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Rollup merge of #28466 - baskerville:trpl-heap-highest-addr, r=steveklabnik

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Steve Klabnik 2015-09-17 17:06:55 -04:00
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src/doc/trpl/the-stack-and-the-heap.md
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@ -217,18 +217,18 @@ on the heap. The actual value of the box is a structure which has a pointer to
it allocates some memory for the heap, and puts `5` there. The memory now looks it allocates some memory for the heap, and puts `5` there. The memory now looks
like this: like this:
| Address | Name | Value | | Address | Name | Value |
|-----------------|------|------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 5 | | (2<sup>30</sup>) - 1 | | 5 |
| ... | ... | ... | | ... | ... | ... |
| 1 | y | 42 | | 1 | y | 42 |
| 0 | x | → 2<sup>30</sup> | | 0 | x | → (2<sup>30</sup>) - 1 |
We have 2<sup>30</sup> in our hypothetical computer with 1GB of RAM. And since We have (2<sup>30</sup>) - 1 in our hypothetical computer with 1GB of RAM. And since
our stack grows from zero, the easiest place to allocate memory is from the our stack grows from zero, the easiest place to allocate memory is from the
other end. So our first value is at the highest place in memory. And the value other end. So our first value is at the highest place in memory. And the value
of the struct at `x` has a [raw pointer][rawpointer] to the place we’ve of the struct at `x` has a [raw pointer][rawpointer] to the place we’ve
allocated on the heap, so the value of `x` is 2<sup>30</sup>, the memory allocated on the heap, so the value of `x` is (2<sup>30</sup>) - 1, the memory
location we’ve asked for. location we’ve asked for.
[rawpointer]: raw-pointers.html [rawpointer]: raw-pointers.html
@ -244,18 +244,18 @@ layout of a program which has been running for a while now:
| Address | Name | Value | | Address | Name | Value |
|----------------------|------|------------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 5 | | (2<sup>30</sup>) - 1 | | 5 |
| (2<sup>30</sup>) - 1 | | |
| (2<sup>30</sup>) - 2 | | | | (2<sup>30</sup>) - 2 | | |
| (2<sup>30</sup>) - 3 | | 42 | | (2<sup>30</sup>) - 3 | | |
| (2<sup>30</sup>) - 4 | | 42 |
| ... | ... | ... | | ... | ... | ... |
| 3 | y | → (2<sup>30</sup>) - 3 | | 3 | y | → (2<sup>30</sup>) - 4 |
| 2 | y | 42 | | 2 | y | 42 |
| 1 | y | 42 | | 1 | y | 42 |
| 0 | x | → 2<sup>30</sup> | | 0 | x | → (2<sup>30</sup>) - 1 |
In this case, we’ve allocated four things on the heap, but deallocated two of In this case, we’ve allocated four things on the heap, but deallocated two of
them. There’s a gap between 2<sup>30</sup> and (2<sup>30</sup>) - 3 which isn’t them. There’s a gap between (2<sup>30</sup>) - 1 and (2<sup>30</sup>) - 4 which isn’t
currently being used. The specific details of how and why this happens depends currently being used. The specific details of how and why this happens depends
on what kind of strategy you use to manage the heap. Different programs can use on what kind of strategy you use to manage the heap. Different programs can use
different ‘memory allocators’, which are libraries that manage this for you. different ‘memory allocators’, which are libraries that manage this for you.
@ -366,29 +366,29 @@ fn main() {
First, we call `main()`: First, we call `main()`:
| Address | Name | Value | | Address | Name | Value |
|-----------------|------|------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 | | (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... | | ... | ... | ... |
| 2 | j | → 0 | | 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> | | 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 | | 0 | h | 3 |
We allocate memory for `j`, `i`, and `h`. `i` is on the heap, and so has a We allocate memory for `j`, `i`, and `h`. `i` is on the heap, and so has a
value pointing there. value pointing there.
Next, at the end of `main()`, `foo()` gets called: Next, at the end of `main()`, `foo()` gets called:
| Address | Name | Value | | Address | Name | Value |
|-----------------|------|-----------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 | | (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... | | ... | ... | ... |
| 5 | z | → 4 | | 5 | z | → 4 |
| 4 | y | 10 | | 4 | y | 10 |
| 3 | x | → 0 | | 3 | x | → 0 |
| 2 | j | → 0 | | 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup>| | 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 | | 0 | h | 3 |
Space gets allocated for `x`, `y`, and `z`. The argument `x` has the same value Space gets allocated for `x`, `y`, and `z`. The argument `x` has the same value
as `j`, since that’s what we passed it in. It’s a pointer to the `0` address, as `j`, since that’s what we passed it in. It’s a pointer to the `0` address,
@ -396,42 +396,42 @@ since `j` points at `h`.
