ANSI C 1999 TC3中的逗号运算符示例

Question

When I was trying to figure out, !0 's result is implementation defined expecting that it shall be unequal to zero I just read something what confused me.

(By the way may it be on some implementations -1 or 1 or is it strict defined? If any one could me tell in comment would be nice)

But my real question is:

in

6.5.17 Comma operator 2

is said:

If an attempt is made to modify the result of a comma operator or to access it after the next sequence point, the behavior is undefined.

In exactly the next line there is an example how to parse a parameter into a function with use of comma operator.

f(a, (t=3, t+2), c);

But the example is in point of my knowledge so far undefined behavior, isn't it? Since t gets assigned 3 and in the next sequence it gets increased by 2 .

But the standard doesn't mention that the example isn't valid.

Or is an assignment to be not understood as modification?

Answer 1

1. !0 evaluates to 1
2. In (t=3, t+2) , there is a sequence point between the assignment and the access to t . The expression is defined, it evaluates to 5 and leaves the value 3 in t . It would be undefined if there was no sequence point between the two, for instance, (t=3)+(t+2) .

Answer 2

I'm not sure what prompted you to ask the question. The section of the standard that you've picked up the example from clearly says:

As indicated by the syntax, the comma operator (as described in this subclause) cannot appear in contexts where a comma is used to separate items in a list (such as arguments to functions or lists of initializers). On the other hand, it can be used within a parenthesized expression or within the second expression of a conditional operator in such contexts. In the function call

  f(a, (t=3, t+2), c) 

the function has three arguments, the second of which has the value 5.

Emphasised the relevant portion to clarify your doubt.

Answer 3

逗号运算符引入了一个序列点，因此应明确定义行为（首先将t设置为3 ，然后将2加到t得到结果5但在t保留3 ）。