Charles R Harris wrote: > Hi All, > > Currently subtract for boolean arrays is defined in > > /**begin repeat > * Arithmetic operators > * > * # OP = ||, ^, &&# > * #kind = add, subtract, multiply# > */ > static void > [EMAIL PROTECTED]@(char **args, intp *dimensions, intp *steps, void *func) > { > register intp i; > intp is1=steps[0],is2=steps[1],os=steps[2], n=dimensions[0]; > char *i1=args[0], *i2=args[1], *op=args[2]; > for(i=0; i<n; i++, i1+=is1, i2+=is2, op+=os) { > *((Bool *)op)=*((Bool *)i1) @OP@ *((Bool *)i2); > } > } > /**end repeat**/ > > > Note that this might yield unexpected results if the boolean value is > not 0 or 1, which is possible using views. Note also that bitwise_xor > converts the boolean to 0 or 1 before using ^. I think subtract should > work the same way, but that would change current behavior. So... do I > make the change and call it a bug fix or leave it as it is?
I think it's a bugfix and so am +1 on the change. -Travis _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion