Define priority_queue with lambda in header file
1
I have a question about defining a priority_queue pointer in header file and initializing it in source file.
header file
std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, ?> *readyQueue;
source file
auto readyQueueComparator = (const std::shared_ptr<Process> &first, const std::shared_ptr<Process> &second) {
return true;
};
readyQueue = new std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, decltype(readyQueueComparator)>;
Basically, I want to implement a structure like this but I couldn't figure out how should I define priority queue in header file in this format.
Thanks.
c++
edited Nov 19 '18 at 21:23
NathanOliver
90.8k15126190
asked Nov 19 '18 at 21:21
Mert AkozcanMert Akozcan
6317
add a comment |
1
I have a question about defining a priority_queue pointer in header file and initializing it in source file.
header file
std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, ?> *readyQueue;
source file
auto readyQueueComparator = (const std::shared_ptr<Process> &first, const std::shared_ptr<Process> &second) {
return true;
};
readyQueue = new std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, decltype(readyQueueComparator)>;
Basically, I want to implement a structure like this but I couldn't figure out how should I define priority queue in header file in this format.
Thanks.
c++
edited Nov 19 '18 at 21:23
NathanOliver
90.8k15126190
asked Nov 19 '18 at 21:21
Mert AkozcanMert Akozcan
6317
I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.
– NathanOliver
Nov 19 '18 at 21:24
That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.
– SergeyA
Nov 19 '18 at 21:28
@SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?
– Mert Akozcan
Nov 19 '18 at 21:58
@MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.
– SergeyA
Nov 19 '18 at 22:00
add a comment |
1
1
1
I have a question about defining a priority_queue pointer in header file and initializing it in source file.
header file
std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, ?> *readyQueue;
source file
auto readyQueueComparator = (const std::shared_ptr<Process> &first, const std::shared_ptr<Process> &second) {
return true;
};
readyQueue = new std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, decltype(readyQueueComparator)>;
Basically, I want to implement a structure like this but I couldn't figure out how should I define priority queue in header file in this format.
Thanks.
c++
edited Nov 19 '18 at 21:23
NathanOliver
90.8k15126190
asked Nov 19 '18 at 21:21
Mert AkozcanMert Akozcan
6317
I have a question about defining a priority_queue pointer in header file and initializing it in source file.
header file
std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, ?> *readyQueue;
source file
auto readyQueueComparator = (const std::shared_ptr<Process> &first, const std::shared_ptr<Process> &second) {
return true;
};
readyQueue = new std::priority_queue<std::shared_ptr<Process>, std::vector<std::shared_ptr<Process>>, decltype(readyQueueComparator)>;
Basically, I want to implement a structure like this but I couldn't figure out how should I define priority queue in header file in this format.
Thanks.
c++
c++
edited Nov 19 '18 at 21:23
NathanOliver
90.8k15126190
asked Nov 19 '18 at 21:21
Mert AkozcanMert Akozcan
6317
edited Nov 19 '18 at 21:23
NathanOliver
90.8k15126190
asked Nov 19 '18 at 21:21
Mert AkozcanMert Akozcan
6317
edited Nov 19 '18 at 21:23
NathanOliver
90.8k15126190
edited Nov 19 '18 at 21:23
NathanOliver
90.8k15126190
edited Nov 19 '18 at 21:23
NathanOliver
90.8k15126190
90.8k15126190
asked Nov 19 '18 at 21:21
Mert AkozcanMert Akozcan
6317
asked Nov 19 '18 at 21:21
Mert AkozcanMert Akozcan
6317
asked Nov 19 '18 at 21:21
Mert AkozcanMert Akozcan
6317
6317
I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.
– NathanOliver
Nov 19 '18 at 21:24
That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.
– SergeyA
Nov 19 '18 at 21:28
@SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?
– Mert Akozcan
Nov 19 '18 at 21:58
@MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.
– SergeyA
Nov 19 '18 at 22:00
add a comment |
I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.
– NathanOliver
Nov 19 '18 at 21:24
That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.
– SergeyA
Nov 19 '18 at 21:28
@SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?
– Mert Akozcan
Nov 19 '18 at 21:58
@MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.
– SergeyA
Nov 19 '18 at 22:00
I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.
– NathanOliver
Nov 19 '18 at 21:24
I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.
– NathanOliver
Nov 19 '18 at 21:24
That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.
– SergeyA
Nov 19 '18 at 21:28
That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.
– SergeyA
Nov 19 '18 at 21:28
@SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?
– Mert Akozcan
Nov 19 '18 at 21:58
@SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?
