Convert Little Endian to Big Endian - Not getting expected result

Question

I have a very small code where I am trying to convert a 16 bit number from little endian to big endian format.

The value of number is 0x8000 and after conversion I am expecting it to be as 0x0080 - but I am getting some different value as mentioned below:

#include <iostream> int main() { int num = 0x8000; int swap = (num>>8) | (num<<8); std::cout << swap << std::endl; return 0; }

The program outputs 8388736 as value which in hex is 0x800080 - I am not sure what wrong am I doing?

On almost all platforms, int is a 32 bit type. You initialize num to 0x00008000, and all bits are significant. — Some programmer dude
– Some programmer dude, Commented Aug 29, 2021 at 19:19
Using just bit operations is valid for unsigned int. For signed type the bit representation is different. — Ripi2
– Ripi2, Commented Aug 29, 2021 at 19:29
Use (unsigned) short int, or better, (u)int16_t, then you will get the result you want. — Remy Lebeau
– Remy Lebeau, Commented Aug 29, 2021 at 20:27

Ted Lyngmo · Accepted Answer · 2021-08-29 19:18:45Z

5

If you do 0x8000 << 8 you'll get 0x800000. If you | that with 0x80 you get the answer you now get. You need to filter away the upper part:

int swap = (num>>8) | (0xFF00 & (num<<8));

Suggestion: Use fixed width types, like uint8_t, uint16_t, uint32_t and uint64_t.

answered Aug 29, 2021 at 19:18

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4 Comments

Thanks - so is this the general way always to swap 16 bits number?

No, general way is to use htons() and ntohs() functions

Thanks - I meant using bit pattern - in many places I see code like - (num << 8 ) | (num >> 8) ?

@Programmer That works with uint16_t. If you use a type wirh more than 16 bits, you need a bitwise AND to remove any set bits above the 16th bit