Explain how we get 2 for below c++ program int x=0; for(int z=0;z<5;z++){if((z>2)&&(++x>2)){x+=7;}} cout<<x;

Question

for z=0 we get x=1.
z=1 then x=2
z=2 then x=3.
z=3 then x=4 then we enter into if block and we get x=11
for z=4 we get x=12 and also we enter into if block there we get x=19
for z=5 we left the if loop and print the x then obviously I think that x is 19 but how we get 2 here
I had a major doubt in this question anyone can explain this question????

Answer 1

&& will evaluate the second expression only if the first expression is true, as "False&&Stmt" will always be "False" irrespective of the value of Stmt

So for z=0,1, and 2, the value of x will be unchanged.

For z=3 it will perform ++x to get x=1, but it still won't go inside the if as 1>2 is false.

For z=4 again ++x gives x=2, 2>2 is false, so it doesn't go into the if statement.

When z=5 it will exit the loop as the condition is z<5.

Answer 2

Hey just follow this
z=0:
nothing happens to x
z=1:
nothing happens to x
z=2:
nothing happens to x
z=3:
here z>2 but ++x (i.e x is 1) is less than 2
z=4:
here z>2 but again ++x (i.e x is 2) is not greater than 2
z=5:
loop is completed

now x is 2

Answer 3

when z>2 is false ++x is not increment
and when z=3,4 and then ++x execute 2 time
now the value of x is 2 and and z executed 5 time
hope you got my point

Answer 4

and(&&) does not check the 2 condition unless 1 condition is true btw the brackets eg-(1st cond.&& 2nd cond.) so at last x=2

Answer 5

if you want to get x = 19 - change IF conditions in some places.

when (z>2) is true - && will not evaluate the second expression