Thread: help with a simple program

Hello guys,
I'm kind of stuck here trying to do some homework...writing a simple program which
1. displays numbers from 1-5.
2. user picks a number (1-5).
3. the programs displays the available numbers except the number the user picked.
4. the program runs until the last number is picked and displays a message that there are no available numbers left.

Could you suggest me how to approach this problem?
So far, I've been able to do 1,2. but not 3 and 4..

We would be glad to help. Show us what you've got so far, and where you're stuck, and we can offer advice from there.

If you'll allow me to be silly...

Originally Posted by holler

Hello guys,
I'm kind of stuck here trying to do some homework...writing a simple program which
1. displays numbers from 1-5.

So far, I've been able to do 1,2. but not 3 and 4..

Code:

printf("%d and %d\n", 3, 4);

Originally Posted by Matticus

If you'll allow me to be silly...

Code:

printf("%d and %d\n", 3, 4);

Sure..
It only works when I enter numbers one by one from 1-5.

Code:

int choice,tries,i;
   int nums[5] = {1,2,3,4,5};


   printf("Available numbers:\n");
   for(i=0; i<5; i++)
   {
       printf("Numbers: %d\n", nums[i]);
   }


   LOOP:
    for(tries=0; tries<5; tries++)
    {
            printf("Enter a number you want to pick: (1-5)\n");
            scanf("%d", &choice);
            if(choice>5)
            {
                printf("wrong number. Try again!\n");
                goto LOOP;
            }
       for(i=0; i<5; i++)
       {
            if(choice==nums[i])
                {
                    nums[i]++;
                    continue;
                }
            printf("Numbers: %d\n", nums[i]);
       }
    }

Originally Posted by Salem

Perhaps start with

Code:

int picked[5] = { 0 };

When the user chooses a number 'num', you do
picked[num] = 1;

When it comes to displaying numbers to choose, you do
if ( picked[n] != 1 ) print n

Thanks for this. Though I'm here to learn, with the limited knowledge that I have, still can't implement your suggestion.
So far, I was able to accomplish only this chunk of code:

Code:

int len,choice,i,j;
   int nums[5] = {1,2,3,4,5};


   printf("Available numbers:\n");
   for(i=0; i<5; i++)
       printf("Numbers: %d\n", nums[i]);


    len = 5;


    do
    {
        printf("\nChoose a number: or press '6' to exit)\n");
        scanf("%d",&choice);


        for(i=0;i<len;i++)
        {
            if(nums[i]==choice)
            {
                for(j=i;j<len;j++)
                nums[j]=nums[j+1];
                len--;
            }
        }


        printf("\nNumbers left:\n");
            for(i=0;i<len;i++)
            {
                printf("Numbers: %d \n",nums[i]);
                if(nums[i]== '\0') // want to check if the array is empty to display the message below
                    printf("none left\n"); 
            }


    }
    while(choice<6);

I understand that the exit condition isn't the best one, could you suggest anything?

Thank you! Sharing my latest progress.
The program executes while there are free numbers left and stops when the array is empty.

Code:

int len,choice,i,j;
   int nums[5] = {1,2,3,4,5};


   printf("Available numbers:\n");
   for(i=0; i<5; i++)
       printf(" %d", nums[i]);


    len = 5;


    do
    {
        LOOP:
        printf("\nChoose a number: \n");
        scanf("%d",&choice);
        if(choice != 1 && choice != 2 && choice != 3 && choice != 4 && choice != 5)
        {
            printf("The number you entered doesn't exist. Try again.\n");
            goto LOOP;
        }
            for(i=0;i<len;i++)
            {
                if(nums[i]==choice)
                {
                    for(j=i;j<len;j++)
                    nums[j]=nums[j+1];
                    len--;
                }
            }
                printf("\nNumbers left:");
                    for(i=0;i<len;i++)
                    {
                        printf(" %d",nums[i]);
                    }
        }
        while(len!=0);
        printf(" none\n");
        getch();

I see. Replaced goto with continue and getch() with getchar().

Code:

    do
    {
        printf("\nChoose a number: \n");
        scanf("%d",&choice);
        if(choice != 1 && choice != 2 && choice != 3 && choice != 4 && choice != 5)
        {
            printf("The number you entered doesn't exist. Try again.\n");
          continue; // instead of goto
        }
            for(i=0;i<len;i++)
            {
                if(nums[i]==choice)
                {
                    for(j=i;j<len;j++)
                    nums[j]=nums[j+1];
                    len--;
                }
            }
                printf("\nNumbers left:");
                    for(i=0;i<len;i++)
                    {
                        printf(" %d",nums[i]);
                    }
        }
        while(len!=0);
        printf(" none\n");
        getchar(); // instead of getch();