| A temporary value is being dropped while a borrow is still in active use. |
| |
| Erroneous code example: |
| |
| ```compile_fail,E0716 |
| fn foo() -> i32 { 22 } |
| fn bar(x: &i32) -> &i32 { x } |
| let p = bar(&foo()); |
| // ------ creates a temporary |
| let q = *p; |
| ``` |
| |
| Here, the expression `&foo()` is borrowing the expression `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. You could imagine that `let p = bar(&foo());` is equivalent to |
| the following, which uses an explicit temporary variable. |
| |
| Erroneous code example: |
| |
| ```compile_fail,E0597 |
| # fn foo() -> i32 { 22 } |
| # fn bar(x: &i32) -> &i32 { x } |
| let p = { |
| let tmp = foo(); // the temporary |
| bar(&tmp) // error: `tmp` does not live long enough |
| }; // <-- tmp is freed as we exit this block |
| let q = p; |
| ``` |
| |
| Whenever a temporary is created, it is automatically dropped (freed) according |
| to fixed rules. Ordinarily, the temporary is dropped at the end of the enclosing |
| statement -- in this case, after the `let p`. This is illustrated in the example |
| above by showing that `tmp` would be freed as we exit the block. |
| |
| To fix this problem, you need to create a local variable to store the value in |
| rather than relying on a temporary. For example, you might change the original |
| program to the following: |
| |
| ``` |
| fn foo() -> i32 { 22 } |
| fn bar(x: &i32) -> &i32 { x } |
| let value = foo(); // dropped at the end of the enclosing block |
| let p = bar(&value); |
| let q = *p; |
| ``` |
| |
| By introducing the explicit `let value`, we allocate storage that will last |
| until the end of the enclosing block (when `value` goes out of scope). When we |
| borrow `&value`, we are borrowing a local variable that already exists, and |
| hence no temporary is created. |
| |
| Temporaries are not always dropped at the end of the enclosing statement. In |
| simple cases where the `&` expression is immediately stored into a variable, the |
| compiler will automatically extend the lifetime of the temporary until the end |
| of the enclosing block. Therefore, an alternative way to fix the original |
| program is to write `let tmp = &foo()` and not `let tmp = foo()`: |
| |
| ``` |
| fn foo() -> i32 { 22 } |
| fn bar(x: &i32) -> &i32 { x } |
| let value = &foo(); |
| let p = bar(value); |
| let q = *p; |
| ``` |
| |
| Here, we are still borrowing `foo()`, but as the borrow is assigned directly |
| into a variable, the temporary will not be dropped until the end of the |
| enclosing block. Similar rules apply when temporaries are stored into aggregate |
| structures like a tuple or struct: |
| |
| ``` |
| // Here, two temporaries are created, but |
| // as they are stored directly into `value`, |
| // they are not dropped until the end of the |
| // enclosing block. |
| fn foo() -> i32 { 22 } |
| let value = (&foo(), &foo()); |
| ``` |