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trpl: Fix off-by-one highest memory address

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Bastien Dejean 2015-09-17 18:53:01 +02:00
parent 2be0d0ad92
commit 6d2cb6c688
1 changed files with 84 additions and 84 deletions
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168
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
like this:
| Address | Name | Value |
|-----------------|------|------------------|
| 2<sup>30</sup> | | 5 |
| ... | ... | ... |
| 1 | y | 42 |
| 0 | x | → 2<sup>30</sup> |
| Address | Name | Value |
|----------------------|------|------------------------|
| (2<sup>30</sup>) - 1 | | 5 |
| ... | ... | ... |
| 1 | y | 42 |
| 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
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
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.
[rawpointer]: raw-pointers.html
@ -244,18 +244,18 @@ layout of a program which has been running for a while now:
| Address | Name | Value |
|----------------------|------|------------------------|
| 2<sup>30</sup> | | 5 |
| (2<sup>30</sup>) - 1 | | |
| (2<sup>30</sup>) - 1 | | 5 |
| (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 |
| 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
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
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.
@ -366,29 +366,29 @@ fn main() {
First, we call `main()`:
| Address | Name | Value |
|-----------------|------|------------------|
| 2<sup>30</sup> | | 20 |
| ... | ... | ... |
| 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> |
| 0 | h | 3 |
| Address | Name | Value |
|----------------------|------|------------------------|
| (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... |
| 2 | j | → 0 |
| 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 |
We allocate memory for `j`, `i`, and `h`. `i` is on the heap, and so has a
value pointing there.
Next, at the end of `main()`, `foo()` gets called:
| Address | Name | Value |
|-----------------|------|-----------------|
| 2<sup>30</sup> | | 20 |
| ... | ... | ... |
| 5 | z | → 4 |
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup>|
| 0 | h | 3 |
| Address | Name | Value |
|----------------------|------|------------------------|
| (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... |
| 5 | z | → 4 |
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 |
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,
@ -396,42 +396,42 @@ since `j` points at `h`.
Next, `foo()` calls `baz()`, passing `z`:
| Address | Name | Value |
|-----------------|------|------------------|
| 2<sup>30</sup> | | 20 |
| ... | ... | ... |
| 7 | g | 100 |
| 6 | f | → 4 |
| 5 | z | → 4 |
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> |
| 0 | h | 3 |
| Address | Name | Value |
|----------------------|------|------------------------|
| (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... |
| 7 | g | 100 |
| 6 | f | → 4 |
| 5 | z | → 4 |
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 |
We’ve allocated memory for `f` and `g`. `baz()` is very short, so when it’s
over, we get rid of its stack frame:
| Address | Name | Value |
|-----------------|------|------------------|
| 2<sup>30</sup> | | 20 |
| ... | ... | ... |
| 5 | z | → 4 |
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> |
| 0 | h | 3 |
| Address | Name | Value |
|----------------------|------|------------------------|
| (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... |
| 5 | z | → 4 |
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 |
Next, `foo()` calls `bar()` with `x` and `z`:
| Address | Name | Value |
|----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 |
| (2<sup>30</sup>) - 1 | | 5 |
| (2<sup>30</sup>) - 1 | | 20 |
| (2<sup>30</sup>) - 2 | | 5 |
| ... | ... | ... |
| 10 | e | → 9 |
| 9 | d | → (2<sup>30</sup>) - 1 |
| 9 | d | → (2<sup>30</sup>) - 2 |
| 8 | c | 5 |
| 7 | b | → 4 |
| 6 | a | → 0 |
@ -439,24 +439,24 @@ Next, `foo()` calls `bar()` with `x` and `z`:
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> |
| 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 |
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.
At the end of `bar()`, it calls `baz()`:
| Address | Name | Value |
|----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 |
| (2<sup>30</sup>) - 1 | | 5 |
| (2<sup>30</sup>) - 1 | | 20 |
| (2<sup>30</sup>) - 2 | | 5 |
| ... | ... | ... |
| 12 | g | 100 |
| 11 | f | → 9 |
| 10 | e | → 9 |
| 9 | d | → (2<sup>30</sup>) - 1 |
| 9 | d | → (2<sup>30</sup>) - 2 |
| 8 | c | 5 |
| 7 | b | → 4 |
| 6 | a | → 0 |
@ -464,7 +464,7 @@ At the end of `bar()`, it calls `baz()`:
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> |
| 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 |
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 |
|----------------------|------|------------------------|
| 2<sup>30</sup> | | 20 |
| (2<sup>30</sup>) - 1 | | 5 |
| (2<sup>30</sup>) - 1 | | 20 |
| (2<sup>30</sup>) - 2 | | 5 |
| ... | ... | ... |
| 10 | e | → 9 |
| 9 | d | → (2<sup>30</sup>) - 1 |
| 9 | d | → (2<sup>30</sup>) - 2 |
| 8 | c | 5 |
| 7 | b | → 4 |
| 6 | a | → 0 |
@ -486,32 +486,32 @@ After `baz()` is over, we get rid of `f` and `g`:
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> |
| 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 |
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 |
|-----------------|------|------------------|
| 2<sup>30</sup> | | 20 |
| ... | ... | ... |
| 5 | z | → 4 |
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> |
| 0 | h | 3 |
| Address | Name | Value |
|----------------------|------|------------------------|
| (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... |
| 5 | z | → 4 |
| 4 | y | 10 |
| 3 | x | → 0 |
| 2 | j | → 0 |
| 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 |
And after that, `foo()` returns:
| Address | Name | Value |
|-----------------|------|------------------|
| 2<sup>30</sup> | | 20 |
| ... | ... | ... |
| 2 | j | → 0 |
| 1 | i | → 2<sup>30</sup> |
| 0 | h | 3 |
| Address | Name | Value |
|----------------------|------|------------------------|
| (2<sup>30</sup>) - 1 | | 20 |
| ... | ... | ... |
| 2 | j | → 0 |
| 1 | i | → (2<sup>30</sup>) - 1 |
| 0 | h | 3 |
And then, finally, `main()`, which cleans the rest up. When `i` is `Drop`ped,
it will clean up the last of the heap too.