Method ProduceInput

src/script/miniscript.h:1247–1492

    template<typename Ctx>
    internal::InputResult ProduceInput(const Ctx& ctx) const {
        using namespace internal;

        // Internal function which is invoked for every tree node, constructing satisfaction/dissatisfactions
        // given those of its subnodes.
        auto helper = [&ctx](const Node& node, std::span<InputResult> subres) -> InputResult {
            switch (node.fragment) {
                case Fragment::PK_K: {
                    std::vector<unsigned char> sig;
                    Availability avail = ctx.Sign(node.keys[0], sig);
                    return {ZERO, InputStack(std::move(sig)).SetWithSig().SetAvailable(avail)};
                }
                case Fragment::PK_H: {
                    std::vector<unsigned char> key = ctx.ToPKBytes(node.keys[0]), sig;
                    Availability avail = ctx.Sign(node.keys[0], sig);
                    return {ZERO + InputStack(key), (InputStack(std::move(sig)).SetWithSig() + InputStack(key)).SetAvailable(avail)};
                }
                case Fragment::MULTI_A: {
                    // sats[j] represents the best stack containing j valid signatures (out of the first i keys).
                    // In the loop below, these stacks are built up using a dynamic programming approach.
                    std::vector<InputStack> sats = Vector(EMPTY);
                    for (size_t i = 0; i < node.keys.size(); ++i) {
                        // Get the signature for the i'th key in reverse order (the signature for the first key needs to
                        // be at the top of the stack, contrary to CHECKMULTISIG's satisfaction).
                        std::vector<unsigned char> sig;
                        Availability avail = ctx.Sign(node.keys[node.keys.size() - 1 - i], sig);
                        // Compute signature stack for just this key.
                        auto sat = InputStack(std::move(sig)).SetWithSig().SetAvailable(avail);
                        // Compute the next sats vector: next_sats[0] is a copy of sats[0] (no signatures). All further
                        // next_sats[j] are equal to either the existing sats[j] + ZERO, or sats[j-1] plus a signature
                        // for the current (i'th) key. The very last element needs all signatures filled.
                        std::vector<InputStack> next_sats;
                        next_sats.push_back(sats[0] + ZERO);
                        for (size_t j = 1; j < sats.size(); ++j) next_sats.push_back((sats[j] + ZERO) | (std::move(sats[j - 1]) + sat));
                        next_sats.push_back(std::move(sats[sats.size() - 1]) + std::move(sat));
                        // Switch over.
                        sats = std::move(next_sats);
                    }
                    // The dissatisfaction consists of as many empty vectors as there are keys, which is the same as
                    // satisfying 0 keys.
                    auto& nsat{sats[0]};
                    CHECK_NONFATAL(node.k != 0);
                    assert(node.k < sats.size());
                    return {std::move(nsat), std::move(sats[node.k])};
                }
                case Fragment::MULTI: {
                    // sats[j] represents the best stack containing j valid signatures (out of the first i keys).
                    // In the loop below, these stacks are built up using a dynamic programming approach.
                    // sats[0] starts off being {0}, due to the CHECKMULTISIG bug that pops off one element too many.
                    std::vector<InputStack> sats = Vector(ZERO);
                    for (size_t i = 0; i < node.keys.size(); ++i) {
                        std::vector<unsigned char> sig;
                        Availability avail = ctx.Sign(node.keys[i], sig);
                        // Compute signature stack for just the i'th key.
                        auto sat = InputStack(std::move(sig)).SetWithSig().SetAvailable(avail);
                        // Compute the next sats vector: next_sats[0] is a copy of sats[0] (no signatures). All further
                        // next_sats[j] are equal to either the existing sats[j], or sats[j-1] plus a signature for the
                        // current (i'th) key. The very last element needs all signatures filled.