制作一個簡單的C語言詞法分析程序_用c語言編寫詞法分析程序-CSDN博客文章瀏覽閱讀322次。C語言的程序中，有很單詞多符號和保留字。一些單詞符號還有對應的左線性文法。所以我們需要先做出一個單詞字符表，給出對應的識別碼，然后跟據對應的表格來寫出程序。_用c語言編寫詞法分析程序https://blog.csdn.net/lijj0304/article/details/134078944

前置程序詞法分析器參考這個帖子??

遞歸下降語法分析程序設計與實現-CSDN博客文章瀏覽閱讀127次。制作一個簡單的C語言詞法分析程序_用c語言編寫詞法分析程序-CSDN博客文章瀏覽閱讀276次。C語言的程序中，有很單詞多符號和保留字。一些單詞符號還有對應的左線性文法。所以我們需要先做出一個單詞字符表，給出對應的識別碼，然后跟據對應的表格來寫出程序。_用c語言編寫詞法分析程序前置程序詞法分析器參考這個帖子??。https://blog.csdn.net/lijj0304/article/details/134331022

遞歸下降實現語法分析可以看這個??

1.程序目標

制作一個LL(1)語法分析程序，程序可以識別詞法分析器的輸出文件中的二元序列，拼湊出用戶輸入。通過表驅動程序，實現賦值語句的LL(1)文法的LL(1)分析過程。算式的語法如下：

G[S]: S→V=E????????E→TE′????????E′→ATE′|ε????????T→FT′????????T′→MFT′|ε????????F→ (E)|i????????A→+|-M→*|/????????V→i

2.程序設計

我根據給定的語法，計算處所需要用到的first集和follow集，接著做select集，然后可以做出LL(1)分析表：

	i	+	-	*	/	(	)	#
S	S→V=E
E	E→TE’					E→TE’
E’		E’→ATE’	E’→ATE’				E’→ ε	E’→ ε
T	T→FT’					T→FT’
T’		T’→ ε	T’→ ε	T’→MFT’	T’→MFT’		T’→ ε	T’→ ε
F	F→i					F→(E)
M				M→*	M→/
A		A→+	A→-
V	V→i

LL(1)的語法分析部分使用棧的思想來實現，定義了一個字符串的棧stack，LL1分析過程中產生的串存儲在棧中，同時對于輸入串有一個定位指針，棧頂元素和指針指向的字符比對，一樣則棧頂元素出棧，指針往后移動。?

下面是程序LL(1)分析流程圖

?

3.完整程序?

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAX_LEN 1000
char str[MAX_LEN];
char stack[MAX_LEN];
int top = 0;int LL1(char *str, char *stack) {int i = 0;while(str[i] != '#') {if(top < 0) return 0;if(stack[top] == str[i]) {printf("%c -> ", stack[top]);top--;i++;}else if(stack[top] == 'S') {if(str[i] == 'i') {stack[top++] = 'E';stack[top++] = '=';stack[top] = 'V';}else return 0;}else if(stack[top] == 'E') {if(str[i] == 'i') {stack[top++] = 'e';stack[top] = 'T';}else if(str[i] == '(') {stack[top++] = 'e';stack[top] = 'T';}else return 0;}else if(stack[top] == 'e') {if(str[i] == '+') {stack[top++] = 'e';stack[top++] = 'T';stack[top] = 'A';}else if(str[i] == '-') {stack[top++] = 'e';stack[top++] = 'T';stack[top] = 'A';}else if(str[i] == ')' || str[i] == '#') {printf("%c -> ", str[i]);top--;}else return 0;}else if(stack[top] == 'T') {if(str[i] == 'i') {stack[top++] = 't';stack[top] = 'F';}else if(str[i] == '(') {stack[top++] = 't';stack[top] = 'F';}else return 0;}else if(stack[top] == 't') {if(str[i] == '*') {stack[top++] = 't';stack[top++] = 'F';stack[top] = 'M';}else if(str[i] == '/') {stack[top++] = 't';stack[top++] = 'F';stack[top] = 'M';}else if(str[i] == '+' || str[i] == '-' || str[i] == ')' || str[i] == '#') {printf("%c -> ", str[i]);top--;}else return 0;}else if(stack[top] == 'F') {if(str[i] == 'i') stack[top] = 'i';else if(str[i] == '(') {stack[top++] = ')';stack[top++] = 'E';stack[top] = '(';}else return 0;}else if(stack[top] == 'A') {if(str[i] == '+')stack[top] = '+';else if(str[i] == '-')stack[top] = '-';else return 0;}else if(stack[top] == 'M') {if(str[i] == '*')stack[top] = '*';else if(str[i] == '/')stack[top] = '/';else return 0;}else if(stack[top] == 'V') {if(str[i] == 'i') stack[top] = 'i';else return 0;}else return 0;}return 1;
}int main() {for(int m = 1; m <= 4; m++) {printf("test%d:\n", m);char txt[] = "./lexical/analyze";char num[6];sprintf(num, "%d.txt", m);strcat(txt, num);FILE *fp = fopen(txt, "r");char buf[MAX_LEN] = "";char input[MAX_LEN] = "";fgets(buf, MAX_LEN, fp);int i = 0, j = 0;for(int k = 0; k < strlen(buf); k++) {if(buf[k] == '1' && buf[k+1] == ',') {str[i++] = 'i';k += 3;while(1) {if(buf[k] == ')' && buf[k+1] == ' ')break;input[j++] = buf[k++];}continue;}if(buf[k] == ',' && buf[k+1] == ' ') {k += 2;while(1) {if(buf[k] == ')' && buf[k+1] == ' ')break;str[i++] = buf[k];input[j++] = buf[k++];}}}printf("Input scentence: %s\n", input);str[i] = '#';fclose(fp);stack[0] = 'S', top = 0;int flag = LL1(str, stack);if(flag == 1) {printf("end\n");printf("Gramma legal: %s\n", str);}else {printf("error\n");printf("Gramma illegal\n");}}return 0;
}

?4.運行測試

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