8 Instructions & Programs

8 指令和程序

Hi, I'm Carrie Anne and this is Crash Course Computer Science!

嗨，我是 Carrie Anne，欢迎收看计算机科学速成课！

Last episode, we combined an ALU, control unit, some memory, and a clock together to

上集我们把 ALU，控制单元， RAM，时钟结合在一起

make a basic, but functional Central Processing Unit – or CPU

做了个基本，但可用的"中央处理单元"，简称 CPU

the beating, ticking heart of a computer.

它是计算机的核心

We've done all the hard work of building many of these components from the electronic

我们已经用电路做了很多组件.

circuits up, and now it's time to give our CPU some actual instructions to process!

这次我们给 CPU 一些指令来运行！

The thing that makes a CPU powerful is the fact that it is programmable –

CPU 之所以强大，是因为它是可编程的 -

if you write a different sequence of instructions, then the CPU will perform a different task.

如果写入不同指令，就会执行不同任务

So the CPU is a piece of hardware which is controlled by easy-to-modify software!

CPU 是一块硬件，可以被软件控制！

Let's quickly revisit the simple program that we stepped through last episode.

我们重新看一下上集的简单程序

The computer memory looked like this.

内存里有这些值

Each address contained 8 bits of data.

每个地址可以存 8 位数据

For our hypothetical CPU, the first four bits specified the operation code, or opcode, and

因为我们的 CPU 是假设的，这里前4位是"操作码"

the second set of four bits specified an address or registers.

后4位指定一个内存地址，或寄存器.

In memory address zero we have 0010 1110.

内存地址 0 是 0010 1110

Again, those first four bits are our opcode which corresponds to a "LOAD_A" instruction.

前 4 位代表 LOAD_A 指令

This instruction reads data from a location of memory specified in those last four bits

意思是：把后 4 位指定的内存地址的值，放入寄存器 A

of the instruction and saves it into Register A. In this case, 1110, or 14 in decimal.

后 4 位是 1110，十进制的 14

So let's not think of this of memory address 0 as "0010 1110", but rather as the instruction

我们来把 0010 1110 看成 "LOAD_A 14" 指令

LOAD_A 14.

我们来把 0010 1110 看成 "LOAD_A 14" 指令

That's much easier to read and understand!

这样更好理解！

And for me to say!

也更方便说清楚

And we can do the same thing for the rest of the data in memory.

可以对内存里剩下的数也这样转换.

In this case, our program is just four instructions long,

这里，我们的程序只有4个指令

and we've put some numbers into memory too, 3 and 14.

还有数字 3 和 14

So now let's step through this program:

现在一步步看

First is LOAD_A 14, which takes the value in address 14, which is the number 3,

LOAD_A 14 是从地址 14 中拿到数字3，

and stores it into Register A.

放入寄存器A

Then we have a "LOAD_B 15" instruction, which takes the value in memory location 15,

LOAD_B 15 是从地址 15 中拿到数字14，

which is the number 14, and saves it into Register B.

放入寄存器B

Okay.

好.

Easy enough.

挺简单的！

But now we have an "ADD" instruction.

下一个是 ADD 指令

This tells the processor to use the ALU to add two registers together,

ADD B A 告诉 ALU，

in this case, B and A are specified.

把寄存器 B 和寄存器 A 里的数字加起来

The ordering is important, because the resulting sum is saved into the second register that's specified.

（B和A的）顺序很重要，因为结果会存在第二个寄存器

So in this case, the resulting sum is saved into Register A.

也就是寄存器 A

And finally, our last instruction is "STORE_A 13", which instructs the CPU to write whatever

最后一条指令是 "STORE_A 13"，

value is in Register A into memory location 13.

把寄存器 A 的值存入内存地址 13

Yesss!

好棒！

Our program adds two numbers together.

我们把 2 个数加在了一起！

That's about as exciting as it gets when we only have four instructions to play with.

毕竟只有4个指令，也只能做这个了.

So let's add some more!

加多一些指令吧！

Now we've got a subtract function, which like ADD, specifies two registers to operate on.

SUB 是减法，和 ADD 一样也要 2 个寄存器来操作.

We've also got a fancy new instruction called JUMP.

还有 JUMP（跳转）

As the name implies, this causes the program to "jump" to a new location.

