Hi!
Here is a small piece of code with the output:
I do understand that given the 2nd initialization (i = 1) , the following condition is not true and hence the program leaves the loop right away without printing out "ABC".
But I do not understand why it would print out
the first initialization (printf("Z")) ?
Has someone an idea?
Regards,
Placebo
edit: Please ignore the 2nd image - somehow I cannot remove it.
Last edited by Placebo; 04-07-2019 at 03:35 AM.
That's not an initialisation; that's a subexpression, the left hand side of a comma expression. So, of course it will be executed: control enters that part, and only exits the loop when it reaches the next part, i.e., the loop condition.
Note that i = 1 is also not an initialisation: it is an assignment, although it does assign the initial value of i. If you wanted to initialise i, you would do that at the point of declaration.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Thanks, helps a lot.
Just did not imagine for an unknown reason that expressions are getting executed when they appear within the brackets of a for-loop.
But in hindsight, of course the expression must get executed, as same as the programm successively executes all other commands, such as value-assignements, increments, etc.
