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Fix the return type of stackalloc #888

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Aug 15, 2023
Merged

Fix the return type of stackalloc #888

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ecb36b2
Update expressions.md
Nigel-Ecma Aug 14, 2023
2a0c176
Update expressions.md
Nigel-Ecma Aug 14, 2023
589fcd4
Fix typo
Nigel-Ecma Aug 15, 2023
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4 changes: 3 additions & 1 deletion standard/expressions.md
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Expand Up @@ -3118,7 +3118,9 @@ When a *stackalloc_initializer* is present:

Each *stackalloc_element_initializer* shall have an implicit conversion to *unmanaged_type* ([§10.2](conversions.md#102-implicit-conversions)). The *stackalloc_element_initializer*s initialize elements in the allocated memory in increasing order, starting with the element at index zero. In the absence of a *stackalloc_initializer*, the content of the newly allocated memory is undefined.

The result of a *stackalloc_expression* is a pointer of type `T *` ([§23.9](unsafe-code.md#239-stack-allocation)), where `T` is the *unmanaged_type*, which is implicitly converted to `Span<T>` in safe code.
The result of a *stackalloc_expression* is an instance of type `Span<T>`, where `T` is the *unmanaged_type* and the instance’s `Length` property returns the number of items allocated.
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> *Note*: When occurring in unsafe code the result of a *stackalloc_expression* may be of a different type, see ([§23.9](unsafe-code.md#239-stack-allocation)). *end note*

Access via an instance of `Span<T>` to the elements of an allocated block is range checked.

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