is this undefined behavior of printf function or not?

Question

 int c=2; printf("%d %d %d",c,c<<=2,c>>=2);

Is this undefined behavior ?

 int c=2;
 printf ("%d %d %d",c,c<<=2,c>>=2);

Yes, `c` is modified multiple times without a sequence point in between. — EOF, Jun 06 '16 at 18:26
It has nothing to do with printf. It would be UB in any function call. — rici, Jun 06 '16 at 18:30
This... _again_? **Why** do the teachers in your country obsess over this meaningless garbage? Mind you, at least they're apparently teaching its pitfalls. — Lightness Races in Orbit, Jun 06 '16 at 18:32
@SouravGhosh sorry actually c is initialize with 2 . i forgot to write . — VikAsh KuMar, Jun 06 '16 at 18:33
@rici I see now that the evaluation of the format is sequenced, not the arguments passed to `printf()`. Ref [Sequence Points in printf](http://stackoverflow.com/q/34637712/2410359) — chux - Reinstate Monica, Jun 06 '16 at 18:42

score 0 · Answer 1 · answered Jun 06 '16 at 18:39

The behaviour is undefined.

[C99: 4/2]: If a ‘‘shall’’ or ‘‘shall not’’ requirement that appears outside of a constraint is violated, the behavior is undefined. Undefined behavior is otherwise indicated in this International Standard by the words ‘‘undefined behavior’’ or by the omission of any explicit definition of behavior. There is no difference in emphasis among these three; they all describe ‘‘behavior that is undefined’’.

[C99: 5.1.2.3/2]: Accessing a volatile object, modifying an object, modifying a file, or calling a function that does any of those operations are all side effects,11) which are changes in the state of the execution environment. Evaluation of an expression may produce side effects. At certain specified points in the execution sequence called sequence points, all side effects of previous evaluations shall be complete and no side effects of subsequent evaluations shall have taken place. (A summary of the sequence points is given in annex C.)

[C99: C/1]: The following are the sequence points described in 5.1.2.3: - The call to a function, after the arguments have been evaluated (6.5.2.2). - The end of the first operand of the following operators: logical AND && (6.5.13); logical OR || (6.5.14); conditional ? (6.5.15); comma , (6.5.17). - The end of a full declarator: declarators (6.7.5); - The end of a full expression: an initializer (6.7.8); the expression in an expression statement (6.8.3); the controlling expression of a selection statement (if or switch) (6.8.4); the controlling expression of a while or do statement (6.8.5); each of the expressions of a for statement (6.8.5.3); the expression in a return statement (6.8.6.4). - Immediately before a library function returns (7.1.4). - After the actions associated with each formatted input/output function conversion specifier (7.19.6, 7.24.2). - Immediately before and immediately after each call to a comparison function, and also between any call to a comparison function and any movement of the objects passed as arguments to that call (7.20.5).

[C99: 6.2.2.2/10]: The order of evaluation of the function designator, the actual arguments, and subexpressions within the actual arguments is unspecified, but there is a sequence point before the actual call.

And, although I can't prove a negative, there is no wording to make an exception for the varargs case.

is this undefined behavior of printf function or not?

1 Answers1