Question

What will be the output of following program int incr (int i) { static int count = 0;
count = count i; return (count); } main ({intij; for (i = 0; i <=4i--)j-incr(i); }​

Answer 1

REQUIRED OUTPUT :

1

2

3

:) STAY HAPPY AND SAFE

Answer 2

Answer:

On successful execution of the given program, the output will be $j=10$

Static Variables:

• Static is a storage class in C++
• The keyword static is used to declare a variable as static.
• When a variable is declared as static, it is initialized only once in the entire program.
• If the value of the variable is altered further in the program, then that value is retained in the next iteration.
• The compiler persists the variable till the end of the program.

incr function:

• The incr function declared at the beginning of the code returns the value of the count variable whenever it is called in the main function.
• The return data type of this function is an integer.
• The incr function includes a variable called count of the integer data type having storage class as static.
• When the incr function is called, it increments the value of count by the value of variable i from the main function, and returns the incremented value of count to the calling function.

Main function:

• The main function consists of two variables i and j, of integer data type.
• In the next line, for loop is declared to perform five iterations with the value of variable i going from zero to four.
• The for loop calls the incr function and assigns the value returned from the incr function to the variable j.

Explanation:

Step 1:

When the incr function is called in the first iteration $i=0$, the value of count will be zero.

This value is returned to the main function and assigned to j.

So, after first iteration $count=0$ and $j=0$

Step 2:

When the incr function is called in the second iteration $i=1$, the value of count will become one, i.e., $count = 0+1$

This value is returned to the main function and assigned to j.

So, after the second iteration $count=1$ and $j=1$

Step 3:

When the incr function is called in the third iteration $i=2$, the value of count will become 3, i.e., $count=1+2$

This value is returned to the main function and assigned to j.

So, after the third iteration $count=3$ and $j=3$

Step 4:

When the incr function is called in the fourth iteration $i=3$, the value of count will become 6, i.e., $count=3+3$

This value is returned to the main function and assigned to j.

So, after the fourth iteration $count=6$ and $j=6$

Step 5:

When the incr function is called in the fifth iteration $i=4$, the value of count will become 10, i.e., $count=6+4$

This value is returned to the main function and assigned to j.

So, after the fifth iteration $count=10$ and $j=10$

Therefore, the output of the program will be $j=10$

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