From 4463087223983c40a6d67beab0513fba7cdb7538 Mon Sep 17 00:00:00 2001
From: Jacques Lucke <jacques@blender.org>
Date: Wed, 16 Dec 2020 15:59:58 +0100
Subject: BLI: remove implicit casts between some span types

Casting pointers from one type to another does change the
value of the pointer in some cases. Therefore, casting a span
that contains pointers of one type to a span that contains
pointers of another type, is not generally safe. In practice, this
issue mainly comes up when dealing with classes that have a
vtable.

There are some special cases that are still allowed. For example,
adding const to the pointer does not change the address.
Also, casting to a void pointer is fine.

In cases where implicit conversion is disabled, but one is sure
that the cast is valid, an explicit call of `span.cast<NewType>()`
can be used.
---
 source/blender/blenlib/BLI_vector.hh | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

(limited to 'source/blender/blenlib/BLI_vector.hh')

diff --git a/source/blender/blenlib/BLI_vector.hh b/source/blender/blenlib/BLI_vector.hh
index 053dcb6faea..fe6d54ae9e5 100644
--- a/source/blender/blenlib/BLI_vector.hh
+++ b/source/blender/blenlib/BLI_vector.hh
@@ -315,13 +315,13 @@ class Vector {
     return MutableSpan<T>(begin_, this->size());
   }
 
-  template<typename U, typename std::enable_if_t<is_convertible_pointer_v<T, U>> * = nullptr>
+  template<typename U, typename std::enable_if_t<is_span_convertible_pointer_v<T, U>> * = nullptr>
   operator Span<U>() const
   {
     return Span<U>(begin_, this->size());
   }
 
-  template<typename U, typename std::enable_if_t<is_convertible_pointer_v<T, U>> * = nullptr>
+  template<typename U, typename std::enable_if_t<is_span_convertible_pointer_v<T, U>> * = nullptr>
   operator MutableSpan<U>()
   {
     return MutableSpan<U>(begin_, this->size());
-- 
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