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refactor(build): remove some code duplication from queue

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This commit is contained in:
Jef Roosens 2022-12-13 18:24:21 +01:00
parent 03f2240ff6
commit d3151863ee
Signed by: Jef Roosens
GPG key ID: B75D4F293C7052DB
3 changed files with 43 additions and 52 deletions
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88
src/build/queue.v
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@ -110,6 +110,21 @@ fn (mut q BuildJobQueue) reschedule(job BuildJob, arch string) ! {
q.queues[arch].insert(new_job)
}
// pop_invalid pops all invalid jobs.
fn (mut q BuildJobQueue) pop_invalid(arch string) {
for {
job := q.queues[arch].peek() or { return }
if job.config.target_id in q.invalidated
&& job.created < q.invalidated[job.config.target_id] {
// This pop *should* never fail according to the source code
q.queues[arch].pop() or {}
} else {
break
}
}
}
// peek shows the first job for the given architecture that's ready to be
// executed, if present.
pub fn (mut q BuildJobQueue) peek(arch string) ?BuildJob {
@ -118,20 +133,11 @@ pub fn (mut q BuildJobQueue) peek(arch string) ?BuildJob {
return none
}
for {
job := q.queues[arch].peek() or { return none }
q.pop_invalid(arch)
job := q.queues[arch].peek()?
// Skip any invalidated jobs
if job.config.target_id in q.invalidated
&& job.created < q.invalidated[job.config.target_id] {
// This pop *should* never fail according to the source code
q.queues[arch].pop() or { return none }
continue
}
if job.timestamp < time.now() {
return job
}
if job.timestamp < time.now() {
return job
}
}
@ -146,27 +152,18 @@ pub fn (mut q BuildJobQueue) pop(arch string) ?BuildJob {
return none
}
for {
mut job := q.queues[arch].peek() or { return none }
q.pop_invalid(arch)
mut job := q.queues[arch].peek()?
// Skip any invalidated jobs
if job.config.target_id in q.invalidated
&& job.created < q.invalidated[job.config.target_id] {
// This pop *should* never fail according to the source code
q.queues[arch].pop() or { return none }
continue
}
if job.timestamp < time.now() {
job = q.queues[arch].pop()?
if job.timestamp < time.now() {
job = q.queues[arch].pop()?
// TODO how do we handle this properly? Is it even possible for a
// cron expression to not return a next time if it's already been
// used before?
q.reschedule(job, arch) or {}
// TODO how do we handle this properly? Is it even possible for a
// cron expression to not return a next time if it's already been
// used before?
q.reschedule(job, arch) or {}
return job
}
return job
}
}
@ -182,28 +179,19 @@ pub fn (mut q BuildJobQueue) pop_n(arch string, n int) []BuildJob {
mut out := []BuildJob{}
outer: for out.len < n {
for {
mut job := q.queues[arch].peek() or { break outer }
for out.len < n {
q.pop_invalid(arch)
mut job := q.queues[arch].peek() or { break }
// Skip any invalidated jobs
if job.config.target_id in q.invalidated
&& job.created < q.invalidated[job.config.target_id] {
// This pop *should* never fail according to the source code
q.queues[arch].pop() or { break outer }
continue
}
if job.timestamp < time.now() {
job = q.queues[arch].pop() or { break }
if job.timestamp < time.now() {
job = q.queues[arch].pop() or { break outer }
// TODO idem
q.reschedule(job, arch) or {}
// TODO idem
q.reschedule(job, arch) or {}
out << job
} else {
break outer
}
out << job
} else {
break
}
}