Zoho Complex and Nested Flows Quiz 1

The value of j is 10 and the loop is iterates from k=1 to k<=5 as its prints the sum of the j+k hence the output of the above code is 11 12 13 14 15

Numbers will be displayed until the signed limit of short and the program will successfully terminate

while(TRUE) or while(true) does not work. true is not a keyword

(++a < 4) first increments and compares afterwards. (b++ < 4) first compares and increments afterwards.

for(anything; anything; anything) is Ok. printf("YELLOW") prints YELLOW and returns 1 as result. So for loop runs forever. Actually break is saving us from quitting the for loop. Only after checking condition and executing the loop statements, third section is executed. break causes the loop to quit without incre/decrement section

You can initialize any number of variables inside for loop

 

while(1) is infinite loop. So we kept if(condition) to break the loop. a++ is equivalent to a=a+1;

Execution of program starts from main. Inside main are two nested for loops.

First for loop is as follows:

Initialization : i = 2

Terminating condition : i <= 5

Increment Value : 1 (i++)

Inner for loop is as follows:

Initialization : j = i

Terminating condition : j <= 5

Increment Value : 1 (j++)

Total number Of Iterations : 4 + 3 + 2 + 1 = 10

Iteration 1 :

Display statement prints i = 2 j = 2

j is incremented to 3.

Iteration 2 :

Display statement prints i = 2 j = 3

j is incremented to 4.

Iteration 3 :

Display statement prints i = 2 j = 4

j is incremented to 5.

Iteration 4 :

Display statement prints i = 2 j = 5

j is incremented to 5 and inner loop terminates. Now i increments and becomes 3. Now inner loop again starts.

Similarly the other iterations.

When i, j becomes 5 loops terminate.

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In this program, sum is initialised to 23 and i is initialized to 2. i is incremented after every completion of the inner loop of j. j starts from 7 and goes up to 9. At the end of the program, sum gets the value 359.