Is there a better way to reverse an array of bytes in memory?

Question

typedef unsigned char Byte;

...

void ReverseBytes( void *start, int size )
{
    Byte *buffer = (Byte *)(start);

    for( int i = 0; i < size / 2; i++ ) {
        std::swap( buffer[i], buffer[size - i - 1] );
    }
}

What this method does right now is it reverses bytes in memory. What I would like to know is, is there a better way to get the same effect? The whole "size / 2" part seems like a bad thing, but I'm not sure.

EDIT: I just realized how bad the title I put for this question was, so I [hopefully] fixed it.

Your example seems flawed, how can you swap two chars without a location? I suspect you need pass in the address. — leppie
– leppie, Commented Feb 25, 2009 at 16:45

kmkaplan · Accepted Answer · 2013-04-23 19:00:13Z

34

The standard library has a std::reverse function:

#include <algorithm>
void ReverseBytes( void *start, int size )
{
    char *istart = start, *iend = istart + size;
    std::reverse(istart, iend);
}

edited Apr 23, 2013 at 19:00

user283145

answered Feb 25, 2009 at 9:16

kmkaplan

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2 Comments

xian Over a year ago

Thanks. I should have looked it up before writing it myself.

Nate Eldredge Apr 10 at 4:00

@JulesGagnon-Marchand: The "as-if" rule means that the description in the standard only specifies the observable effects; reverse does not actually have to be implemented that way. It may be implemented in a more optimized way, perhaps including hand-written optimizations that the compiler won't do for OP's manual version. So I would say it's certainly cleaner, no worse in terms of performance, and at least potentially better.

Michael Manner · Accepted Answer · 2012-07-07 14:30:13Z

28

A performant solution without using the STL:

void reverseBytes(void *start, int size) {
    unsigned char *lo = start;
    unsigned char *hi = start + size - 1;
    unsigned char swap;
    while (lo < hi) {
        swap = *lo;
        *lo++ = *hi;
        *hi-- = swap;
    }
}

Though the question is 3 ½ years old, chances are that someone else will be searching for the same thing. That's why I still post this.

answered Jul 7, 2012 at 14:30

Michael Manner

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2 Comments

Aviad P. Over a year ago

The chance has realized :)

Ilan Over a year ago

To complete the code, just add the casting: (unsigned char*)start + size - 1;

Mykola Golubyev · Accepted Answer · 2009-02-25 10:56:34Z

2

If you need to reverse there is a chance that you can improve your algorithms and just use reverse iterators.

answered Feb 25, 2009 at 10:56

Mykola Golubyev

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2 Comments

xian Over a year ago

It's for reading data from files that uses different endiannesses.

Hosam Aly Over a year ago

@kitchen, this seems more like reversing the bytes of an integer, not the bytes of a whole array...

Rob K · Accepted Answer · 2009-02-25 16:35:28Z

1

If you're reversing binary data from a file with different endianness you should probably use the ntoh* and hton* functions, which convert specified data sizes from network to host order and vice versa. ntohl for instance converts a 32 bit unsigned long from big endian (network order) to host order (little endian on x86 machines).

answered Feb 25, 2009 at 16:35

Rob K

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2 Comments

Xofo Over a year ago

This question concerns swapping an array, so the solutions provided above specifically answer than need.

davidbak Over a year ago

Or see endian.h if you're using glib.

phuclv · Accepted Answer · 2025-04-10 03:31:19Z

0

I would review the stl::swap and make sure it's optimized; after that I'd say you're pretty optimal for space. I'm reasonably sure that's time-optimal as well.

edited Apr 10 at 3:31

phuclv

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answered Feb 25, 2009 at 16:43

Paul Nathan

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1 Comment

Head Geek Over a year ago

Nowhere near time-optimal. The size/2 calculation might be optimized out of being run every loop, but the size-i-1 calculation wouldn't be, nor would the cost of the array indexing. That said, a perfectly optimized loop wouldn't be that much faster than what he's got.

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