Mixing
Can static generic function interfere with each other (c#)?
Suppose there is a static extension function for caching the results like this:
public static Func<T, R> Memoize<T, R>(this Func<T, R> func)
{
var cache = new Dictionary<T, R>();
R result = default(R);
return x => cache.TryGetValue(x, out result) ? result : cache[x] = func(x);
}
If in the main method I declare
Func<int, int> func1;
func1 = (x) => x + 1; // one method with <int, int>
func1 = func1.Memoize();
Func<int, long> func2;
func2 = (x) => x - 1; // other method with <int, long>
func2 = func2.Memoize();
they will work fine. But what if I declare both delegates with identical type signature? Will they corrupt the results for each other if they are called?
c# static
asked Jan 1 at 15:17
TimofeusTimofeus
747
add a comment |
Suppose there is a static extension function for caching the results like this:
public static Func<T, R> Memoize<T, R>(this Func<T, R> func)
{
var cache = new Dictionary<T, R>();
R result = default(R);
return x => cache.TryGetValue(x, out result) ? result : cache[x] = func(x);
}
If in the main method I declare
Func<int, int> func1;
func1 = (x) => x + 1; // one method with <int, int>
func1 = func1.Memoize();
Func<int, long> func2;
func2 = (x) => x - 1; // other method with <int, long>
func2 = func2.Memoize();
they will work fine. But what if I declare both delegates with identical type signature? Will they corrupt the results for each other if they are called?
c# static
asked Jan 1 at 15:17
TimofeusTimofeus
747
1
This is an extension method so while technically it is a static mehod, you are using an instance (calledfuncin this case) so no, they should not collide.
– Zohar Peled
Jan 1 at 15:27
Thanks. They also do not collide even if non-extension. I suppose they would if the dictionary was separate static field.
– Timofeus
Jan 1 at 16:09
add a comment |
Suppose there is a static extension function for caching the results like this:
public static Func<T, R> Memoize<T, R>(this Func<T, R> func)
{
var cache = new Dictionary<T, R>();
R result = default(R);
return x => cache.TryGetValue(x, out result) ? result : cache[x] = func(x);
}
If in the main method I declare
Func<int, int> func1;
func1 = (x) => x + 1; // one method with <int, int>
func1 = func1.Memoize();
Func<int, long> func2;
func2 = (x) => x - 1; // other method with <int, long>
func2 = func2.Memoize();
they will work fine. But what if I declare both delegates with identical type signature? Will they corrupt the results for each other if they are called?
c# static
asked Jan 1 at 15:17
TimofeusTimofeus
747
Suppose there is a static extension function for caching the results like this:
public static Func<T, R> Memoize<T, R>(this Func<T, R> func)
{
var cache = new Dictionary<T, R>();
R result = default(R);
return x => cache.TryGetValue(x, out result) ? result : cache[x] = func(x);
}
If in the main method I declare
Func<int, int> func1;
func1 = (x) => x + 1; // one method with <int, int>
func1 = func1.Memoize();
Func<int, long> func2;
func2 = (x) => x - 1; // other method with <int, long>
func2 = func2.Memoize();
they will work fine. But what if I declare both delegates with identical type signature? Will they corrupt the results for each other if they are called?
c# static
c# static
asked Jan 1 at 15:17
TimofeusTimofeus
747
asked Jan 1 at 15:17
TimofeusTimofeus
747
edited Jan 1 at 16:13
Timofeus
asked Jan 1 at 15:17
TimofeusTimofeus
747
asked Jan 1 at 15:17
TimofeusTimofeus
747
asked Jan 1 at 15:17
TimofeusTimofeus
747
747
1
This is an extension method so while technically it is a static mehod, you are using an instance (calledfuncin this case) so no, they should not collide.
– Zohar Peled
Jan 1 at 15:27
Thanks. They also do not collide even if non-extension. I suppose they would if the dictionary was separate static field.
– Timofeus
Jan 1 at 16:09
add a comment |
1
This is an extension method so while technically it is a static mehod, you are using an instance (calledfuncin this case) so no, they should not collide.
– Zohar Peled
Jan 1 at 15:27
Thanks. They also do not collide even if non-extension. I suppose they would if the dictionary was separate static field.
