, 148 6.4.2 États mémoire symboliques abstraits, p.154
, alors : si n x = puits alors retourner segf, si n x = null retourner le même ems Sinon on transforme G de façon légèrement différente du cas concret. Les noeuds n tels que s(n) = n x vérifient alors s(n) = puits. Soit k x = c(n x ) et k ? c(N ) un nouveau compteur. Alors c (n x ) = k , et ? = ?k x .(? ? k = k x ? 1) Il faut normaliser l'arc (n x , s(n x )) Le résultat f a (ems) est minimal
, ? si g est de la forme IsNull(x) alors ems = f a ((G, c, ?| IsNull(x) )), ? si g est de la forme ¬IsNull(x) alors ems = f a ((G, c, ?| ¬IsNull(x) )), ? si ems = segf ou ems = ? alors ems = ? ; ? pour les ems généraux
, on veut décider s'il existe un graphe mémoire G ? EM S tel que G soit une fuite mémoire
, La concrétisation d'un ems atomique minimal ems = (G, c, ?) contient des fuites mémoire si et seulement si G est une fuite mémoire
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