Author
Hey,
Just a quick question for you all. Do the two following lines of code result in the same?
1: assert( (block->prevsize & 0x1) == 0x1 && (block->size & 0x1) == 0x1 );
2: assert( ((block->prevsize & block->size) & 0x1) == 0x1 );
If so, which of the above do you consider the most elegant and "clean"?
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;
}
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;
}
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.This topic is closed to new replies.
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