Why does increment operation like “a[i] = i++;” re

Question

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Closed 8 years ago.

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Undefined Behavior and Sequence Points

#include <iostream>
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 1

Clearly explained here: C-Faq

Why doesn't this code: a[i] = i++; work?

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.)

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.

Answer 2

You are modifying a variable and using its value without an intervening sequence point. What do you expect is the value of i when x[i] appears? Because whatever you expect, you'd be wrong to expect that.