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What does stack.pop() return?


What does stack.pop() return?

By : RichardM
Date : October 25 2020, 07:10 AM
may help you . Yes, as Willem has mentioned pop removes top element from the stack and returns its value. So how are you obtaining the index? You need to note what the element removed represents. In this case, it represents the index from the temperatures array:
code :
stack.push(i);


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Return the stack element at a given index without modifying the original Stack in Java

Return the stack element at a given index without modifying the original Stack in Java


By : slammerhead
Date : March 29 2020, 07:55 AM
To fix the issue you can do If you cannot use another stack, you can cheat and abuse a local variable on the call stack for the same purpose by making a recursive method:
code :
public static <T> T getStackElement(Stack<T> stack, int index) {
  if (index == 0) {
    return stack.peek();
  }

  T x = stack.pop();
  try {
    return getStackElement(stack, index - 1);
  } finally {
    stack.push(x);
  }
}
Method to copy a stack into a new one and return it without changing the original stack using a single queue using stand

Method to copy a stack into a new one and return it without changing the original stack using a single queue using stand


By : Jesse Gillmore
Date : March 29 2020, 07:55 AM
help you fix your problem Stack is a Collection, and collections can addAll(Collection)
code :
Stack<Integer> intStackCopy = new Stack<Integer>();
intStackCopy.addAll(origStack);
Iterator<Integer> intItr = origStack.iterator();
while(intItr.hasNext())  {
   intStackCopy.push(intItr.next());
}
public Stack<Integer> copyStack(Stack<Integer> orig)  {
   Queue<Integer> q = new Queue<Integer>();
   q.addAll(orig);
   Stack<Integer> intStackCopy = new Stack<Integer>();
   intStackCopy.addAll(q);
   return  intStackCopy;
}
Return address prediction stack buffer vs stack-stored return address?

Return address prediction stack buffer vs stack-stored return address?


By : Amitoz deol
Date : March 29 2020, 07:55 AM
Does that help Predictors are normally part of the fetch stage, in order to determine which instructions to fetch next. This takes place before the processor has decoded the instructions, and therefore doesn't even know with certainty that a branch instruction exists. Like all predictors, the intent of the return address predictor is to get the direction / target of the branch faster. A return instruction is a branch, and so it would normally have a branch predictor entry to determine whether it is taken and where the target is. The return address predictor is consulted in lieu of the normal branch target buffer.
So perhaps 50 instructions before the return statement is actually "executed", the fetch stage predicts a return instruction and the instruction address to fetch next. Later, when the return is executed, the address is read from the stack and compared with where the return was predicted to go. If they are the same, execution continues, else execution is rolled back to use the proper return address.
Simple stack example, how to return if stack is empty?

Simple stack example, how to return if stack is empty?


By : David Simonelli
Date : March 29 2020, 07:55 AM
Hope that helps This is exactly the type of thing that optionals are for. Optionals are like regular variables, except they have a special case to say, "no value is assigned", so instead of returning type "T", return type "optional T":
code :
mutating func pop() -> T? {
    if !elements.isEmpty {
        return elements.removeLast() // return the last item on the stack
    } else {
        return nil // indicate there was nothing on the stack
    }
}
if let item = myStack.pop() { // this implicitly unwraps the optional if present
    // Do something. "item" is of type T, not T?
else {
    // Nothing was on the stack. "item" is not defined
}
What is the return type of stack.top() and stack.pop() using stack<string> in cpp?

What is the return type of stack.top() and stack.pop() using stack<string> in cpp?


By : Mukesh
Date : March 29 2020, 07:55 AM
it helps some times A std::string is a different type from a char*, the following won't work:
code :
ch1 = cards.top(); // top() returns a string, not a char*
int main(){

  stack <string> cards;
  string ch1; // std::strings
  string ch2;
  ...

  ch1 = cards.top(); // Correct

  printf("Top of the Stack : %s \n", ch1.c_str()); // c_str() needed
  return 0;
}
std::cout << "Top of the Stack : " << ch1;
#include <iostream>
#include <stack>
#include <string>

int main() {
    std::stack <std::string> cards;
    std::string ch1;

    cards.push("Ace");
    cards.push("King");
    cards.push("Queen");
    cards.push("Jack in the Jack");

    std::cout << "Number of cards : " << cards.size() << std::endl;

    ch1 = cards.top(); // Get the top card

    std::cout << "Top of the Stack : " << ch1;
}
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