Use new unique_append function signature

master
Aadhavan Srinivasan 2 months ago
parent aee04cd8fe
commit aee24644e9

@ -55,7 +55,7 @@ func verifyLastStates(start []*State) {
func concatenate(s1 *State, s2 *State) *State {
for i := range s1.output {
s1.output[i].transitions[s2.content] = unique_append(s1.output[i].transitions[s2.content], s2)
s1.output[i].transitions[s2.content], _ = unique_append(s1.output[i].transitions[s2.content], s2)
}
s1.output = s2.output
return s1
@ -69,9 +69,9 @@ func kleene(s1 State) *State {
toReturn.isKleene = true
toReturn.output = append(toReturn.output, toReturn)
for i := range s1.output {
s1.output[i].transitions[toReturn.content] = unique_append(s1.output[i].transitions[toReturn.content], toReturn)
s1.output[i].transitions[toReturn.content], _ = unique_append(s1.output[i].transitions[toReturn.content], toReturn)
}
toReturn.transitions[s1.content] = unique_append(toReturn.transitions[s1.content], &s1)
toReturn.transitions[s1.content], _ = unique_append(toReturn.transitions[s1.content], &s1)
return toReturn
}
@ -85,8 +85,8 @@ func alternate(s1 *State, s2 *State) *State {
// For example, given the transition 'a', the state 's1' can only be mentioned once.
// This would lead to multiple instances of the same set of match indices, since both
// 's1' states would be considered to match.
toReturn.transitions[s1.content] = unique_append(toReturn.transitions[s1.content], s1)
toReturn.transitions[s2.content] = unique_append(toReturn.transitions[s2.content], s2)
toReturn.transitions[s1.content], _ = unique_append(toReturn.transitions[s1.content], s1)
toReturn.transitions[s2.content], _ = unique_append(toReturn.transitions[s2.content], s2)
toReturn.content = EPSILON
toReturn.isEmpty = true

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