int lkbitwise (
kobject i1,
kobject i2,
long sval,
char *function,
kobject o)
The supported functions are:
.TS center tab(:) ; l l . and : o = i1 & (i2 or sval) or : o = i1 | (i2 or sval) xor : o = i1 ^ (i2 or sval) nand : o = (~i1) | (~(i2 or sval)) nor : o = (~i1) & (~(i2 or sval)) andrev : o = i1 & (~(i2 or sval)) andinv : o = (~i1) & (i2 or sval) orrev : o = i1 | (~(i2 or sval)) orinv : o = (~i1) | (i2 or sval) rs : o = i1 >> by (i2 or sval) ls : o = i1 << by (i2 or sval) srs : Signed Right Shift by (i2 or sval) sls : Signed Left Shift by (i2 or sval) rr : Right Rotate by (i2 or sval) lr : Left Rotate by (i2 or sval) .TETherefore if lkbitwise() were called with two input data objects:
lkbitwise(i1, i2, 0, "and", o);
then the following would be performed:
o = i1 & i2
If lkbitwise were called with a single data object and a constant:
lkbitwise(i1, NULL, 10, "and", o);
then the following would be performed:
o = i1 & 10
none
none
$DATAMANIP/objects/library/kdatamanip/src/lkbitwise.c