cout print order?

[rvangend]

I have a function that I am passing two arguments, one of them by value and one by reference.

int problem_001(int,int& t);

Before calling "problem_001", "t" has a known value (let's say 0).
Inside "problem_001", I assign "t" a new value (let's say 10).

In my main program, I have this statement:

cout << "Problem 01 = " << problem_001(1000,t) << " , " << t <<endl;

In this case "t" still prints with the original value of "0".

but, if I change my statement to:

cout << "Problem 01 = " << problem_001(1000,t);
cout << " , " << t <<endl;

"t" now prints the value I expect i.e. "10".

Is there a explanation for this please.

Thanks
Rob

[vlad from moscow]

According to the C++ Standard "

The order of evaluation of function arguments is unspecified.

"
So the first statement has an undefined behavior. According to C conventions usually arguments are evaluated from right to left but you should not rely upon that.
In the next statements there is no such problem because statements are executed sequentially.

[cire]

@OP - You may find http://stackoverflow.com/questions/10778627/why-is-i-i-1-undefined-behavior-in-c11 interesting reading. The principals involved are the same.