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Page 591 if((tok_type == DELIMITER) && *token=='+' || *token == '-') { op = *token; get_token(); } eval_exp6(answer); if(op == '-') *answer = -(*answer); } /* Process a parenthesized expression */ void eval_exp6(double *answer) { if((*token == '(')) { get_token(); eval_exp2(answer); if(*token != ')') serror(1); get_token(); } else atom(answer); } /* Get the value of a number */ void atom(double *answer) { if(tok_type == NUMBER) { *answer = atof(token); get_token(); return; } serror(0); /* otherwise syntax error in expression */ } /* Return a token to the input stream */ void putback(void) { char *t; t = token; for(; *t; t++) prog--; }
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Page 592 /* Display a syntax error */ void serror(int error) { static char *e[]= { ''Syntax Error", "Unbalanced Parentheses", "No Expression Present", "Division by Zero" }; printf("%s\n", e[error]); } /* Return the next token */ void get_token(void) { register char *temp; tok_type = 0; temp = token; *temp = '\0'; if(!*prog) return; /* at end of expression */ while(isspace(*prog)) ++prog; /* skip over white space */ if(strchr("+-*/%^=()", *prog)){ tok_type = DELIMITER; /* advance to next char */ *temp++ = *prog++; } else if(isalpha(*prog)) { while(!isdelim(*prog)) *temp++ = *prog++; tok_type = VARIABLE; } else if(isdigit(*prog)) { while(!isdelim(*prog)) *temp++ = *prog++; tok_type = NUMBER; } *temp = '\0'; } /* Return true if c is a delimiter */
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Page 593 int isdelim(char c) { if(strchr(" +-/*%^=()", c) || c==9 || c=='\r' || c==0) return 1; return 0; }
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The parser as it is shown can handle the following operators: +, *, /, % In addition, it can handle , integer exponentiation (^) and the unary minus The parser can also deal with parentheses correctly Notice that it has six levels as well as the atom( ) function, which returns the value of a number As discussed, the two globals token and tok_type return, respectively, the next token and its type from the expression string The pointer prog points to the string that holds the expression The simple main( ) function that follows demonstrates the use of the parser:
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/* Parser demo program */ #include <stdlibh> #include <ctypeh> #include <stdioh> #include <stringh> char *prog; void eval_exp(double *answer); int main(void) { double answer; char *p; p = (char *) malloc(100); if(!p) { printf(''Allocation failure\n"); exit(1); } /* Process expressions until a blank line is entered */ do { prog = p;
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Page 594 printf("Enter expression: "); gets(prog); if(!*prog) break; eval_exp(&answer); printf(''Answer is: %2f\n", answer); } while(*p); return 0; }
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To understand exactly how the parser evaluates an expression, work through the following expression (Assume that prog points to the start of the expression) 10 3 * 2 When eval_exp( ), the entry point into the parser, is called, it gets the first token If the token is null, the function prints the message No Expression Present and returns However, in this case, the token contains the number 10 Since the token is not null, eval_exp2( ) is called As a result, eval_exp2( ) calls eval_exp3( ), and eval_exp3( ) calls eval_exp4( ) , which in turn calls eval_exp5 ( ) Then eval_exp5( ) checks whether the token is a unary plus or minus, which in this case, it is not, so eval_exp6( ) is called At this point eval_exp6( ) recursively calls either eval_exp2( ) (in the case of a parenthesized expression) or atom( ) to find the value of a number Since the token is not a left parentheses, atom( ) is executed, and *answer is assigned the value 10 Next, another token is retrieved, and the functions begin to return up the chain The token is now the operator and the , functions return up to eval_exp2( ) What happens next is very important Because the token is it is saved in op The parser then gets , the next token, which is 3, and the descent down the chain begins again As before, atom( ) is entered The value 3 is returned in *answer, and the token * is read This causes a return back up the chain to eval_exp3( ), where the final token 2 is read At this point, the first arithmetic operation occurs the multiplication of 2 and 3 The result is returned to eval_exp2( ) , and the subtraction is performed The subtraction yields the answer 4 Although this process may seem complicated at first, working through some other examples on your own will clarify the parser's operation This parser would be suitable for use by a desktop calculator, as illustrated by the previous program It also could be used in a limited database Before it could be used in a computer language or in a sophisticated calculator, however, it would need the ability to handle variables This is the subject of the next section
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