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Nothing about lambdas requires heap allocation. See also: C++, Rust

If the lambda captures some value, and also outlives the current scope, then that captured value has to necessarily be heap allocated.

That would be a bug, so just... don't do that?

If you return a pointer to a local variable that outlives the scope, the pointer would be dangling. Does that mean we should ban pointers?

If you close over a pointer to a local variable that outlives the scope, the closure would be dangling. Does that mean we should ban closures?

No, (in C++) the lambda can capture the the variable by value, and the lambda itself can be passed around by value. If you capture a variable by reference or pointer that your lambda outlives, your code got a serious bug.

And in Rust, it will enforce correct usage via the borrow checker - the outlive case simply will not compile.

If you do want it, you have the option to, say, heap allocate.

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