var c = (a,b); The above uses the comma operator. It evaluates as the value of its right-hand side (i.e. b). var c = a,b; This does not use the comma operator. The comma character here forms part of the var expression which takes a comma-separated list of variables to create in the current scope, … Read more
The first one is equivalent to: (true ? (++x, ++y) : (–x)), –y; The second one is equivalent to: (false ? (++x, ++y) : (–x)), –y; Thus the –y is always executed. In the first line, the increments are executed first so x = 1, y = 0 is expected. In the second line, the … Read more
The , operator evaluates a series of expressions and returns the value of the last. c=a,b is the same as (c=a),b. That is why c is 10 c=(a,b) will assign the result of a,b, which is 20, to c. As Mike points out in the comments, assignment (=) has higher precedence than comma
A common idiom is to use the comma operator which evaluates both operands, and returns the second operand. Thus: for(int i = 0; i != 5; ++i,++j) do_something(i,j); But is it really a comma operator? Now having wrote that, a commenter suggested it was actually some special syntactic sugar in the for statement, and not … Read more
In your example it serves no reason at all. It is on occasion useful when written as if(cond) perror(“an error occured”), exit(1) ; — then you don’t need curly braces. But it’s an invitation to disaster. The comma operator is to put two or more expressions in a position where the reference only allows one. … Read more
In the expression (i, ++i, 1), the comma used is the comma operator the comma operator (represented by the token ,) is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type). Because it discards its first operand, it is generally … Read more
Section 6.6/3, “Constant expressions”, of the ISO C99 standard is the section you need. It states: Constant expressions shall not contain assignment, increment, decrement, function-call, or comma operators, except when they are contained within a subexpression that is not evaluated. In the C99 rationale document from ISO, there’s this little snippet: An integer constant expression … Read more
The comma operator evaluates both of its operands (from left to right) and returns the value of the second operand. Source: https://developer.mozilla.org/en/JavaScript/Reference/Operators/Special_Operators/Comma_Operator For example, the expression 1,2,3,4,5 evaluates to 5. Obviously the comma operator is useful only for operations with side-effects. console.log(1,2,3,4,5); console.log((1,2,3,4,5));
The comma operator evaluates both of its operands (from left to right) and returns the value of the second operand. Source: https://developer.mozilla.org/en/JavaScript/Reference/Operators/Special_Operators/Comma_Operator For example, the expression 1,2,3,4,5 evaluates to 5. Obviously the comma operator is useful only for operations with side-effects. console.log(1,2,3,4,5); console.log((1,2,3,4,5));
The following is probably not very useful as you don’t write it yourself, but a minifier can shrink code using the comma operator. For example: if(x){foo();return bar()}else{return 1} would become: return x?(foo(),bar()):1 The ? : operator can be used now, since the comma operator (to a certain extent) allows for two statements to be written … Read more