An expensive jump with GCC 5.4.0
Solution 1:
The logical AND operator (&&
) uses short-circuit evaluation, which means that the second test is only done if the first comparison evaluates to true. This is often exactly the semantics that you require. For example, consider the following code:
if ((p != nullptr) && (p->first > 0))
You must ensure that the pointer is non-null before you dereference it. If this wasn't a short-circuit evaluation, you'd have undefined behavior because you'd be dereferencing a null pointer.
It is also possible that short circuit evaluation yields a performance gain in cases where the evaluation of the conditions is an expensive process. For example:
if ((DoLengthyCheck1(p) && (DoLengthyCheck2(p))
If DoLengthyCheck1
fails, there is no point in calling DoLengthyCheck2
.
However, in the resulting binary, a short-circuit operation often results in two branches, since this is the easiest way for the compiler to preserve these semantics. (Which is why, on the other side of the coin, short-circuit evaluation can sometimes inhibit optimization potential.) You can see this by looking at the relevant portion of object code generated for your if
statement by GCC 5.4:
movzx r13d, WORD PTR [rbp+rcx*2]
movzx eax, WORD PTR [rbx+rcx*2]
cmp r13w, 478 ; (curr[i] < 479)
ja .L5
cmp ax, 478 ; (l[i + shift] < 479)
ja .L5
add r8d, 1 ; nontopOverlap++
You see here the two comparisons (cmp
instructions) here, each followed by a separate conditional jump/branch (ja
, or jump if above).
It is a general rule of thumb that branches are slow and are therefore to be avoided in tight loops. This has been true on virtually all x86 processors, from the humble 8088 (whose slow fetch times and extremely small prefetch queue [comparable to an instruction cache], combined with utter lack of branch prediction, meant that taken branches required the cache to be dumped) to modern implementations (whose long pipelines make mispredicted branches similarly expensive). Note the little caveat that I slipped in there. Modern processors since the Pentium Pro have advanced branch prediction engines that are designed to minimize the cost of branches. If the direction of the branch can be properly predicted, the cost is minimal. Most of the time, this works well, but if you get into pathological cases where the branch predictor is not on your side, your code can get extremely slow. This is presumably where you are here, since you say that your array is unsorted.
You say that benchmarks confirmed that replacing the &&
with a *
makes the code noticeably faster. The reason for this is evident when we compare the relevant portion of the object code:
movzx r13d, WORD PTR [rbp+rcx*2]
movzx eax, WORD PTR [rbx+rcx*2]
xor r15d, r15d ; (curr[i] < 479)
cmp r13w, 478
setbe r15b
xor r14d, r14d ; (l[i + shift] < 479)
cmp ax, 478
setbe r14b
imul r14d, r15d ; meld results of the two comparisons
cmp r14d, 1 ; nontopOverlap++
sbb r8d, -1
It is a bit counter-intuitive that this could be faster, since there are more instructions here, but that is how optimization works sometimes. You see the same comparisons (cmp
) being done here, but now, each is preceded by an xor
and followed by a setbe
. The XOR is just a standard trick for clearing a register. The setbe
is an x86 instruction that sets a bit based on the value of a flag, and is often used to implement branchless code. Here, setbe
is the inverse of ja
. It sets its destination register to 1 if the comparison was below-or-equal (since the register was pre-zeroed, it will be 0 otherwise), whereas ja
branched if the comparison was above. Once these two values have been obtained in the r15b
and r14b
registers, they are multiplied together using imul
. Multiplication was traditionally a relatively slow operation, but it is darn fast on modern processors, and this will be especially fast, because it's only multiplying two byte-sized values.
You could just as easily have replaced the multiplication with the bitwise AND operator (&
), which does not do short-circuit evaluation. This makes the code much clearer, and is a pattern that compilers generally recognize. But when you do this with your code and compile it with GCC 5.4, it continues to emit the first branch:
movzx r13d, WORD PTR [rbp+rcx*2]
movzx eax, WORD PTR [rbx+rcx*2]
cmp r13w, 478 ; (curr[i] < 479)
ja .L4
cmp ax, 478 ; (l[i + shift] < 479)
setbe r14b
cmp r14d, 1 ; nontopOverlap++
sbb r8d, -1
There is no technical reason it had to emit the code this way, but for some reason, its internal heuristics are telling it that this is faster. It would probably be faster if the branch predictor was on your side, but it will likely be slower if branch prediction fails more often than it succeeds.
Newer generations of the compiler (and other compilers, like Clang) know this rule, and will sometimes use it to generate the same code that you would have sought by hand-optimizing. I regularly see Clang translate &&
expressions to the same code that would have been emitted if I'd have used &
. The following is the relevant output from GCC 6.2 with your code using the normal &&
operator:
movzx r13d, WORD PTR [rbp+rcx*2]
movzx eax, WORD PTR [rbx+rcx*2]
cmp r13d, 478 ; (curr[i] < 479)
jg .L7
xor r14d, r14d ; (l[i + shift] < 479)
cmp eax, 478
setle r14b
add esi, r14d ; nontopOverlap++
Note how clever this is! It is using signed conditions (jg
and setle
) as opposed to unsigned conditions (ja
and setbe
), but this isn't important. You can see that it still does the compare-and-branch for the first condition like the older version, and uses the same setCC
instruction to generate branchless code for the second condition, but it has gotten a lot more efficient in how it does the increment. Instead of doing a second, redundant comparison to set the flags for a sbb
operation, it uses the knowledge that r14d
will be either 1 or 0 to simply unconditionally add this value to nontopOverlap
. If r14d
is 0, then the addition is a no-op; otherwise, it adds 1, exactly like it is supposed to do.
GCC 6.2 actually produces more efficient code when you use the short-circuiting &&
operator than the bitwise &
operator:
movzx r13d, WORD PTR [rbp+rcx*2]
movzx eax, WORD PTR [rbx+rcx*2]
cmp r13d, 478 ; (curr[i] < 479)
jg .L6
cmp eax, 478 ; (l[i + shift] < 479)
setle r14b
cmp r14b, 1 ; nontopOverlap++
sbb esi, -1
The branch and the conditional set are still there, but now it reverts back to the less-clever way of incrementing nontopOverlap
. This is an important lesson in why you should be careful when trying to out-clever your compiler!
But if you can prove with benchmarks that the branching code is actually slower, then it may pay to try and out-clever your compiler. You just have to do so with careful inspection of the disassembly—and be prepared to re-evaluate your decisions when you upgrade to a later version of the compiler. For example, the code you have could be rewritten as:
nontopOverlap += ((curr[i] < 479) & (l[i + shift] < 479));
There is no if
statement here at all, and the vast majority of compilers will never think about emitting branching code for this. GCC is no exception; all versions generate something akin to the following:
movzx r14d, WORD PTR [rbp+rcx*2]
movzx eax, WORD PTR [rbx+rcx*2]
cmp r14d, 478 ; (curr[i] < 479)
setle r15b
xor r13d, r13d ; (l[i + shift] < 479)
cmp eax, 478
setle r13b
and r13d, r15d ; meld results of the two comparisons
add esi, r13d ; nontopOverlap++
If you've been following along with the previous examples, this should look very familiar to you. Both comparisons are done in a branchless way, the intermediate results are and
ed together, and then this result (which will be either 0 or 1) is add
ed to nontopOverlap
. If you want branchless code, this will virtually ensure that you get it.
GCC 7 has gotten even smarter. It now generates virtually identical code (excepting some slight rearrangement of instructions) for the above trick as the original code. So, the answer to your question, "Why does the compiler behave this way?", is probably because they're not perfect! They try to use heuristics to generate the most optimal code possible, but they don't always make the best decisions. But at least they can get smarter over time!
One way of looking at this situation is that the branching code has the better best-case performance. If branch prediction is successful, skipping unnecessary operations will result in a slightly faster running time. However, branchless code has the better worst-case performance. If branch prediction fails, executing a few additional instructions as necessary to avoid a branch will definitely be faster than a mispredicted branch. Even the smartest and most clever of compilers will have a hard time making this choice.
And for your question of whether this is something programmers need to watch out for, the answer is almost certainly no, except in certain hot loops that you are trying to speed up via micro-optimizations. Then, you sit down with the disassembly and find ways to tweak it. And, as I said before, be prepared to revisit those decisions when you update to a newer version of the compiler, because it may either do something stupid with your tricky code, or it may have changed its optimization heuristics enough that you can go back to using your original code. Comment thoroughly!
Solution 2:
One important thing to note is that
(curr[i] < 479) && (l[i + shift] < 479)
and
(curr[i] < 479) * (l[i + shift] < 479)
are not semantically equivalent! In particular, the if you ever have the situation where:
-
0 <= i
andi < curr.size()
are both true -
curr[i] < 479
is false -
i + shift < 0
ori + shift >= l.size()
is true
then the expression (curr[i] < 479) && (l[i + shift] < 479)
is guaranteed to be a well-defined boolean value. For example, it does not cause a segmentation fault.
However, under these circumstances, the expression (curr[i] < 479) * (l[i + shift] < 479)
is undefined behavior; it is allowed to cause a segmentation fault.
This means that for the original code snippet, for example, the compiler can't just write a loop that performs both comparisons and does an and
operation, unless the compiler can also prove that l[i + shift]
will never cause a segfault in a situation it's required not to.
In short, the original piece of code offers fewer opportunities for optimization than the latter. (of course, whether or not the compiler recognizes the opportunity is an entirely different question)
You might fix the original version by instead doing
bool t1 = (curr[i] < 479);
bool t2 = (l[i + shift] < 479);
if (t1 && t2) {
// ...