Question

push(item) { if (top < SIZE) { stack[top] = item; top++; } else ERROR } pop() { if (!is_empty()) { top--; return stack[top]; } else ERROR } is_empty() { if (top == 0) return true; else return false; }

          push(item) {
if (top < SIZE) {
stack[top] = item;
top++;
}
else
ERROR
}
pop() {
if (!is_empty()) {
top--;
return stack[top];
}
else
ERROR
}
is_empty() {
if (top == 0)
return true;
else
return false;
}

Added by Christopher K.

Discrete Mathematics and its Applications

Kenneth Rosen 8th Edition

Chapter 5

Instant Answer

Solved by Expert Oswaldo Jiménez

Step 1

- push: the if condition is reversed. It should allow push when top < SIZE; the code uses if (top == SIZE) which is the overflow case. Correct intent: if (top != SIZE) { stack[top] = item; top++; } else ERROR - pop: the if is reversed. It should return an error Show more…

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The pseudocode of Figure 1 illustrates the basic push() and pop() operations of an array-based stack. Assuming that this algorithm could be used in a concurrent environment, answer the following questions: push(item) if (top == SIZE) stack[top] = item; top++; else ERROR pop() if (is_empty()) { top--; return stack[top]; } else ERROR is_empty() { if (top == 0) return true; else return false; }