列表中只有单元素时 reduce 的处理方式?

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#1 Created at ... [Delete] [Delete and Lock User]

def add(x,y): print(x); print(y); print('----------'); return x+y;

print(reduce(add, [1])); # 只输出1; 没有 ------- 这个是为什么?

#2 Created at ... [Delete] [Delete and Lock User]

def add(x,y): print(x); print(y); print('----------'); return (x+y) * 2;

print(reduce(add, [1])); # 还是输出1; 说明对于只有一个元素的list, reduce 直接返回了这个元素; 而不会去调用传入函数的处理逻辑;

#3 Created at ... [Delete] [Delete and Lock User]

那就有一个问题了, 对于 "1" 这样的字符串, 前面那个字符串转换整数估计不能成功, 还是会返回字符串;

#4 Created at ... [Delete] [Delete and Lock User]

奇怪了, str2int("9") 还是转换成为了整数 9 , 我琢磨一下.

#5 Created at ... [Delete] [Delete and Lock User]

额, map(char2num, s) 这儿已经转换成为了数字;