英文:
Why are there problems using map in an IF statement?
问题
为什么以下代码无法运行？
if (map [ [a b c] -> a + b = c ] [1] [2] [3]) [show 100]
以下代码的输出结果为'true'：
show (map [ [a b c] -> a + b = c ] [1] [2] [3])
所以我期望上面的第一个语句与下面的语句相同：
if true [show 100]
（附注：在我的完整版本中，列表更长，但使用reduce折叠成单个true/false。）
谢谢。
英文:
Why won't the following run?
if (map [ [a b c] -> a + b = c ] [1] [2] [3]) [show 100]
The following produces 'true' as an output:
show (map [ [a b c] -> a + b = c ] [1] [2] [3])
so I expected the first statement above to be the same as:
if true [show 100]
(P.S. in my full version the lists are longer but are collapsed into a single true/false using reduce.)
Thanks.
答案1
得分: 1
为了详细说明ThomasC的评论，map函数总是生成一个列表，即使它只有一个元素。
因此，以下代码：
(map [ [a b c] -> a + b = c ] [1 2] [2 3] [3 5])
会生成[true true]。在这里，reduce函数很有帮助：
(reduce AND (map [ [a b c] -> a + b = c ] [1 2] [2 3] [3 5]))
通过"与"所有map输出的元素，会生成一个简单的true。
而以下代码：
(reduce AND (map [ [a b c] -> a + b = c ] [1 2] [2 3] [3 6]))
会生成一个简单的false。
英文:
To elaborate on ThomasC's comment, map always produces a list, even if it only has one element. So
(map [ [a b c] -> a + b = c ] [1] [2] [3])
does produce [true]. Thus
(map [ [a b c] -> a + b = c ] [1 2] [2 3] [3 5])
will produce [true true]. reduce is helpful here,
reduce AND (map [ [a b c] -> a + b = c ] [1 2] [2 3] [3 5])
will produce a simple true by "anding" all the elements of the map output, and
reduce AND (map [ [a b c] -> a + b = c ] [1 2] [2 3] [3 6])
will produce a simple false.
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