#include
using namespace std;
int main()
{
int x[3] = {};
int i=0;
x[i] = i++;
cout << x[0] << " " << x[1] << endl;
return 0;
}
Codepad is giving me this: Line 9: warning: operation on 'i' may be undefined
Why is the operation undefined?
Answer
Clearly explained here: C-Faq
Why doesn't this code: a[i] = i++;
work?
The subexpression
i++
causes a side effect--it modifiesi'
s value--which leads to undefined behavior sincei
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.)
Relevant Standard Quotation is as follows:
C++03 5 Expressions [expr]:
Para 4:
....
Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be accessed only to determine the value to be stored. The requirements of this paragraph shall be met for each allowable ordering of the subexpressions of a full
expression; otherwise the behavior is undefined.
No comments:
Post a Comment