output of this c program is weird ,instead of value -4 at second position third line it shows -5 how?

Question

#include<stdio.h>
int main()
{
    int a=5,b=9;
    char ch='c';
    printf("%c %c\n",ch,++ch);
    printf("%d %d\n",a,!!a);
    printf("%d %d %d\n",a=a+2,-a + ++b%a,a<<2);//
    printf("%d",a);
    return 0;
}



/* 
    op
    d d
    5 1
    7 -5 20
    7
*/

output of this c program is weird ,instead of value -4 at second position third line it shows -5 how? i am using devc++IDE

Answer 1

The program could actually show any value or even crash (undefined behaviour), since you are modifying a and using it a second time within the same sequence point. You will have to move the a = a+2 instruction before or after the printf invocation. Same should be done with the ++ch instruction. See this question for further info.

EDIT While the linked question covers the case up to C99, the C11 standard was worded differently. Now the relevant section is §6.5/2.

"If a side effect on a scalar object is unsequenced relative to either a different side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined. If there are multiple allowable orderings of the subexpressions of an expression, the behavior is undefined if such an unsequenced side effect occurs in any of the orderings."

Here the expression is the whole printf invocation, and the "value computation using the value of the same scalar object" is the use of ch in an additional place in the parameter list, other than the place in which it is changed. Since the evaluations of the parameters are unsequenced (§6.5/3), the "multiple allowable orderings" clause applies.

EDIT 2 Because of §6.5.2.2/10, what precedes might be not appliable if one called f(&ch) (with f altering the stored value) instead of ++ch. I am however not totally sure about this.

Answer 2

Different from the first thoughts, this program does actually not show undefined behaviour, as each expression doe not violate (C11, 6.5/2). In brief: each expression only has one side-effect and each expression per se is clearly defined (each would very well work as expression statements or the right side of an assignment).

Suspect are actually just these lines:

printf("%c %c\n",ch,++ch);
printf("%d %d %d\n",a=a+2,-a + ++b%a,a<<2);//

After each sequence point (roughly: function call), all variables are well-defined.

However, the actual values passed to printf vary, as the sequence the expressions for the arguments are evaluated is not defined (5.1.2.3/2 - indeterminately sequenced). So, for the first line, either ch or ch+1 is passed as the first argument, depending which argument expression is evaluated first. The second argument is, however always ch+1.

Comments welcome, please cite the standard, no second/3rd source. Yes, that is a gauntlet thrown down.

Disclaimer: this code is of course crap. This I will not discuss. It is, however, a good exercise for students exactly to enforce the question as asked. (wereas I wonder if there is no other like this already - the ones given in the comments are actually not, as they do exhibit UB). It is vital actually to sensitize programmers for this problem, which is very specific to C-like languages as those do allow side-effects for expressions.

Edit:

After further thinking, I am not that sure anymore. As my chain of prove is missing a link which I cannot find (lack of understanding the english phrases or just being too stupid), I will accept to treat this as UB now. While still in doubt, the behavior is bad however it would be called and has no application beyond as an example for - at least - ambivalent coding. I leave it here in hope, someone might stumble around and show up the final solution.

Thanks @AlbertoM for watering the seed of doubt.