2.3 节的练习
2.3.1
构建一个语法制导翻译方案，该方案把算数表达式从中缀表达式翻译成前缀表达式。
解答
产生式：
expr -> expr + term
| expr - term
| term
term -> term * factor
| term / factor
| factor
factor -> digit | (expr)
翻译方案：
expr -> {print("+")} expr + term
| {print("-")} expr - term
| term
term -> {print("*")} term * factor
| {print("/")} term / factor
| factor
factor -> digit {print(digit)}
| (expr)
2.3.2
构建一个语法制导翻译方案，该方案把算数表达式从后缀表达式翻译成中缀表达式。
解答
产生式：
expr -> expr expr +
| expr expr -
| expr expr *
| expr expr /
| digit
翻译方案：
expr -> expr {print("+")} expr +
| expr {print("-")} expr -
| {print("(")} expr {print(")*(")} expr {print(")")} *
| {print("(")} expr {print(")/(")} expr {print(")")} /
| digit {print(digit)}
疑问
参考答案是：
E -> {print("(")} E {print(op)} E {print(")"}} op | digit {print(digit)}
起初我的答案也是这样，但发现括号冗余比较严重，才改成现在的解答。虽然在多级优先级运算中，其实只能减少最低优先级运算的括号，不过看起来在这样的情况下还是能节约比较多括号的。这个解法正确否？
2.3.3
构建一个将整数翻译成罗马数字的语法制导翻译方案。
解答
辅助方法：
repeat(sign, times) // 将 sign 重复 times 次输出，如 repeat('a',2) = 'aa'
翻译方案：
num -> thousand hundred ten digit
{ num.roman = thousand.roman || hundred.roman || ten.roman || digit.roman;
print(num.roman)}
thousand -> low {thousand.roman = repeat('M', low.v)}
hundred -> low {hundred.roman = repeat('C', low.v)}
| 4 {hundred.roman = 'CD'}
| high {hundred.roman = 'D' || repeat('X', high.v - 5)}
| 9 {hundred.roman = 'CM'}
ten -> low {ten.roman = repeat('X', low.v)}
| 4 {ten.roman = 'XL'}
| high {ten.roman = 'L' || repeat('X', high.v - 5)}
| 9 {ten.roman = 'XC'}
digit -> low {digit.roman = repeat('I', low.v)}
| 4 {digit.roman = 'IV'}
| high {digit.roman = 'V' || repeat('I', high.v - 5)}
| 9 {digit.roman = 'IX'}
low -> 0 {low.v = 0}
| 1 {low.v = 1}
| 2 {low.v = 2}
| 3 {low.v = 3}
high -> 5 {high.v = 5}
| 6 {high.v = 6}
| 7 {high.v = 7}
| 8 {high.v = 8}
2.3.4
构建一个将罗马数字翻译成整数的语法制导方案。
解答
产生式：
romanNum -> thousand hundred ten digit
thousand -> M | MM | MMM | ε
hundred -> smallHundred | C D | D smallHundred | C M
smallHundred -> C | CC | CCC | ε
ten -> smallTen | X L | L smallTen | X C
smallTen -> X | XX | XXX | ε
digit -> smallDigit | I V | V smallDigit | I X
smallDigit -> I | II | III | ε
翻译方案：
romanNum -> thousand hundred ten digit {romanNum.v = thousand.v || hundred.v || ten.v || digit.v; print(romanNun.v)}
thousand -> M {thousand.v = 1}
| MM {thousand.v = 2}
| MMM {thousand.v = 3}
| ε {thousand.v = 0}
hundred -> smallHundred {hundred.v = smallHundred.v}
| C D {hundred.v = smallHundred.v}
| D smallHundred {hundred.v = 5 + smallHundred.v}
| C M {hundred.v = 9}
smallHundred -> C {smallHundred.v = 1}
| CC {smallHundred.v = 2}
| CCC {smallHundred.v = 3}
| ε {hundred.v = 0}
ten -> smallTen {ten.v = smallTen.v}
| X L {ten.v = 4}
| L smallTen {ten.v = 5 + smallTen.v}
| X C {ten.v = 9}
smallTen -> X {smallTen.v = 1}
| XX {smallTen.v = 2}
| XXX {smallTen.v = 3}
| ε {smallTen.v = 0}
digit -> smallDigit {digit.v = smallDigit.v}
| I V {digit.v = 4}
| V smallDigit {digit.v = 5 + smallDigit.v}
| I X {digit.v = 9}
smallDigit -> I {smallDigit.v = 1}
| II {smallDigit.v = 2}
| III {smallDigit.v = 3}
| ε {smallDigit.v = 0}
疑问
参考答案和我的解法有一个显著的区别，如下：
LowHundreds → ε { LowHundreds.val = 0; }
| C { LowHundreds.val = 100; }
| CC { LowHundreds.val = 200; }
| CCC { LowHundreds.val = 300; }
所以他的最后一步获得翻译后的数字是用的加法：
RomanNumeral → Thousands Hundreds Tens Ones
{ RomanNumeral.val = Thousands.val + Hundreds.val + Tens.val + Ones.val;
printf(“%d
”, RomanNumeral.val;) }
看起来我的解法因为使用了更少的0而节约了一些空间，其他也没看出什么坏处，是这样？
2.3.5
构建一个将后缀表达式翻译成前缀表达式的语法制导翻译方案。
解答
产生式：
expr -> expr expr op | digit
翻译方案：
expr -> {print(op)} expr expr op | digit {print(digit)}