前提条件–解析|设置2(自下而上或Shift减少解析器)
Shift Reduce解析器尝试以与自底向上解析相同的方式来构建解析，即，解析树是从叶子(底部)到根(上)构建的。移位减少解析器的一种更通用的形式是LR解析器。
该解析器需要一些数据结构，即
- 用于存储输入字符串的输入缓冲区。
- 用于存储和访问生产规则的堆栈。
基本操作–
- 移位：这涉及将符号从输入缓冲区移到堆栈上。
- 减少：如果手柄出现在堆栈顶部，则通过使用适当的生产规则进行减少，即从堆栈中弹出生产规则的RHS并将生产规则的LHS推入堆栈。
- 接受：如果堆栈中仅存在开始符号且输入缓冲区为空，则解析操作称为接受。当获得接受动作时，意味着成功解析。
- 错误：在这种情况下，解析器既不能执行shift操作也不能执行reduce操作，甚至无法接受操作。
示例1 –考虑语法
S –> S + S
S –> S * S
S –> id
对输入字符串“ id + id + id”执行Shift Reduce解析。
示例2 –考虑语法
E –> 2E2
E –> 3E3
E –> 4
对输入字符串“ 32423”执行Shift Reduce解析。
以下是C-中的实现
C++
// Including Libraries
#include
using namespace std;
// Global Variables
int z = 0, i = 0, j = 0, c = 0;
// Modify array size to increase
// length of string to be parsed
char a[16], ac[20], stk[15], act[10];
// This Function will check whether
// the stack contain a production rule
// which is to be Reduce.
// Rules can be E->2E2 , E->3E3 , E->4
void check()
{
// Coping string to be printed as action
strcpy(ac,"REDUCE TO E -> ");
// c=length of input string
for(z = 0; z < c; z++)
{
// checking for producing rule E->4
if(stk[z] == '4')
{
printf("%s4", ac);
stk[z] = 'E';
stk[z + 1] = '\0';
//pinting action
printf("\n$%s\t%s$\t", stk, a);
}
}
for(z = 0; z < c - 2; z++)
{
// checking for another production
if(stk[z] == '2' && stk[z + 1] == 'E' &&
stk[z + 2] == '2')
{
printf("%s2E2", ac);
stk[z] = 'E';
stk[z + 1] = '\0';
stk[z + 2] = '\0';
printf("\n$%s\t%s$\t", stk, a);
i = i - 2;
}
}
for(z = 0; z < c - 2; z++)
{
//checking for E->3E3
if(stk[z] == '3' && stk[z + 1] == 'E' &&
stk[z + 2] == '3')
{
printf("%s3E3", ac);
stk[z]='E';
stk[z + 1]='\0';
stk[z + 1]='\0';
printf("\n$%s\t%s$\t", stk, a);
i = i - 2;
}
}
return ; // return to main
}
// Driver Function
int main()
{
printf("GRAMMAR is -\nE->2E2 \nE->3E3 \nE->4\n");
// a is input string
strcpy(a,"32423");
// strlen(a) will return the length of a to c
c=strlen(a);
// "SHIFT" is copied to act to be printed
strcpy(act,"SHIFT");
// This will print Lables (column name)
printf("\nstack \t input \t action");
// This will print the initial
// values of stack and input
printf("\n$\t%s$\t", a);
// This will Run upto length of input string
for(i = 0; j < c; i++, j++)
{
// Printing action
printf("%s", act);
// Pushing into stack
stk[i] = a[j];
stk[i + 1] = '\0';
// Moving the pointer
a[j]=' ';
// Printing action
printf("\n$%s\t%s$\t", stk, a);
// Call check function ..which will
// check the stack whether its contain
// any production or not
check();
}
// Rechecking last time if contain
// any valid production then it will
// replace otherwise invalid
check();
// if top of the stack is E(starting symbol)
// then it will accept the input
if(stk[0] == 'E' && stk[1] == '\0')
printf("Accept\n");
else //else reject
printf("Reject\n");
}
// This code is contributed by Shubhamsingh10 C
//Including Libraries
#include
#include
#include
//Global Variables
int z = 0, i = 0, j = 0, c = 0;
// Modify array size to increase
// length of string to be parsed
char a[16], ac[20], stk[15], act[10];
// This Function will check whether
// the stack contain a production rule
// which is to be Reduce.
// Rules can be E->2E2 , E->3E3 , E->4
void check()
{
// Coping string to be printed as action
strcpy(ac,"REDUCE TO E -> ");
// c=length of input string
for(z = 0; z < c; z++)
{
//checking for producing rule E->4
if(stk[z] == '4')
{
printf("%s4", ac);
stk[z] = 'E';
stk[z + 1] = '\0';
//pinting action
printf("\n$%s\t%s$\t", stk, a);
}
}
for(z = 0; z < c - 2; z++)
{
//checking for another production
if(stk[z] == '2' && stk[z + 1] == 'E' &&
stk[z + 2] == '2')
{
printf("%s2E2", ac);
stk[z] = 'E';
stk[z + 1] = '\0';
stk[z + 2] = '\0';
printf("\n$%s\t%s$\t", stk, a);
i = i - 2;
}
}
for(z=0; z3E3
if(stk[z] == '3' && stk[z + 1] == 'E' &&
stk[z + 2] == '3')
{
printf("%s3E3", ac);
stk[z]='E';
stk[z + 1]='\0';
stk[z + 1]='\0';
printf("\n$%s\t%s$\t", stk, a);
i = i - 2;
}
}
return ; //return to main
}
//Driver Function
int main()
{
printf("GRAMMAR is -\nE->2E2 \nE->3E3 \nE->4\n");
// a is input string
strcpy(a,"32423");
// strlen(a) will return the length of a to c
c=strlen(a);
// "SHIFT" is copied to act to be printed
strcpy(act,"SHIFT");
// This will print Lables (column name)
printf("\nstack \t input \t action");
// This will print the initial
// values of stack and input
printf("\n$\t%s$\t", a);
// This will Run upto length of input string
for(i = 0; j < c; i++, j++)
{
// Printing action
printf("%s", act);
// Pushing into stack
stk[i] = a[j];
stk[i + 1] = '\0';
// Moving the pointer
a[j]=' ';
// Printing action
printf("\n$%s\t%s$\t", stk, a);
// Call check function ..which will
// check the stack whether its contain
// any production or not
check();
}
// Rechecking last time if contain
// any valid production then it will
// replace otherwise invalid
check();
// if top of the stack is E(starting symbol)
// then it will accept the input
if(stk[0] == 'E' && stk[1] == '\0')
printf("Accept\n");
else //else reject
printf("Reject\n");
}
// This code is contributed by Ritesh Aggarwal 输出
GRAMMAR is -
E->2E2
E->3E3
E->4
stack input action
$ 32423$ SHIFT
$3 2423$ SHIFT
$32 423$ SHIFT
$324 23$ REDUCE TO E -> 4
$32E 23$ SHIFT
$32E2 3$ REDUCE TO E -> 2E2
$3E 3$ SHIFT
$3E3 $ REDUCE TO E -> 3E3
$E $ Accept