Which is faster: while(1) or while(2)?

Both loops are infinite, but we can see which one takes more instructions/resources per iteration.

Using gcc, I compiled the two following programs to assembly at varying levels of optimization:

int main(void) {
    while(1) {}
    return 0;
}

int main(void) {
    while(2) {}
    return 0;
}

Even with no optimizations (-O0), the generated assembly was identical for both programs. Therefore, there is no speed difference between the two loops.

For reference, here is the generated assembly (using gcc main.c -S -masm=intel with an optimization flag):

With -O0:

    .file   "main.c"
    .intel_syntax noprefix
    .def    __main; .scl    2;  .type   32; .endef
    .text
    .globl  main
    .def    main;   .scl    2;  .type   32; .endef
    .seh_proc   main
main:
    push    rbp
    .seh_pushreg    rbp
    mov rbp, rsp
    .seh_setframe   rbp, 0
    sub rsp, 32
    .seh_stackalloc 32
    .seh_endprologue
    call    __main
.L2:
    jmp .L2
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

With -O1:

    .file   "main.c"
    .intel_syntax noprefix
    .def    __main; .scl    2;  .type   32; .endef
    .text
    .globl  main
    .def    main;   .scl    2;  .type   32; .endef
    .seh_proc   main
main:
    sub rsp, 40
    .seh_stackalloc 40
    .seh_endprologue
    call    __main
.L2:
    jmp .L2
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

With -O2 and -O3 (same output):

    .file   "main.c"
    .intel_syntax noprefix
    .def    __main; .scl    2;  .type   32; .endef
    .section    .text.startup,"x"
    .p2align 4,,15
    .globl  main
    .def    main;   .scl    2;  .type   32; .endef
    .seh_proc   main
main:
    sub rsp, 40
    .seh_stackalloc 40
    .seh_endprologue
    call    __main
.L2:
    jmp .L2
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

In fact, the assembly generated for the loop is identical for every level of optimization:

 .L2:
    jmp .L2
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

The important bits being:

.L2:
    jmp .L2

I can't read assembly very well, but this is obviously an unconditional loop. The jmp instruction unconditionally resets the program back to the .L2 label without even comparing a value against true, and of course immediately does so again until the program is somehow ended. This directly corresponds to the C/C++ code:

L2:
    goto L2;

Edit:

Interestingly enough, even with no optimizations, the following loops all produced the exact same output (unconditional jmp) in assembly:

while(42) {}

while(1==1) {}

while(2==2) {}

while(4<7) {}

while(3==3 && 4==4) {}

while(8-9 < 0) {}

while(4.3 * 3e4 >= 2 << 6) {}

while(-0.1 + 02) {}

And even to my amazement:

#include<math.h>

while(sqrt(7)) {}

while(hypot(3,4)) {}

Things get a little more interesting with user-defined functions:

int x(void) {
    return 1;
}

while(x()) {}

#include<math.h>

double x(void) {
    return sqrt(7);
}

while(x()) {}

At -O0, these two examples actually call x and perform a comparison for each iteration.

First example (returning 1):

.L4:
    call    x
    testl   %eax, %eax
    jne .L4
    movl    $0, %eax
    addq    $32, %rsp
    popq    %rbp
    ret
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

Second example (returning sqrt(7)):

.L4:
    call    x
    xorpd   %xmm1, %xmm1
    ucomisd %xmm1, %xmm0
    jp  .L4
    xorpd   %xmm1, %xmm1
    ucomisd %xmm1, %xmm0
    jne .L4
    movl    $0, %eax
    addq    $32, %rsp
    popq    %rbp
    ret
    .seh_endproc
    .ident  "GCC: (tdm64-2) 4.8.1"

However, at -O1 and above, they both produce the same assembly as the previous examples (an unconditional jmp back to the preceding label).

TL;DR

Under GCC, the different loops are compiled to identical assembly. The compiler evaluates the constant values and doesn't bother performing any actual comparison.

The moral of the story is:

There exists a layer of translation between C++ source code and CPU instructions, and this layer has important implications for performance.

Therefore, performance cannot be evaluated by only looking at source code.

The compiler should be smart enough to optimize such trivial cases. Programmers should not waste their time thinking about them in the vast majority of cases.

---

Yes, while(1) is much faster than while(2), for a human to read! If I see while(1) in an unfamiliar codebase, I immediately know what the author intended, and my eyeballs can continue to the next line.

