Output of the c program [closed] - c++

Question

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 4 years ago.
Improve this question
When running the following code, the output is 2 4 6. I expected it to be 2 5 and so on...
Why the difference?
#include <iostream>
using namespace std;
int main() {
static int i;
for (i++; ++i; i++) {
printf("%d ", i);
if (i == 6)
break;
}
return 0;
}
I did the same thing here, without the for loop, from 2nd iteration. The result is 5.
Why is that?
#include <iostream>
using namespace std;
int main() {
static int i=2;
i++;
i++;
cout<<++i;
return 0;
}

Answer 1

Each time the loop is traversed you increase i by two, not three (I guess you expected the latter to happen).
In a general case
for ( init_statement ; condition ; iteration_expression ) {
loop_statements
}
is equivalent to
init_statement
for ( ; condition ; ) {
loop_statements
iteration_expression
}
and
init_statement
while ( condition ) {
loop_statements
iteration_expression
}
See, e.g., this.
In your case,
for (i++; ++i; i++) {
...
}
is equivalent to
i++
for ( ; ++i; ) {
...
i++
}
and
i++
while ( ++i ) {
...
i++ ;
}
As for your second piece of code, you mention that "The result is 5". It appears you expect otherwise, perhaps 4. If so, try replacing
cout<<++i;
with
cout<<i++;
(incrementing prior vs. after other operations in the sentence).

Answer 2

When you see the syntax of for loop which is
for(initialize;condition;inc/decrement)
the statement in initialize block only executes once at the starting of for loop that's why it is increamenting i by 2.
Hopefully that help

Answer 3

In a for loop you've got 4 parts. I forget the "official" name for the individual pieces but I typically refer to them as:
Initialization
Anchor Statement
Body
Follow-up
effectively:
for(Initialization (1); Anchor Statement (2); Follow-up (4))
{
Body (3)
}
It's going to hit in the following order:
1, 2, 3, 4, [2, 3, 4]... until step 2 evaluates to false or a break statement is hit. Then it exits.
So i starts as 0.
You hit step 1 and 2, making i evaluate to 2 in step 3 (the body).
Then you hit step 4 and 2, making i evaluate to 4 in step 3.
Then you hit step 4 and 2, making i evaluate to 6 in step 3.
Then you have a break statement that will exit the loop.
The important part that seems to be the confusion, is that the initialization step only runs once.