两种风格的代码有何不同啊？

Question

winSSQ count noRed noBlue = do {
    let yesRed =  [1..33] \ noRed;
    let yesBlue = [1..16] \ noBlue;
    bracket (openFile "ssqNum.txt" WriteMode) (hClose) (hd1 -> pickSSQ count yesRed yesBlue hd1);
}
将报错: parse error on input `let' !

但下面代码却正确！

winSSQ count noRed noBlue = do
    let yesRed =  [1..33] \ noRed
    let yesBlue = [1..16] \ noBlue
    bracket (openFile "ssqNum.txt" WriteMode) (hClose) (hd1 -> pickSSQ count yesRed yesBlue hd1)

这是什么原因呢？

Answer 1

winSSQ count noRed noBlue = do {
    let {yesRed=[1..33] \ noRed; yesBlue=[1..16] \ noBlue};
    bracket (openFile "ssqNum.txt" WriteMode) (hClose) (hd1 -> pickSSQ count yesRed yesBlue hd1);
}
这样就可以了！

Answer 2

帮flw发的~

Answer 3

因为这里本来是有语法糖的，你用了花括号之后，语法糖不见了，然后就有语法错误了。

写成这样就对了（这也是第二种写法的展开形式）：

winSSQ count noRed noBlue = do {
    let yesRed =  [1..33] \ noRed;
    in let yesBlue = [1..16] \ noBlue;
        in bracket (openFile "ssqNum.txt" WriteMode) (hClose) (hd1 -> pickSSQ count yesRed yesBlue hd1);
}