cout, operator<<, sequence points

[kalki70]

Hello,

I have a doubt about this operator, and I haven't found the answer in
internet (yet).
When calling the operator << several times in an expression, are all
parameters evaluated first and then passed to cout::<<, or are being
passed when they are evaluated ?
An example to clarify my question :

int foo()
{
cout << "calling foo";
return 1;
}

......
cout << "i = " << foo() << endl;

Should I see this:

calling fooi = 1

or

i = calling foo1 ??

Or is it undefined behaviour?

So going to the original question.... is "i=" passed to cout::<< first
and then foo() is evaluated and passed to cout::<<, or is foo()
evaluated and just then "i = " and foo() are passed to cout::<< ?? Or
undefined behaviour?

Thanks in advance,

Luis

[kalki70]

Victor, Jack, I really appreciate your answers, now it is also clear
to me.

Thanks a lot,

Luis