Q: Why doesn't this code:
a[i] = i++;work?
A: The subexpression i++ causes a side effect--it modifies i's value--which leads to undefined behavior since i is also referenced elsewhere in the same expression. There is no way of knowing whether the reference will happen before or after the side effect--in fact, neither obvious interpretation might hold; see question 3.9. (Note that although the language in K&R suggests that the behavior of this expression is unspecified, the C Standard makes the stronger statement that it is undefined--see question 11.33.)
References:
K&R1 Sec. 2.12
K&R2 Sec. 2.12
ISO Sec. 6.3
H&S Sec. 7.12 pp. 227-9
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