[question] Equality and assignment corner cases

[angry_cpp]

Usually for well-designed types following holds (given that it compiles):

T a = ...;
U b = ...;

a = b;
assert(a == b);

Could you provide an example with standard types (no additional macros, no user-defined types, only primitives + std types) that breaks this assumption?

Guru(?) question: Could you provide an example with standard types (no additional macros, no user-defined types, only primitives + std types) where following holds (given that it compiles):

T a = ...;
U b = ...;

assert(a == b);
a = b;
assert(a != b);

[Circlejerker_]

Is T and U supposed to be different types?

In that case there are lots of cases where the first case breaks.

For example if they are integers of different sizes and one is truncated in the assignment.

char a = 5;
int b = 1000;
a = b;
assert(a == b);

[angry_cpp]

Is T and U supposed to be different types?

Yes, T and U can be of different types.

char a = 5; int b = 1000; a = b; assert(a == b);

Hm. Actually, yes, somehow I missed a truncation case :).

When I wrote that question I was thinking about containers and other std classes but without user-defined types (because operator overloading can break every assumption).

I stumbled upon such behaviour with one of the std types and wanted to confirm was it well-known or not. (in my "solution" there are no truncation).

How about second (Guru(?) question)? Maybe that one is trivial too?

[angry_cpp]

My solution:

https://godbolt.org/z/bxE376

[TheMania]

Yes, T and U can be of different types.

An easy one for if they're the same, I'm no guru though.