How about something like
int getbit(bits, n) {
return (bits[n>>4] & (n & 0x0F)) == (n & 0x0F);
}
This just tests the bit(s) rather than shifting it all the way to the end.
The result is a Boolean (well as close as C comes :).
-----Original Message-----
From: mspgcc-users-ad...@lists.sourceforge.net
[mailto:mspgcc-users-ad...@lists.sourceforge.net] On Behalf Of Kelly Murray
Sent: Tuesday, June 15, 2004 4:08 AM
To: mspgcc-users@lists.sourceforge.net
Subject: [Mspgcc-users] bit arrays
Given the small amount of ram (128 bytes) in the x1101, I need to use
bit arrays for a 64bit value. Is there a good/best way to do this?
I've done roughly the code below:
unsigned int bits[4]; // 16bit words x 4 = 64bits
getbit(bits, n) { return( (bits[n>>4] >> 15-(n&0xF)) & 0x1) ); }
However, it's really quite ugly compared to just " bits[n] ",
I've also looked at the code generated, and it's quite bloated,
and given the ROM of a 1101 (1K), I can't really afford that either,
the shifting is not a single instruction on a MSP430.
The bits[n] generates just 8 bytes of code.
(bits[n>>4] >> 15-(n&0xF)) & 0x1) generates 50 bytes
My most recent idea is to use something similiar with somewhat
less shifting, at least not the loop used above.
unsigned char bits[16];
getbit(bits, n) {
int bits4 = bits[n>>2];
switch (n & 0x3) {
case 11: bits4 >>= 1;
case 01: bits4 >>= 1;
case 10: bits4 >>= 1;
case 11: return bits4 & 1;
}}
This generates 88 bytes...
there must be a simple way to do this that I'm just overlooking.
Perhaps some simple assembly code is the best approach?
thanks,
Kelly Murray
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