If I see while(2), I'll probably halt in my tracks and try to figure out why the author didn't write while(1). Did the author's finger slip on the keyboard? Do the maintainers of this codebase use while(n) as an obscure commenting mechanism to make loops look different? Is it a crude workaround for a spurious warning in some broken static analysis tool? Or is this a clue that I'm reading generated code? Is it a bug resulting from an ill-advised find-and-replace-all, or a bad merge, or a cosmic ray? Maybe this line of code is supposed to do something dramatically different. Maybe it was supposed to read while(w) or while(x2). I'd better find the author in the file's history and send them a "WTF" email... and now I've broken my mental context. The while(2) might consume several minutes of my time, when while(1) would have taken a fraction of a second!

I'm exaggerating, but only a little. Code readability is really important. And that's worth mentioning in an interview!

---

The existing answers showing the code generated by a particular compiler for a particular target with a particular set of options do not fully answer the question -- unless the question was asked in that specific context ("Which is faster using gcc 4.7.2 for x86_64 with default options?", for example).

As far as the language definition is concerned, in the abstract machine while (1) evaluates the integer constant 1, and while (2) evaluates the integer constant 2; in both cases the result is compared for equality to zero. The language standard says absolutely nothing about the relative performance of the two constructs.

I can imagine that an extremely naive compiler might generate different machine code for the two forms, at least when compiled without requesting optimization.

On the other hand, C compilers absolutely must evaluate some constant expressions at compile time, when they appear in contexts that require a constant expression. For example, this:

int n = 4;
switch (n) {
    case 2+2: break;
    case 4:   break;
}

requires a diagnostic; a lazy compiler does not have the option of deferring the evaluation of 2+2 until execution time. Since a compiler has to have the ability to evaluate constant expressions at compile time, there's no good reason for it not to take advantage of that capability even when it's not required.

The C standard (N1570 6.8.5p4) says that

An iteration statement causes a statement called the loop body to be executed repeatedly until the controlling expression compares equal to 0.

So the relevant constant expressions are 1 == 0 and 2 == 0, both of which evaluate to the int value 0. (These comparison are implicit in the semantics of the while loop; they don't exist as actual C expressions.)

A perversely naive compiler could generate different code for the two constructs. For example, for the first it could generate an unconditional infinite loop (treating 1 as a special case), and for the second it could generate an explicit run-time comparison equivalent to 2 != 0. But I've never encountered a C compiler that would actually behave that way, and I seriously doubt that such a compiler exists.

Most compilers (I'm tempted to say all production-quality compilers) have options to request additional optimizations. Under such an option, it's even less likely that any compiler would generate different code for the two forms.

If your compiler generates different code for the two constructs, first check whether the differing code sequences actually have different performance. If they do, try compiling again with an optimization option (if available). If they still differ, submit a bug report to the compiler vendor. It's not (necessarily) a bug in the sense of a failure to conform to the C standard, but it's almost certainly a problem that should be corrected.

Bottom line: while (1) and while(2) almost certainly have the same performance. They have exactly the same semantics, and there's no good reason for any compiler not to generate identical code.

And though it's perfectly legal for a compiler to generate faster code for while(1) than for while(2), it's equally legal for a compiler to generate faster code for while(1) than for another occurrence of while(1) in the same program.

(There's another question implicit in the one you asked: How do you deal with an interviewer who insists on an incorrect technical point. That would probably be a good question for the Workplace site).

---

Wait a minute. The interviewer, did he look like this guy?

https://i.stack.imgur.com/L6P1H.png

It's bad enough that the interviewer himself has failed this interview, what if other programmers at this company have "passed" this test?

No. Evaluating the statements 1 == 0 and 2 == 0 should be equally fast. We could imagine poor compiler implementations where one might be faster than the other. But there's no good reason why one should be faster than the other.

Even if there's some obscure circumstance when the claim would be true, programmers should not be evaluated based on knowledge of obscure (and in this case, creepy) trivia. Don't worry about this interview, the best move here is to walk away.

Disclaimer: This is NOT an original Dilbert cartoon. This is merely a mashup.

---

Your explanation is correct. This seems to be a question that tests your self-confidence in addition to technical knowledge.

By the way, if you answered

Both pieces of code are equally fast, because both take infinite time to complete

the interviewer would say

But while (1) can do more iterations per second; can you explain why? (this is nonsense; testing your confidence again)

So by answering like you did, you saved some time which you would otherwise waste on discussing this bad question.

