This was an interview question asked by a senior manager.
Which is faster?
while(1) {
// Some code
}
or
while(2) {
//Some code
}
I said that both have the same execution speed, as the expression inside while
should finally evaluate to true
or false
. In this case, both evaluate to true
and there are no extra conditional instructions inside the while
condition. So, both will have the same speed of execution and I prefer while (1).
But the interviewer said confidently:
"Check your basics. while(1)
is faster than while(2)
."
(He was not testing my confidence)
Is this true?
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).
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: