What will be the output of the following code? int main () { int x = 4, y = 0;…

Concept: sequence points and unsequenced modifications

Between two sequence points, the C standard allows a scalar object to be modified at most once; if its old value is also read in the same span, that read must only be used to compute the new value. Ordinary operators such as + do not create a sequence point between their two operands — the compiler is free to evaluate the left or right operand first, and their side effects (increments/decrements) are not ordered relative to each other. Only a small set of operators (&&, ||, the comma operator, and the ?: conditional) impose an evaluation order together with a sequence point.

Application: tracing z = (x++ + ++y + y++ , x++);

1. The parentheses hold a single comma expression: a left operand E1 = x++ + ++y + y++, and a right operand E2 = x++. The comma is one of the operators that DOES insert a sequence point, so all of E1's side effects are guaranteed to complete before E2 begins.

2. Look inside E1 itself: it is (x++) + (++y) + (y++). This is two applications of + chained together, and C leaves the order in which the three operands are evaluated unspecified — there is no sequence point between them.

3. Both ++y and y++ write to the same object, y, and both sit as operands of + inside E1, with no sequence point separating those two writes. That is exactly the case the standard forbids: y is modified more than once between sequence points.

4. Because that constraint is violated, evaluating E1 is undefined behaviour. A compiler is free to compute the value of E1 (and even the internal state of x and y afterward) in more than one legitimate way — nothing about the code guarantees a single answer for what z ends up holding once E2 is evaluated and assigned.

Cross-check

Contrast this with z = (x++, y++, x + y);, where every side effect is separated from the next by an explicit comma — that version IS well-defined, because each operand of the comma operator is fully sequenced before the next one starts. The difference here is that the two writes to y (++y and y++) sit on the SAME side of the only comma in this expression, inside one additive sub-expression, so nothing sequences them against each other. Mainstream compilers (e.g. GCC with -Wall) flag lines like this with an 'operation on y may be undefined' warning for precisely this reason — confirming that no single fixed output like 5 can be relied upon.