用c语言编写一个LL1语法分析器

要编写一个 LL(1) 语法分析器，需要先进行语法分析，然后生成一张预测分析表，最后使用该表进行语法分析。

下面是一个简单的 LL(1) 语法分析器的 C 语言代码示例：

#include <stdio.h>
#include <ctype.h>
#include <string.h>

#define MAX_PROD 10
#define MAX_TERM 10

char grammar[MAX_PROD][MAX_TERM];
char first_set[MAX_TERM][MAX_TERM];
char follow_set[MAX_TERM][MAX_TERM];
char predict_set[MAX_PROD][MAX_TERM];
char terminals[MAX_TERM];
char non_terminals[MAX_TERM];
int no_of_productions, no_of_terminals, no_of_non_terminals;

void first(char symbol) {
    int i, j, k;
    for (i = 0; i < no_of_productions; i++) {
        if (grammar[i][0] == symbol) {
            if (islower(grammar[i][2])) {
                for (k = 0; k < no_of_terminals; k++) {
                    if (grammar[i][2] == terminals[k]) {
                        first_set[symbol - 'A'][k] = '1';
                        break;
                    }
                }
            } else {
                first(grammar[i][2]);
                for (j = 0; j < no_of_terminals; j++) {
                    if (first_set[grammar[i][2] - 'A'][j] == '1') {
                        first_set[symbol - 'A'][j] = '1';
                    }
                }
            }
        }
    }
}

void follow(char symbol) {
    int i, j, k;
    if (symbol == grammar[0][0]) {
        follow_set[symbol - 'A'][no_of_terminals - 1] = '1';
    }
    for (i = 0; i < no_of_productions; i++) {
        for (j = 2; j < strlen(grammar[i]); j++) {
            if (grammar[i][j] == symbol) {
                if (grammar[i][j + 1] != '\0') {
                    if (islower(grammar[i][j + 1])) {
                        for (k = 0; k < no_of_terminals; k++) {
                            if (grammar[i][j + 1] == terminals[k]) {
                                follow_set[symbol - 'A'][k] = '1';
                                break;
                            }
                        }
                    } else {
                        int l;
                        for (l = 0; l < no_of_terminals; l++) {
                            if (first_set[grammar[i][j + 1] - 'A'][l] == '1') {
                                follow_set[symbol - 'A'][l] = '1';
                            }
                        }
                        if (first_set[grammar[i][j + 1] - 'A'][no_of_terminals - 2] == '1') {
                            follow(grammar[i][0]);
                        }
                    }
                } else if (grammar[i][j + 1] == '\0') {
                    follow(grammar[i][0]);
                }
            }
        }
    }
}

void predict() {
    int i, j, k, l;
    for (i = 0; i < no_of_productions; i++) {
        for (j = 0; j < no_of_terminals; j++) {
            if (first_set[grammar[i][2] - 'A'][j] == '1') {
                predict_set[i][j] = grammar[i][2];
            }
        }
        if (first_set[grammar[i][2] - 'A'][no_of_terminals - 2] == '1') {
            for (k = 0; k < no_of_terminals; k++) {
                if (follow_set[grammar[i][0] - 'A'][k] == '1') {
                    if (predict_set[i][k] != grammar[i][2]) {
                        predict_set[i][k] = grammar[i][2];
                    } else {
                        for (l = 0; l < no_of_terminals; l++) {
                            if (predict_set[i][l] == grammar[i][2]) {
                                predict_set[i][l] = '\0';
                            }
                        }
                    }
                }
            }
        }
    }
}

int main() {
    int i, j;
    printf("Enter the number of productions: ");
    scanf("%d", &no_of_productions);
    printf("Enter the productions:\n");
    for (i = 0; i < no_of_productions; i++) {
        scanf("%s", grammar[i]);
        non_terminals[i] = grammar[i][0];
    }
    no_of_non_terminals = i;
    printf("Enter the number of terminals: ");
    scanf("%d", &no_of_terminals);
    printf("Enter the terminals:\n");
    for (i = 0; i < no_of_terminals; i++) {
        scanf(" %c", &terminals[i]);
    }
    strcpy(terminals + i, "$");
    no_of_terminals++;
    for (i = 0; i < no_of_non_terminals; i++) {
        for (j = 0; j < no_of_terminals; j++) {
            first_set[non_terminals[i] - 'A'][j] = '0';
            follow_set[non_terminals[i] - 'A'][j] = '0';
        }
    }
    for (i = 0; i < no_of_non_terminals; i++) {
        first(non_terminals[i]);
    }
    follow(grammar[0][0]);
    predict();
    printf("Predictive Parsing Table:\n");
    printf("Non-terminal\tTerminal\tProduction\n");
    for (i = 0; i < no_of_non_terminals; i++) {
        for (j = 0; j < no_of_terminals; j++) {
            if (predict_set[i][j] != '\0') {
                printf("%c\t\t%c\t\t%c->%s\n", non_terminals[i], terminals[j], non_terminals[i], predict_set[i]);
            }
        }
    }
    return 0;
}

该程序可以读入文法和终结符号，然后计算出每个符号的 FIRST 集和 FOLLOW 集，并使用它们构建预测分析表。最后，程序输出预测分析表，以便进行语法分析。

这只是一个简单的 LL(1) 语法分析器实现，如果需要更完整的实现，可以使用更复杂的数据结构和算法，以及更精细的错误处理机制。