合肥工业大学编译原理实验一词法分析设计

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实验一 构造词法分析器

实验题目：识别字符串的标识符，关键字，关系运算符，分界符, 常数，算术运算符以及语法错误
实验示例：If i=0 then n++;
a﹤= 3b %)
实验输出：

算法复杂度:$O(n^2)$
部分代码展示：

while (fgets(string1, 100, fp) != nullptr) {
        int len = strlen(string1);
        int column = 0;
        int flag = 1;
        int first, flaga = 1;
        i = 0;
        while (i < len) {
            first = i;
            if (( string1[first] >= '0' && string1[first] <= '9' )) { //常数
                while (( string1[i] != ' ' ) && ( string1[i] != '\n' )) {
                    buffer[k++] = string1[i++];
                }
                column++;
                number_judge(buffer, column, row);
                memset(buffer, 0, sizeof(buffer));
                k = 0;
            } else if (( string1[first] >= 'a' && string1[first] <= 'z' ) ||
                       ( string1[first] >= 'A' && string1[first] <= 'Z' )) { //标识符或者关键字
                while ((( string1[i] != ' ' ) && ( string1[i] != '\n' ) && ( string1[i] >= 'a' && string1[i] <= 'z' ) ||
                        ( string1[i] >= 'A' && string1[i] <= 'Z' ) || ( string1[i] >= '0' && string1[i] <= '9' ))) {
                    buffer[k++] = string1[i++];
                }
                column++;
                alphabet_judge(buffer, column, row);
                memset(buffer, 0, sizeof(buffer));
                k = 0;
            } else if (relation_operator_judge(string1 + first, i, column, row)) { //关系运算符
                k = 0;
                i++;
            } else if(arithmetical_operator_judge(string1+first,&i,column,row)){
                i++;
            }else if(division_operator_judge(string1+first,i,column,row)){
                i++;
            }else if (string1[i] == ' '||string1[i] == '\n') {
                i++;
            }else{
                struct error er(string1[i], row, column);
                i++;
            } 
        }

完整代码链接：pick me!
总结：个人认为代码写得比较冗杂，应该可以通过string类中的字符串处理、切割等函数，进一步简化。