Answer 1

@jefferyE 匿名函数是可以递归调用的，只不过在严格模式下会报错

(function(counter){
	if (counter) {
		return;
	}
	return arguments.callee(1);
})()

Answer 2

all(list) {
        return new Promise((resolve, reject) => {
            let resValues = [];
            let counts = 0;
            for (let [i, p] of list) {
                resolve(p).then(res => {
                    counts++;
                    resValues[i] = res;
                    if (counts === list.length) {
                        resolve(resValues)
                    }
                }, err => {
                    reject(err)
                })
            }
        })
    }

你这里直接resolve(p)不对吧