Output of a simple c++ code?

[xpertbyshishir]

void main()
{
int x=5;
int y;

y=++x*++x;

printf("%d,%d" ,x,y);
}

This was a simple ques i was asked in the Written test.. But i couldnt get exact answer..

[Banfa]

This was a simple ques i was asked in the Written test.. But i couldnt get exact answer..

Unless it was a fairly advanced test this is a very poor question.

The answer is that this program invokes undefined behaviour on 2 counts

1. main MUST return int

2. You can not preincrement x twice like that. Specifically it is undefined behaviour to access a variable more than once to read and write it's value more than once between sequence points.


Since undefiined behaviour is invoked the output could be pretty much anything (or nothing for that matter).

[ashitpro]

void main()
{
int x=5;
int y;

y=++x*++x;

printf("%d,%d" ,x,y);
}

This was a simple ques i was asked in the Written test.. But i couldnt get exact answer..

Answer should be 7,49

Because x is incremented twice so x should be 7.
And for y statement becomes
y=7*7;
Read about 'sequence points in C'
here, '=' is sequence point where updated values will be put into all variables right to it .

[ilikesuresh]

The answer should be
x = 7
y =36;
The execution starts from right to left not left to right.
so x++ * ++x
becomes 6 * 6(pre incremented)
immediately after x++ the x value becomes 7
so
x = 7
y =36;

[r035198x]

Did you guys read post#2 above?