Unit #4- Loops and Nested Loops

Front

Consider the following code segment. int count = 0;    for (int i = 10; i > 0; i--) {       for (int j = 0; j < 20; j += 2) {             count++;    } } System.out.println("count = " + count); What is the output of the code segment?

Back

count  = 100

Answer explanation

The outer loop will execute 10 times, and the inner loop will execute 10 times (i.e., from 0 to 19 by 2’s), so the count would be incremented 100 times, so count = 100 is the correct answer.

Front

Consider the following code segment. for (int i = 0; i <= 4; i++) { String printLine = "";      for (int j = 0; j <= i; j++) {          printLine += String.valueOf(j);      } System.out.println(printLine); } What will be the output of the code segment?

Back

0 01 012 0123 01234

Front

Consider the following code segment. int printNum = 0; for (int i = 0; i < 4; i++) { String printLine = "";     for (int j = 0; j <= i; j++) {                                      printLine += String.valueOf(printNum);              printNum++;         } System.out.println(printLine); } What will be the output of the code segment?

Back

0 12 345 6789

Answer explanation

The outer loop iterates four times so there will be four lines of output. On each iteration of the inner loop, the string format of the integer printNum is appended to printLine (the line to be printed at the bottom of the outer loop). The integer printNum starts at 0, and it iterates i times (the loop control variable), so 0 is the first line printed. At the end of the inner loop, printNum is incremented by 1. On the second iteration of the outer loop, the loop control variable i  = 1, so the inner loop will be executed twice, and 12 (the values of printNum) will be printed to the screen. This continues until finally 6789 is printed and the code segment ends.

Front

Consider the following code segment. for (int i = 0; i < 5; i++) { String printLine = "";      for (int j = 0; j <= i; j++) {          printLine = "*" + printLine;          }           for (int j = 0; j <= (4 - i); j++) {          printLine = " " + printLine;          } System.out.println(printLine); } What is the output of the following code segment?

Back

 * ** ***  **** *****

Answer explanation

The outer loop for (int i = 0; i < 5; i++) will execute five times, which corresponds to the number of pattern rows that will be printed. At the top of the outer loop, the string that will be printed to the screen printLine is set to empty so that individual characters can be added to it. We then have two inner loops that follow. The first one, for (int j = 0; j <= i; j++), will add the proper number of "*" characters to printLine based on the current row. The second inner loop, for (int j = 0; j <= (4 - i); j++), will add the proper number of " " (space) characters to printLine based on the current row. At the end of the outer loop, the current contents of printLine are printed to the screen resulting in the following pattern:

*

**

***

****

*****

Front

Which statements will loop every character of String myString?

Back

I and II

Answer explanation

Statements I and II model the for loop to visit every character of myString: for (int i = 0;i < myString.length();i++). Option III will stop at myString.length() - 1, which means that the last character of myString will not be visited.

Front

Consider the following code. int count = 0; for (int x = 1; x <= 3; x++) { /* missing loop header */ { count++; } } System.out.println(count); Which loop header will make the code print 6 as the value of count? Options:
for (int y = 0; y <= 2; y++), for (int y = 0; y < 3; y++), for (int y = 2; y >= 0; y--),
for (int y = 3; y > 0; y--),
for (int y = 0; y < x; y++)

Back

for (int y = 0; y < x; y++)

Front

Consider the following code segment. String str = "Self-belief"; String reverse = ""; String ithLetter; for(int i=0; i < str.length(); i++) { ithLetter = str.substring(i,i+1); reverse = ithLetter + reverse; } System.out.println(reverse); What will be the output of the above code segment?

Back

feileb-fleS

Answer explanation

The for loop is designed to start at the first letter of string str and to add that to reverse. After the first iteration, reverse = "S". On the second iteration of the for loop, i = 1, so append the first "e" in str to the beginning of reverse, so reverse now equals "eS".

And the loop will continue until the end of the string str index.

Front

Consider the code segment. int count = 0; for (int i = 0; i < 10; i++) {   for (int j = 0; j < 5; j++) {        count++;    } } System.out.println("count = " + count); What will be output as a result of running the code segment?

Back

count  = 50