Answer / venu

i don't know the answer.
but here is the proble.
int i=5;
j=++i + ++i + ++i;
printf("%d",j);This code gives the answer 22.

How and What is the process?

Answer / raghu

In the first case
initially j=15 is assigned that is (5+5+5).then i gets incremented thrice.if we try to print i then i will be
8 (5 is incremented thrice).

In the second case
as we know that printf gets evaluated 4m right to left
that is pf("%d", 5++ + 5++ +5++);
that means 4m right to left 7+6+5=18
firstly i's value is 5 then it gets incremented 6 then 7.

Answer / gireesh

I have tried the second case in linux with gcc. But, the
output is 15 and not 18.

Answer / divakar

in the 1st case post increment will be done at the end of
the statement that is 15 is assigned to 'j' then 'i' will
increment 3 times bcoz 3 post increments r there.
in the later case first 15 assign to 'i' then increment
3 times and assign to 'i'

Answer / prasoon

both the codes possess undefined behaviour because the value
of i is changing more than once between two sequence points...

hence my answer is undefined behaviour

Answer / satyasaran

Remember increment operator is compiler depended.
1.
if u compiled in TC++ then u will get j=18
if u compile in VC++ then u will get j=15.
2.
if u compiled in TC++ then u will get j=18
if u compile in VC++ then u will get j=15.

Answer / amey

in the 1st case post increment will be done at the end of
the statement that is 15 is assigned to 'j' then 'i' will
increment 3 times bcoz 3 post increments r there.
in the later case first 15 assign to 'i' then increment
3 times and assign to 'i'

Answer / keerthan

Here, in first case, i=5, j=(i++)+ (i++) + (i++)............ i is added three times without incrementing because it is post incremented so... 5 is added with 5 and again with 5 answer is 15.
where as in second case... execution from right to left when it is done... initially i=5, first i++ this value is 5 then incremented so that 6 is added to 5... now is incremented again so 7 now 7+6+5 =18.

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