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# Comma operator and order or precedence

Ah hello.

I am sure I am having a silly moment here, but I am trying to understand the comma operator in C++ and "operator precedence" rules (which I believe are a general maths thing, hence me posting this in the Maths TA too:)).

From what I have read about the comma operator, we have two facts

1) The result of it is the right hand value, so for example a,b would yield b.
2) It has the lowest operator precedence.  I read this means "it's always the last one to bind to an expression", but I don't quite get that bit.

So, if we have

a = (b , c)

the evaulation sequence would first be (b,c) which would yield c, hence a would be assigned c.  But I am struggling with

a = b , c

This is apparently equivalent to

(a = b) , c

The result of this expression is b.

Now, I know we have to evaluate the brackets first, which results in b, so we then have

b, c

Q1) From rule 1 above, why is the result hence not c?  Clearly something is incorrect in what I deduced above, but I cannot see it...

Q2) Can someone clarify fact 2) above and what "it's always the last one to bind to an expression" means?

TIA
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1 Solution

Commented:

1. This is because the comma seperates the two statements, so the expression a = b , c is equivalent to ( a = b ) , c

One maybe important fact: In C++ it's possible to override comma-operators which may lead to confusion since if it is overridden it starts acting like a function call which means there's no order defined in which left- and rightside expressions are executed. For this you can find a sample at i.e. http://madebyevan.com/obscure-cpp-features/

Hope that helps,

ZOPPO
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Author Commented:
Thanks ZOPPO, but what you say I have already observed.  The result of

( a = b ) , c

is (and this is where I might be wrong!)

b, c

which is why I cannot see why the overall answer is not c!
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Commented:
Yes, that's correct, but the c is never assigned to anything. You have two different statements, "a = b" and "c", so a is equal to b afterward, no matter in which order the statements are evaluated.

Maybe it's clear when you check this:

x = ( a = b, c );

This results in 'a = b' and 'x = c'.
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Author Commented:

"but the c is never assigned to anything"

plus the lovely little example perfectly explain this.

Thank you!
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Commented:
You're welcome, I'm glad I could help :o)

Have a nice day,

best regards,

ZOPPO
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