< 比 <= 快吗？

Question

Is if (a < 901) faster than if (a <= 900) ?

Not exactly as in this simple example, but there are slight performance changes on loop complex code. I suppose this has to do something with generated machine code in case it's even true.

Answer 1

No, it will not be faster on most architectures. You didn't specify, but on x86, all of the integral comparisons will be typically implemented in two machine instructions:

• A test or cmp instruction, which sets EFLAGS
• And a Jcc (jump) instruction , depending on the comparison type (and code layout):
• jne - Jump if not equal --> ZF = 0
• jz - Jump if zero (equal) --> ZF = 1
• jg - Jump if greater --> ZF = 0 and SF = OF
• (etc...)

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Example (Edited for brevity) Compiled with $ gcc -m32 -S -masm=intel test.c

    if (a < b) {
        // Do something 1
    }

Compiles to:

    mov     eax, DWORD PTR [esp+24]      ; a
    cmp     eax, DWORD PTR [esp+28]      ; b
    jge     .L2                          ; jump if a is >= b
    ; Do something 1
.L2:

And

    if (a <= b) {
        // Do something 2
    }

Compiles to:

    mov     eax, DWORD PTR [esp+24]      ; a
    cmp     eax, DWORD PTR [esp+28]      ; b
    jg      .L5                          ; jump if a is > b
    ; Do something 2
.L5:

So the only difference between the two is a jg versus a jge instruction. The two will take the same amount of time.

---

I'd like to address the comment that nothing indicates that the different jump instructions take the same amount of time. This one is a little tricky to answer, but here's what I can give: In the Intel Instruction Set Reference , they are all grouped together under one common instruction, Jcc (Jump if condition is met). The same grouping is made together under the Optimization Reference Manual , in Appendix C. Latency and Throughput.

Latency — The number of clock cycles that are required for the execution core to complete the execution of all of the μops that form an instruction.

Throughput — The number of clock cycles required to wait before the issue ports are free to accept the same instruction again. For many instructions, the throughput of an instruction can be significantly less than its latency

The values for Jcc are:

      Latency   Throughput
Jcc     N/A        0.5

with the following footnote on Jcc :

7. Selection of conditional jump instructions should be based on the recommendation of section Section 3.4.1, “Branch Prediction Optimization,” to improve the predictability of branches. When branches are predicted successfully, the latency of jcc is effectively zero.

So, nothing in the Intel docs ever treats one Jcc instruction any differently from the others.

If one thinks about the actual circuitry used to implement the instructions, one can assume that there would be simple AND/OR gates on the different bits in EFLAGS , to determine whether the conditions are met. There is then, no reason that an instruction testing two bits should take any more or less time than one testing only one (Ignoring gate propagation delay, which is much less than the clock period.)

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Edit: Floating Point

This holds true for x87 floating point as well: (Pretty much same code as above, but with double instead of int .)

        fld     QWORD PTR [esp+32]
        fld     QWORD PTR [esp+40]
        fucomip st, st(1)              ; Compare ST(0) and ST(1), and set CF, PF, ZF in EFLAGS
        fstp    st(0)
        seta    al                     ; Set al if above (CF=0 and ZF=0).
        test    al, al
        je      .L2
        ; Do something 1
.L2:

        fld     QWORD PTR [esp+32]
        fld     QWORD PTR [esp+40]
        fucomip st, st(1)              ; (same thing as above)
        fstp    st(0)
        setae   al                     ; Set al if above or equal (CF=0).
        test    al, al
        je      .L5
        ; Do something 2
.L5:
        leave
        ret

Answer 2

Historically (we're talking the 1980s and early 1990s), there were some architectures in which this was true. The root issue is that integer comparison is inherently implemented via integer subtractions. This gives rise to the following cases.

Comparison     Subtraction
----------     -----------
A < B      --> A - B < 0
A = B      --> A - B = 0
A > B      --> A - B > 0

Now, when A < B the subtraction has to borrow a high-bit for the subtraction to be correct, just like you carry and borrow when adding and subtracting by hand. This "borrowed" bit was usually referred to as the carry bit and would be testable by a branch instruction. A second bit called the zero bit would be set if the subtraction were identically zero which implied equality.

There were usually at least two conditional branch instructions, one to branch on the carry bit and one on the zero bit.

Now, to get at the heart of the matter, let's expand the previous table to include the carry and zero bit results.

Comparison     Subtraction  Carry Bit  Zero Bit
----------     -----------  ---------  --------
A < B      --> A - B < 0    0          0
A = B      --> A - B = 0    1          1
A > B      --> A - B > 0    1          0

So, implementing a branch for A < B can be done in one instruction, because the carry bit is clear only in this case, , that is,

;; Implementation of "if (A < B) goto address;"
cmp  A, B          ;; compare A to B
bcz  address       ;; Branch if Carry is Zero to the new address

But, if we want to do a less-than-or-equal comparison, we need to do an additional check of the zero flag to catch the case of equality.

