Why is a big loop within a small loop faster than

问题:

This is my Perl code

$big=10_000_000;
#A:big loop outside
my $begin_time = time;
foreach my $i (1..$big) {
        foreach my $p (1..10){
        }
}
my $end_time = time;
my $t1=$end_time-$begin_time; 

#B:small loop outside
my $begin_time = time;
foreach my $i (1..10){
     foreach my $p (1..$big){
     }
}
my $end_time = time;
my $t2=$end_time-$begin_time; 

#output
print $t1;
print "\n";
print $t2;

t1=8 seconds
t2=3 seconds

And the mathematica code:

Timing[Do[2, {i, 1, 10}, {j, 2*1, 10^7}]]
output:{14.328, Null}
Timing[Do[2, {j, 1, 2*10^7}, {i, 1, 10}]]
output:{30.937, Null}

Why does the big loop outside take more time?

回答1:

There's a certain amount of overhead in executing the inner loop (initialising the variable; making the checks to see if it should end) and in the first case, you are losing this overhead 10,000,000 times; in the second, you are only doing it 10 times.

EDIT: Let s be the time to setup a loop (e.g. initialise a variable) and i the time to iterate a loop (e.g. test the end condition). Then:

Big Inner Loop

T = s1 + 10 * ( i1 + s2 + 10,000,000*i2 )
  = s1 + 10*i1 + 10*s2 + 100,000,000*i2

Big Outer Loop

T = s1 + 10,000,000 * ( i1 + s2 + 10*i2 )
  = s1 + 10,000,000*i1 + 10,000,000*s2 + 100,000,000*i2

Difference

diff = 9,999,990*i1 + 9,999,990*s2

So the iteration time of the outer loop (i1) and the set-up time of the inner-loop (s2) are both performed 9,999,990 times more with the big outer loop than with the big inner loop.