Why is modulus operator slow?

Question

Paraphrasing from in \"Programming Pearls\" book (about c language on older machines, since book is from the late 90\'s):

Integer arithmetic operations (+

Answer 1

The modulus/modulo operation is usually understood as the integer equivalent of the remainder operation - a side effect or counterpart to division.

Except for some degenerate cases (where the divisor is a power of the operating base - i.e. a power of 2 for most number formats) this is just as expensive as integer division!

So the question is really, why is integer division so expensive?

I don't have the time or expertise to analyze this mathematically, so I'm going to appeal to grade school maths:

Consider the number of lines of working out in the notebook (not including the inputs) required for:

• Equality: (Boolean operations) essentially none - in computer "big O" terms this is known a O(1)
• addition: two, working left to right, one line for the output and one line for the carry. This is an O(N) operation
• long multiplication: n*(n+1) + 2: two lines for each of the digit products (one for total, one for carry) plus a final total and carry. So O(N^2) but with a fixed N (32 or 64), and it can be pipelined in silicon to less than that
• long division: unknown, depends upon the argument size - it's a recursive descent and some instances descend faster than others (1,000,000 / 500,000 requires less lines than 1,000 / 7). Also each step is essentially a series of multiplications to isolate the closest factors. (Although multiple algorithms exist). Feels like an O(N^3) with variable N

So in simple terms, this should give you a feel for why division and hence modulo is slower: computers still have to do long division in the same stepwise fashion tha you did in grade school.

If this makes no sense to you; you may have been brought up on school math a little more modern than mine (30+ years ago).

---

The Order/Big O notation used above as O(something) expresses the complexity of a computation in terms of the size of its inputs, and expresses a fact about its execution time. http://en.m.wikipedia.org/wiki/Big_O_notation

O(1) executes in constant (but possibly large) time. O(N) takes as much time as the size of its data-so if the data is 32 bits it takes 32 times the O(1) time of the step to calculate one of its N steps, and O(N^2) takes N times N (N squared) the time of its N steps (or possibly N times MN for some constant M). Etc.

---

In the above working I have used O(N) rather than O(N^2) for addition since the 32 or 64 bits of the first input are calculated in parallel by the CPU. In a hypothetical 1 bit machine a 32 bit addition operation would be O(32^2) and change. The same order reduction applies to the other operations too.