让程序跳转到新位置

This is useful if we want to change the order of instructions, or choose to skip some instructions.

如果想改变指令顺序，或跳过一些指令，这个很实用

For example, a JUMP 0, would cause the program to go back to the beginning.

举例， JUMP 0 可以跳回开头

At a low level, this is done by writing the value specified in the last four bits into

JUMP 在底层的实现方式是，把指令后 4 位代表的内存地址的值

the instruction address register, overwriting the current value.

覆盖掉 "指令地址寄存器" 里的值

We've also added a special version of JUMP called JUMP_NEGATIVE.

还有一个特别版的 JUMP 叫 JUMP_NEGATIVE

This only jumps the program if the ALU's negative flag is set to true.

它只在 ALU 的负数标志" 为真时，进行 JUMP

As we talked about in Episode 5, the negative flag is only set

第5集讲过，算术结果为负，

when the result of an arithmetic operation is negative.

负数标志才是真

If the result of the arithmetic was zero or positive, the negative flag would not be set.

结果不是负数时， "负数标志"为假

So the JUMP NEGATIVE won't jump anywhere, and the CPU will just continue on to the next instruction.

如果是假，JUMP_NEGATIVE 就不会执行，程序照常进行

Our previous program really should have looked like this to be correct,

我们之前的例子程序，其实应该是这样，才能正确工作.

otherwise the CPU would have just continued on after the STORE instruction, processing all those 0's.

否则跑完 STORE_A 13 之后，CPU 会不停运行下去，处理后面的 0

But there is no instruction with an opcode of 0, and so the computer would have crashed!

因为 0 不是操作码，所以电脑会崩掉！

It's important to point out here that we're storing

我还想指出一点，

both instructions and data in the same memory.

指令和数据都是存在同一个内存里的.

There is no difference fundamentally -it's all just binary numbers.

它们在根本层面上毫无区别都是二进制数

So the HALT instruction is really important because it allows us to separate the two.

HALT 很重要，能区分指令和数据

Okay, so let's make our program a bit more interesting, by adding a JUMP.

好，现在用 JUMP 让程序更有趣一些.

We'll also modify our two starting values in memory to 1 and 1.

我们还把内存中 3 和 14 两个数字，改成 1 和 1

Lets step through this program just as our CPU would.

现在来从 CPU 的视角走一遍程序

First, LOAD_A 14 loads the value 1 into Register A.

首先 LOAD_A 14，把 1 存入寄存器A ，（因为地址 14 里的值是 1）

Next, LOAD_B 15 loads the value 1 into Register B.

然后 LOAD_B 15，把 1 存入寄存器B，（因为地址 15 里的值也是 1）

As before, we ADD registers B and A together, with the sum going into Register A. 1+1 = 2,

然后 ADD B A 把寄存器 B 和 A 相加，结果放到寄存器 A 里

so now Register A has the value 2 in it (stored in binary of course)

现在寄存器 A 的值是 2，(当然是以二进制存的）

Then the STORE instruction saves that into memory location 13.

然后 STORE_A 13 指令，把寄存器 A 的值存入内存地址 13

Now we hit a "JUMP 2" instruction.

现在遇到 JUMP 2 指令

This causes the processor to overwrite the value in the instruction address register,

CPU 会把"指令地址寄存器"的值，

which is currently 4, with the new value, 2.

现在是 4，改成 2

Now, on the processor's next fetch cycle, we don't fetch HALT,

因此下一步不再是 HALT

instead we fetch the instruction at memory location 2, which is ADD B A.

而是读内存地址 2 里的指令，也就是 ADD B A

We've jumped!

我们跳转了！

寄存器 A 里是 2，寄存器 B 里是 1

So 1+2 = 3, so now Register A has the value 3.

1+2=3，寄存器 A 变成 3

We store that into memory.

存入内存

And we've hit the JUMP again, back to ADD B A.

又碰到 JUMP 2，又回到 ADD B A.

1+3=4

So now register A has the value 4.

现在寄存器 A 是 4

See what's happening here?

发现了吗？

Every loop, we're adding one.

每次循环都+1

Its counting up!

不断增多

Cooooool.

酷

But notice there's no way to ever escape.

