ref: bb16b5a70ddf77d03f1e985cb761ad5aa6c19b2c
dir: /tdq.c/
/* * tdq.c: implement a 'to-do queue', a simple de-duplicating to-do * list mechanism. */ #include <assert.h> #include "puzzles.h" /* * Implementation: a tdq consists of a circular buffer of size n * storing the integers currently in the queue, plus an array of n * booleans indicating whether each integer is already there. * * Using a circular buffer of size n to store between 0 and n items * inclusive has an obvious failure mode: if the input and output * pointers are the same, how do you know whether that means the * buffer is full or empty? * * In this application we have a simple way to tell: in the former * case, the flags array is all 1s, and in the latter case it's all * 0s. So we could spot that case and check, say, flags[0]. * * However, it's even easier to simply determine whether the queue is * non-empty by testing flags[buffer[op]] - that way we don't even * _have_ to compare ip against op. */ struct tdq { int n; int *queue; int ip, op; /* in pointer, out pointer */ bool *flags; }; tdq *tdq_new(int n) { int i; tdq *tdq = snew(struct tdq); tdq->queue = snewn(n, int); tdq->flags = snewn(n, bool); for (i = 0; i < n; i++) { tdq->queue[i] = 0; tdq->flags[i] = false; } tdq->n = n; tdq->ip = tdq->op = 0; return tdq; } void tdq_free(tdq *tdq) { sfree(tdq->queue); sfree(tdq->flags); sfree(tdq); } void tdq_add(tdq *tdq, int k) { assert((unsigned)k < (unsigned)tdq->n); if (!tdq->flags[k]) { tdq->queue[tdq->ip] = k; tdq->flags[k] = true; if (++tdq->ip == tdq->n) tdq->ip = 0; } } int tdq_remove(tdq *tdq) { int ret = tdq->queue[tdq->op]; if (!tdq->flags[ret]) return -1; tdq->flags[ret] = false; if (++tdq->op == tdq->n) tdq->op = 0; return ret; } void tdq_fill(tdq *tdq) { int i; for (i = 0; i < tdq->n; i++) tdq_add(tdq, i); }