evaluate it

[ashim.seth]

hey people i was just testin this:
int i=5,j;
j=i++ + ++i;
cout<<j;
and then...

int i=5;
int j= i++ + ++i;
cout<<j;
both givin different results
y?
plz help

[Victor Bazarov]

as********@gmail.com wrote:

hey people i was just testin this:
int i=5,j;
j=i++ + ++i;
cout<<j;
and then...

int i=5;
int j= i++ + ++i;
cout<<j;
both givin different results
y?
plz help

Read the FAQ, section 39

V

[Peter Julian]

<as********@gmail.com> wrote in message
news:11**********************@g44g2000cwa.googlegr oups.com...

hey people i was just testin this:
int i=5,j;
j=i++ + ++i;
cout<<j;
and then...

int i=5;
int j= i++ + ++i;
cout<<j;
both givin different results
y?
plz help

The results are irrelevent. Since you are refering to the same variable in
that compound statement, the standard does not guarentee when, how or
whether the variable i gets incremented. In fact, the standard has a
requirement that the said variable be only modified once in between sequence
points.

Try:

#include <iostream>

int main()
{
int i = 5;

std::cout << "i = " << i << "\n";

int j(i++);

std::cout << "i = " << i << "\n";
std::cout << "j = " << j << "\n";

j += (++i);

std::cout << "i = " << i << "\n";
std::cout << "j = " << j << "\n";

return 0;
}

/*
i = 5
i = 6
j = 5
i = 7
j = 12
*/