-
编译原理语法分析器编程
编译原理语法分析器编程- The translation principle grammar analyzer programs
- 2022-05-10 20:22:50下载
- 积分：1
-
pascals compiler source code and explanations
pascals编译程序源代码及说明-pascals compiler source code and explanations
- 2022-01-26 19:59:33下载
- 积分：1
-
avr 程序开发中可以用到BOOTLOADER程序、ISP汇编程序ISP C程序
avr 程序开发中可以用到BOOTLOADER程序、ISP汇编程序ISP C程序-avr program development process can be used BootLoader, ISP assembler procedures ISP C
- 2022-10-03 17:50:03下载
- 积分：1
-
编译原理 实现多个表达式的识别和判断出错
编译原理 实现多个表达式的识别和判断出错-Compilation Principle of multiple expressions of recognition and assessment of error
- 2022-03-26 03:32:58下载
- 积分：1
-
Libboost_regex regular expression
基于libboost_regex的正则表达式测试程序，可用于入门者的参考代码。-Libboost_regex regular expression-based test procedure can be used for a reference code for beginners.
- 2022-03-21 21:31:53下载
- 积分：1
-
object pascal编译器源码
object pascal编译器源码-object pascal compiler source
- 2022-05-30 22:16:49下载
- 积分：1
-
这是关于小Python的实现
this is about tiny python implementation
- 2022-06-02 23:04:16下载
- 积分：1
-
词法分析程序，可对以下的C源程序进行分析
词法分析程序，可对以下的C源程序进行分析：main() {int a[12] ,sum for(i=1 i=1 j--){ if(i==j&&i+j==13)sum+=a[i][j] } } printf("%c",sum) }-lexical analysis procedures, right below the C source code analysis : main () (int a [12] [12], the sum for (i = 1 ilt; I = 12) (for (j = 1 JLT; J = 12) Scanf ( "% d", a [i] [j])) for (i = 12 IGT; a = i--) (for (j = 12 IGT; = 1 j--) (if (i == ji j == 13) sum = a [i] [j])) printf ( "% c", sum
- 2022-07-19 15:44:58下载
- 积分：1
-
1、构造该算符优先文法的优先关系矩阵或优先函数；
2、输入串应是词法分析的输出二元式序列，即某算术表达式“实验项目一”的输出结果。输出为输入串是否为该文法定
1、构造该算符优先文法的优先关系矩阵或优先函数；
2、输入串应是词法分析的输出二元式序列，即某算术表达式“实验项目一”的输出结果。输出为输入串是否为该文法定义的算术表达式的判断结果。
3、算符优先分析过程应能发现输入串出错。
-1, construction of the operator the priority of the priority relation matrix grammar or priority function 2, input strings should be the output of lexical analysis of binary sequence, that is, the arithmetic expression of a
- 2023-06-11 00:45:03下载
- 积分：1
-
(1)构造文法G的LR(0)项目
(2)构造文法G的LR(0)项目集规范族及识别活前缀的DFA
(3)证明文法G不是LR(0)文法而是SLR(1)文法，...
(1)构造文法G的LR(0)项目
(2)构造文法G的LR(0)项目集规范族及识别活前缀的DFA
(3)证明文法G不是LR(0)文法而是SLR(1)文法，并构造SLR(1)分析表
(4)设计LR语法分析程序，且能输出分析过程
(5)列举两个例子测试语法分析程序（识别失败一例，识别成功一例，后者推导步骤不得少于10步）
-(1) construct grammar G of the LR (0) item (2) construct grammar G of the LR (0) item sets standards and identification of family living prefix DFA (3) prove that the grammar G is not LR (0) grammar, but the SLR (1 ) grammar, and constructs SLR (1) analysis table (4) design LR parser, and can output analysis process (5) cited two examples of parsing test procedures (identification fails an example, identify the success of a case, the latter shall not be derived step Less than 10 steps)
- 2022-05-16 14:03:06下载
- 积分：1
