Converting a numerical char array to a long?

I'm working with some C code on an embedded processor (8051)in which I have an array:

Code:

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unsigned char x[3];        // unsigned char is a 8-bit value on 8051
x[0] = 0x02;
x[1] = 0xFF;
x[2] = 0xFF;

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I then have an unsigned long variable in which I would like to equal 0x02FFFF (in the example above). On the 8051, the unsigned long variable is 32-bits.

Code:

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unsigned long i;

i = ???

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So how would this be accomplished?

Thanks in Advance
Philip

How about:

Code:

---

unsigned char x[3];        // unsigned char is a 8-bit value on 8051
x[0] = 0x02;
x[1] = 0xFF;
x[2] = 0xFF;

unsigned long i = 0;
for (int index = 0; index < 3; index++)
{
    i <<= 8;
    i |= x[index];
}

// or

unsigned long i = x[0];
i = i << 8;
i = i | x[1];
i = i << 8;
i = i | x[2];

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Is there a reason you declare a 4 element array but only use 3?

unsigned char x[3];
declares a three element array, right?

Quote:

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Originally posted by jlou
unsigned char x[3];
declares a three element array, right?

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No, [3] is a 4 element array.

0 is the first number

I think when you are declaring an array, the number inside is the size, in this example, 3.

Then when you are accessing the array, the number inside is the index, starting with 0. Indexes 0, 1, and 2 are valid in an array of size 3.

So:
int myArray[3]; // declares an array of size 3.
int x = myArray[2]; // set x to the 3rd element in myArray.

and:
int myArray[4]; // declares an array of size 4.
int x = myArray[3]; // set x to the 4th element in myArray.

no

Code:

---

#include <iostream>
using namespace std;

int main()
{
        int myarray[3];
        myarray[0] = 1;
        myarray[1] = 2;
        myarray[2] = 3;
        myarray[3] = 4;

        cout << myarray[0] << myarray[1] << myarray[2] << myarray[3] << endl;
}

---

This will output 1234

myarray[3] holds 4 elements.

I would also like to recommend the use of vectors, arrays suck compared to vectors :D

Oh God, NOOOOO!

type name[size];
Declares an array with size elements, not size+1. No arguing.

Quote:

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This will output 1234

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Only that this means nothing. Try this code:

Code:

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#include <iostream>
using namespace std;

int main()
{
        int previous;
        int myarray[3];
        int next;
        myarray[0] = 1;
        myarray[1] = 2;
        myarray[2] = 3;
        myarray[3] = 4;
        previous = 0;
        next = 0;

        cout << myarray[0] << myarray[1] << myarray[2] << myarray[3] << endl;
}

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Now try again. As you can see, with myarray[3] you are accessing memory that does not belong to the array!

Quote:

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Originally posted by kasracer
I would also like to recommend the use of vectors, arrays suck compared to vectors :D

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Now there is something I very much agree with!

In other words, yes...it's only a 3 element array..hehe.

Thanks for the idea jlou. Both the shifting will do the job or casting a pointer. I could also use a union to access the individual bytes, but I had too many variables to redefine.

Thanks
P