Next, `foo()` calls `baz()`, passing `z`: Next, `foo()` calls `baz()`, passing `z`:
| Address | Name | Value | | Address | Name | Value |
|-----------------|------|------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 | | (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... | | ... | ... | ... |
| 7 | g | 100 | | 7 | g | 100 |
| 6 | f | → 4 | | 6 | f | → 4 |
| 5 | z | → 4 | | 5 | z | → 4 |
| 4 | y | 10 | | 4 | y | 10 |
| 3 | x | → 0 | | 3 | x | → 0 |
| 2 | j | → 0 | | 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> | | 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 | | 0 | h | 3 |
We’ve allocated memory for `f` and `g`. `baz()` is very short, so when it’s We’ve allocated memory for `f` and `g`. `baz()` is very short, so when it’s
over, we get rid of its stack frame: over, we get rid of its stack frame:
| Address | Name | Value | | Address | Name | Value |
|-----------------|------|------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 | | (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... | | ... | ... | ... |
| 5 | z | → 4 | | 5 | z | → 4 |
| 4 | y | 10 | | 4 | y | 10 |
| 3 | x | → 0 | | 3 | x | → 0 |
| 2 | j | → 0 | | 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> | | 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 | | 0 | h | 3 |
Next, `foo()` calls `bar()` with `x` and `z`: Next, `foo()` calls `bar()` with `x` and `z`:
| Address | Name | Value | | Address | Name | Value |
|----------------------|------|------------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 | | (2<sup>30</sup>) - 1 | | 20 |
| (2<sup>30</sup>) - 1 | | 5 | | (2<sup>30</sup>) - 2 | | 5 |
| ... | ... | ... | | ... | ... | ... |
| 10 | e | → 9 | | 10 | e | → 9 |
| 9 | d | → (2<sup>30</sup>) - 1 | | 9 | d | → (2<sup>30</sup>) - 2 |
| 8 | c | 5 | | 8 | c | 5 |
| 7 | b | → 4 | | 7 | b | → 4 |
| 6 | a | → 0 | | 6 | a | → 0 |
@ -439,24 +439,24 @@ Next, `foo()` calls `bar()` with `x` and `z`:
| 4 | y | 10 | | 4 | y | 10 |
| 3 | x | → 0 | | 3 | x | → 0 |
| 2 | j | → 0 | | 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> | | 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 | | 0 | h | 3 |
We end up allocating another value on the heap, and so we have to subtract one We end up allocating another value on the heap, and so we have to subtract one
from 2<sup>30</sup>. It’s easier to just write that than `1,073,741,823`. In any from (2<sup>30</sup>) - 1. It’s easier to just write that than `1,073,741,822`. In any
case, we set up the variables as usual. case, we set up the variables as usual.
At the end of `bar()`, it calls `baz()`: At the end of `bar()`, it calls `baz()`:
| Address | Name | Value | | Address | Name | Value |
|----------------------|------|------------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 | | (2<sup>30</sup>) - 1 | | 20 |
| (2<sup>30</sup>) - 1 | | 5 | | (2<sup>30</sup>) - 2 | | 5 |
| ... | ... | ... | | ... | ... | ... |
| 12 | g | 100 | | 12 | g | 100 |
| 11 | f | → 9 | | 11 | f | → 9 |
| 10 | e | → 9 | | 10 | e | → 9 |
| 9 | d | → (2<sup>30</sup>) - 1 | | 9 | d | → (2<sup>30</sup>) - 2 |
| 8 | c | 5 | | 8 | c | 5 |
| 7 | b | → 4 | | 7 | b | → 4 |
| 6 | a | → 0 | | 6 | a | → 0 |
@ -464,7 +464,7 @@ At the end of `bar()`, it calls `baz()`:
| 4 | y | 10 | | 4 | y | 10 |
| 3 | x | → 0 | | 3 | x | → 0 |
| 2 | j | → 0 | | 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> | | 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 | | 0 | h | 3 |
With this, we’re at our deepest point! Whew! Congrats for following along this With this, we’re at our deepest point! Whew! Congrats for following along this
@ -474,11 +474,11 @@ After `baz()` is over, we get rid of `f` and `g`:
| Address | Name | Value | | Address | Name | Value |
|----------------------|------|------------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 | | (2<sup>30</sup>) - 1 | | 20 |
| (2<sup>30</sup>) - 1 | | 5 | | (2<sup>30</sup>) - 2 | | 5 |
| ... | ... | ... | | ... | ... | ... |
| 10 | e | → 9 | | 10 | e | → 9 |
| 9 | d | → (2<sup>30</sup>) - 1 | | 9 | d | → (2<sup>30</sup>) - 2 |
| 8 | c | 5 | | 8 | c | 5 |
| 7 | b | → 4 | | 7 | b | → 4 |
| 6 | a | → 0 | | 6 | a | → 0 |
@ -486,32 +486,32 @@ After `baz()` is over, we get rid of `f` and `g`:
| 4 | y | 10 | | 4 | y | 10 |
| 3 | x | → 0 | | 3 | x | → 0 |
| 2 | j | → 0 | | 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> | | 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 | | 0 | h | 3 |
Next, we return from `bar()`. `d` in this case is a `Box<T>`, so it also frees Next, we return from `bar()`. `d` in this case is a `Box<T>`, so it also frees
what it points to: (2<sup>30</sup>) - 1. what it points to: (2<sup>30</sup>) - 2.
| Address | Name | Value | | Address | Name | Value |
|-----------------|------|------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 | | (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... | | ... | ... | ... |
| 5 | z | → 4 | | 5 | z | → 4 |
| 4 | y | 10 | | 4 | y | 10 |
| 3 | x | → 0 | | 3 | x | → 0 |
| 2 | j | → 0 | | 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> | | 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 | | 0 | h | 3 |
And after that, `foo()` returns: And after that, `foo()` returns:
| Address | Name | Value | | Address | Name | Value |
|-----------------|------|------------------| |----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 | | (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... | | ... | ... | ... |
| 2 | j | → 0 | | 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> | | 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 | | 0 | h | 3 |
And then, finally, `main()`, which cleans the rest up. When `i` is `Drop`ped, And then, finally, `main()`, which cleans the rest up. When `i` is `Drop`ped,
it will clean up the last of the heap too. it will clean up the last of the heap too.