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?

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.

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 ???

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 .

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];

}

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...

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?

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;
}

---

How about

Code:

---

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

---