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| author | Andrew Banchich <andrewbanchich@gmail.com> | 2019-04-11 16:12:04 -0400 |
|---|---|---|
| committer | GitHub <noreply@github.com> | 2019-04-11 16:12:04 -0400 |
| commit | aefc1581b125876f37c652f84fddc81b0b25fe12 (patch) | |
| tree | 217e398d6ccc1fa1300b17043a037dfe138d962a | |
| parent | 3de0106789468b211bcc3a25c09c0cf07119186d (diff) | |
| download | rust-aefc1581b125876f37c652f84fddc81b0b25fe12.tar.gz rust-aefc1581b125876f37c652f84fddc81b0b25fe12.zip | |
Update diagnostics.rs
Add `a` and other minor text improvements
| -rw-r--r-- | src/librustc_mir/diagnostics.rs | 4 |
1 files changed, 2 insertions, 2 deletions
diff --git a/src/librustc_mir/diagnostics.rs b/src/librustc_mir/diagnostics.rs index e1b66312da2..c8836fe5193 100644 --- a/src/librustc_mir/diagnostics.rs +++ b/src/librustc_mir/diagnostics.rs @@ -2307,10 +2307,10 @@ let q = *p; ``` Here, the expression `&foo()` is borrowing the expression -`foo()`. As `foo()` is call to a function, and not the name of +`foo()`. As `foo()` is a call to a function, and not the name of a variable, this creates a **temporary** -- that temporary stores the return value from `foo()` so that it can be borrowed. -So you might imagine that `let p = bar(&foo())` is equivalent +You could imagine that `let p = bar(&foo());` is equivalent to this: ```compile_fail,E0597 |
