博文
词法分析器（C语言版）(2007-06-06 22:59:00)
摘要：这是我用C语言编写的能识别C语言的关键字之类的一个词法分析器，需要有一个程序文件，然后在程序的第一行把#define Filaname “文件名”的那个文件名换成要识别的文件的准确路径，之后就运行就可以了，这个程序是在TurboC 2.0和本网站提供的TC for windows 3.1中调试成功的，如有问题请及时联系我，谢谢！（自认为不足之处就是没有调用函数，一个main弄到底，还有就是没有注释，看不懂的话千万不要硬撑啊！）
#define Filename "1.txt"#include<string.h>#include<stdio.h>
main(){char keyword[32][9]={"auto","break","case","char","const","continue","default","double","do","else","enum","extern","float","for","goto","if","int","long","register","return","short","signed","sizeof","static","struct","switch","typedef","union","unsigned","void","volatile","while"}; char sing[40][5]={"->","++","--","<<",">>","<=",">=","==","!=","&&","||","?:","+=","-=","*=","/=","%=",">>=","<<=","&=","|=","^=","=",".","!","~","-","*","&'","/","%","+","<",">","^","|'","?",":",",",";"}; char *s; char temp[10][9],ci[9]; char *go,*now; int i,j,n,k,out,tag1=0,tag2=0,tag3=0; FILE *p; if((p=fopen(Filename,"r"))=......
用C语言编写的模拟进程控制的程序(2006-10-05 11:47:00)
摘要：本程序是在Turbo C 2.0环境下编写的，预计在其他C程序编译系统中也没问题。
#define null 0#define M " Menu\n\n1.Creat a process\n2.Destroy a
process\n3.Blockaprocess\n4.Wakeup a process\n5.Suspend a process\n6.Active a
process\n7.Observe the processes\n\nPlease select the item you want to realize:"
struct PCB{char name; int ko; int cha;}
struct PCB pcb[10];int i=0;
void creat(){char a; int b; printf("Please input the name of the process:"); scanf("%c",&a); printf("\nPlease input the ko of the process:"); scanf("%d",&b); i++; pcb[i].name=a; pcb[i].ko=b; pcb[i].cha=1; printf(M);}
void destroy(){char a; int j; printf("Select a process you want to destroy:"); scanf("%c",&a); printf("\n"); for(j=1;j<=10;j++) {if(pcb[j].name==a) {pcb[j].name=0; pcb[j].ko=0; pcb[j].cha=0; printf("Process %c has been destroied."); } ......
FIFO算法模拟(2006-07-04 12:18:00)
摘要：#define M 4
#define N 10
#include<stdio.h>
typedef struct Queue
{char *head;
char *tail:
int=length;
}Que;
Que create()
{Que Q;
Q.head=(Queue)malloc(sizeof(Que));
Q.tail=Q.head;
Q.length=0;
return(Q);
}
Que push(Que Q,char c)
{*++Q.tail=c;
Q.length++;
if(Q.length>M)printf("Overflow!");
else return(Q);
}
char pop(Que Q)
{char c;
if(Q.length==0)exit(NULL);
Q.length++;
else return(c);
}
void visit(Que Q)
{int count;
char e;
for(count=0;count<M;count++)
{e=pop(Q);
printf("%c",e);
push(Q,pop(Q));
}
}
main()
{Que Q,p;
char c,e;
c=getchar();
while(c!=NULL)
{for(p=Q.head;p<=Q.tail;p++)
{if(*p==c)visit(Q);
else
{e=pop(Q);
push(Q,c);
}
}
}
}......
用链式存储实现的选择排序(2006-07-02 08:06:00)
摘要：#include<stdio.h>#define N 10typedef struct Seq{int data; struct Seq *next;}Seq,*Sq;
Sq create(){Sq L,p; int i; L=(Sq)malloc(sizeof(Seq)); L->next=NULL; for(i=N;i>0;i--) {p=(Sq)malloc(sizeof(Seq)); scanf("%d",&p->data); p->next=L->next; L->next=p; } return(L);}
int minkey(Sq L){Sq p,q; int n; p=L->next; n=p->data; for(;p!=NULL;p=p->next) {if(p->data<n) n=p->data; } return(n);}
void selectsort(Sq L){Sq p,q,s; for(p=L;p->next!=NULL;p=p->next) {for(q=p->next;q!=NULL;q=q->next) {s=q->next; if(s->data==minkey(p)) {q->next=s->next; s->next=p->next; p->next=s; } } }}
void main(){Sq L,p; L=create(); selectsort(L); for(p=L->next;p!=NULL;p=p->next) printf("%d ",p->data); getch();}
&nbs......
