make never_loop applicability more flexible

Author: lapla-cogito

clippy_lints/src/loops/never_loop.rs

@@ -4,11 +4,13 @@ use clippy_utils::diagnostics::span_lint_and_then;
 use clippy_utils::higher::ForLoop;
 use clippy_utils::macros::root_macro_call_first_node;
 use clippy_utils::source::snippet;
+use clippy_utils::visitors::{Descend, for_each_expr_without_closures};
 use rustc_errors::Applicability;
 use rustc_hir::{Block, Destination, Expr, ExprKind, HirId, InlineAsmOperand, Pat, Stmt, StmtKind, StructTailExpr};
 use rustc_lint::LateContext;
 use rustc_span::{Span, sym};
 use std::iter::once;
+use std::ops::ControlFlow;
 
 pub(super) fn check<'tcx>(
     cx: &LateContext<'tcx>,
@@ -27,14 +29,19 @@ pub(super) fn check<'tcx>(
                     ..
                 }) = for_loop
                 {
-                    // Suggests using an `if let` instead. This is `Unspecified` because the
-                    // loop may (probably) contain `break` statements which would be invalid
-                    // in an `if let`.
+                    let app = if let Some(block_expr) = block.expr
+                        && !contains_any_break_or_continue(block_expr)
+                    {
+                        Applicability::MachineApplicable
+                    } else {
+                        Applicability::Unspecified
+                    };
+
                     diag.span_suggestion_verbose(
                         for_span.with_hi(iterator.span.hi()),
                         "if you need the first element of the iterator, try writing",
                         for_to_if_let_sugg(cx, iterator, pat),
-                        Applicability::Unspecified,
+                        app,
                     );
                 }
             });
@@ -43,6 +50,15 @@ pub(super) fn check<'tcx>(
     }
 }
 
+fn contains_any_break_or_continue(expr: &Expr<'_>) -> bool {
+    for_each_expr_without_closures(expr, |e| match e.kind {
+        ExprKind::Break(..) | ExprKind::Continue(..) => ControlFlow::Break(()),
+        ExprKind::Loop(_, None, ..) => ControlFlow::Continue(Descend::No),
+        _ => ControlFlow::Continue(Descend::Yes),
+    })
+    .is_some()
+}
+
 /// The `never_loop` analysis keeps track of three things:
 ///
 /// * Has any (reachable) code path hit a `continue` of the main loop?

tests/ui/never_loop_fixable.fixed

@@ -0,0 +1,20 @@
+#![allow(clippy::iter_next_slice, clippy::needless_return)]
+
+fn no_break_or_continue_loop() {
+    if let Some(i) = [1, 2, 3].iter().next() {
+        //~^ never_loop
+        return;
+    }
+}
+
+fn no_break_or_continue_loop_outer() {
+    if let Some(i) = [1, 2, 3].iter().next() {
+        //~^ never_loop
+        return;
+        loop {
+            if true {
+                continue;
+            }
+        }
+    }
+}

tests/ui/never_loop_fixable.rs

@@ -0,0 +1,20 @@
+#![allow(clippy::iter_next_slice, clippy::needless_return)]
+
+fn no_break_or_continue_loop() {
+    for i in [1, 2, 3].iter() {
+        //~^ never_loop
+        return;
+    }
+}
+
+fn no_break_or_continue_loop_outer() {
+    for i in [1, 2, 3].iter() {
+        //~^ never_loop
+        return;
+        loop {
+            if true {
+                continue;
+            }
+        }
+    }
+}

tests/ui/never_loop_fixable.stderr

@@ -0,0 +1,35 @@
+error: this loop never actually loops
+  --> tests/ui/never_loop_fixable.rs:4:5
+   |
+LL | /     for i in [1, 2, 3].iter() {
+LL | |
+LL | |         return;
+LL | |     }
+   | |_____^
+   |
+   = note: `#[deny(clippy::never_loop)]` on by default
+help: if you need the first element of the iterator, try writing
+   |
+LL -     for i in [1, 2, 3].iter() {
+LL +     if let Some(i) = [1, 2, 3].iter().next() {
+   |
+
+error: this loop never actually loops
+  --> tests/ui/never_loop_fixable.rs:11:5
+   |
+LL | /     for i in [1, 2, 3].iter() {
+LL | |
+LL | |         return;
+LL | |         loop {
+...  |
+LL | |     }
+   | |_____^
+   |
+help: if you need the first element of the iterator, try writing
+   |
+LL -     for i in [1, 2, 3].iter() {
+LL +     if let Some(i) = [1, 2, 3].iter().next() {
+   |
+
+error: aborting due to 2 previous errors
+