Name: CS50 2019 - リーディング5 - データ構造 (CS50 2019 - Lecture 5 - Data Structures)
Uploaded: 2021-01-14T10:20:55.000Z
Duration: 1 h 45 min 3 s
Description: VoiceTubeの動画で発音を聞きながら英語表現を覚えよう！学べる英語：

And this is our look today at data structures.

You'll recall last week that we gave ourselves

a few new ingredients and some new syntax in C.

We introduced pointers, the ability to address chunks of memory

And we introduced the star notation and a few other functions,

malloc among them, free among them, so that you can actually

Just to hammer this home, let's take a look

at just this small example that from the outset is buggy.

But let's just walk through it step by step.

This first line highlighted in yellow in English

If you're a little rusty since last week, what's that first line of code

AUDIENCE: It's creating a point to an int named x.

It's creating a pointer to an integer and calling

And let me propose that the next line is doing the same thing giving us

another pointer to an integer but this time calling it y.

The third line of code, someone else, what's

happening in English with this line here?

AUDIENCE: It's creates a memory of size of int and assigns it to x,

but it's more difficult to execute if it's not a pointer, I don't know.

malloc is the function we introduced last week

It takes one argument, the number of bytes that you want to allocate.

Even if you don't recall how many bytes you need for an integer,

you can call this other operator, sizeof,

that we saw briefly last week, that will return, in this case, 4 most likely

depending on the computer that you're on.

So this is saying, hey, computer give me 4 bytes of memory.

And it returns that chunk of memory to conventionally

by way of the first address, so ox something,

And then it stores that address in x, which is in fact OK,

because as you noted initially, x is in fact a pointer.

So all this is doing is it's declaring a variable called

x and storing in it ultimately the address

You wouldn't typically allocate an int like this.

You would allocate an int with just int and semicolon just like in week 1.

But now that we have the ability to allocate addresses and allocate memory,

This line here now, the fourth line of code, says what in English?

AUDIENCE: It goes to the address in the x and then sets it to 42.

DAVID MALAN: Good goes to the address in x and sets it to 42.

So the star operator is the dereference operator,

which is a fancy way of saying go to that address.

Well, per week 1 when we discussed the assignment operator,

So wherever malloc found 4 bytes of available memory for me,

this fourth line of code says, go there and put the number 42

by putting the appropriate zeros and ones.

and here's the bug, if we finally reveal it here--

AUDIENCE: You haven't allocated memory for that variable yet.

I haven't allocated memory for that variable yet.

And because up here I've just said int star y semicolon.

You can only assume safely that has some garbage value, some unknown value,

maybe remnants from some other part of the program,

that might not necessarily be true here at the beginning of the program.

But for safety's sake assume that if you don't give a variable

It's got some bogus address, such that if you

say star y go to that bogus address something bad is going to happen.

And maybe you experienced this already in P Set 4 or prior,

some kind of memory problem with your code,

or a segmentation fault or seg fault, bad things

happen when you go to addresses that don't exist

or that you don't even know where they are.

What if instead I do something like this?

So that just says put in y the same address that's in x.

And then with this last line of code, what if I now say star y equals 13?

AUDIENCE: Now there's memory allocated for y.

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CS50 2019 - リーディング5 - データ構造 (CS50 2019 - Lecture 5 - Data Structures)

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