I am writng a program that a user enters the amount of sentences he would like to enter ,then the computer ask for a sentence but i can't seem to get it to loop rString numberSentences = JOptionPane.showInputDialog("How Many Senteces To Be Entered");. It
The "break" only breaks out of the for loop, not the while loop, which is the infinite one. You need a labeled break, or just a return instead. In addition to what Greg suggests, another possibility to exit the infinite loop is to use a Boolean variable in the while condition e.g....
There’s two problems but to make this less cluttered I’ll focus on the first. ‘films’ refers to an array containing different movie titles. ‘numberSelection’ is for the user to input which movie they’d like to select based on the array. /// private static Film filmSelection()...
create_loop: LOOP IF @__interval > CREATE_NEXT_INTERVALSTHEN LEAVE create_loop;ENDIF; SETLESS_THAN_TIMESTAMP = CUR_TIME + (HOURLY_INTERVAL * @__interval * 3600);SETPARTITION_NAME = FROM_UNIXTIME(CUR_TIME + HOURLY_INTERVAL * (@__interval - 1) * 3600,'p%Y%m%d%H00'); CALL partitio...
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Seeing this error "Matlab has encountered an internal problem and needs to close error" at the if and for loop. Suspecting it is because of the memory issue. In the command window, I typed memory. I saw the below Maximum possible array: 18594 MB (1.950e+01...
Unlike traditional approaches, RL in QSMODE/DQSMODE is not merely a parameter adjustment method but is fully utilized to generate paths based on the accumulated search experience to enhance path quality. QSMODE/DQSMODE introduces a unique self-training update mechanism for the Q-table and DQN ...
If you need to parse it (and extract data) you will need either a for or a while loop. If you have the the whole text as a string you will need first to split it by a token (I guess by the newline character), then iterate over the elements (lines) and then perform actions on...
It is easy to get the solution for k based on the solution for k - 1. This leads to a greedy solution to the original problem — solve for consecutive values of k starting from 1 until we reach the maximal value for which the solution exists. → Reply indweller 7 years ago,...