;; Implementation of "if (A <= B) goto address;"
cmp A, B           ;; compare A to B
bcz address        ;; branch if A < B
bzs address        ;; also, Branch if the Zero bit is Set

So, on some machines, using a "less than" comparison might save one machine instruction . This was relevant in the era of sub-megahertz processor speed and 1:1 CPU-to-memory speed ratios, but it is almost totally irrelevant today.

Answer 3

Assuming we're talking about internal integer types, there's no possible way one could be faster than the other. They're obviously semantically identical. They both ask the compiler to do precisely the same thing. Only a horribly broken compiler would generate inferior code for one of these.

If there was some platform where < was faster than <= for simple integer types, the compiler should always convert <= to < for constants. Any compiler that didn't would just be a bad compiler (for that platform).

Answer 4

I see that neither is faster. The compiler generates the same machine code in each condition with a different value.

if(a < 901)
cmpl  $900, -4(%rbp)
jg .L2

if(a <=901)
cmpl  $901, -4(%rbp)
jg .L3

My example if is from GCC on x86_64 platform on Linux.

Compiler writers are pretty smart people, and they think of these things and many others most of us take for granted.

I noticed that if it is not a constant, then the same machine code is generated in either case.

int b;
if(a < b)
cmpl  -4(%rbp), %eax
jge   .L2

if(a <=b)
cmpl  -4(%rbp), %eax
jg .L3

Answer 5

For floating point code, the <= comparison may indeed be slower (by one instruction) even on modern architectures. Here's the first function:

int compare_strict(double a, double b) { return a < b; }

On PowerPC, first this performs a floating point comparison (which updates cr , the condition register), then moves the condition register to a GPR, shifts the "compared less than" bit into place, and then returns. It takes four instructions.

Now consider this function instead:

int compare_loose(double a, double b) { return a <= b; }

This requires the same work as compare_strict above, but now there's two bits of interest: "was less than" and "was equal to." This requires an extra instruction ( cror - condition register bitwise OR) to combine these two bits into one. So compare_loose requires five instructions, while compare_strict requires four.

You might think that the compiler could optimize the second function like so:

int compare_loose(double a, double b) { return ! (a > b); }

However this will incorrectly handle NaNs. NaN1 <= NaN2 and NaN1 > NaN2 need to both evaluate to false.

Answer 6

Maybe the author of that unnamed book has read that a > 0 runs faster than a >= 1 and thinks that is true universally.

But it is because a 0 is involved (because CMP can, depending on the architecture, replaced eg with OR ) and not because of the < .

Answer 7

至少，如果这是真的，编译器可以简单地将 a <= b 优化为 !(a > b)，因此即使比较本身实际上更慢，除了最天真的编译器之外，您不会注意到任何差异.

Answer 8

TL;DR answer

For most combinations of architecture, compiler and language, < will not be faster than <= .

Full answer

Other answers have concentrated on x86 architecture, and I don't know the ARM architecture (which your example assembler seems to be) well enough to comment specifically on the code generated, but this is an example of a micro-optimisation which is very architecture specific, and is as likely to be an anti-optimisation as it is to be an optimisation .

As such, I would suggest that this sort of micro-optimisation is an example of cargo cult programming rather than best software engineering practice.

Counterexample

There are probably some architectures where this is an optimisation, but I know of at least one architecture where the opposite may be true. The venerable Transputer architecture only had machine code instructions for equal to and greater than or equal to , so all comparisons had to be built from these primitives.

Even then, in almost all cases, the compiler could order the evaluation instructions in such a way that in practice, no comparison had any advantage over any other. Worst case though, it might need to add a reverse instruction (REV) to swap the top two items on the operand stack . This was a single byte instruction which took a single cycle to run, so had the smallest overhead possible.

Summary

Whether or not a micro-optimisation like this is an optimisation or an anti-optimisation depends on the specific architecture you are using, so it is usually a bad idea to get into the habit of using architecture specific micro-optimisations, otherwise you might instinctively use one when it is inappropriate to do so, and it looks like this is exactly what the book you are reading is advocating.

Answer 9

They have the same speed. Maybe in some special architecture what he/she said is right, but in the x86 family at least I know they are the same. Because for doing this the CPU will do a substraction (a - b) and then check the flags of the flag register. Two bits of that register are called ZF (zero Flag) and SF (sign flag), and it is done in one cycle, because it will do it with one mask operation.

