[noselasd]

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.

[gf000]

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.