CODEWITHSUNDEEP
c++
Joaquin
Joaquin

Reputation: 3

Unify 2 bytes to Int

I have a function that allows me to separate a INT value in 2 Bytes (e.g: INT "123123" results in E0F3; highByte= 0xF3 and lowByte=0xE0)

using this:

void int2bytes(unsigned char dest[2],int val){
             int hByte=0;
             int lByte=0;

             hByte=val&0XFF;
             lByte=val>>8;

             dest[0]= (char)hByte;
             dest[1]= (char)lByte;
            }

my question is:

How do I convert(unify) those 2 bytes to one INT equals to "123123"?

Upvotes: 0

Views: 293

Answers (2)

TrueY
TrueY

Reputation: 7610

Try

(dest[1] << 8) & dest[0]

IMHO the hByte and lByte is swapped... And what is more 123123 is not a two byte value. Int is signed, so on two bytes the max value is 32767, but int (usually) is 4 bytes.

I might suggest to use a union in this case, because you can spare all the arithmetics and you do not need to use explicit conversion at all.

An example code to use union:

#include <iostream>
using std::cout;
using std::hex;
using std::dec;
using std::endl;

int main() {
   union int2bytes {
       unsigned char byte[sizeof(int)];
       int val;
   };

   int2bytes i;
   // Convert int to byte
   i.val = 123123;
   cout << i.val << " : " << hex << i.val << dec << endl;
   for (int j = 0; j < sizeof(int); ++j)
       cout << "Byte#" << j << " : " << hex << (int)i.byte[j] << dec << endl;

   // Convert byte to int
    i.byte[1]--;
    cout << i.val << " : " << hex << i.val << dec << endl;
}

The output:

123123 : 1e0f3
Byte#0 : f3
Byte#1 : e0
Byte#2 : 1
Byte#3 : 0
122867 : 1dff3

Union can be improved a little bit

union int2bytes {
    unsigned char byte[sizeof(int)];
    int val;
    int2bytes (const int2bytes& i = 0) : val(i.val) {};
    int2bytes (int i) : val(i) {};
};

Now these work

int2bytes i = 123123;
int2bytes j; // j.val == 0
int2bytes k = i;
j = i;

Upvotes: 0

David Heffernan
David Heffernan

Reputation: 612894

To perform the reverse of your function int2bytes you simply do this:

int val = (lByte << 8) | hByte;

Although, I think you have the names of your variables back-to-front. I'd call val&0xff the low byte, and val >> 8 the high byte.

You are also mistaken in supposing that you can fit 123123 into 2 bytes. Remember that 2^16 is 65536.

I have a function that allows me to separate a int value into 2 bytes (e.g: int "123123" results in E0F3; highByte= 0xF3 and lowByte=0xE0)

Well, E0F3 represented in decimal is equal to 57587. Which is the value you report in the comment you made to my answer.

Now, 123123 represented in hexadecimal is 0x1E0F3 and you need at least three bytes to store that value.

Upvotes: 3

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