Here is an example code generated by the compiler on my system (MS Visual Studio 2012), with optimizations turned off:

yyy:
    xor eax, eax
    cmp eax, 1     (or 2, depending on your code)
    je xxx
    jmp yyy
xxx:
    ...

With optimizations turned on:

xxx:
    jmp xxx

So the generated code is exactly the same, at least with an optimizing compiler.

---

The most likely explanation for the question is that the interviewer thinks that the processor checks the individual bits of the numbers, one by one, until it hits a non-zero value:

1 = 00000001
2 = 00000010

If the "is zero?" algorithm starts from the right side of the number and has to check each bit until it reaches a non-zero bit, the while(1) { } loop would have to check twice as many bits per iteration as the while(2) { } loop.

This requires a very wrong mental model of how computers work, but it does have its own internal logic. One way to check would be to ask if while(-1) { } or while(3) { } would be equally fast, or if while(32) { } would be even slower.

---

Of course I do not know the real intentions of this manager, but I propose a completely different view: When hiring a new member into a team, it is useful to know how he reacts to conflict situations.

They drove you into conflict. If this is true, they are clever and the question was good. For some industries, like banking, posting your problem to Stack Overflow could be a reason for rejection.

But of course I do not know, I just propose one option.

---

I think the clue is to be found in "asked by a senior manager". This person obviously stopped programming when he became a manager and then it took him/her several years to become a senior manager. Never lost interest in programming, but never wrote a line since those days. So his reference is not "any decent compiler out there" as some answers mention, but "the compiler this person worked with 20-30 years ago".

At that time, programmers spent a considerable percentage of their time trying out various methods for making their code faster and more efficient as CPU time of 'the central minicomputer' was so valueable. As did people writing compilers. I'm guessing that the one-and-only compiler his company made available at that time optimized on the basis of 'frequently encountered statements that can be optimized' and took a bit of a shortcut when encountering a while(1) and evaluated everything else, including a while(2). Having had such an experience could explain his position and his confidence in it.

The best approach to get you hired is probably one that enables the senior manager to get carried away and lecture you 2-3 minutes on "the good old days of programming" before YOU smoothly lead him towards the next interview subject. (Good timing is important here - too fast and you're interrupting the story - too slow and you are labelled as somebody with insufficient focus). Do tell him at the end of the interview that you'd be highly interested to learn more about this topic.

---

You should have asked him how did he reached to that conclusion. Under any decent compiler out there, the two compile to the same asm instructions. So, he should have told you the compiler as well to start off. And even so, you would have to know the compiler and platform very well to even make a theoretical educated guess. And in the end, it doesn't really matter in practice, since there are other external factors like memory fragmentation or system load that will influence the loop more than this detail.

---

For the sake of this question, I should that add I remember Doug Gwyn from C Committee writing that some early C compilers without the optimizer pass would generate a test in assembly for the while(1) (comparing to for(;;) which wouldn't have it).

I would answer to the interviewer by giving this historical note and then say that even if I would be very surprised any compiler did this, a compiler could have:

without optimizer pass the compiler generate a test for both while(1) and while(2)

with optimizer pass the compiler is instructed to optimize (with an unconditional jump) all while(1) because they are considered as idiomatic. This would leave the while(2) with a test and therefore makes a performance difference between the two.

I would of course add to the interviewer that not considering while(1) and while(2) the same construct is a sign of low-quality optimization as these are equivalent constructs.

---

Another take on such a question would be to see if you got courage to tell your manager that he/she is wrong! And how softly you can communicate it.

My first instinct would have been to generate assembly output to show the manager that any decent compiler should take care of it, and if it's not doing so, you will submit the next patch for it :)

---

To see so many people delve into this problem, shows exactly why this could very well be a test to see how quickly you want to micro-optimize things.

My answer would be; it doesn't matter that much, I rather focus on the business problem which we are solving. After all, that's what I'm going to be paid for.

Moreover, I would opt for while(1) {} because it is more common, and other teammates would not need to spend time to figure out why someone would go for a higher number than 1.

Now go write some code. ;-)

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If you're that worried about optimisation, you should use

for (;;)

because that has no tests. (cynic mode)

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It seems to me this is one of those behavioral interview questions masked as a technical question. Some companies do this - they will ask a technical question that should be fairly easy for any competent programmer to answer, but when the interviewee gives the correct answer, the interviewer will tell them they are wrong.

