Why is a = (a+b) - (b=a) a bad choice for swapping two integers?

Question

I stumbled into this code for swapping two integers without using a temporary variable or the use of bitwise operators.

int main(){

    int a=2,b=3;
    printf("a=%d,b=%d",a,b);
    a=(a+b)-(b=a);
    printf("\na=%d,b=%d",a,b);
    return 0;
}

But I think this code has undefined behavior in the swap statement a = (a+b) - (b=a); as it does not contain any sequence points to determine the order of evaluation.

My question is: Is this an acceptable solution to swap two integers?

Answer 1

No. This is not acceptable. This code invokes Undefined behavior. This is because of the operation on b is not defined. In the expression

a=(a+b)-(b=a);  

it is not certain whether b gets modified first or its value gets used in the expression (a+b) because of the lack of the sequence point.
See what standard syas:

C11: 6.5 Expressions:

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.84)₁.

Read C-faq- 3.8 and this answer for more detailed explanation of sequence point and undefined behavior.

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_{1. Emphasis is mine.}