CSE340Spring2009HW1_Solutions

CSE340Spring2009HW1_Solutions - Arizona State University...

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Arizona State University Fulton School of Engineering Computer Science and Engineering CSE 340 Fall 2009 Homework 1 Solution Chapter 4 4.12 expr -> expr + term | term term -> term * factor | term % factor | factor factor -> base ^ factor | base base -> ( expr ) | number number -> number digit | digit digit -> 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 4.22 This eliminates the possibility of having expressions with many terms added together. For example, it will be impossible to generate 2+3+4 using this grammar. In fact the only way to parse this input is to use term+term, so either the 2 matches a term and 3+4 matches the other term or 2+3 matches a term and 4 matches the other term. Both cases are not possible because the rule for term has no addition operator. 4.41 a. Consider the number 11.11 which is of the form number 1 . number 2 . The value for number1 is 11 (eleven). The value associated with the other number is1/10+1/100. Since the rule for numbers is of the form number -> number digit | digit it is easy to calculate the value of number 1 because for each added digit we multiple the current value by 10 and add the digit. This is not the case for number 2 because the value associated with a new digit depends on it position and that information is not available when we parse from left to right. When a scanner recognizes the number, we get a string that we could process from one end to get the integer value and from the other end to get the fractional value. b. The new rule looks like frac_number -> digit frac_number | digit number -> number digit | digit decimal_number -> number [.frac_number]
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This is clearly better for the fractional part because we can express the value of the fractional part as vf(number) = v(digit)/10+vf(number)/10 where vf stands for the fractional value and v(digit) is the value of a digit. So, the value of the 11 that is in the fractional part is v(10)/10+ vf(1)/10 = 0.1 +0.1/10 = 0.11 For the integer part of the number, we can express the decimal value vd as vd(number) = vd(number)*10+v(digit) c) 4.42 a) expr-> ( list ) | a list-> expr { , expr } b)
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c) #include <stdio.h> #include <stdlib.h> #include <ctype.h> int token; void expr(void); void list(void); void match(char); void getToken(void) { token = getChar(); } void error(void) { printf("\nAn error occured on input.\n"); } void expr (void)
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{ switch(token) { case 'a': match('a'); break; case '(': match('('); list(); match(')'); break; default: error(); break; } } void list (void) { expr (); while (token == ‘,’ ) { match (‘,’) ; expr () ; } } void match(char c) { if (token == c) getToken(); else error(); } int command(void)
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CSE340Spring2009HW1_Solutions - Arizona State University...

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