编译原理的词法分析器设计与实现

https://download.csdn.net/download/xuptacm/10383733
完整项目 保证可以运行
运行环境要求低
DEV_C++ 可以运行

int wordanalysis()//词法分析，分为3部分，完整项目在下面
{
    if ((a[i] >= 'A'&&a[i] <= 'Z')||(a[i]>='a'&&a[i]<='z'))  //分析标识符和保留字
    {
        char word[10];
        char pro[100][100] = { "PROGRAM", "BEGIN", "END", "VAR", "INTEGER", "WHILE", "IF", "THEN", "ELSE", "DO", "PROCEDURE" ,
                                "char","int","if","else","var" ,"return","break","do","while","for","double","float","short"}; //保留字表

        int n = 0;
        word[n++] = a[i++];
        //若字符为A~Z或0~9，则继续读取
        while ((a[i] >= 'A'&&a[i] <= 'Z') || (a[i] >= '0' && a[i] <= '9')||(a[i]>='a'&&a[i]<='z'))
        {
            word[n++] = a[i++];
        }
        word[n] = '\0';
        i--;

        //判断该标识符是否为保留字
        for (n = 0; n < 100; n++)
        {
            if (strcmp(word, pro[n]) == 0)
            {
                printf("%s\t(%d,-) 保留字\n", pro[n], n + 1);
                return 3;
            }
        }

        //判断标识符长度是否超出规定
        if (strlen(word)>10)
        {
            printf("%s\t Lexical analyzer read Unlawful identifier",word);
            return 0;
        }

        //判断该标识符是否存在标识符表中
        int m = 0;
        if (line != 0)
        {
            int q = 0;
            while (qif (strcmp(word, mark[q++]) == 0)
                {
                    printf("%s\t(12,%d) 标识符\n", word, q);
                    return 3;
                }
            }

        }

        //将该标识符保存到标识符表中
        strcpy(mark[line], word);

        printf("%s\t(12, %d) 标识符\n", word, line + 1);
        line++;
        return 3;

    }

    else if (a[i] >= '0' && a[i] <= '9')  //分析常数
    {
        char x[100];
        int n = 0, sum;
        x[n++] = a[i++];
     //判断字符是否是0~9
        while (a[i] >= '0' && a[i] <= '9')
        {
            x[n++] = a[i++];
        }
        x[n] = '\0';
        i--;
        int num = atoi(x); //将字符串转换成int型

        //判断该常数是否存在于常数表中
        if (row != 0)
        {   
            int y; 
            for (y = 0; y < 1000; y++)
            {
                int w = number[y][0];
                sum = 0;
                int d;
                for (d = 1; d <= number[y][0]; d++)
                {
                    w = w - 1;
                    sum = sum + number[y][d] * pow(2, w);
                }
                if (num == sum)
                {
                    printf("%d\t(13,%d)\n", num, y + 1);
                    return 3;
                }
            }
        }
        int z = num, c = num;
        int m = 0;
        do        //计算是几位二进制数
        {
            z = z / 2;
            m++;
        } while (z != 0);

        for (n = m; n > 0; n--)  //将二进制保存于常数表中
        {
            number[row][n] = c % 2;
            c = c / 2;
        }
        number[row][0] = m;

        int line = row;
        printf("%d\t(13,%d)\n", num, line + 1);
        row++;

        return 3;
    }

    else                      //分析符号
        switch (a[i])
    {
        case ' ':
        case '\n':
            return -1;
        case '#': return 0;
        case '=':printf("=\t(14,-)\n"); return 3;
        case '<':
            i++;
            if (a[i] == '=')
            {
                printf("<= \t(16,-)\n");
                return 3;
            }
            else if (a[i] == '>')
            {
                printf("<>\t(19,-)\n");
                return 3;
            }
            else
            {
                i--;
                printf("<\t(15,-)\n");
                return 3;
            }
        case '>':
            i++;
            if (a[i] == '=')
            {
                printf(">=\t(18,-)\n");
                return 3;
            }
            else
            {
                i--;
                printf(">\t(17,-)\n");
                return 3;
            }
        case '+': printf("+\t(20,-)\n"); return 3;
        case '-': printf("-\t(21,-)\n"); return 3;
        case '*': printf("*\t(22,-)\n"); return 3;
        case '/': 
            i++;
            if(a[i]!='/'){
                i--;
                printf("/\t(23,-)\n"); return 3;
            }

            else{

                while(1){
                    if(a[i++]=='\n')
                        return -1;
                }
                printf("//\t(35,-)\n");return 3;

            }
    }
}

}

https://download.csdn.net/download/xuptacm/10383733