– Timofeus
Jan 1 at 16:09
1
1
This is an extension method so while technically it is a static mehod, you are using an instance (called
func in this case) so no, they should not collide.– Zohar Peled
Jan 1 at 15:27
This is an extension method so while technically it is a static mehod, you are using an instance (called
func in this case) so no, they should not collide.– Zohar Peled
Jan 1 at 15:27
Thanks. They also do not collide even if non-extension. I suppose they would if the dictionary was separate static field.
– Timofeus
Jan 1 at 16:09
Thanks. They also do not collide even if non-extension. I suppose they would if the dictionary was separate static field.
– Timofeus
Jan 1 at 16:09
add a comment |
1 Answer
1
active
oldest
votes
I'm not sure why you think the type signatures would have any impact on your Memoize function. Each invocation of Memoize instantiates its own dictionary. Therefore func1 and func2 would operate independenty of one another regardless of the types involved.
To clarify a bit further, the function you return from Memoize captures the dictionary in a closure and thus has a reference to the instance created on the first line. Since you call Memoize twice, there are two separate dictionaries in play.
answered Jan 1 at 15:48
Kirk WollKirk Woll
61.6k16159173
I was under the impression that static methods are created once on first calling, so if one Memoize<int, int> was present the program would use it if called from another delegate.
– Timofeus
Jan 1 at 16:11
Static methods are not "created" -- they are just methods that do not require an instance of the surrounding class to operate (i.e. they have no access tothis). However, each invocation of a static method operates independently -- you are creating a dictionary upon each invocation, so each function returned has its own unique copy.
– Kirk Woll
Jan 1 at 16:13
Yes, I meant static instances, not methods.
– Timofeus
Jan 1 at 16:16
There is no such thing as "static instances" in C#. What you have is a static method.
– Kirk Woll
Jan 1 at 16:16
add a comment |
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I'm not sure why you think the type signatures would have any impact on your Memoize function. Each invocation of Memoize instantiates its own dictionary. Therefore func1 and func2 would operate independenty of one another regardless of the types involved.
To clarify a bit further, the function you return from Memoize captures the dictionary in a closure and thus has a reference to the instance created on the first line. Since you call Memoize twice, there are two separate dictionaries in play.
answered Jan 1 at 15:48
Kirk WollKirk Woll
61.6k16159173
I was under the impression that static methods are created once on first calling, so if one Memoize<int, int> was present the program would use it if called from another delegate.
– Timofeus
Jan 1 at 16:11
Static methods are not "created" -- they are just methods that do not require an instance of the surrounding class to operate (i.e. they have no access tothis). However, each invocation of a static method operates independently -- you are creating a dictionary upon each invocation, so each function returned has its own unique copy.
– Kirk Woll
Jan 1 at 16:13
Yes, I meant static instances, not methods.
– Timofeus
Jan 1 at 16:16
There is no such thing as "static instances" in C#. What you have is a static method.
– Kirk Woll
Jan 1 at 16:16
add a comment |
I'm not sure why you think the type signatures would have any impact on your Memoize function. Each invocation of Memoize instantiates its own dictionary. Therefore func1 and func2 would operate independenty of one another regardless of the types involved.
To clarify a bit further, the function you return from Memoize captures the dictionary in a closure and thus has a reference to the instance created on the first line. Since you call Memoize twice, there are two separate dictionaries in play.
answered Jan 1 at 15:48
Kirk WollKirk Woll
61.6k16159173
I was under the impression that static methods are created once on first calling, so if one Memoize<int, int> was present the program would use it if called from another delegate.
– Timofeus
Jan 1 at 16:11
Static methods are not "created" -- they are just methods that do not require an instance of the surrounding class to operate (i.e. they have no access tothis). However, each invocation of a static method operates independently -- you are creating a dictionary upon each invocation, so each function returned has its own unique copy.
– Kirk Woll
Jan 1 at 16:13
Yes, I meant static instances, not methods.
– Timofeus
Jan 1 at 16:16
There is no such thing as "static instances" in C#. What you have is a static method.
