Answer:
The method definition to this question can be given as:
Method definition:
double max(double x, double y) //define method with double parameter
{
if (x>=y) //check condition.
return x; //return value
else
return y; //return value
}
double max(int x, int y) //define method with integer parameter
{
if (x>=y) //check condition
return x; //return value
else
return y; //return value
}
double max(char x, char y) //define method with char parameter
{
if (x>=y) //check condition
return x; //return value
else
return y; //return value
}
Explanation:
The above method definition can be described as below:
- In the first method definition first, we define a method that is "max()". In this method we pass two variables as a parameter that is "x and y" and the datatype of this is double. Then we use a conditional statement. In the if block we check if variable x is greater then equal to y then it will return x else it will return y.
- In the second method definition, we define a method that is same as the first method name but in this method, we pass two integer variable that is "x and y". Then we use a conditional statement. In the if block we check if variable x is greater then equal to y then it will return x else it will return y.
- In the third method definition, we define a method that is same as the first and second method name but in this method, we pass two char variable that is "x and y". Then we use a conditional statement. In the if block we check if variable x is greater then equal to y then it will return x else it will return y.
B.Teamwork is a soft skill involves the ability to work harmoniously with your
colleagues and improve productivity. It is the collaborative effort of a team to
complete a task or to achieve a common goal in the most effective and efficient
way.
Answer:
Option 1: May crash at runtime because it can input more elements than the array can hold
Explanation:
Given the code as follows:
- int[] a = {1, 3, 7, 0, 0, 0};
- int size = 3, capacity = 6;
- int value = cin.nextInt();
- while (value > 0)
- {
- a[size] = value;
- size++;
- value = cin.nextInt();
- }
From the code above, we know the <em>a</em> is an array with six elements (Line 1). Since the array has been initialized with six elements, the capacity of the array cannot be altered in later stage.
However, a while loop is created to keep prompting for user input an integer and overwrite the value in the array started from index 3 (Line 4- 9). In every round of loop, the index is incremented by 1 (Line 7). If the user input for variable <em>value</em> is always above zero, the while loop will persist. This may reach a point where the index value is out of bound and crash the program. Please note the maximum index value for the array is supposedly be 5.
Answer:
- low = 10
- high = 50
- count = 0
-
- for i in range(low, high + 1):
- if(i % 3 == 0 and i % 5 == 0):
- count += 1
- print(count)
Explanation:
The solution code is written in Python.
We can create low and high variables to store the lower bound and upper bound in the range (Line 1-2)
Next create a counter variable, count (Line 3).
Use a for loop to traverse through the number between lower bound and upper bound and check if the current number-i is divisible by 3 and by 5, increment the count by one.
After the loop, print the count and we can get the number of ideal integers within the range (Line 8).
