Does the statement `int val = (++i > ++j) ? ++i : ++j;` invoke undefined behavior?

Given the following program:

#include <stdio.h>
int main(void)
{
    int i = 1, j = 2;
    int val = (++i > ++j) ? ++i : ++j;
    printf("%d\n", val); // prints 4
    return 0;
}

The initialization of val seems like it could be hiding some undefined behavior, but I don't see any point at which an object is either modified more than once or modified and used without a sequence point in between. Could someone either correct or corroborate me on this?

The behavior of this code is well defined.

The first expression in a conditional is guaranteed to be evaluated before either the second expression or the third expression, and only one of the second or third will be evaluated. This is described in section 6.5.15p4 of the C standard:

The first operand is evaluated; there is a sequence point between its evaluation and the evaluation of the second or third operand (whichever is evaluated). The second operand is evaluated only if the first compares unequal to 0; the third operand is evaluated only if the first compares equal to 0; the result is the value of the second or third operand (whichever is evaluated), converted to the type described below.

In the case of your expression:

int val = (++i > ++j) ? ++i : ++j;

++i > ++j is evaluated first. The incremented values of i and j are used in the comparison, so it becomes 2 > 3. The result is false, so then ++j is evaluated and ++i is not. So the (again) incremented value of j (i.e. 4) is then assigned to val.