Comments (1)
Expression 2
If you remove pipe operator from your second expression
list2 |> Seq.iter (fun x -> list1 |> should contain x)
you will get
list2 |> Seq.iter (fun x -> should contain x list1)
Expression 1
In the same time, if you add param to the function from first expression
list2 |> Seq.iter (list1 |> (should contain) )
you will get
list2 |> Seq.iter (fun x -> (list1 |> (should contain)) x)
that is equal to
list2 |> Seq.iter (fun x -> (should contain list1) x)
that is equal to
list2 |> Seq.iter (fun x -> should contain list1 x)
Conclusion
Now it should be obvious that they are two different expressions:
list2 |> Seq.iter (fun x -> should contain list1 x)
list2 |> Seq.iter (fun x -> should contain x list1)
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Related Issues (20)
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