Implementation of LEXICAL ANALYZER for IF STATEMENT with the help of Lex Tool-
C Program-
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<ctype.h>
int isKeyword(char buffer[]) {
char keywords[32][10] = {
"auto",
"break",
"case",
"char",
"const",
"continue",
"default",
"do",
"double",
"else",
"enum",
"extern",
"float",
"for",
"goto",
"if",
"int",
"long",
"register",
"return",
"short",
"signed",
"sizeof",
"static",
"struct",
"switch",
"typedef",
"union",
"unsigned",
"void",
"volatile",
"while"
};
int i, flag = 0;
for (i = 0; i < 32; ++i) {
if (strcmp(keywords[i], buffer) == 0) {
flag = 1;
break;
}
}
return flag;
}
int main() {
char ch, buffer[15], operators[] = "+-*/%=";
FILE * fp;
int i, j = 0;
fp = fopen("program.txt", "r");
if (fp == NULL) {
printf("error while opening the file\n");
exit(0);
}
while ((ch = fgetc(fp)) != EOF) {
for (i = 0; i < 6; ++i) {
if (ch == operators[i])
printf("%c is operator\n", ch);
}
if (isalnum(ch)) {
buffer[j++] = ch;
} else if ((ch == ' ' || ch == '\n') && (j != 0)) {
buffer[j] = '\0';
j = 0;
if (isKeyword(buffer) == 1)
printf("%s is keyword\n", buffer);
else
printf("%s is indentifier\n", buffer);
}
}
fclose(fp);
return 0;
}
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