Bitwise

Yes, they do evaluate to the same thing. Which you can check using:

assert( ((block->prevsize & 0x1) == 0x1 && (block->size & 0x1) == 0x1 ) == ( ((block->prevsize & block->size) & 0x1) == 0x1) );

The most readable version is:

assert( block->ValidSize( ) );

Block::ValidSize( ) {
return (size & prevsize & 0x1) == 0x1;
}

And then make it even more readable by making a IS_BIT_SET macro.

Quote:

Original post by Andrew Russell And then make it even more readable by making a IS_BIT_SET macro.

I usually mandate readable code but really "& 1" is every bit as readable as IS_BIT_SET...

some people thinks diffrently though...

Also note that: assert( size & prevsize & 1) is the same as assert( (size & prevsize & 1) == 1)

Quote:

Original post by ToohrVyk Yes, they do evaluate to the same thing. Which you can check using: assert( ((block->prevsize & 0x1) == 0x1 && (block->size & 0x1) == 0x1 ) == ( ((block->prevsize & block->size) & 0x1) == 0x1) ); The most readable version is: assert( block->ValidSize( ) ); Block::ValidSize( ) { return (size & prevsize & 0x1) == 0x1; }

and mark that method as const so that people feel comfortable using it in an assert:)

Cheers
Chris

Or, organized another way around, you could do this:

// Useful for other situations too :/inline bool is_set(int value, int bit) {  return (value & (1 << bit)) != 0;}// now instead of writing a bunch of asserts that call a function,// we just write a bunch of calls to the function:inline void Block::debug_checkSize() const {  assert(is_set(size, 0));  assert(is_set(prevsize, 0));}// which in release mode should still disappear completely.