stdin buffer after your call to scanf.

    after your call to fgets tin contains only the newline character (since it was available in the buffer), and
    the rest of what you typed is once again left in the buffer.

    You need to flush the buffer after your call to scanf.
    There are many posts on these boards on how to do that. You can search for fflush, but do NOT use it to flush stdin.
    (to be lazy you can just put a call to getchar() before fgets to remove the newline, this will only help you in a perfect world where the user enters what you want them to).

    Also, when using fgets the newline character gets stored in your string
    (well not necessairly in your case since you only allow 4 characters with fgets).
    It's something you should be aware of, as it may affect your comparison. ("no\n" will not match "no")
    Last edited by spydoor; 09-07-2005 at 02:38 PM.

  8. #8
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    here ur advice is good but i dont think i know enough to do it, i changed it even more but it still dont work

    Code:
    #include <stdio.h>
    #include <string.h>
    
    int main()
    {
    char tin[4];
    char yes[4];
    char buffer[4];
    int x,flag=0;
    
    while(flag == 0 ){
    strcpy(yes,"yes");
    
    printf("enter a ascii value to see its corresponding character - ");
    scanf("%3d",&x);
    printf(" \n  the ascii character of %d is %c \n",x,x);
    printf(" \t Would you like to start over?(yes or no)");
    
    fgets(buffer, sizeof(buffer) , stdin);
    sscanf(buffer, "%s", &tin);
    fflush(stdout);
    getchar();
    flag = strcmp(tin,yes);
    }
    }

  9. #9
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    sorry, I probably should have not mentioned fflush, it was just a keyword for you to search with.

    try this:

    Code:
    #include <stdio.h>
    #include <string.h>
    
    int main()
    {
        char tin[4];
        // no need to store "yes" in a variable... it never changes
        int x,flag,ch;
        flag =0;
    
        while(flag == 0 )
        {
            printf("enter a ascii value to see its corresponding character - ");
            scanf("%3d",&x);
            while ((ch = getchar()) != '\n' && ch != EOF);  //flush stdin
    
            printf(" \n  the ascii character of %d is %c \n",x,x);
            printf(" \t Would you like to start over?(yes or no)");
    
            fgets(tin, 4 , stdin);
            while ((ch = getchar()) != '\n' && ch != EOF);  //flush stdin 
    
            flag = strcmp(tin, "yes"); //compare should be okay since we fgets(,4,) otherwise we'd have to worry about newline char in tin
        }
    }
    Last edited by spydoor; 09-07-2005 at 02:56 PM.

  10. #10
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    what does this do?
    -
    Code:
    while ((ch = getchar()) != '\n' && ch != EOF);  //flush stdin
    -
    (btw program works great, but i just need to find out how it works)
    Last edited by rodrigorules; 09-07-2005 at 02:58 PM.

  11. #11
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    EOF stands for End of File.

    getchar() returns EOF (an int) when there is nothing more to read or when there is an error reading.

    The while loop is calling getchar() as many times as needed to read (and essentially discard as we never use it again) everything that is sitting in the stdin buffer up to a newline character.

  12. #12
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    why is the comment - // flush stdin
    in there?

  13. #13
    ATH0 quzah's Avatar
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    Quote Originally Posted by rodrigorules
    why is the comment - // flush stdin
    in there?
    Probably because getchar is often a macro for getc( stdin ). At any rate, it's just a comment, explaining to the reader what that line of code does. You know, kinda what comments were made for.


    Quzah.
    Hope is the first step on the road to disappointment.

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