Answer 1

Nah, this one is not undefined behaviour, just when you use x twice in the same expression.

The meaning of ++x is:
First increment x, then return the new value of x.

x++ means:
Make a copy of x, increment the original value of x and return the old copy of x.

Using parenthesis doesn't make a difference here.

y = 2*(++x); -> y = 2*3, x = 3.
z = 2*(x++); -> z = 2*3, x = 4.

Answer 2

Using parenthesis is not going to change the behaviour of x++. It is going to return the old value of x, with or without them.
x will only be changed after the fact.

Answer 3

In z, the x is incremented in post-increment form so the value of x is firstly used then increment.
In y, there is pre-increment in x so firstly increase the value then use it.
Example: x=2;
y=2*(++x); --- i.e. 2*(3) because first increment then use
z=2*(x++); --- i.e. 2*(3) because first use then increment

Answer 4

https://code.sololearn.com/c4nc6FYgE7w9/?ref=app

Answer 5

After the operation of z variable x's value will be 4. I have done some more changes in the code just check it

Answer 6

I suppose this is an undefined behavior.

https://stackoverflow.com/questions/949433/why-are-these-constructs-using-pre-and-post-increment-undefined-behavior

Answer 7

It is pre-increment and post-increment concept
In variable y ,x value(2) is pre-incremented by 1 (i.e; 3) and assigned to variable y so it becomes 2*3=6
........
After that
x value(i.e; 3 ) is assigned to variable z so it becomes 2*3=6 and then x post-incremented by 1 then the value becomes 4
......
Hopes it helps you

Answer 8

x++ is a post-increment, in which the value of the variable 'x' is first used prior to the increment.

++x is a pre-increment, in which the value is incremented first prior to use of the variable.

In the second line, y = 2 * (++x), the value of x is 3, because the expression uses a pre-increment operator and thus the value of y is 6.

However in the third line, z = 2 * (x++), the value of x is still 3, since the expression uses a post-increment operator and hence the value is z is also 6.

So those are the reasons why the output shows 6,6 and not 6,8.

Answer 9

Hope this helps you

Answer 10

CarrieForle Is my answer incorrect?

Answer 11

Prince I suppose that the behavior of post increment is always the same in an expression, regardless of the usage of parentheses around it. The increment is done only after the expression is evaluated.

Answer 12

Y=2*(++x) here it generate 6 and x=3 becoz of pre increment of x
Now
Y=2*(x++) =2*3=6 again it will generate the same as above due to post increment of x .In post increment first assign the value than increment it so after assigning it will remain same as 3 but if it ask about the value of x then it become 4..

Answer 13

Nah, this one is not undefined behaviour, just when you use x twice in the same expression.

The meaning of ++x is:
First increment x, then return the new value of x.

x++ means:
Make a copy of x, increment the original value of x and return the old copy of x.

Using parenthesis doesn't make a difference here.

y = 2*(++x); -> y = 2*3, x = 3.
z = 2*(x++); -> z = 2*3, x = 4.