CODEWITHSUNDEEP
c++c
David Trevor
David Trevor

Reputation: 989

Swapping lower byte (0-7) with the higher one (8-15) one

I now know how it's done in one line, altough I fail to realise why my first draft doesn't work aswell. What I'm trying to do is saving the lower part into a different variable, shifting the higher byte to the right and adding the two numbers via OR. However, it just cuts the lower half of the hexadecimal and returns the rest.

short int method(short int number) {


short int a = 0;
for (int x = 8; x < 16; x++){
    if ((number & (1 << x)) == 1){
        a = a | (1<<x);
    }
}

    number = number >> 8;

short int solution = number | a;
return solution;

Upvotes: 5

Views: 1433

Answers (2)

QuestionC
QuestionC

Reputation: 10064

if ((number & (1 << x)) == 1)

This is only going to return true if x is 0. Since 1 in binary is 00000000 00000001, and 1 << x is going to set all but the x'th bit to 0.

You don't care if it's 1 or not, you just care if it's non-zero. Use

if (number & (1 << x))

Upvotes: 4

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726509

You are doing it one bit at a time; a better approach would do it with a single operation:

uint16_t method(uint16_t number) {
    return (number << 8) | (number >> 8);
}

The code above specifies 16-bit unsigned type explicitly, thus avoiding issues related to sign extension. You need to include <stdint.h> (or <cstdint> in C++) in order for this to compile.

Upvotes: 8

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