Does "(f(x))+g(y)" can make sure call f(x) first in C++?

Question

And does f(x)+(g(y)) can make sure call g(y) first? I know the order in expression is undefined in many case, but in this case does parentheses work?

possible duplicate of [Order of expression evaluation in C](http://stackoverflow.com/questions/9437683/order-of-expression-evaluation-in-c) — devnull, Jul 05 '13 at 11:36
If not close, can somebody at least quote the relevant parts of the standard from the links above. — devnull, Jul 05 '13 at 11:39

score 24 · Accepted Answer · answered Jul 05 '13 at 11:27

24

Parentheses exist to override precedence. They have no effect on the order of evaluation.

answered Jul 05 '13 at 11:27

R. Martinho Fernandes

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I would like to add a reference here http://en.cppreference.com/w/cpp/language/eval_order – DRC Jul 05 '13 at 11:28
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@musefan No. The order in which operations are done has nothing to do with the order in which the parameters for those operations are themselves evaluated. If you compare `(a() + b()) + c()` to `a() + (b() + c())`, the parenthesis change the order of the additions, but you can still *evaluate* `a()`, `b()`, and `c()` in any order you want. – David Schwartz Jul 05 '13 at 11:30
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Talk about [lazy reputation day](http://stackoverflow.com/a/17485851/596781)... :-) – Kerrek SB Jul 05 '13 at 11:33
@DavidSchwartz I wouldn't phrase it as "changes the order of the additions". It doesn't. I would phrase it as "changes *what* gets added to *what*". (a better example would use subtraction which is not associative) – R. Martinho Fernandes Jul 05 '13 at 11:39

score 12 · Answer 2 · answered Jul 05 '13 at 11:27

12

Look ma, two lines!

auto r = g(y);
f(x) + r;

This introduces the all-important sequence point between the two function calls. There may be other ways to do it, but this way seems straightforward and obvious. Note that your parentheses do not introduce a sequence point, so aren't a solution.

answered Jul 05 '13 at 11:27

John Zwinck

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This does not actually answer the question that was asked. – David Heffernan Jul 05 '13 at 11:28
You were right, so I expanded a bit to explain and include a link (which funny enough, on Wikipedia shows this exact case as lacking a sequence point for C++). – John Zwinck Jul 05 '13 at 11:31
Shouldn't this be `auto &&r = g(y);`? Otherwise you might introduce an additional copy, depending what type `g` returns. – Steve Jessop Jul 05 '13 at 13:13

score 2 · Answer 3 · edited Jul 05 '13 at 11:31

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No. Unless the + operator is redefined, things like that are evaluated left to right. Even if you were able to influence the precedence in the operator, it wouldn't necessarily mean that f and g were evaluated in the same order. If you need f to be evaluated before g, you can always do:

auto resultOfF = f(x);
auto resultOfG = g(x);
resultOfF + resultOfG;

edited Jul 05 '13 at 11:31

rubenvb

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answered Jul 05 '13 at 11:28

Thorsten Dittmar

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You're welcome. It's a C++11 feature, but a handy one at that, especially in exactly this case :) – rubenvb Jul 05 '13 at 11:37

Does "(f(x))+g(y)" can make sure call f(x) first in C++?

3 Answers3