simplify slice expr (avoid len)

[notJoon]

Simplify expression when specifying lengths with the len function inside a slice.

Handling these cases:

slice[:len(slice)] → slice[:]
slice[n:len(slice)] → slice[n:] // n: number literal
slice[a:len(slice)] → slice[a:] // a: variable

Example

package slice

var slice = []int{1, 2, 3}

func foo() {
	_ = slice[1:len(slice)]
}

func bar() {
	_ = slice[:len(slice)]
}

func baz() {
	a := 1
	_ = slice[a:len(slice)]
}

Linter result:

error: simplify-slice-range
 --> testdata/slice0.gno
  |
6 |         _ = slice[1:len(slice)]
  |      ^ unnecessary use of len() in slice expression, can be simplified

Suggestion:
6 | slice[1:]

Note: unnecessary use of len() in slice expression, can be simplified
Here, `slice[1:len(slice)]` can be simplified to `slice[1:]`. When slicing to the end of a slice, using len() is unnecessary.

error: simplify-slice-range
 --> testdata/slice0.gno
   |
10 |         _ = slice[:len(slice)]
   |      ^ unnecessary use of len() in slice expression, can be simplified

Suggestion:
10 | slice[:]

Note: unnecessary use of len() in slice expression, can be simplified
In this case, `slice[:len(slice)]` is equivalent to `slice[:]`. The full length of the slice is already implied when omitting both start and end indices.

error: simplify-slice-range
 --> testdata/slice0.gno
   |
15 |         _ = slice[a:len(slice)]
   |      ^ unnecessary use of len() in slice expression, can be simplified

Suggestion:
15 | slice[a:]

Note: unnecessary use of len() in slice expression, can be simplified
In this instance, `slice[a:len(slice)]` can be written as `slice[a:]`. The len() function is redundant when slicing to the end, regardless of the start index.