please walk me through this code

This is a discussion on please walk me through this code within the C Programming forums, part of the General Programming Boards category; Can somebody please walk me through this problem and show me how you get an output of: [0] [14] [28] ...

  1. #1
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    Question please walk me through this code

    Can somebody please walk me through this problem and show me how you get an output of:
    [0]
    [14]
    [28]

    I've been looking at it for the past 30 minutes and I still can't figure out how this code is supposed to work? Thanks in advance!

    Code:
    What is the output of the following program?
    --------------------------------------------------------
    	#include <stdio.h>
    	
    	int main()
    	{
    		int x[20];
    		int count1, count2;
    
    		for( count1 = 0; count1 < 20; count1++ )
    		{
    			x[count1] = count1 * 2;
    		}
    
    		for( count2 = 0; count2 < 20; count2 += 7 )
    		{
    			printf( "%d\n", x[count2] ); 
    		}
    		return 0;
    	}

  2. #2
    The superhaterodyne twomers's Avatar
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    What's the problem with it?

    count1 is being increased by one each loop, and count two is being increased by seven. the zeroth element of x is 2*0=0. The seventh is 2*7=14 and the 14th is 2*14=28. The next iteration for the second loop gives a value of 21 for count2 which is greater than 20 so the loop terminates and the program completes.

  3. #3
    Registered User Noir's Avatar
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    The first loop puts all of the even numbers:
    Code:
    0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38
    into the array. The second loop prints every 7th number after 0, so it prints the red ones before count2 is bigger than 20:
    Code:
    0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38

  4. #4
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    Oh I see, I was confused as to what the relationship between count1 and count2 was. So for this code, the first part sets freq[1]-freq[4] equal to zero? Then I'm not sure what the second part is supposed to do. The output should be (3 1 2 0 1)


    main()
    {
    int i;
    int freq[5];
    int a[] = {0,1,4,0,2,2,0};

    for(i=0;i<5;i++) {
    freq[i]=0;
    }

    for(i=0;i<7;i++) {
    freq[a[i]]++;
    }

    for(i=0;i<5;i++) {
    printf("&#37;d ",freq[i]);
    }
    printf("\n");
    }
    Last edited by cboard; 04-11-2007 at 04:00 PM.

  5. #5
    Registered User whiteflags's Avatar
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    > the first part sets freq[1]-freq[4] equal to zero?
    No, both count variables are assigned zero so that we can count indices.
    Something in pseudocode might be more legible:
    Code:
    Let A[] be an array of 20 items
    
    count = 0
    for each item in A[]:
        A[item] = count * 2
        count++
    
    for each seventh item in A[]:
       print A[item]
    Does it make sense now?

  6. #6
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    I forgot to put the second code in my last post sorry! But yeah that makes perfect sense, do you mine writing me a psuedo code for that second code? If you don't mind of course

  7. #7
    Registered User whiteflags's Avatar
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    Basically
    Code:
    Let A[] be an array of some number of items
    Let B[] be an array of 5 items
    
    for each item in B[]:
       B[item] = 0
    
    for each item in A[]:
        save = A[item]
        access B[save] then increment B[save]
    
    for each item in B[]:
       print B[item]
    Skipping the pseudocode step is reserved for expert programmers.

  8. #8
    Lean Mean Coding Machine KONI's Avatar
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    I'm being picky here but your pseudocode is misleading. In fact, the for each operation is used incorrectly:

    If you were to write "for each item in B[]", then item already refers to the value, not to the index. You can output all the values of an array like this:

    Code:
    for each item in B[]
        print item
    The correct way of writing is to say:
    Code:
    Let A[] be an array of some number of items
    Let B[] be an array of 5 items
    
    for each index in 0 to length(B[]) -1:
       B[index] = 0
    
    for each item in A[]:
        B[item] = B[item] + 1
    
    for each item in B[]:
       print item
    I know it's not "that" important to write correct pseudocode, but your construction of for each is against every syntax of every language implementing the foreach sugar.

  9. #9
    Registered User whiteflags's Avatar
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    I know that Koni. Thanks for the concern anyways.

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