Though in practice, the biggest variability comes from memory access — accessing an inner cache is dramatically faster than accessing RAM. Add to that that on modern CPUs an access to RAM can cost the same as hundreds of instructions, so previous memory accesses will be crucial cache effects. Multiplication is one instruction, in most instruction sets, for 64 or 32 bit integers. You can find reference manuals for various instruction sets by Googling -- look for "MIPS" for a relatively easy-to-understand version.
Good algorithms exist. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Is it possible to accurately determine the number of instructions required to multiply or add two integers in a modern processor?
Ask Question. Asked 7 years, 6 months ago. Active 5 years, 6 months ago. Viewed 2k times. Addition I've found Wiki's explanation of an adder-subtractor to be far more advanced, in description, not operation, surely , and I've had less luck interpreting it thus far. Here is an example:. But with other compilers, other compiler arguments, differently written inner loops, the results can vary and I cannot even get an approximation. Multiplication of two n -bit numbers can in fact be done in O log n circuit depth , just like addition.
Addition in O log n is done by splitting the number in half and recursively adding the two parts in parallel , where the upper half is solved for both the "0-carry" and "1-carry" case. Once the lower half is added, the carry is examined, and its value is used to choose between the 0-carry and 1-carry case. Multiplication in O log n depth is also done through parallelization , where every sum of 3 numbers is reduced to a sum of just 2 numbers in parallel, and the sums are done in some manner like the above.
I won't explain it here, but you can find reading material on fast addition and multiplication by looking up "carry-lookahead" and "carry-save" addition. So from a theoretical standpoint, since circuits are obviously inherently parallel unlike software , the only reason multiplication would be asymptotically slower is the constant factor in the front, not the asymptotic complexity.
This is an even more complex answer than simply multiplication versus addition. Multiplication, electronically, is a much more complicated circuit. Most of the reasons why, is that multiplication is the act of a multiplication step followed by an addition step, remember what it was like to multiply decimal numbers prior to using a calculator. The other thing to remember is that multiplication will take longer or shorter depending on the architecture of the processor you are running it on.
This may or may not be simply company specific. While an AMD will most likely be different than an Intel, even an Intel i7 may be different from a core 2 within the same generation , and certainly different between generations especially the farther back you go. This is more an exercise in understanding your architecture, and electronics.
To do this would get rid of the inherent performance gains we are able to get when adding pipelining into a processor. Pipelining is the idea of taking a task and breaking it down into smaller sub-tasks that can be performed much quicker.
By storing and forwarding the results of each sub-task between sub-tasks, we can now run a faster clock rate that only needs to allow for the longest latency of the sub-tasks, and not from the overarching task as a whole. In the above diagram, the non-pipelined circuit takes 50 units of time. In the pipelined version, we have split the 50 units into 5 steps each taking 10 units of time, with a store step in between.
For an operation to be completed, it must move through all 5 steps in order but another of the same operation with operands can be in step 2 as one is in step 1, 3, 4, and 5. With all of this being said, this pipelined approach allows us to continuously fill the operator each clock cycle, and get a result out on each clock cycle IF we are able to order our operations such that we can perform all of one operation before we switch to another operation, and all we take as a timing hit is the original amount of clocks necessary to get the FIRST operation out of the pipeline.
Mystical brings up another good point. It is also important to look at the architecture from a more systems perspective. It is true that the newer Haswell architectures was built to better the Floating Point multiply performance within the processor. For this reason as the System level, it was architected to allow multiple multiplies to occur in simultaneity versus an add which can only happen once per system clock.
Ryzen is similar. Bulldozer-family has much lower integer throughput and not-fully-pipelined multiply, including extra slow for bit operand-size multiply.
But a good compiler could auto-vectorize your loops. Or simply constant-propagate through them to just print out the answer! See Why does clang produce inefficient asm with -O0 for this simple floating point sum?
Store-forwarding latency on a modern x86 like Sandybridge-family including Haswell and Skylake is about 3 to 5 cycles, depending on timing of the reload. So with a 1-cycle latency ALU add in there, too, you're looking at about two 6-cycle latency steps in the critical path for this loop. See also Adding a redundant assignment speeds up code when compiled without optimization for another no-optimization benchmark. In that one, store-forwarding latency is actually reduced by having more independent work in the loop, delaying the reload attempt.
Or on Bulldozer-family, not fully pipelined with 1 per 2-clock throughput. Not so with the multiply. I came to this thread to get an idea of what the modern processors are doing in regard to integer math and the number of cycles required to do them.
I worked on this problem of speeding up bit integer multiplies and divides on the 65c processor in the 's. So the idea that multiplies are faster than adds is simply not the case except rarely but like people said it depends upon how the architecture is implemented.
If there are enough steps being performed available between clock cycles, yes a multiply could effectively be the same speed as an add based on the clock, but there would be a lot of wasted time.
One can dream. On the 65c processor, there were no multiply or divide instructions. Mult and Div were done with shifts and adds. To perform a 16 bit add, you would do the following:. If dealing with a call like from C, you would have additional overhead of dealing with pushing and pulling values off the stack. Creating routines that would do two multiples at once would save overhead for example.
The traditional way of doing the multiply is shifts and adds through the entire value of the one number. Each time the carry became a one as it is shifted left would mean you needed to add the value again. Since modern processors are super scalar and can execute out of order, you can often get total instructions per cycle that exceed 1. Calculating the efficiency of assembly code is not the best way to go in these days of Out of Order Execution Super Scalar pipelines.
It'll vary by processor type. It'll vary on instructions both before and after you can add extra code and have it run faster sometimes! Some operations division notably can have a range of execution times even on older more predictable chips. Actually timing of lots of iterations is the only way to go. You can find information on intel cpu at intel software developer manuals.
For instance the latency is 1 cycle for an integer addition and 3 cycles for an integer multiplication. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Asked 8 years, 2 months ago.
Active 4 years, 9 months ago. I am speaking about cost per occurrence — and this cost is indeed damn high for exceptions. However, if somebody is careless enough to raise them 1M times per second — well, they will be in trouble. So then, if your data structure contains tens of thousands or more items, the accumulated savings can easily exceed the cost of a throw.
I think that this is a great chart, and every software engineer should have a rough idea of these costs in their head when they are designing code. One thing that might be a useful addition to your diagram is to add the cost of acquiring uncontended locks e.
Please compare the exception cost to realistic alternatives. You throw an exception and catch it, say, three function calls above. The alternative is at least 4 tests and alternatives. That leads to more code. Also, the stack unraveling code is ad hoc and not easily optimized. I should say that comparing things as such is beyond the scope of this exercise comparing all the pairs which might be of interest, would take way too long.
Basically, the cost of return-and-check seems to consist of three components: 1. For the Exceptions vs.
The error-case happens after a bunch of non-error-cases, so the branch predictor would continue to predict non-error. If there were a bunch of error-case paths before this, this is not an exceptional error at all. As it stands right now, it looks like you are directly comparing the two, and that comparison is currently misleading. You have a point — and I added a note about it being a normal case. Going into further details will be probably even more misleading Thanks for the update.
Note that the answers decreased significantly a few generations ago on Intel hardware, though those operations still take significantly longer than a normal instruction. Let me guess — it happened around the time when FSB was abolished and therefore bus lock was no longer necessary for sync ;-. Your email address will not be published.
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