Even or odd, count the set bits, multiply by shifting

Four questions about one number, none of them answered with arithmetic. Is 182 odd, what is it times eight, what are its bottom four bits, and how many of its 32 bits are ones. The answers land in $t1 to $t4.

The instructions of the Example before this one treat a register as a number. These four treat the same register as 32 bits side by side, which is the other way to read one and often the cheaper way.

You need to know: the "Arithmetic, logic and bits" lecture. What is new here is that a masked bit is already a 0 or a 1, so counting one costs an add and no branch at all.

andi $t1, $t0, 1 is C's n & 1, and the answer is the value: $t1 comes out at 0 because 182 is even, and it would be 1 for an odd number, with nothing else to read and no flag anywhere. The M68K tests that bit with btst, which sets Z to 1 when the bit was 0, and then needs an sne to turn the flag back into a number.

Shifting left by three multiplies by eight, since every place a bit moves left doubles what it is worth. $t2 comes out at 000005B0, which is 1456. The shift amount is five bits, so 0 to 31, and sllv takes it from a register when the program worked it out.

andi $t3, $t0, 0xF keeps the four bits the mask has set and clears everything else, so $t3 is 6, the 6 of 0xB6. That is how any field is taken out of a packed value: mask what you want, then shift it down to the bottom if it was not there already. The constant of andi is 16 bits, so a mask that reaches into the top half of a register goes through li and a register first.

The loop runs 32 times, once per bit, and does C's count += n & 1; n >>= 1;. srl is the shift that brings zeroes in at the top. sra copies the sign bit down instead, which is what divides a signed number by two, and here the register is a row of bits to take apart. $t4 comes out at 5, the number of ones in 10110110. The M68K writes the same loop around its carry flag, shifting the bottom bit into C and branching on it; here the masked bit is a number and add $t4, $t4, $t7 counts it without a branch.

clz $t8, $t0 counts the leading zeroes, the run of 0 bits from the top down, and comes out at 24: the highest set bit of 182 is bit 7, and there are 24 bits above it. That is how a program finds the position of the top bit of a number in one instruction.

Try changing li $t0, 182 to li $t0, 183, one more. $t1 becomes 1 because the number is now odd, $t3 becomes 7, and $t4 becomes 6.