Re: [android-developers] Re: help me urgent
sree wrote:
After takeing two one dimensional arrays, i [sic] prepared one two dimensional array.like this wayString [ ][ ] taskoutput=new String[prioritynamearray.length][taskvaluesarray. length];
Please follow the Java naming conventions. It will benefit you, too.
Also, "array" (or variant) in a variable name is generally not useful. Variables usually should not be
named for their type.
then Loop is rotating like this way,for (int s = 0; s <prioritynamearray.length; s++) {
Watch your crazy indentation, please.
DO NOT USE TAB CHARACTERS TO INDENT USENET POSTS!
for (int t = 0; t <taskvaluesarray.length; t++) {taskoutput[s][t]=here how can i add above two values into 2d arrays.
What do you want here? You have not told us.
You are, in effect, defining a transformation from the cross between the value sets represented by
(I'll rename your variables) 'priorities' and 'tasks'. Let me draw it:
\ priorities
| [0] | [1] | [2] | ...
tasks |--- ---|--- ---|--- ---| ...
[0] | ? | ? | ? | ...
--- ---|--- ---|--- ---|--- ---| ...
[1] | ? | ? | ? | ...
--- ---|--- ---|--- ---|--- ---| ...
[2] | ? | ? | ? | ...
--- ---|--- ---|--- ---|--- ---| ...
[3] | ? | ? | ? | ...
--- ---|--- ---|--- ---|--- ---| ...
You will need 'priorities.length' times 'tasks.length' values to fill in those question-marked
matrix entries.
What goes there?
Please tell us.
Foo results = new Foo [tasks.length] [priorities.length];
for (int tx = 0; tx < tasks.length; ++tx)
{
for (int px = 0; ix < priorities.length; ++px)
{
Task task = tasks[tx];
Priority priority = priorities[px];
results [px][tx] = whatDoYouWantHerePleaseTellUs(task, priority);
}
}
You need to tell us what you want in that box.
Obviously it's a value that depends on the two indexing values. You likely would want to factor
that out as a method on the two arguments as I showed here.
So what do you want in that box?
--
Lew
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