Answer 10

This would be highly dependent on the underlying architecture that the C is compiled to. Some processors and architectures might have explicit instructions for equal to, or less than and equal to, which execute in different numbers of cycles.

That would be pretty unusual though, as the compiler could work around it, making it irrelevant.

Answer 11

You should not be able to notice the difference even if there is any. Besides, in practice, you'll have to do an additional a + 1 or a - 1 to make the condition stand unless you're going to use some magic constants, which is a very bad practice by all means.

Answer 12

When I wrote the first version of this answer, I was only looking at the title question about < vs. <= in general, not the specific example of a constant a < 901 vs. a <= 900 . Many compilers always shrink the magnitude of constants by converting between < and <= , eg because x86 immediate operand have a shorter 1-byte encoding for -128..127.

For ARM, being able to encode as an immediate depends on being able to rotate a narrow field into any position in a word. So cmp r0, #0x00f000 would be encodeable, while cmp r0, #0x00efff would not be. So the make-it-smaller rule for comparison vs. a compile-time constant doesn't always apply for ARM. AArch64 is either shift-by-12 or not, instead of an arbitrary rotation, for instructions like cmp and cmn , unlike 32-bit ARM and Thumb modes.

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< vs. <= in general, including for runtime-variable conditions

In assembly language on most machines, a comparison for <= has the same cost as a comparison for < . This applies whether you're branching on it, booleanizing it to create a 0/1 integer, or using it as a predicate for a branchless select operation (like x86 CMOV). The other answers have only addressed this part of the question.

But this question is about the C++ operators, the input to the optimizer. Normally they're both equally efficient; the advice from the book sounds totally bogus because compilers can always transform the comparison that they implement in asm. But there is at least one exception where using <= can accidentally create something the compiler can't optimize.

As a loop condition, there are cases where <= is qualitatively different from < , when it stops the compiler from proving that a loop is not infinite. This can make a big difference, disabling auto-vectorization.

Unsigned overflow is well-defined as base-2 wrap around, unlike signed overflow (UB). Signed loop counters are generally safe from this with compilers that optimize based on signed-overflow UB not happening: ++i <= size will always eventually become false. ( What Every C Programmer Should Know About Undefined Behavior )

void foo(unsigned size) {
    unsigned upper_bound = size - 1;  // or any calculation that could produce UINT_MAX
    for(unsigned i=0 ; i <= upper_bound ; i++)
        ...

Compilers can only optimize in ways that preserve the (defined and legally observable) behaviour of the C++ source for all possible input values , except ones that lead to undefined behaviour.

(A simple i <= size would create the problem too, but I thought calculating an upper bound was a more realistic example of accidentally introducing the possibility of an infinite loop for an input you don't care about but which the compiler must consider.)

In this case, size=0 leads to upper_bound=UINT_MAX , and i <= UINT_MAX is always true. So this loop is infinite for size=0 , and the compiler has to respect that even though you as the programmer probably never intend to pass size=0. If the compiler can inline this function into a caller where it can prove that size=0 is impossible, then great, it can optimize like it could for i < size .

Asm like if(!size) skip the loop; do{...}while(--size); is one normally-efficient way to optimize a for( i<size ) loop, if the actual value of i isn't needed inside the loop ( Why are loops always compiled into "do...while" style (tail jump)? ).

But that do{}while can't be infinite: if entered with size==0 , we get 2^n iterations. ( Iterating over all unsigned integers in a for loop C makes it possible to express a loop over all unsigned integers including zero, but it's not easy without a carry flag the way it is in asm.)

With wraparound of the loop counter being a possibility, modern compilers often just "give up", and don't optimize nearly as aggressively.

Example: sum of integers from 1 to n

Using unsigned i <= n defeats clang's idiom-recognition that optimizes sum(1 .. n) loops with a closed form based on Gauss's n * (n+1) / 2 formula.

unsigned sum_1_to_n_finite(unsigned n) {
    unsigned total = 0;
    for (unsigned i = 0 ; i < n+1 ; ++i)
        total += i;
    return total;
}

x86-64 asm from clang7.0 and gcc8.2 on the Godbolt compiler explorer

 # clang7.0 -O3 closed-form
    cmp     edi, -1       # n passed in EDI: x86-64 System V calling convention
    je      .LBB1_1       # if (n == UINT_MAX) return 0;  // C++ loop runs 0 times
          # else fall through into the closed-form calc
    mov     ecx, edi         # zero-extend n into RCX
    lea     eax, [rdi - 1]   # n-1
    imul    rax, rcx         # n * (n-1)             # 64-bit
    shr     rax              # n * (n-1) / 2
    add     eax, edi         # n + (stuff / 2) = n * (n+1) / 2   # truncated to 32-bit
    ret          # computed without possible overflow of the product before right shifting
.LBB1_1:
    xor     eax, eax
    ret

But for the naive version, we just get a dumb loop from clang.

unsigned sum_1_to_n_naive(unsigned n) {
    unsigned total = 0;
    for (unsigned i = 0 ; i<=n ; ++i)
        total += i;
    return total;
}

# clang7.0 -O3
sum_1_to_n(unsigned int):
    xor     ecx, ecx           # i = 0
    xor     eax, eax           # retval = 0
.LBB0_1:                       # do {
    add     eax, ecx             # retval += i
    add     ecx, 1               # ++1
    cmp     ecx, edi
    jbe     .LBB0_1            # } while( i<n );
    ret

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GCC doesn't use a closed-form either way, so the choice of loop condition doesn't really hurt it ; it auto-vectorizes with SIMD integer addition, running 4 i values in parallel in the elements of an XMM register.

# "naive" inner loop
.L3:
    add     eax, 1       # do {
    paddd   xmm0, xmm1    # vect_total_4.6, vect_vec_iv_.5
    paddd   xmm1, xmm2    # vect_vec_iv_.5, tmp114
    cmp     edx, eax      # bnd.1, ivtmp.14     # bound and induction-variable tmp, I think.
    ja      .L3 #,       # }while( n > i )

 "finite" inner loop
  # before the loop:
  # xmm0 = 0 = totals
  # xmm1 = {0,1,2,3} = i
  # xmm2 = set1_epi32(4)
 .L13:                # do {
    add     eax, 1       # i++
    paddd   xmm0, xmm1    # total[0..3] += i[0..3]
    paddd   xmm1, xmm2    # i[0..3] += 4
    cmp     eax, edx
    jne     .L13      # }while( i != upper_limit );

     then horizontal sum xmm0
     and peeled cleanup for the last n%3 iterations, or something.
     

It also has a plain scalar loop which I think it uses for very small n , and/or for the infinite loop case.

BTW, both of these loops waste an instruction (and a uop on Sandybridge-family CPUs) on loop overhead. sub eax,1 / jnz instead of add eax,1 /cmp/jcc would be more efficient. 1 uop instead of 2 (after macro-fusion of sub/jcc or cmp/jcc). The code after both loops writes EAX unconditionally, so it's not using the final value of the loop counter.

Answer 13

You could say that line is correct in most scripting languages, since the extra character results in slightly slower code processing. However, as the top answer pointed out, it should have no effect in C++, and anything being done with a scripting language probably isn't that concerned about optimization.

Answer 14

Only if the people who created the computers are bad with boolean logic. Which they shouldn't be.

Every comparison ( >= <= > < ) can be done in the same speed.

What every comparison is, is just a subtraction (the difference) and seeing if it's positive/negative.
(If the msb is set, the number is negative)

How to check a >= b ? Sub ab >= 0 Check if ab is positive.
How to check a <= b ? Sub 0 <= ba Check if ba is positive.
How to check a < b ? Sub ab < 0 Check if ab is negative.
How to check a > b ? Sub 0 > ba Check if ba is negative.

Simply put, the computer can just do this underneath the hood for the given op:

a >= b == msb(ab)==0
a <= b == msb(ba)==0
a > b == msb(ba)==1
a < b == msb(ab)==1

and of course the computer wouldn't actually need to do the ==0 or ==1 either.
for the ==0 it could just invert the msb from the circuit.

Anyway, they most certainly wouldn't have made a >= b be calculated as a>b || a==b a>b || a==b lol

Answer 15

In C and C++, an important rule for the compiler is the “as-if” rule: If doing X has the exact same behavior as if you did Y, then the compiler is free to choose which one it uses.

In your case, “a < 901” and “a <= 900” always have the same result, so the compiler is free to compile either version. If one version was faster, for whatever reason, then any quality compiler would produce code for the version that is faster. So unless your compiler produced exceptionally bad code, both versions would run at equal speed.

Now if you had a situation where two bits of code will always produce the same result, but it is hard to prove for the compiler, and/or it is hard for the compiler to prove which if any version is faster, then you might get different code running at different speeds.

PS The original example might run at different speeds if the processor supports single byte constants (faster) and multi byte constants (slower), so comparing against 255 (1 byte) might be faster than comparing against 256 (two bytes). I'd expect the compiler to do whatever is faster.

Answer 16

Only if computation path depends on data:

a={1,1,1,1,1000,1,1,1,1}
while (i<=4)
{
     for(j from 0 to a[i]){ do_work(); }
     i++;
}

will compute 250 times more than while(i<4)

Real-world sample would be computing mandelbrot-set. If you include a pixel that iterates 1000000 times, it will cause a lag but the coincidence with <= usage probability is too low.