byte to long

This is a discussion on byte to long within the C++ Programming forums, part of the General Programming Boards category; ok i have a byte buffer byte buf[16]; in which i have stored longs. now how do i add the ...

  1. #1
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    byte to long

    ok i have a byte buffer

    byte buf[16]; in which i have stored longs.

    now how do i add the longs in it and get the total in another long?
    or in other words how would i read a long out of it?

  2. #2
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    You can't have longs in a byte buffer unless you use 4 byte slots for every long. In your case, you could store 4 longs in the array. Add them together using:

    long total=0;
    for(long i = 0; i < 4; i++)
    {
    total += (long*)(buf)[i];
    }

    I think I got it right.
    // Gliptic

  3. #3
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    i came across this but i wasn't sure. becuase

    lets say i have a long thats something like this

    00010010 01011010 10010010 10010010

    when i save it to a
    byte buf[4]
    it then splites into four parts. if i add each byte that is buf[0] +buf[1] +buf[2] +buf[3] would not the answer be different?

    like:
    buf[0] = 00010010
    buf[0] = 01011010 +
    buf[0] = 10010010 +
    buf[0] = 10010010 +
    =================
    11001 0000 ??? not the same ???

  4. #4
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    Of course it would be different. That's because you are adding all the bytes in the long together. If you want to read the long you just type: *(long*)buf .
    // Gliptic

  5. #5
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    sorry could you give me an example?

    lets say i have a


    byte buf[16]; with four longs in it.

    to add should i use

    long total=0;
    for(long i = 0; i < 4; i+=4 )
    {

    total += * (long*)(buf)[i];

    }

  6. #6
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    No, this one:

    long total=0;
    for(long i = 0; i < 4; i++)
    {
    total += (long*)(buf)[i];
    }

    Think of the byte array as what it is, a byte array!
    There are 16 bytes in it. In this code, we are treating the array just as it was a long array with four elements.

    Sorry, I don't have time to write more now...
    // Gliptic

  7. #7
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    thanks for your help, but i get this error

    '+=' : illegal, right operand has type 'long *'

    when using above, and if i remove the * the results still wrong?

  8. #8
    zen
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    Here's how you'd do it with a byte array holding two longs -

    Code:
    #include <iostream>
    
    using namespace std;
    
    typedef unsigned char byte;
    
    int main ()
    {
       
    	byte buf[8]; 
    	
    	long a=10,b=25;
    
    	//Put two longs in byte array
    	*(long*)buf = a;
    	*(long*)&buf[4]=b;
    
    		
    	long total=0; 
    	for(int i = 0; i < 8; i+=4) 
    	{ 
    	total += *(long*)&buf[i]; 
    	} 
    
    	cout << total;
    	
    	return 0;
    }
    zen

  9. #9
    and the hat of wrongness Salem's Avatar
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    How about

    Code:
    byte buf[16];
    long *lbuff = (long *)buff;
    for ( i = 0 ; i < 4 ; i++ ) {
        sum += lbuff[i];
    }

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