Call an expensive function periodically inside a loop
I call a computationally expensive function inside a loop:
for( int j = 0; j < Max; ++j ) {
// ...
processQueuedEvents(); // Computationally expensive call
// ...
}
However, I don't need to run the expensive function on every single loop iteration, so I want to call it periodically:
for( int j = 0; j < Max; ++j ) {
// ...
if ( /* The condition I'm talking about */ )
processQueuedEvents(); // Computationally expensive call
// ...
}
At this point, I need to develop a proper condition for my periodic call. The condition should correlate to Max, I mean, if Max is larger, the expensive call is less-frequent and if Max is smaller the expensive call is more-frequent.
Does anybody have any suggestions or hints? For some reason, I have a hard time coming up with a proper condition.
c++ loops for-loop periodicity periodic-processing
asked Nov 19 '18 at 10:28
user3405291user3405291
1,87911937
add a comment |
I call a computationally expensive function inside a loop:
for( int j = 0; j < Max; ++j ) {
// ...
processQueuedEvents(); // Computationally expensive call
// ...
}
However, I don't need to run the expensive function on every single loop iteration, so I want to call it periodically:
for( int j = 0; j < Max; ++j ) {
// ...
if ( /* The condition I'm talking about */ )
processQueuedEvents(); // Computationally expensive call
// ...
}
At this point, I need to develop a proper condition for my periodic call. The condition should correlate to Max, I mean, if Max is larger, the expensive call is less-frequent and if Max is smaller the expensive call is more-frequent.
Does anybody have any suggestions or hints? For some reason, I have a hard time coming up with a proper condition.
c++ loops for-loop periodicity periodic-processing
asked Nov 19 '18 at 10:28
user3405291user3405291
1,87911937
2
Sounds to me like you want a fixed number of calls across the whole loop, i.e. check whetherj >= Max / numberOfCalls(potential off-by-one error here).
– Quentin
Nov 19 '18 at 10:31
@Quentin Thanks! I'm going to try it
– user3405291
Nov 19 '18 at 10:33
1
j % period == 0, and period should increase as Max increases, e.g.j % (Max / numberOfCalls) == 0.
– felix
Nov 19 '18 at 10:40
@felix Right! That's the relationship :)
– user3405291
Nov 19 '18 at 10:45
add a comment |
I call a computationally expensive function inside a loop:
for( int j = 0; j < Max; ++j ) {
// ...
processQueuedEvents(); // Computationally expensive call
// ...
}
However, I don't need to run the expensive function on every single loop iteration, so I want to call it periodically:
for( int j = 0; j < Max; ++j ) {
// ...
if ( /* The condition I'm talking about */ )
processQueuedEvents(); // Computationally expensive call
// ...
}
At this point, I need to develop a proper condition for my periodic call. The condition should correlate to Max, I mean, if Max is larger, the expensive call is less-frequent and if Max is smaller the expensive call is more-frequent.
Does anybody have any suggestions or hints? For some reason, I have a hard time coming up with a proper condition.
c++ loops for-loop periodicity periodic-processing
asked Nov 19 '18 at 10:28
user3405291user3405291
1,87911937
I call a computationally expensive function inside a loop:
for( int j = 0; j < Max; ++j ) {
// ...
processQueuedEvents(); // Computationally expensive call
// ...
}
However, I don't need to run the expensive function on every single loop iteration, so I want to call it periodically:
for( int j = 0; j < Max; ++j ) {
// ...
if ( /* The condition I'm talking about */ )
processQueuedEvents(); // Computationally expensive call
// ...
}
At this point, I need to develop a proper condition for my periodic call. The condition should correlate to Max, I mean, if Max is larger, the expensive call is less-frequent and if Max is smaller the expensive call is more-frequent.
Does anybody have any suggestions or hints? For some reason, I have a hard time coming up with a proper condition.
c++ loops for-loop periodicity periodic-processing
c++ loops for-loop periodicity periodic-processing
asked Nov 19 '18 at 10:28
user3405291user3405291
1,87911937
asked Nov 19 '18 at 10:28
user3405291user3405291
1,87911937
asked Nov 19 '18 at 10:28
user3405291user3405291
1,87911937
asked Nov 19 '18 at 10:28
user3405291user3405291
1,87911937
asked Nov 19 '18 at 10:28
user3405291user3405291
1,87911937
1,87911937
2
Sounds to me like you want a fixed number of calls across the whole loop, i.e. check whetherj >= Max / numberOfCalls(potential off-by-one error here).
– Quentin
Nov 19 '18 at 10:31
@Quentin Thanks! I'm going to try it
– user3405291
Nov 19 '18 at 10:33
1
j % period == 0, and period should increase as Max increases, e.g.j % (Max / numberOfCalls) == 0.
– felix
Nov 19 '18 at 10:40
@felix Right! That's the relationship :)
– user3405291
Nov 19 '18 at 10:45
add a comment |
2
Sounds to me like you want a fixed number of calls across the whole loop, i.e. check whetherj >= Max / numberOfCalls(potential off-by-one error here).
