How are arguments passed in small brackets processed in C?

Question

For the following code:

int main(void) {
    int x = 1000, y = 5000;
    printf ("%d\n", x, y);
    printf ("%d\n", (x, y));
    return 0;
}

Output: 1000 5000

Can somebody please explain this?

@AntonH: How is it undefined behavior? Passing to few arguments to `printf` causes undefined behavior. If you pass to many arguments, the extra ones are evaluated and then discarded. — Keith Thompson, Mar 10 '14 at 18:02

LearningC · Accepted Answer · 2014-03-10T18:32:35.920

2

its about operator precedence.
in case of (x,y) first statement inside () is evaluated so the last value y is taken as result from (). without () all comma operators have equal precedence so evaluation takes from left to right so x value is taken for printf()

edited Mar 10 '14 at 18:32

answered Mar 10 '14 at 17:59

LearningC

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score 2 · Answer 2 · answered Mar 10 '14 at 18:00

Parentheses make (x, y) a single expression, composed of x and y with a comma operator. The operator evaluates x, throws the value away, evaluates y, and make it the value of the expression.

Since evaluating variable x has no side effect, the expression (x, y) in this case is equivalent to y passed by itself:

printf ("%d\n", y);

Note: if your first printf used a "%d %d" format string, you would see both x and y.

score 1 · Answer 3 · answered Mar 10 '14 at 18:00

1

For the first printf, you are just passing a second argument that is being ignored.

For the second, you are invoking the comma operator which evaluates both it's arguments, and returns the value of the second.

answered Mar 10 '14 at 18:00

Linuxios

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score 1 · Answer 4 · answered Mar 10 '14 at 18:01

1

What you're seeing here is the evaluation of the C comma operator.
http://en.wikipedia.org/wiki/Comma_operator

The little used comma operator is often used within the third part of a for() loop to take multiple actions on increment, but technically it's a valid expression anywhere.

answered Mar 10 '14 at 18:01

erik258

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3

It's also commonly used in macro definitions. – Keith Thompson Mar 10 '14 at 18:03

score 0 · Answer 5 · answered Mar 10 '14 at 18:11

In the first call of function printf the number of arguments exceeds the number of format specifiers in the format string

printf ("%d\n", x, y);

In this case the second argument that is y will be ignored. Only x will be outputed. So the output of the call is

In the second call of function printf there is only one argument that is an expression enclosed in parentheses.

printf ("%d\n", (x, y));

This expression is an expression of the comma operator. At first the first operand that is x is evaluated . Its value is ignored. And then the second operand of the expression that is y is evaluated. Its value is used as the result of the full expression. So this function call outputs

How are arguments passed in small brackets processed in C?

5 Answers5