SICP : The Elements of Programming

好的计算机编程语言应具备的三个特性

基础单元表达式，计算机编程语言最最最基础单元，理应具备的表达式
组合的能力，能够通过基础单元表达式组合成更复杂的元素
抽象的能力，能通过复杂的元素抽象成更高层的单元

基础单元表达式

加减乘除。
其实加减乘除才是最牛逼的接口，左边和右边提供输入参数，最后返回计算结果，又是最基础的可以组合成更复杂单元的基础单元表达式接口。

SICP javascript版本，可以在网页版解释运行环境中写代码

在线快乐地编写javascript代码并运行

137 + 349; 
486

1000 - 334; 
666

5 * 99; 
495

10 / 4; 
2.5

2.7 + 10; 
12.7

3 * 2 * (3 - 5 + 4) + 27 / 6 * 10;

Naming and the Environment

const pi = 3.14159;
const radius = 10;

pi * radius * radius;
314.159

const circumference = 2 * pi * radius;
circumference;
62.8318

Evaluating Operator Combinations

One of our goals in this chapter is to isolate issues about thinking procedurally.

(2 + 4 * 6) * (3 + 12);

在这里插入图片描述
== In general, we shall see that recursion is a very powerful technique for dealing with
hierarchical, treelike objects.==

Compound Functions

Numbers and arithmetic operations are primitive data and functions.
Nesting of combinations provides a means of combining operations.
Constant declarations that associate names with values provide a limited means of abstraction.

function square(x)
{
    return x*x;
}

square(21);

square(2 + 5);

function square(x)
{
    return x*x;
}

square(21);

square(2 + 5);

square(square(3));

// x² + y² can be expressed as
// square(x) + square(y)

function sum_of_squares(x,y)
{
    return square(x) + square(y);
}

sum_of_squares(3,4);
25

// now we can use sum_of_squares as a building block in constructing further functions.
function f(a)
{
    return sum_of_squares(a + 1,a * 2);
}

f(5);
136

上述计算模型 square及sum_of_squares都可以是外部领域模型，所以一个大型系统可以由很多外部领域模型组合成复杂功能。

在这里插入图片描述
突然间恍然大悟上面这个计算机中的精灵图：EVAL计算出规则APPLY的结果，计算得到的结果又再次成为规则APPLY的输入，然后EVAL又再次计算得到规则APPLY的结果

Conditional Expressions and Predicates


function abs(x)
{
    return x >= 0 ? x : -x;
}

abs(3);

abs(-43);

function greater_or_equal(x,y)
{
    return x > y || x === y;
}

greater_or_equal(3,4);

greater_or_equal(4,4);

greater_or_equal(6,2);


function greater_or_equal(x,y)
{
    return !( x < y);
}

练习题

10;

5 + 3 + 4;

9 - 1;

6 / 2;

2 * 4 + (4 - 6);

const a  = 3;
const b = a + 1;

a + b + a * b;

a === b;

b > a && b < a * b ? b : a;

a === 4 ? 6 : b === 4 ? 6 + 7 + a : 25;

exercise 1.1 exercise 1.2 exercise 1.3

(5 + 4 + (2 - (3 - (6 + 4 / 5)))) / (3 * (6 - 2) * (2 - 7));

function square(x)
{
    return x * x;
}

function square_of_the_larger_two(x,y,z)
{
    return x > y ? 
    (y > z ? (square(x) + square(y)) : (square(x) + square(z))) 
    :
    (x > z ? (square(x) + square(y)) : (square(y) + square(z)));
}

square_of_the_larger_two(2,3,4);

square_of_the_larger_two(2,1,3);