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copied!<p>A common idiom is to use the <a href="http://en.wikipedia.org/wiki/Comma_operator" rel="noreferrer">comma operator</a> which evaluates both operands, and returns the second operand. Thus:</p> <pre><code>for(int i = 0; i != 5; ++i,++j) do_something(i,j); </code></pre> <h2>But is it really a comma operator?</h2> <p>Now having wrote that, a commenter suggested it was actually some special syntactic sugar in the for statement, and not a comma operator at all. I checked that in GCC as follows:</p> <pre><code>int i=0; int a=5; int x=0; for(i; i<5; x=i++,a++){ printf("i=%d a=%d x=%d\n",i,a,x); } </code></pre> <p>I was expecting x to pick up the original value of a, so it should have displayed 5,6,7.. for x. What I got was this</p> <pre><code>i=0 a=5 x=0 i=1 a=6 x=0 i=2 a=7 x=1 i=3 a=8 x=2 i=4 a=9 x=3 </code></pre> <p>However, if I bracketed the expression to force the parser into really seeing a comma operator, I get this</p> <pre><code>int main(){ int i=0; int a=5; int x=0; for(i=0; i<5; x=(i++,a++)){ printf("i=%d a=%d x=%d\n",i,a,x); } } i=0 a=5 x=0 i=1 a=6 x=5 i=2 a=7 x=6 i=3 a=8 x=7 i=4 a=9 x=8 </code></pre> <p>Initially I thought that this showed it wasn't behaving as a comma operator at all, but as it turns out, this is simply a precedence issue - the comma operator has the <a href="http://www.cppreference.com/wiki/operator_precedence" rel="noreferrer">lowest possible precedence</a>, so the expression x=i++,a++ is effectively parsed as (x=i++),a++</p> <p>Thanks for all the comments, it was an interesting learning experience, and I've been using C for many years!</p>

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