– Mert Akozcan
Nov 19 '18 at 21:58
@MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.
– SergeyA
Nov 19 '18 at 22:00
@MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.
– SergeyA
Nov 19 '18 at 22:00
add a comment |
1 Answer
1
active
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3
The short answer is don't use a lambda for this. Define a custom callable type instead. For example:
struct OrderByProcessPriority
{
bool operator()(
std::shared_ptr<Process> const& first,
std::shared_ptr<Process> const& second) const;
}
using ProcessQueue = std::priority_queue<
std::shared_ptr<Process>,
std::vector<std::shared_ptr<Process>>,
OrderByProcessPriority>;
ProcessQueue *readyQueue;
Then, in your cpp file, create the instance like this:
readyQueue = new ProcessQueue{};
answered Nov 19 '18 at 21:31
Peter RudermanPeter Ruderman
10.2k2352
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
The short answer is don't use a lambda for this. Define a custom callable type instead. For example:
struct OrderByProcessPriority
{
bool operator()(
std::shared_ptr<Process> const& first,
std::shared_ptr<Process> const& second) const;
}
using ProcessQueue = std::priority_queue<
std::shared_ptr<Process>,
std::vector<std::shared_ptr<Process>>,
OrderByProcessPriority>;
ProcessQueue *readyQueue;
Then, in your cpp file, create the instance like this:
readyQueue = new ProcessQueue{};
answered Nov 19 '18 at 21:31
Peter RudermanPeter Ruderman
10.2k2352
add a comment |
3
The short answer is don't use a lambda for this. Define a custom callable type instead. For example:
struct OrderByProcessPriority
{
bool operator()(
std::shared_ptr<Process> const& first,
std::shared_ptr<Process> const& second) const;
}
using ProcessQueue = std::priority_queue<
std::shared_ptr<Process>,
std::vector<std::shared_ptr<Process>>,
OrderByProcessPriority>;
ProcessQueue *readyQueue;
Then, in your cpp file, create the instance like this:
readyQueue = new ProcessQueue{};
answered Nov 19 '18 at 21:31
Peter RudermanPeter Ruderman
10.2k2352
add a comment |
3
3
3
The short answer is don't use a lambda for this. Define a custom callable type instead. For example:
struct OrderByProcessPriority
{
bool operator()(
std::shared_ptr<Process> const& first,
std::shared_ptr<Process> const& second) const;
}
using ProcessQueue = std::priority_queue<
std::shared_ptr<Process>,
std::vector<std::shared_ptr<Process>>,
OrderByProcessPriority>;
ProcessQueue *readyQueue;
Then, in your cpp file, create the instance like this:
readyQueue = new ProcessQueue{};
answered Nov 19 '18 at 21:31
Peter RudermanPeter Ruderman
10.2k2352
The short answer is don't use a lambda for this. Define a custom callable type instead. For example:
struct OrderByProcessPriority
{
bool operator()(
std::shared_ptr<Process> const& first,
std::shared_ptr<Process> const& second) const;
}
using ProcessQueue = std::priority_queue<
std::shared_ptr<Process>,
std::vector<std::shared_ptr<Process>>,
OrderByProcessPriority>;
ProcessQueue *readyQueue;
Then, in your cpp file, create the instance like this:
readyQueue = new ProcessQueue{};
answered Nov 19 '18 at 21:31
Peter RudermanPeter Ruderman
10.2k2352
answered Nov 19 '18 at 21:31
Peter RudermanPeter Ruderman
10.2k2352
answered Nov 19 '18 at 21:31
Peter RudermanPeter Ruderman
10.2k2352
answered Nov 19 '18 at 21:31
Peter RudermanPeter Ruderman
10.2k2352
10.2k2352
add a comment |
add a comment |
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I wouldn't use a lambda here. I'd just write a functor. I'd also avoid having a global variable if it can be helped.
– NathanOliver
Nov 19 '18 at 21:24
That won't be possible with lambda, since it's type is unknown. On a side note, I am not sure if there is really a reason to have a pointer to priority_queue in the program.
– SergeyA
Nov 19 '18 at 21:28
@SergeyA Let's say that I won't use a pointer to priority_queue, but I still want that as a member variable in a class. (I didn't put the whole code, but ready_queue is actually a member variable.) How should I define priority_queue in header file in this case?
– Mert Akozcan
Nov 19 '18 at 21:58
@MertAkozcan pointer note was a side note. The actual answer is that you can't use lambda's type before you define the lambda. You can use a functor like shown in the answer.
– SergeyA
Nov 19 '18 at 22:00