– Kirk Woll
Jan 1 at 16:16
add a comment |
I'm not sure why you think the type signatures would have any impact on your Memoize function. Each invocation of Memoize instantiates its own dictionary. Therefore func1 and func2 would operate independenty of one another regardless of the types involved.
To clarify a bit further, the function you return from Memoize captures the dictionary in a closure and thus has a reference to the instance created on the first line. Since you call Memoize twice, there are two separate dictionaries in play.
answered Jan 1 at 15:48
Kirk WollKirk Woll
61.6k16159173
I'm not sure why you think the type signatures would have any impact on your Memoize function. Each invocation of Memoize instantiates its own dictionary. Therefore func1 and func2 would operate independenty of one another regardless of the types involved.
To clarify a bit further, the function you return from Memoize captures the dictionary in a closure and thus has a reference to the instance created on the first line. Since you call Memoize twice, there are two separate dictionaries in play.
answered Jan 1 at 15:48
Kirk WollKirk Woll
61.6k16159173
answered Jan 1 at 15:48
Kirk WollKirk Woll
61.6k16159173
answered Jan 1 at 15:48
Kirk WollKirk Woll
61.6k16159173
answered Jan 1 at 15:48
Kirk WollKirk Woll
61.6k16159173
61.6k16159173
I was under the impression that static methods are created once on first calling, so if one Memoize<int, int> was present the program would use it if called from another delegate.
– Timofeus
Jan 1 at 16:11
Static methods are not "created" -- they are just methods that do not require an instance of the surrounding class to operate (i.e. they have no access tothis). However, each invocation of a static method operates independently -- you are creating a dictionary upon each invocation, so each function returned has its own unique copy.
– Kirk Woll
Jan 1 at 16:13
Yes, I meant static instances, not methods.
– Timofeus
Jan 1 at 16:16
There is no such thing as "static instances" in C#. What you have is a static method.
– Kirk Woll
Jan 1 at 16:16
add a comment |
I was under the impression that static methods are created once on first calling, so if one Memoize<int, int> was present the program would use it if called from another delegate.
– Timofeus
Jan 1 at 16:11
Static methods are not "created" -- they are just methods that do not require an instance of the surrounding class to operate (i.e. they have no access tothis). However, each invocation of a static method operates independently -- you are creating a dictionary upon each invocation, so each function returned has its own unique copy.
– Kirk Woll
Jan 1 at 16:13
Yes, I meant static instances, not methods.
– Timofeus
Jan 1 at 16:16
There is no such thing as "static instances" in C#. What you have is a static method.
– Kirk Woll
Jan 1 at 16:16
I was under the impression that static methods are created once on first calling, so if one Memoize<int, int> was present the program would use it if called from another delegate.
– Timofeus
Jan 1 at 16:11
I was under the impression that static methods are created once on first calling, so if one Memoize<int, int> was present the program would use it if called from another delegate.
– Timofeus
Jan 1 at 16:11
Static methods are not "created" -- they are just methods that do not require an instance of the surrounding class to operate (i.e. they have no access to
this). However, each invocation of a static method operates independently -- you are creating a dictionary upon each invocation, so each function returned has its own unique copy.– Kirk Woll
Jan 1 at 16:13
Static methods are not "created" -- they are just methods that do not require an instance of the surrounding class to operate (i.e. they have no access to
this). However, each invocation of a static method operates independently -- you are creating a dictionary upon each invocation, so each function returned has its own unique copy.– Kirk Woll
Jan 1 at 16:13
Yes, I meant static instances, not methods.
– Timofeus
Jan 1 at 16:16
Yes, I meant static instances, not methods.
– Timofeus
Jan 1 at 16:16
There is no such thing as "static instances" in C#. What you have is a static method.
– Kirk Woll
Jan 1 at 16:16
There is no such thing as "static instances" in C#. What you have is a static method.
– Kirk Woll
Jan 1 at 16:16
add a comment |
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1
This is an extension method so while technically it is a static mehod, you are using an instance (called
funcin this case) so no, they should not collide.– Zohar Peled
Jan 1 at 15:27
Thanks. They also do not collide even if non-extension. I suppose they would if the dictionary was separate static field.
– Timofeus
Jan 1 at 16:09