The company wants to see how you will react in this situation. Do you sit there quietly and don't push that your answer is correct, due to either self-doubt or fear of upsetting the interviewer? Or are you willing to challenge a person in authority who you know is wrong? They want to see if you are willing to stand up for your convictions, and if you can do it in a tactful and respectful manner.

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I used to program C and Assembly code back when this sort of nonsense might have made a difference. When it did make a difference we wrote it in Assembly.

If I were asked that question I would have repeated Donald Knuth's famous 1974 quote about premature optimization and walked if the interviewer didn't laugh and move on.

---

Maybe the interviewer posed such dumb question intentionally and wanted you to make 3 points:

Basic reasoning. Both loops are infinite, it's hard to talk about performance. Knowledge about optimisation levels. He wanted to hear from you if you let the compiler do any optimisation for you, it would optimise the condition, especially if the block was not empty. Knowledge about microprocessor architecture. Most architectures have a special CPU instruction for comparision with 0 (while not necessarily faster).

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Here's a problem: If you actually write a program and measure its speed, the speed of both loops could be different! For some reasonable comparison:

unsigned long i = 0;
while (1) { if (++i == 1000000000) break; }

unsigned long i = 0;
while (2) { if (++i == 1000000000) break; }

with some code added that prints the time, some random effect like how the loop is positioned within one or two cache lines could make a difference. One loop might by pure chance be completely within one cache line, or at the start of a cache line, or it might to straddle two cache lines. And as a result, whatever the interviewer claims is fastest might actually be fastest - by coincidence.

Worst case scenario: An optimising compiler doesn't figure out what the loop does, but figures out that the values produced when the second loop is executed are the same ones as produced by the first one. And generate full code for the first loop, but not for the second.

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They are both equal - the same.

According to the specifications anything that is not 0 is considered true, so even without any optimization, and a good compiler will not generate any code for while(1) or while(2). The compiler would generate a simple check for != 0.

---

Judging by the amount of time and effort people have spent testing, proving, and answering this very straight forward question Id say that both were made very slow by asking the question.

And so to spend even more time on it ...

"while (2)" is ridiculous, because,

"while (1)", and "while (true)" are historically used to make an infinite loop which expects "break" to be called at some stage inside the loop based upon a condition that will certainly occur.

The "1" is simply there to always evaluate to true and therefore, to say "while (2)" is about as silly as saying "while (1 + 1 == 2)" which will also evaluate to true.

And if you want to be completely silly just use: -

while (1 + 5 - 2 - (1 * 3) == 0.5 - 4 + ((9 * 2) / 4.0)) {
    if (succeed())
        break;
}

Id like to think your coder made a typo which did not effect the running of the code, but if he intentionally used the "2" just to be weird then sack him before he puts weird sh!t all through your code making it difficult to read and work with.

---

That depends on the compiler.

If it optimizes the code, or if it evaluates 1 and 2 to true with the same number of instructions for a particular instruction set, the execution speed will be the same.

In real cases it will always be equally fast, but it would be possible to imagine a particular compiler and a particular system when this would be evaluated differently.

I mean: this is not really a language (C) related question.

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Since people looking to answer this question want the fastest loop, I would have answered that both are equally compiling into the same assembly code, as stated in the other answers. Nevertheless you can suggest to the interviewer using 'loop unrolling'; a do {} while loop instead of the while loop.

Cautious: You need to ensure that the loop would at least always run once.

The loop should have a break condition inside.

Also for that kind of loop I would personally prefer the use of do {} while(42) since any integer, except 0, would do the job.

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The obvious answer is: as posted, both fragments would run an equally busy infinite loop, which makes the program infinitely slow.

Although redefining C keywords as macros would technically have undefined behavior, it is the only way I can think of to make either code fragment fast at all: you can add this line above the 2 fragments:

#define while(x) sleep(x);

it will indeed make while(1) twice as fast (or half as slow) as while(2).

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The only reason I can think of why the while(2) would be any slower is:

The code optimizes the loop to cmp eax, 2 When the subtract occurs you're essentially subtracting a. 00000000 - 00000010 cmp eax, 2 instead of b. 00000000 - 00000001 cmp eax, 1

cmp only sets flags and does not set a result. So on the least significant bits we know if we need to borrow or not with b. Whereas with a you have to perform two subtractions before you get a borrow.

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