顺序链表的简单选择排序(2006-06-25 07:55:00)
摘要：#include<stdio.h>#define N 10typedef struct Seq{int data; struct Seq *next;}Seq,*Sq;
Sq create(){Sq L,p; int i; L=(Sq)malloc(sizeof(Seq)); L->next=NULL; for(i=N;i>0;i--) {p=(Sq)malloc(sizeof(Seq)); scanf("%d",&p->data); p->next=L->next; L->next=p; } return(L);}
int minkey(Sq L){Sq p,q; int n; p=L->next; n=p->data; for(;p!=NULL;p=p->next) {if(p->data<n) n=p->data; } return(n);}
void selectsort(Sq L){Sq p,q,s; for(p=L;p->next!=NULL;p=p->next) {for(q=p->next;q!=NULL;q=q->next) {s=q->next; if(s->data==minkey(p)) {q->next=s->next; s->next=p->next; p->next=s; } } }}
void main(){Sq L,p; L=create(); selectsort(L); for(p=L->next;p!=NULL;p=p->next) printf("%d ",p->data); getch();}
&nbs......
LRU算法实现(2006-05-31 14:17:00)
摘要：这是模拟计算机中高速缓存分组读取的程序，有意者请帮忙改下程序，有错之处，请多指教。
#define M 4
#define N 20
main()
{char row[M],s[N],t;
int i,j;
printf("You are allowed to input a maximum of N objects.");
for(i=0;i<N;i++)
{for(j=0;j<M;j++)
{if(s[i]!='\0'&&s[i]==row[j])
{for(;j>0;j--)
{t=row[i];
row[j]=row[j-1];
row[j-1]=t;
}
}
else if(s[i]!='\0'&&s[i]!=row[j])
{for(j=M-1;j>=1;j--)
row[j]=row[j-1];
row[0]=s[i];
}
}
}
} ......
用栈做的数制转换（十进制转换为八进制）(2006-05-12 16:04:00)
摘要：这是我用栈实现的十进制数转换为八进制数，其他进制转换同理，由于是在Turbo C++ 3.0上编写的，所以在Turbo C 2.0 或其他版本的C编译工具上可能不能通过编译而成为可执行文件。
#define maxlength 20#include<malloc.h>typedef struct stacknode{int *base; int *top; int length;}stack;
stack *initstack(){stack *S; (*S).base=(stack*)malloc(maxlength*sizeof(stack)); if(!(*S).base)exit(0); (*S).top=(*S).base; return(S);}
stack *push(stack *S,int r){*(*S).top++=r; if((*S).top-(*S).base>=maxlength) {(*S).top=(int*)realloc((*S).base,sizeof(int)); *(*S).top++=r; } return(S);}
int pop(stack *S){int r; if((*S).top==(*S).base)exit(0); r=*(--(*S).top); return(r);}
void main(){stack *S; int n,m; printf("Please input the number you want to convert:"); scanf("%d",&n); printf("\n"); S=initstack(); while(n!=0) {push(S,n%8); n=n/8; } while(S!=NULL) {m=pop(S); printf("%d",m); }}
......
Hanoi塔问题(2006-05-04 17:18:00)
摘要：解决汉诺塔问题。[printf("(++c).move disk n from x to z\n",);即move(x,n,z)]
void hanoi(int n,int x,int y,int z)
{int c=0;
if(n==1)printf("%d".move disk %d from %d to %d\n",++c,n,x,z);
else
{hanoi(n-1,x,z,y);
printf("%d.move disk %d from %d to %d\n",++c,n,x,z);
hanoi(n-1,y,x,z);
}
}
main()
{int n,x,y,z;
scanf(%d %d %d %d",&n,&x,&y,&z);
hanoi(n,x,y,z);
}
......
Fabs(2006-05-04 17:05:00)
摘要：这是一个关于fabs的程序，输入t，输出大于t的最小fabs值。
main()
{int f1=0,f2=1,fn,t;
fn=f1+f2;
scanf("%d",&t);
while(fn<=t)
{f1=f2;
f2=fn;
fn=f1+f2;
}
printf("The number you want is:%d\n",fn);
}......
统计一个字符串在另一个字符串中出现次数(2006-05-04 16:45:00)
摘要：这是一个在一个字符串中统计出另一个子串出现的次数，并返回次数n，保存在文件out.dat中。
＃include<stdio.h>
Void ReadWrite(char *str1,char *str2)
{FILE *fp;
fp=fopen("C:\\out.dat","w+");
if(fp)
{if(*str1)
{fprintf(fp,"%s",*str1);
str1++;
}
if(*str2)
{fprintf(fp,%s",*str2);
str2++;
}
}
else printf("Can't find or something is wrong to the file!");
}
int findstr(char *str1,char *str2)
{char *p,*r;
int n=0;
while(*str1)
{p=str1;
r=str2;
while(*r)
{if(*r==*p)
{r++;
p++;
}
&n......