4

I have read that comma operator is used to assign expression, and the right expression is supplied to lvalue.

But why does this program assign left expression to lvalue when not using parenthesis? I am using turbo c compiler.

int b=2;
int a;
a=(b+2,b*5);  // prints 10 as expected
a=b+2,b*5;    // prints 4 when not using parenthesis

Also the following works:

int a =(b+2,b*5);

But this generates an error:

int a =b+2,b*5;   // Error

I can't understand why.

BSMP
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Sanjana
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    The comma operator has lower precence than the assignment operator, and has a special meaning within type declarations, which is why your last example fails. – fuz Sep 06 '13 at 20:31
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    The comma operator is almost never needed and almost always used incorrectly. Do not use it. – Adam Burry Sep 07 '13 at 03:12
  • read [example](http://stackoverflow.com/questions/17785206/why-does-this-c-code-compile-what-does-it-do/17785232#17785232) – Grijesh Chauhan Sep 07 '13 at 07:44

1 Answers1

8

Because precedence of , operator is lower than of = one, this...

a=b+2,b*5;

... will actually be evaluated as...

a = b + 2;
b * 5;

With int i = b + 2, b * 5; is a bit different, because comma has different meaning in declaration statements, separating different declarations from each other. Consider this:

int a = 3, b = 4;

You still have comma here, but now it separates two variable assignment-on-declarations. And that's how the compiler attempts to treat that line from your example - but fails to get any meaning from b * 5 line (it's neither assignment nor declaration).

Now, int a = (b + 2, b * 5) is different: you assign a value of b + 2, b * 5 expression to a variable a of type int. The first sub-expression is discarded, leaving you just with b * 5.

raina77ow
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