但没法结束啊

We're never.. ever.. going to get to that halt instruction,

永远不会碰到 HALT

because we're always going to hit that JUMP.

总是会碰到 JUMP

This is called an infinite loop – a program that runs forever… ever… ever… ever…

这叫无限循环这个程序会永远跑下去. 下去. 下去. 下去

ever

下去

To break the loop, we need a conditional jump.

为了停下来，我们需要有条件的 JUMP

A jump that only happens if a certain condition is met.

只有特定条件满足了，才执行 JUMP.

Our JUMP_NEGATIVE is one example of a conditional jump,

比如 JUMP NEGATIVE 就是条件跳转的一个例子

but computers have other types too like JUMP IF EQUAL and JUMP IF GREATER.

还有其他类型的条件跳转，比如， JUMP IF EQUAL（如果相等），JUMP IF GREATER（如果更大）

So let's make our code a little fancier and step through it.

现在把代码弄花哨一点，再过一遍代码

Like before, the program starts by loading values from memory into registers A and B.

就像之前，程序先把内存值放入寄存器 A 和 B.

In this example, the number 11 gets loaded into Register A, and 5 gets loaded into Register B.

寄存器 A 是 11，寄存器 B 是 5

Now we subtract register B from register A. That's 11 minus 5, which is 6,

SUB B A，用 A 减 B，11-5=6

and so 6 gets saved into Register A.

6 存入寄存器 A

Now we hit our JUMP NEGATIVE.

JUMP NEGATIVE 出场

The last ALU result was 6.

上一次 ALU 运算的结果是 6

That's a positive number, so the the negative flag is false.

是正数，所以 "负数标志" 是假

That means the processor does not jump.

因此处理器不会执行 JUMP

So we continue on to the next instruction...

继续下一条指令

...which is a JUMP 2.

JUMP 2

No conditional on this one, so we jump to instruction 2 no matter what.

JUMP 2 没有条件，直接执行！

Ok, so we're back at our SUBTRACT Register B from Register A. 6 minus 5 equals 1.

又回到寄存器 A-B，6-5=1

So 1 gets saved into register A.

A 变成 1

Next instruction.

下一条指令

We're back again at our JUMP NEGATIVE.

又是 JUMP NEGATIVE

1 is also a positive number, so the CPU continues on to the JUMP 2, looping back around again

因为 1 还是正数，因此 JUMP NEGATIVE 不会执行，来到下一条指令，JUMP 2

to the SUBTRACT instruction.

又来减一次

This time is different though.

这次就不一样了

1 minus 5 is negative 4.

1-5=-4

And so the ALU sets its negative flag to true for the first time.

这次ALU的 "负数标志" 是真

Now, when we advance to the next instruction,

现在下一条指令

JUMP_NEGATIVE 5, the CPU executes the jump to memory location 5.

JUMP NEGATIVE 5， CPU 的执行跳到内存地址 5

We're out of the infinite loop!

跳出了无限循环！

Now we have a ADD B to A. Negative 4 plus 5, is positive 1, and we save that into Register A.

现在的指令是 ADD B A，-4+5=1，1 存入寄存器 A

Next we have a STORE instruction that saves Register A into memory address 13.

下一条指令 STORE_A 13，把 A 的值存入内存地址 13

Lastly, we hit our HALT instruction and the computer rests.

最后碰到 HALT 指令，停下来.

So even though this program is only 7 instructions long, the CPU ended up executing 13 instructions,

虽然程序只有 7 个指令，但 CPU 执行了 13 个指令，

and that's because it looped twice internally.

因为在内部循环了 2 次.

This code calculated the remainder if we divide 5 into 11, which is one.

这些代码其实是算余数的，11除5余1

With a few extra lines of code, we could also keep track of how many loops we did, the count

如果加多几行指令，我们还可以跟踪循环了多少次

of which would be how many times 5 went into 11… we did two loops, so that means 5 goes

11除5，循环2次

into 11 two times... with a remainder of 1.

余1

And of course this code could work for any two numbers, which we can just change in memory

当然，我们可以用任意2个数，7和81，18和54，什么都行

to whatever we want: 7 and 81, 18 and 54, it doesn't matter

当然，我们可以用任意2个数，7和81，18和54，什么都行

that's the power of software!

这就是软件的强大之处

Software also allowed us to do something our hardware could not.

