Rename images to be in nostarch's convention

And switch images 3 and 4, they appear in the opposite order
This commit is contained in:
Carol (Nichols || Goulding) 2016-09-06 17:39:56 -04:00
parent 6d617aba2a
commit ed557992cb
9 changed files with 5 additions and 5 deletions

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@ -167,7 +167,7 @@ You might say “copy the `String`!” This is both correct and incorrect at the
same time. It does a _shallow_ copy of the `String`. Whats that mean? Well, same time. It does a _shallow_ copy of the `String`. Whats that mean? Well,
lets take a look at what `String` looks like under the covers: lets take a look at what `String` looks like under the covers:
<img alt="string" src="img/foo1.png" class="center" style="width: 50%;" /> <img alt="string" src="img/trpl04-01.png" class="center" style="width: 50%;" />
A `String` is made up of three parts: a pointer to the memory that holds the A `String` is made up of three parts: a pointer to the memory that holds the
contents of the string, a length, and a capacity. The length is how much memory contents of the string, a length, and a capacity. The length is how much memory
@ -180,11 +180,11 @@ When we assign `s1` to `s2`, the `String` itself is copied, meaning we copy the
pointer, the length, and the capacity. We do not copy the data that the pointer, the length, and the capacity. We do not copy the data that the
`String`'s pointer refers to. In other words, it looks like this: `String`'s pointer refers to. In other words, it looks like this:
<img alt="s1 and s2" src="img/foo2.png" class="center" style="width: 50%;" /> <img alt="s1 and s2" src="img/trpl04-02.png" class="center" style="width: 50%;" />
_Not_ this: _Not_ this:
<img alt="s1 and s2 to two places" src="img/foo4.png" class="center" style="width: 50%;" /> <img alt="s1 and s2 to two places" src="img/trpl04-03.png" class="center" style="width: 50%;" />
Theres a problem here. Both data pointers are pointing to the same place. Why Theres a problem here. Both data pointers are pointing to the same place. Why
is this a problem? Well, when `s2` goes out of scope, it will free the memory is this a problem? Well, when `s2` goes out of scope, it will free the memory
@ -240,7 +240,7 @@ also invalidates the first binding, instead of calling this a shallow copy,
it's called a _move_. Here we would read this by saying that `s1` was _moved_ it's called a _move_. Here we would read this by saying that `s1` was _moved_
into `s2`. So what actually happens looks like this: into `s2`. So what actually happens looks like this:
<img alt="s1 and s2 to the same place" src="img/foo3.png" class="center" style="width: 50%;" /> <img alt="s1 and s2 to the same place" src="img/trpl-04-04.png" class="center" style="width: 50%;" />
That solves our problem! With only `s2` valid, when it goes out of scope, it That solves our problem! With only `s2` valid, when it goes out of scope, it
alone will free the memory, and were done. alone will free the memory, and were done.
@ -279,7 +279,7 @@ println!("{}", s1);
This will work just fine. Remember our diagram from before? In this case, This will work just fine. Remember our diagram from before? In this case,
it _is_ doing this: it _is_ doing this:
<img alt="s1 and s2 to two places" src="img/foo4.png" class="center" style="width: 50%;" /> <img alt="s1 and s2 to two places" src="img/trpl04-03.png" class="center" style="width: 50%;" />
When you see a call to `clone()`, you know that some arbitrary code is being When you see a call to `clone()`, you know that some arbitrary code is being
executed, and that code may be expensive. Its a visual indicator that something executed, and that code may be expensive. Its a visual indicator that something

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