本文整理汇总了TypeScript中@jonggrang/parsing.option函数的典型用法代码示例。如果您正苦于以下问题：TypeScript option函数的具体用法？TypeScript option怎么用？TypeScript option使用的例子？那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了option函数的5个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的TypeScript代码示例。
示例1:
export const uri: PS.Parser<URI> = PS.co(function* () {
const us: string = yield PS.attempt(uscheme);
const [ua, up]: [Maybe<URIAuth>, string] = yield hierPart;
const uq: string = yield PS.option('', PS.char('?').chain(() => uquery));
const uf: string = yield PS.option('', PS.char('#').chain(() => ufragment));
return PS.pure(mkURI(us, ua, up, uq, uf));
});示例2: notMatching
export const relativeRef: PS.Parser<URI> = PS.co(function* () {
yield notMatching(uscheme);
const [ua, path]: [Maybe<URIAuth>, string] = yield relativePart;
const uq: string = yield PS.option('', PS.char('?').chain(() => uquery));
const uf: string = yield PS.option('', PS.char('#').chain(() => ufragment));
return PS.pure(mkURI('', ua, path, uq, uf));
});示例3: optNH4cH4
function optNH4cH4(n: number): PS.Parser<string> {
return PS.option('', PS.co(function* () {
const a1: string = yield countMinMax(0, n, h4c).map(joinStr);
const a2: string = yield h4;
return PS.pure(a1 + a2);
}));
}示例4: countMinMax
function countMinMax<A>(m: number, n: number, p: PS.Parser<A>): PS.Parser<L.List<A>> {
if (m > 0) {
return p.chain(x => countMinMax(m - 1, n - 1, p).map(ar => L.cons(x, ar)));
} else if (n <= 0) {
return PS.pure(L.nil);
}
return PS.option(L.nil, p.chain(x => countMinMax(0, n - 1, p).map(ar => L.cons(x, ar))));
}示例5: digits
const time: PS.Parser<[number, number, number]> = PS.co(function* () {
let h = yield digits(2, 2);
yield PS.char(':');
let m = yield digits(2, 2);
// the seconds fields is optional
let s = yield PS.option(0, PS.attempt(PS.char(':').chain(() => digits(2, 2))));
if (h > 23 || m > 59 || s > 59) {
return PS.fail('hour, minute and seconds must be in valid range');
}
return PS.pure([h, m, s]);
});