4 ms·

> All that happens is the following: > - An integer is allocated (and initialized to zero). > - The address of this integer is returned. That is not all that

> All that happens is the following: > - An integer is allocated (and initialized to zero). > - The address of this integer is returned. That is not all that happens, at least down at the C/assembler level. Let me illustrate what I mean. Consider this function, which also does both of these things (cobbled together from Google searches so please excuse incorrect syntax): func foo() uintptr{ var x int return uintptr(unsafe.Pointer(&x)) } All that function does is allocate an integer (and initialise to zero) and return the address of that integer. Exactly the same as your function, right? Except it's obviously not - it doesn't extend the lifetime of the integer variable. So why not? The GC somehow knows to ignore the number returned from my function, even though, under the hood, it's still stored in a register or stack location or whatever in exactly the same way as the address returned from your function. So how does the GC know to ignore it? Is that number somehow marked in a way that says "GC, when you're scanning memory looking for address-like numbers, don't pay attention to this one"? No. It doesn't look at the number in the first place because it hasn't been told to look at it. In contrast, in your example, the memory address is not just returned from the function (in the C sense that it's put in a register for the caller to receive). It, additionally, somehow registers that memory address with the GC to let it know that there's another reference to that variable location. That is the extra thing that your function does that mine doesn't. And that magic happens (or at least starts) at the moment you use the & operator.

foldr 4y ago
Yes, Go has a precise (i.e. non-conservative) garbage collector. It seems odd to me to think about that as some kind of special feature of the & operator. Even if one does, it's certainly not a surprising feature. Knowing that Go has a precise GC, one certainly expects the GC to know that the value of &x references x. If it didn't that would be a major bug. The Go GC isn't a reference counting implementation. It traces the values of variables on the stack and it knows their types (because it knows which function any given stack frame corresponds to and it knows which variables that function allocates). Thus it knows that if a variable is of type *int and has a non-nil value then its value references an int. (And so on for fields of structs that are stored in stack variables, etc.) The & operator does not need to do anything special. The & operator merely takes the address of the object. When that address is stored in a pointer variable (or array member, or struct field...), that's when it becomes visible to the GC as a reference.

foldr 4y ago

Yes, Go has a precise (i.e. non-conservative) garbage collector. It seems odd to me to think about that as some kind of special feature of the & operator. Even if one does, it's certainly not a surprising feature. Knowing that Go has a precise GC, one certainly expects the GC to know that the value of &x references x. If it didn't that would be a major bug. The Go GC isn't a reference counting implementation. It traces the values of variables on the stack and it knows their types (because it knows which function any given stack frame corresponds to and it knows which variables that function allocates). Thus it knows that if a variable is of type *int and has a non-nil value then its value references an int. (And so on for fields of structs that are stored in stack variables, etc.) The & operator does not need to do anything special. The & operator merely takes the address of the object. When that address is stored in a pointer variable (or array member, or struct field...), that's when it becomes visible to the GC as a reference.