– Quentin
Nov 19 '18 at 10:31
@Quentin Thanks! I'm going to try it
– user3405291
Nov 19 '18 at 10:33
1
j % period == 0, and period should increase as Max increases, e.g.j % (Max / numberOfCalls) == 0.
– felix
Nov 19 '18 at 10:40
@felix Right! That's the relationship :)
– user3405291
Nov 19 '18 at 10:45
2
2
Sounds to me like you want a fixed number of calls across the whole loop, i.e. check whether
j >= Max / numberOfCalls (potential off-by-one error here).– Quentin
Nov 19 '18 at 10:31
Sounds to me like you want a fixed number of calls across the whole loop, i.e. check whether
j >= Max / numberOfCalls (potential off-by-one error here).– Quentin
Nov 19 '18 at 10:31
@Quentin Thanks! I'm going to try it
– user3405291
Nov 19 '18 at 10:33
@Quentin Thanks! I'm going to try it
– user3405291
Nov 19 '18 at 10:33
1
1
j % period == 0, and period should increase as Max increases, e.g. j % (Max / numberOfCalls) == 0.– felix
Nov 19 '18 at 10:40
j % period == 0, and period should increase as Max increases, e.g. j % (Max / numberOfCalls) == 0.– felix
Nov 19 '18 at 10:40
@felix Right! That's the relationship :)
– user3405291
Nov 19 '18 at 10:45
@felix Right! That's the relationship :)
– user3405291
Nov 19 '18 at 10:45
add a comment |
1 Answer
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You didn't provide enough details about the increment function you want to use. If you are considering a liner one you can reason with factors of 10 in the following manner:
0 <= Max < 10- invoke the expensive call each execution of the loop.
10 <= Max < 100- invoke the expensive call each10-th (j % 10 == 0)execution of the loop.
100 <= Max < 1000- invoke the expensive call each100-th (j % 100 == 0)execution of the loop.
and so on.
answered Nov 19 '18 at 10:44
NiVeRNiVeR
6,87641930
add a comment |
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You didn't provide enough details about the increment function you want to use. If you are considering a liner one you can reason with factors of 10 in the following manner:
0 <= Max < 10- invoke the expensive call each execution of the loop.
10 <= Max < 100- invoke the expensive call each10-th (j % 10 == 0)execution of the loop.
100 <= Max < 1000- invoke the expensive call each100-th (j % 100 == 0)execution of the loop.
and so on.
answered Nov 19 '18 at 10:44
NiVeRNiVeR
6,87641930
add a comment |
You didn't provide enough details about the increment function you want to use. If you are considering a liner one you can reason with factors of 10 in the following manner:
0 <= Max < 10- invoke the expensive call each execution of the loop.
10 <= Max < 100- invoke the expensive call each10-th (j % 10 == 0)execution of the loop.
100 <= Max < 1000- invoke the expensive call each100-th (j % 100 == 0)execution of the loop.
and so on.
answered Nov 19 '18 at 10:44
NiVeRNiVeR
6,87641930
add a comment |
You didn't provide enough details about the increment function you want to use. If you are considering a liner one you can reason with factors of 10 in the following manner:
0 <= Max < 10- invoke the expensive call each execution of the loop.
10 <= Max < 100- invoke the expensive call each10-th (j % 10 == 0)execution of the loop.
100 <= Max < 1000- invoke the expensive call each100-th (j % 100 == 0)execution of the loop.
and so on.
answered Nov 19 '18 at 10:44
NiVeRNiVeR
6,87641930
You didn't provide enough details about the increment function you want to use. If you are considering a liner one you can reason with factors of 10 in the following manner:
0 <= Max < 10- invoke the expensive call each execution of the loop.
10 <= Max < 100- invoke the expensive call each10-th (j % 10 == 0)execution of the loop.
100 <= Max < 1000- invoke the expensive call each100-th (j % 100 == 0)execution of the loop.
and so on.
answered Nov 19 '18 at 10:44
NiVeRNiVeR
6,87641930
answered Nov 19 '18 at 10:44
NiVeRNiVeR
6,87641930
answered Nov 19 '18 at 10:44
NiVeRNiVeR
6,87641930
answered Nov 19 '18 at 10:44
NiVeRNiVeR
6,87641930
6,87641930
add a comment |
add a comment |
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2
Sounds to me like you want a fixed number of calls across the whole loop, i.e. check whether
j >= Max / numberOfCalls(potential off-by-one error here).– Quentin
Nov 19 '18 at 10:31
@Quentin Thanks! I'm going to try it
– user3405291
Nov 19 '18 at 10:33
1
j % period == 0, and period should increase as Max increases, e.g.j % (Max / numberOfCalls) == 0.– felix
Nov 19 '18 at 10:40
@felix Right! That's the relationship :)
– user3405291
Nov 19 '18 at 10:45