how to use delegate to call function and introduce function as content?
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smallkubi wrote:
I think maybe paint problem.
Why?
smallkubi wrote:
When i ask before, i get the answer is point number is too big.
The more points you need to draw, the longer it takes. If there had been a useless delay in the chart-control, people would have noticed.
Bastard Programmer from Hell :suss: If you can't read my code, try converting it here[^][](X-Clacks-Overhead: GNU Terry Pratchett)
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I mean Dalay is not that Chart generating process is slow, when i finished plot, i want to operate on Chart, like mouse moving, Scroll... is slow.
Meaning it will have to recalculate which part it should show and repaint it. That takes time. Smaller charts with less points will be faster. If you find that the Chart-control is "too slow" for your needs, you may have to implement your own.
Bastard Programmer from Hell :suss: If you can't read my code, try converting it here[^][](X-Clacks-Overhead: GNU Terry Pratchett)
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Meaning it will have to recalculate which part it should show and repaint it. That takes time. Smaller charts with less points will be faster. If you find that the Chart-control is "too slow" for your needs, you may have to implement your own.
Bastard Programmer from Hell :suss: If you can't read my code, try converting it here[^][](X-Clacks-Overhead: GNU Terry Pratchett)
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oh, do you know how to stop control auto refresh? i think if i can refresh it in mannual?
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Now i want to have the function of Zooming,scrolling... Anything changes, screen don't update automatically,I want to updated data on screen mannully, can it be possible?
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So something like this:
static void Main()
{
int x = 1;
int z = 3;
fun1(x, y => fun2(y));
fun1(x, y => fun3(y, z));
}static void fun1(int x, Action<int> call)
{
call(x);
}static void fun2(int y) { ... }
static void fun3(int y, int z) { ... }
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
Hi, i have another question... If i want to use multi-thread to call fun1, how can i input the variable x, y,z? for example: static void Main() { int x = 1; int z = 3; Thread th=new Thread(fun1( ? )) //how can i do ? th.start(); // fun1(x, y => fun2(y)); // fun1(x, y => fun3(y, z)); } static void fun1(int x, Action call) { call(x); } static void fun2(int y) { ... } static void fun3(int y, int z) { ... }
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Hi, i have another question... If i want to use multi-thread to call fun1, how can i input the variable x, y,z? for example: static void Main() { int x = 1; int z = 3; Thread th=new Thread(fun1( ? )) //how can i do ? th.start(); // fun1(x, y => fun2(y)); // fun1(x, y => fun3(y, z)); } static void fun1(int x, Action call) { call(x); } static void fun2(int y) { ... } static void fun3(int y, int z) { ... }
Something like this:
Thread th = new Thread(() => fun1(x, y => fun3(y, z)));
Breaking it down:
Action<int> call = y => fun3(y, z);
ThreadStart start = () => fun1(x, call);
Thread th = new Thread(start);
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
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Something like this:
Thread th = new Thread(() => fun1(x, y => fun3(y, z)));
Breaking it down:
Action<int> call = y => fun3(y, z);
ThreadStart start = () => fun1(x, call);
Thread th = new Thread(start);
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
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Good Idea,i meet it. And if i have a button at main Thread , it can abort Thread "th". How to code button click function? And, if whether Thread th end or abort, i want to run" MessageBox.Show("Thread end")", how can i do it?
smallkubi wrote:
it can abort Thread "th"
Don't. How To Stop a Thread in .NET (and Why Thread.Abort is Evil)[^] If you're using .NET 4.5 or above, you'd probably have better luck using
asyncandawait. Asynchronous Programming with Async and Await (C# and Visual Basic)[^] Otherwise, use something like theBackgroundWorkercomponent[^].
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
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Something like this:
Thread th = new Thread(() => fun1(x, y => fun3(y, z)));
Breaking it down:
Action<int> call = y => fun3(y, z);
ThreadStart start = () => fun1(x, call);
Thread th = new Thread(start);
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
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hi, If i use Func call= y=>fun3(y,z) , fun3 can be one output param function. But if my function need 2 or more output, how can i do it?
Sorry, I'm not following you. Do you mean you want multiple return values from your function? If so, you'll need to create a class or structure to contain them:
public struct Fun3ReturnValues
{
public int SomeValue { get; set; }
public string SomeOtherValue { get; set; }
}
...
public static Fun3ReturnValues Fun3(int y, int z)
{
return new Fun3ReturnValues
{
SomeValue = 42,
SomeOtherValue = "Hello",
};
}Alternatively, you could use the
Tupleclass[^], but the meaning of the values wouldn't be as clear.
"These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer
