Yes, it does have the distinction. From your link:

> When possible, the Go compilers will allocate variables that are local to a function in that function's stack frame. However, if the compiler cannot prove that the variable is not referenced after the function returns, then the compiler must allocate the variable on the garbage-collected heap to avoid dangling pointer errors.

As you can see, Go uses both the stack and the heap; however, unlike Rust, it doesn't require users to know where their variables live to write correct programs.

I believe, the distinction that orclev mentioned was that in Rust, you specifically have to be aware when something is allocated to the stack or heap, but in Go, you don't have to since the compiler deals with it. Obviously Go allocates to both stack and heap, it's just that the user doesn't need to know which.

Fair enough. I simply meant that Go's semantics let you reason about what lives on the stack vs what lives on the heap much like you can in Rust (though in Rust it's clearer still because there is no escape analysis at all). While you don't need to know from a correctness perspective, you can reliably reason about what will live on the stack vs heap in a way you can't in, say, Java.

No, you can't really reason properly about what lives on the stack. As the Go FAQ say, you can safely assume a variable will be allocated on the stack if you never take a reference of it, but that's not always clear as it seems. For instance, if you call a method on your variable, you can't immediately know by looking at the method whether the receiver is by-reference (pointer receiver) or by value. If you're using a pointer receiver, the variable's address may be taken, so the compiler has to perform escape analysis.

Once you take the address and escape analysis is performed, you've got no way to easily reason about what's on the stack vs. what's on the heap, without delving into the gory implementation details which may be changed in future versions.

Since Go isn't C++ and doesn't have a huge and complicated standard that guarantees compiler optimizations, you can't really reason about what's going on.

This is essentially the same situation as Java, although the Hotspot JVM's escape analysis is probably less 'basic'.