logic

问题 As a whole my program is for inserting and deleting nodes in a sorted doubly linked list. Insertion works and deleting the first node from the linked list works fine, except deleting the last node. Also, deleting nodes in the middle and end don't work. If I try to delete the last node, I am led back to main(); If I try deleting a node in the middle, the program crashes. Your help is appreciated! void remove(int n){ struct node* help = head; if(head->data == n){ if (head->next) help->next-

问题 i have a problem with my syntax below, i want to output the division of two numbers in decimal,but if i input 4 and 5, my output is 0 not 0.8 when i divide all others work fine. case $ch in 1)sum=`expr $n1 + $n2` echo "Sum ="$sum;; 2)sum=`expr $n1 - $n2` echo "Sub = "$sum;; 3)sum=`expr $n1 \* $n2` echo "Mul = "$sum;; 4)sum=`expr $n1 / $n2` echo "Div = "$sum;; *)echo "Invalid choice";; esac 回答1: Expr only implements integer arithmetic. Look into using something else, like dc. 回答2: Use bc . sum

问题 The following code System.out.println("1 0 0: " + (true ^ false ^ false)); System.out.println("1 0 1: " + (true ^ false ^ true)); System.out.println("1 1 0: " + (true ^ true ^ false)); System.out.println("1 1 1: " + (true ^ true ^ true)); System.out.println("0 0 0: " + (false ^ false ^ false)); System.out.println("0 0 1: " + (false ^ false ^ true)); System.out.println("0 1 0: " + (false ^ true ^ false)); System.out.println("0 1 1: " + (false ^ true ^ true)); outputs: 1 0 0: true 1 0 1: false

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . What does the term mocking mean in the context of C#/WPF ? 回答1: In the context of unit testing, mocking is providing a fake implementation of a type that your code under test interacts with. See http://en.wikipedia.org/wiki/Mock_object The book "The Art of Unit Testing" by Roy Osherove has good explanations and

问题 Right now I have an idea for a project and I would like to know if anyone can help me on the same logic. As such I need to create or generate a number of EditText according to amount you enter, ie, to select or enter a number such as 5, show me 5 EditText layout for type 5 values​​. They know that the form could accomplish this? Any ideas please? I guess it must be a way to do it with a loop, but not like carrying this calculation Java to XML. Thank you. 回答1: Here is an example of generating

问题 The problem is Brown, Clark, Jones and Smith are four substantial citizens who serve the community as architect, banker, doctor and lawyer, though not necessarily respectively. Brown who is more conservative than Jones but more liberal than Smith, is a better golfer than the men who are older than he is and has a larger income than the men who are younger than Clark The banker who earns more than the architect, is neither the youngest or the oldest. The doctor, who is a poorer golfer than the

问题 Paul here. I have 2 forms: a simple 'Signup for news' input for email, and button. a 'Contact' form with name, email, message. I also have a checkbox to allow sign up to news from this form. You can see them here on a page called test: http://butterflyepidemic.com/test/ I'm not sure how to set up the logic for the checkbox so that both forms are behaving themselves. If a subscriber uses the signup form, and then wants to send us a message, the contact form simply won't let them progress:

问题 Question: I have a method that Instantiates a new Scanner Class with a BufferedReader, and returns the Scanner Class itself. As far as I understand it, Java will return the Scanner class - as a "copy" of the object (pass-by-value), as opposed to what we would call in C++ a Reference or Pointer. Provided that logic is correct - does that then mean, the Scanner instantiated in the method I call, will continue to stay OPEN and thus never be garbage collected? For example: (bare bone) private

问题 I have a binary chunk of data 512 bytes long, I was wondering what the most efficient way would be if I wanted to shift it once to the right. My best guess right now (very new to assembly) would be that I would have to first check a chunk (probably int) to see what it would shift out, shift, then carry in whatever the previous int would have shifted out and proceed carrying this shift down the data. Is there an easier way? If I have to use this carry technique, what's the largest chunk I can

问题 FYI, the logic program I use cannot do contradiction introductions. This point is most likely irrelevant, for I highly doubt I would need to use any form of contradiction for this proof. In my attempt to solve this, I started off with assuming (p ⇒ q) ⇒ p) Is this correct? If so, what next? Forgive me if the solution seems so obvious. 回答1: (p ⇒ q) ⇒ p ((p ⇒ q) ⇒ p) ∨ (p ⇒ p) ; (X ⇒ X) and Or introduction ((p ⇒ q) ∨ p) ⇒ p ; (X ⇒ Z) ∨ (Y ⇒ Z) |- (X ∨ Y ⇒ Z) ((¬p ∨ q) ∨ p) ⇒ p ; (p ⇒ q) ⇔ (¬p ∨