软件还让我们做到硬件做不到的事

Remember, our ALU didn't have the functionality to divide two numbers,

ALU 可没有除法功能

instead it's the program we made that gave us that functionality.

是程序给了我们这个功能.

And then other programs can use our divide program to do even fancier things.

别的程序也可以用我们的除法程序，来做其他事情

And you know what that means.

这意味着

New levels of abstraction!

一层新抽象！

So, our hypothetical CPU is very basic – all of its instructions are 8 bits long,

我们这里假设的 CPU 很基础，所有指令都是 8 位，

with the opcode occupying only the first four bits.

操作码只占了前面 4 位

So even if we used every combination of 4 bits, our CPU would only be able to support

即便用尽 4 位，

a maximum of 16 different instructions.

也只能代表 16 个指令

On top of that, several of our instructions used the last 4 bits to specify a memory location.

而且我们有几条指令，是用后 4 位来指定内存地址

But again, 4 bits can only encode 16 different values,

因为 4 位最多只能表示 16 个值，

meaning we can address a maximum of 16 memory locations that's not a lot to work with.

所以我们只能操作 16 个地址，这可不多.

For example, we couldn't even JUMP to location 17,

我们甚至不能 JUMP 17

because we literally can't fit the number 17 into 4 bits.

因为 4 位二进制无法表示数字 17

For this reason, real, modern CPUs use two strategies.

因此，真正的现代 CPU 用两种策略

The most straightforward approach is just to have bigger instructions, with more bits,

最直接的方法是用更多位来代表指令，

like 32 or 64 bits.

比如 32 位或 64 位

This is called the instruction length.

这叫指令长度

Unsurprisingly.

毫不意外

The second approach is to use variable length instructions.

第二个策略是 "可变指令长度"

For example, imagine a CPU that uses 8 bit opcodes.

举个例子，比如某个 CPU 用 8 位长度的操作码

When the CPU sees an instruction that needs no extra values, like the HALT instruction,

如果看到 HALT 指令，HALT 不需要额外数据

it can just execute it immediately.

那么会马上执行.

However, if it sees something like a JUMP instruction, it knows it must also fetch

如果看到 JUMP，它得知道位置值

the address to jump to, which is saved immediately behind the JUMP instruction in memory.

这个值在 JUMP 的后面

This is called, logically enough, an Immediate Value.

这叫 "立即值"

In such processor designs, instructions can be any number of bytes long,

这样设计，指令可以是任意长度

which makes the fetch cycle of the CPU a tad more complicated.

但会让读取阶段复杂一点点

Now, our example CPU and instruction set is hypothetical,

要说明的是，我们拿来举例的 CPU 和指令集都是假设的，

designed to illustrate key working principles.

是为了展示核心原理

So I want to leave you with a real CPU example.

所以我们来看个真的 CPU 例子.

In 1971, Intel released the 4004 processor.

1971年，英特尔发布了 4004 处理器.

It was the first CPU put all into a single chip

这是第一次把 CPU 做成一个芯片，

and paved the path to the intel processors we know and love today.

给后来的英特尔处理器打下了基础

It supported 46 instructions, shown here.

它支持 46 个指令

Which was enough to build an entire working computer.

足够做一台能用的电脑

And it used many of the instructions we've talked about like JUMP ADD SUBTRACT and LOAD.

它用了很多我们说过的指令，比如 JUMP ADD SUB LOAD

It also uses 8-bit immediate values, like we just talked about, for things like JUMPs,

它也用 8 位的"立即值"来执行 JUMP，

in order to address more memory.

以表示更多内存地址.

And processors have come a long way since 1971.

处理器从 1971 年到现在发展巨大.

A modern computer processor, like an Intel Core i7, has thousands of different instructions

现代 CPU，比如英特尔酷睿 i7，有上千个指令和指令变种

and instruction variants, ranging from one to fifteen bytes long.

长度从1到15个字节.

For example, there's over a dozens different opcodes just for variants of ADD!

举例，光 ADD 指令就有很多变种！

And this huge growth in instruction set size is due in large part to extra bells and whistles

指令越来越多，是因为给 CPU 设计了越来越多功能

that have been added to processor designs overtime, which we'll talk about next episode.

下集我们会讲

See you next week!

下周见