Starting from a simple case of "fold" (I used (+) but can be anything else):

```
Prelude.foldl (+) 0 [10,20,30]
```

is it possible apply an inline transformation similar to (that doesn't work):

```
Prelude.foldl ((+) . (\x -> read x :: Int)) 0 ["10","20","30"]
```

In case not, is there an alternative to fold, to apply a generic function and inline transformation (apart from using specific functions like 'sum', 'max' etc)?

Answer:

The `read`

lambda applies to the first argument and the first argument to the function given to `foldl`

is the accumulator. Those two arguments are the opposite for `foldr`

. So, expanded, it looks like this:

```
foldl (\acc element -> (read acc :: Int) + element) 0 ["10", "20", "30"]
```

Since `acc`

is an `Int`

, this doesn't work.

So, with this information in hand, you can do this with `foldr`

since it has the opposite argument order:

```
foldr ((+) . (read :: String -> Int)) 0 ["10","20","30"]
```

If you want an inline `foldl`

version, you can use `flip`

to achieve this.

You could also use `map`

first (everything else being equal, this would be the solution I'd prefer):

```
foldl (+) 0 $ map (read :: String -> Int) ["10","20","30"]
```

Also, you probably want `foldl'`

instead of `foldl`

.

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