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need help in some code

This is a discussion on need help in some code within the C Programming forums, part of the General Programming Boards category; i have a code assignment which require a few code to be writen (i m new to c programming) anyway ...

  1. #1
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    Jan 2013
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    7

    need help in some code

    i have a code assignment which require a few code to be writen (i m new to c programming)
    anyway heres the question, need to perfect it

    first one,

    Code:
    write a c program to count the number of words in its input, where a"ward" is defined as any continuous sequence of non-whitespace characters
    
    example:
        input:"+++ --- hi bye 879+3"
        output:" the string has 5 words.
    
    ("+++","---","bye", and "879+3")
    
    heres my code
    
    Code:
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    int main()
    {
      char str[40];
      int c;
      int i,num=0;
      int space;
      gets(str);
      c=strlen(str);
      
      for(i=0;str[i]!='\0';i++)
      {
         if(str[i]==' '&&(str[i+1]!=' '))
         {
            num++;
         }
      }
      if(str[0]==' ')
      {
           printf("The string has %d words ",num);
           system("pause");
      }
      else
      {
          printf("The string has %d words",num);
          printf("\n\n");
          system("pause");
      }
      
    }
    2 more code posted below
    the 3rd program i have no idea how to write at all
    please help me asap
    i'm using dev c++ to write the code
    thankyou v. much!

  2. #2
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    7
    second one

    Code:
    write a C program that reads in a string from user and outputs  the middle item of the string. stop the program if -1 is read.
    
    Example:
       input: "i am busy today"
    output: "s"
    
    heres my code
    
    
    Code:
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<conio.h>
    
    int main()
    {
     char str[40];
      int c,mid;
      int i;
      gets(str);
      c=strlen(str);
      for(i=0;i<c;i++)
      {
        if(c%2==0)
        mid=str[c/2];
        else
        mid=str[(c-1)/2];
    }
    printf("%c",mid);
    printf("\n\n");
    system("pause");
    }
    third ones

    Code:
    write a code which work like this
     
    input : 12345
    output: 12543
    
    the output is the middle character of the string  & last character  of the string switched
    input shouldnt be limited.

  3. #3
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    Jun 2011
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    2,348
    First Program:

    Your program is returning an incorrect number of words, so there is something wrong with your logic. This program can be accomplished much more easily with a variable that is used to indicate whether you are in a word or not (let's call it "inWord"). Make "inWord" equal to one if you are in a word (i.e. the current character is not a space) and make "inWord" equal to zero if you are not in a word (i.e. the current character is a space). See if developing the logic based around this idea helps you with the program.

    Also, I don't understand why you're using the last "if/else" statement to print the output. The code in each branch is pretty much the same, so there's no use in an "if/else." Just print the output.

    Some other notes:
    - You should explicitly "return 0;" at the end of "main()"
    - You shouldn't use "system("pause")" to keep the console open - "getchar()" would be better
    - "gets()" is unsafe (see here) - consider "fgets()" as a safer alternative

    --------

    Second Program:

    Your loop seems to be virtually useless here. Note that you are not using the value of "i" in the loop, so nothing is going to be different for any iteration. What if you tried this instead?

    Code:
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<conio.h>
    
    int main()
    {
        char str[40];
        int c,mid;
        int i;
        gets(str);
        c=strlen(str);
    //    for(i=0;i<c;i++)
    //    {
    //        if(c%2==0)
                mid=str[c/2];
    //        else
    //            mid=str[(c-1)/2];
    //    }
        printf("%c",mid);
        printf("\n\n");
        system("pause");
    }
    Also, the same suggestions go for this program as the last one (see "Some other notes" above).

    --------

    For the third program, this doesn't seem too hard if you got the first two working yourself. You simply have to identify the middle character and the last character, and swap them. You already worked out how to identify the middle character. Finding the last character should be just as easy. And swapping them is just a matter of reassigning those indices of the character array (hint: a temporary variable is need to perform this simple swap).

    Try out the third one and post your attempt if you can't solve it.

    It will definitely help to sketch it out on paper by hand, to get an idea of the steps needed to solve that problem.
    Last edited by Matticus; 01-24-2013 at 11:00 AM.
    Ash Mo Naiming likes this.

  4. #4
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    Jan 2013
    Posts
    7
    thanks so much!, i hav successfully solve it

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