Converting decimal to binary

This is a discussion on Converting decimal to binary within the C Programming forums, part of the General Programming Boards category; Hello. I'm encountering some errors in my code. Please help. Code: void binary(int number) { int i, count; for(count = ...

  1. #1
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    Join Date
    Mar 2012
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    Converting decimal to binary

    Hello. I'm encountering some errors in my code. Please help.

    Code:
    void binary(int number)
    {
    	int i, count;
    	for(count = 0; number > 0; count++)
    	{
    		number /= 2;
    	}
    	int remainder[count];
    	for(i = 0; i < count; i++)
    	{
    		
    		remainder[i] = number % 2;
    		printf("%d\t", remainder);
    		number /= 2;
    		printf("%d\t", number);
    		printf("%d\n\n", i);
    	}
    	for(i = count; i >= 0; i--)
    		printf("%d", remainder[1]);
    }

    Errors:
    1. expected constant expression
    2. cannot allocate an array of constant size 0
    3. 'remainder' : unknown size

  2. #2
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    The 1989 C standard did not support variable size arrays (a-la "int remainder [count];" in your code). The 1999 C standard did.

    This means, depending on age of your C compiler, that your code is invalid.

    Either update your compiler, or look up use of malloc() to dynamically allocate memory, and free() to release it.
    Right 98% of the time, and don't care about the other 3%.

    If I seem grumpy or unhelpful in reply to you, or tell you you need to demonstrate more effort before you can expect help, it is likely you deserve it. Suck it up, Sunshine, and read this, this, and this before posting again.

  3. #3
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    But I'm using Visual Studio 2005 Express Edition.

  4. #4
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    It typically takes up to ten years after a standard is released before you can be confident that new compilers implement it reasonably completely.

    Visual Studio only partly implemented features of of the 1999 C standard until at least 2008. 2005 was before 2008, so Visual Studio 2005 editions would also have only had partial support for C99.
    Right 98% of the time, and don't care about the other 3%.

    If I seem grumpy or unhelpful in reply to you, or tell you you need to demonstrate more effort before you can expect help, it is likely you deserve it. Suck it up, Sunshine, and read this, this, and this before posting again.

  5. #5
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    can try this..

    Code:
    void binary(int number)
    {
        int i, count, temp = number;
        for(count = 0; number > 0; count++)
        {
            number /= 2;
        }
    number =temp; // try saving the input.. else you would have lost it in the first loop.
    int * remainder =  calloc(count, sizeof(int)); //to get around your compiler //incompatibility. should use free though as stated.
        for(i = 0; i < count; i++)
        {
             
            remainder[i] = number % 2;
            printf("%d\t",remainder[i]);
            number /= 2;
            printf("%d\t", number);
            printf("%d\n\n", i);
        }
        for(i = count - 1; i >= 0; i--) // introduce a -1 here prints an extra zero.. 
            printf("%d", remainder[i]);
    }
    hope this helps..

    Nitin

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