6982: Remove parentheses when inverting `!(cond)` r=matklad a=Jesse-Bakker

Followup to #6894

When inverting a composite condition twice, the parentheses were left. This also removes those unnecessary parentheses when applying the invert-if assist.

Co-authored-by: Jesse Bakker <github@jessebakker.com>
This commit is contained in:
bors[bot] 2020-12-21 16:57:49 +00:00 committed by GitHub
commit c8a73fe655
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2 changed files with 16 additions and 1 deletions

View file

@ -77,6 +77,15 @@ mod tests {
)
}
#[test]
fn invert_if_remove_not_parentheses() {
check_assist(
invert_if,
"fn f() { i<|>f !(x == 3 || x == 4 || x == 5) { 3 * 2 } else { 1 } }",
"fn f() { if x == 3 || x == 4 || x == 5 { 1 } else { 3 * 2 } }",
)
}
#[test]
fn invert_if_remove_inequality() {
check_assist(

View file

@ -232,7 +232,13 @@ fn invert_special_case(expr: &ast::Expr) -> Option<ast::Expr> {
};
Some(make::expr_method_call(receiver, method, arg_list))
}
ast::Expr::PrefixExpr(pe) if pe.op_kind()? == ast::PrefixOp::Not => pe.expr(),
ast::Expr::PrefixExpr(pe) if pe.op_kind()? == ast::PrefixOp::Not => {
if let ast::Expr::ParenExpr(parexpr) = pe.expr()? {
parexpr.expr()
} else {
pe.expr()
}
}
// FIXME:
// ast::Expr::Literal(true | false )
_ => None,