How Can I Trace A Java Program . Please Give Me Step by Step
//consider the follwing program part
for(int k = 0; k < 10; k++);
int k = 9
while(k < 10)
//now lets trace the program as follows
start, at the for loop, k is initialized to 0 with a scope
local to the for loop itself.
k = 0
Is it less than 10? yup, so we enter the loop.
Next, we have output,
Since there is no brackets, the for loop applies only to the
very next line. In this case, the output. So, return to the
top of the loop, k increments by 1
k = 1
and we proceed again since 1 < 10.
The output "hi" is printed again.
Go back to the top of the loop, increment k, and recheck the
loop condition of k < 10. The above cycle repeats until k is
incremented and equals 10. Since 10 is not less than 10,
exit the for loop.
k = 10
Now, declare an integer k to be equal to 9.
k = 9
Note it has local scope to this part of the program but,
since the for for loop previous one is gone from memory, so
too is the integer k that it (the loop) declared.
Thus, there is no conflict. So, an integer variable k (one
different from the last) is now = 9
Check the while boolean condition and k is less than 10 so
enter the while loop.
Next, print the output
Next, increment k by 1 ("k++") so
k = 10
Check the while loop conditional again but, this time, it's
false. k is not less than 10 anymore. So, exit the while loop.
|Is This Answer Correct ?||3 Yes||5 No|
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Financial: credit card number validation) Credit card numbers follow certain patterns. A credit card number must have between 13 and 16 digits. It must start with: ■ 4 for Visa cards ■ 5 for Master cards ■ 37 for American Express cards ■ 6 for Discover cards In 1954, Hans Luhn of IBM proposed an algorithm for validating credit card numbers. The algorithm is useful to determine whether a card number is entered correctly or whether a credit card is scanned correctly by a scanner. All credit card numbers are generated following this validity check, commonly known as the Luhn check or the Mod 10 check, which can be described as follows (for illustration, consider the card number 4388576018402626): 1. Double every second digit from right to left. If doubling of a digit results in a two-digit number, add up the two digits to get a single-digit number. 2 * 2 = 4 2 * 2 = 4 4 * 2 = 8 1 * 2 = 2 6 * 2 = 12 (1 + 2 = 3) 5 * 2 = 10 (1 + 0 = 1) 8 * 2 = 16 (1 + 6 = 7) 4 * 2 = 8 2. Now add all single-digit numbers from Step 1. 4 + 4 + 8 + 2 + 3 + 1 + 7 + 8 = 37 3. Add all digits in the odd places from right to left in the card number. 6 + 6 + 0 + 8 + 0 + 7 + 8 + 3 = 38 4. Sum the results from Step 2 and Step 3. 37 + 38 = 75 5. If the result from Step 4 is divisible by 10, the card number is valid; otherwise, it is invalid. For example, the number 4388576018402626 is invalid, but the number 4388576018410707 is valid. Write a program that prompts the user to enter a credit card number as a long integer. Display whether the number is valid or invalid. Design your program to use the following methods: /** Return true if the card number is valid */ public static boolean isValid(long number) /** Get the result from Step 2 */ public static int sumOfDoubleEvenPlace(long number) /** Return this number if it is a single digit, otherwise, return * the sum of the two digits */ public static int getDigit(int number) /** Return sum of odd place digits in number */ public static int sumOfOddPlace(long number) /** Return true if the digit d is a prefix for number */ public static boolean prefixMatched(long number, int d) /** Return the number of digits in d */ public static int getSize(long d) /** Return the first k number of digits from number. If the * number of digits in number is less than k, return
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