Outputting a string to the screen

This is a discussion on Outputting a string to the screen within the C++ Programming forums, part of the General Programming Boards category; If you declare a string 32 bits long after you put thirty two bits in it it should only print ...

  1. #1
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    Outputting a string to the screen

    If you declare a string 32 bits long after you put thirty two bits in it it should only print those 32 bits correct???? It shouldn't print all these null characters then another 32 bit string??? sorry just venting a little

  2. #2
    RoD
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    >>sorry just venting a little

    Is there a specific program you are having trouble with? If so post the code and we'll be more then happy to look it over...

  3. #3
    Code Goddess Prelude's Avatar
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    >It shouldn't print all these null characters then another 32 bit string???
    Not unless you are using a C string that you forgot to terminate with a nul character.

    -Prelude
    My best code is written with the delete key.

  4. #4
    I lurk
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    Yes, if you declare a string 32 elements long, you may access elements 0-31 (effectively 32 elements) You cannot access the 32nd element, as, in essence it's actually the 33rd element of your string. Remember, counting is zero-based.

    char *pszString = { "Hello" };

    pszString[0] now contains 'H'
    pszString[1] = 'e'
    pszString[2] = 'l'
    pszString[3] = 'l'
    pszString[4] = 'o'
    pszString[5] = '\0' Null terminator, overwrite this and you'll overflow when attemping to print the string

  5. #5
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    OK here goes I am making a simulator and trying to make an and gate work in the ALU(I have the same problems with or gates, nor, etc)

    Code:
    void main()
    {
    
    	//ALU
    	char CRYOUT = '0';			//Carry out flag
    	char OVF = '0';				//Overflow flag
    	char ZERO = '0';				//Zero flag
    	char SGN = '0';				//Sign bit that comes of ALU, MSB
    	//ALU controls
    	char CTLALU[5];			//Control ALU
    	char CTLCIN = '0';			//Carry in
    
    	char AluInputA[33];		//Output saved from	MUXALUA and is input to ALU A
    	char AluInputB[33];		//Output saved from	MUXALUB and is input to ALU B
    	char AluOutput[33];		//Output saved from the ALU
    
    
    
    cout << "Testing ALU function." << endl;
    	ALU(AluInputA, AluInputB, CTLALU, CTLCIN, CRYOUT, OVF, ZERO, SGN, AluOutput);
    
    	cout << "And Output = " << AluOutput << endl;
    }
    
    
    //--------------------AND--------------------
    void and(char InputA[33], char InputB[33], char Output[33])
    {
    	for(int i = 31; i >= 0; i--)
    	{
    		if(InputA[i] == '1' && InputB[i] == '1')
    			Output[i] = '1';
    		else
    			Output[i] = '0';
    	}
    
    }
    The Output looks like this:

    The binary number in decimal is 0
    4 to 1 Multiplexer = 00000000000000000000000000000010
    MUXALUB
    The binary number in decimal is 3
    4 to 1 Multiplexer = 00000000000000000000000000001111
    Enter in ALU Control: 0001
    Testing ALU function.
    The binary number in decimal is 1
    Output: 00000000000000000000000000000010dddd0000000000 0000000000000000001111

    Press any key to continue


    so as you see the first 32 bits is right but the other characters then InputB from the 4-1 mux is also in the string. I hope someone can help. Thanks.

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