For an expression expr of the form expr-first && expr-second: The definite assignment state of v before expr-first is the same as the definite assignment state of v before expr. The definite assignment state of v before expr-second is definitely assigned if the state of v after expr-first is either definitely assigned or "definitely assigned after true expression". Otherwise, it is not definitely assigned. The definite assignment state of v after expr is determined by: If the state of v after expr-first is definitely assigned, then the state of v after expr is definitely assigned. Otherwise, if the state of v after expr-second is definitely assigned, and the state of v after expr-first is "definitely assigned after false expression", then the state of v after expr is definitely assigned. Otherwise, if the state of v after expr-second is definitely assigned or "definitely assigned after true expression", then the state of v after expr is "definitely assigned after true expression". Otherwise, if the state of v after expr-first is "definitely assigned after false expression", and the state of v after expr-second is "definitely assigned after false expression", then the state of v after expr is "definitely assigned after false expression". Otherwise, the state of v after expr is not definitely assigned. [Example: In the example = 0 && (i = y) >= 0) { // i definitely assigned } else { // i not definitely assigned } // i not definitely assigned } } ]]>the variable i is considered definitely assigned in one of the embedded statements of an if statement but not in the other. In the if statement in method F, the variable i is definitely assigned in the first embedded statement because execution of the expression (i = y) always precedes execution of this embedded statement. In contrast, the variable i is not definitely assigned in the second embedded statement, since x >= 0 might have tested false, resulting in the variable i's being unassigned. end example]