Text in a function printed before it is called

Dear All

I have two C City & Guilds (4240, 4250) and I have decided to continue and learn C++. So I have purchased a book with compiler on CD, Bloodshed Dev-C++. I have written a programme found in the book and the output is as the book describes. However, I do not understand it. I read on to see if the book explined my question but it did not. See if you can.

Code:

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#include <iostream>

int Add (int x, int y)
{
    std:: cout << "In Add(), received " << x << " and " << y  << "\n";
    return (x+y);
}

int main()
{
    std:: cout << "The value returned is : " << Add(3,4);
    return 0;
}

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Code:

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Output to screen
 In Add(), received 3 and 4
 The value returned is : 7

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And the problem:
Why does the text in the function get printed to screen before
the function is called?

Code:

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// Because this function returns

Add(3,4);

//before this function returns

cout << Add(3,4);

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This actually led me to a question myself, though. What's the deal with

Code:

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std::cout.operator<<("The value returned is : ");

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returning an address?

the << operator is right associative. Which means it reads from its right operand and assigns the result to its left operand. As so we must evaluate the right-most operand first...

Quote:

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std:: cout << "The value returned is : " << Add(3,4);

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Add() is evaluated first. The cout inside gets then printed to screen because a function termination flushes any local stream. The result meanwhile is sent to the stream in main().

Quote:

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std:: cout << "The value returned is : " << 7;

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main() then finishes and the stream is flushed, producing the 2nd line of the output

Your cout call in main is equivalent to:

Code:

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operator<<(operator<<(std::cout, "The value returned is : "), Add(3,4));

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The first operator<< function call has two arguments, both of which are the return values of function calls themselves (the first is operator<<(std::cout, "The value returned is : "), the second is Add(3,4)). The order that the two arguments are evaluated isn't specified by the standard, so you don't know which of those two functions will be evaluated first. On your machine with your compiler, the second function was evaluated first, which is why the stuff inside of it got printed first.

>> What's the deal with std::cout.operator<<("The value returned is : "); returning an address?

It returns the stream, not an address, but the stream can be converted to void* so that it can be used in a boolean expression.

Quote:

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Originally Posted by Daved

The order that the two arguments are evaluated isn't specified by the standard, so you don't know which of those two functions will be evaluated first. On your machine with your compiler, the second function was evaluated first, which is why the stuff inside of it got printed first.

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In fact the output operator has an associative rule. It is right associative. The right operands are evaluated first.

Quote:

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Originally Posted by Daved

>> What's the deal with std::cout.operator<<("The value returned is : "); returning an address?

It returns the stream, not an address, but the stream can be converted to void* so that it can be used in a boolean expression.

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That makes sense, I wasn't thinking straight. Thank you.

>> It is right associative.
I don't think that's correct, but I don't have any references handy to prove it. The bitshift operators are left associative, and so is the overloaded version used with streams. All that the associativity means is that you will get:

Code:

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operator<<(operator<<(std::cout, "The value returned is : "), Add(3,4));

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and not:

Code:

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operator<<(operator<<(std::cout, Add(3,4)), "The value returned is : ");

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The order of evaluation of the parameters should still be unspecified.

Thanks to all who replied

A very useful lesson

Quote:

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Originally Posted by Daved

I don't think that's correct, but I don't have any references handy to prove it.

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Actually you are right. I'm sorry. They are left associative. Couldn't be any other way in fact.