Question

what will be the output of the following code
int k=5,j=9
k+=k++-++j+k;
system.out.println("k="+k);​

Answer 1

QUESTION:-

➣

What will be the Output:-

ANSWER:-

➣

Output:- 6

WORKING:-

➣

Given:-

➣ int k = 5 and j = 9

Method:-

➣ 5 = 5 - 10 + 6

➣ 5 + 5 - 10 + 6

➣ 10 - 10 +6

➣ 6

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Explanation:-

➣

Let's understand the Concept of above Output:-

Case 1:

➣ When the sign of increment (++) is present before any variable the default value increases by 1 and the increased value becomes the Default Value then the Value is printed.

➣ This is called Pre-increment

• For eg:-

➣ ++ a [int a = 5]

In this case the default value of a will 1st increase and replace the default value and at last print the Incremented value.

Output:- a = 6

Default value:- a = 6

Case 2:

➣ When the sign of increment (++) is present after any variable the default value is 1st Printed then the increased value increases by 1 and the increased value becomes the Default Value.

➣This is called Post-increment

• For eg:-

➣ a++[int a = 5]

In this case the default value of a will be printed then its value will increase by 1.

Output:- a = 5

Default value:- a = 6

_________________________________________

• We can Simply understand this Concept like this:-

• In Pre-increment(++a) the default value stored in a 1st increases by 1 and then the value is Printed or this changed value is used further in a function if the the variable repeats itself.

Also after Printing the value the Incremented value becomes the Default Value for a.

• In Post-increment(a++) the default value stored in a 1st is printed and then increases by 1 and the increased value is implemented further in expression if a is present one more time.

➣ ++a = Add before

➣ a++ = Add after

Answer 2

Answer:

k = 6

Explanation:

k = 5+(5-10+6)=5+1=6