Local Static variables and cout

[tavianator]

could someone please explain to me why this code produces

1: 3
2: 2
3: 1

as-is, and

1: 1
2: 2
3: 3

when uncommented?

int foo()
{
	static int i = 0;
	return ++i;
}

int main(int argc, char* argv[])
{
	std::cout     << "1: " << foo() << std::endl//;
	/*std::cout*/ << "2: " << foo() << std::endl//;
	/*std::cout*/ << "3: " << foo() << std::endl;
	return 0;
}

 

[cpjust]

Apparently the order of execution is right to left, so the 3rd foo() gets run first...
I wouldn't always count of that being the case though. Different compilers might execute statements from left to right.

 

[tavianator]

ok, it just seems wierd that the operator is evaluated left-to-right, but the functions aren't evaluated in that order.

 

[cpjust]

You can see what it's doing a little better like this:

#include <iostream>

int foo( char c )
{
    static int i = 0;
	std::cout << c << std::endl;
    return ++i;
}

int main(int argc, char* argv[])
{
    std::cout     << "1: " << foo( 'a' ) << std::endl//;
    /*std::cout*/ << "2: " << foo( 'b' ) << std::endl//;
    /*std::cout*/ << "3: " << foo( 'c' ) << std::endl;
    return 0;
}

I get the following output:
c
b
a
1: 3
2: 2
3: 1

If you put a breakpoint on the cout line in main(), then step into each function call, you can see it execute the foo()'s from right to left, then the << insertions from left to right.
It's weird, but that's what happens when you try to combine too many things into a single line in C/C++. ;-)

 

[Salem]

> could someone please explain to me why this code produces
Because the first version invokes undefined behaviour.

http://c-faq.com/expr/seqpoints.html

Getting 3 2 1 is just as likely as getting 1 2 3 or indeed other possible answers (as befits undefined behaviour).

Since your original code does [tt]++i[/tt] three times between sequence points, the result seems at odds with expectation.

> It's weird, but that's what happens when you try to combine too many things into a single line
It's only weird when the thing in question has multiple side effects on the same object.

> it just seems wierd that the operator is evaluated left-to-right, but the functions aren't evaluated in that order.
Given [tt]f1() + f2()[/tt], there is nothing in the standard which says which should be called first.

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