I have a problem I couldn;t read my textfile ?? #include<stdio.h> #include<strin
ID: 3665386 • Letter: I
Question
I have a problem I couldn;t read my textfile ??
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include <ctype.h>
void removeduplicate();
void final();
int Isiden(char ch);
int Isop(char ch);
int Isdel(char ch);
int Iskey(char * str);
void removeduplicate();
char op[8]={'+','-','*','/','=','<','>','%'};
char del[8]={'}','{',';','(',')','[',']',','};
char *key[]={"int","void","main","char","float"};
//char *operato[]={"+","-","/","*","<",">","=","%","<=",">=","++"};
int idi=0,idj=0,k,opi=0,opj=0,deli=0,uqdi=0,uqidi=0,uqoperi=0,kdi=0,liti=0,ci=0;
int uqdeli[20],uqopi[20],uqideni[20],l=0,j;
char uqdel[20],uqiden[20][20],uqop[20][20],keyword[20][20];
char iden[20][20],oper[20][20],delem[20],litral[20][20],lit[20],constant[20][20];
void lexanalysis(char *str)
{
int i=0;
while(str[i]!='')
{
if(Isiden(str[i])) //for identifiers
{
while(Isiden(str[i]))
{
iden[idi][idj++]=str[i++];
}
iden[idi][idj]='';
idi++;idj=0;
}
else
if(str[i]=='"') //for literals
{
lit[l++]=str[i];
for(j=i+1;str[j]!='"';j++)
{
lit[l++]=str[j];
}
lit[l++]=str[j];lit[l]='';
strcpy(litral[liti++],lit);
i=j+1;
}
else
if(Isop(str[i])) // for operators
{
while(Isop(str[i]))
{
oper[opi][opj++]=str[i++];
}
oper[opi][opj]='';
opi++;opj=0;
}
else
if(Isdel(str[i])) //for delemeters
{
while(Isdel(str[i]))
{
delem[deli++]=str[i++];
}
}
else
{
i++;
}
}
removeduplicate();
final();
}
int Isiden(char ch)
{
if(isalpha(ch)||ch=='_'||isdigit(ch)||ch=='.')
return 1;
else
return 0;
}
int Isop(char ch)
{
int f=0,i;
for(i=0;i<8&&!f;i++)
{
if(ch==op[i])
f=1;
}
return f;
}
int Isdel(char ch)
{
int f=0,i;
for(i=0;i<8&&!f;i++)
{
if(ch==del[i])
f=1;
}
return f;
}
int Iskey(char * str)
{
int i,f=0;
for(i=0;i<5;i++)
{
if(!strcmp(key[i],str))
f=1;
}
return f;
}
void removeduplicate()
{
int i,j;
for(i=0;i<20;i++)
{
uqdeli[i]=0;
uqopi[i]=0;
uqideni[i]=0;
}
for(i=1;i<deli+1;i++) //removing duplicate delemeters
{
if(uqdeli[i-1]==0)
{
uqdel[uqdi++]=delem[i-1];
for(j=i;j<deli;j++)
{
if(delem[i-1]==delem[j])
uqdeli[j]=1;
}
}
}
for(i=1;i<idi+1;i++) //removing duplicate identifiers
{
if(uqideni[i-1]==0)
{
strcpy(uqiden[uqidi++],iden[i-1]);
for(j=i;j<idi;j++)
{
if(!strcmp(iden[i-1],iden[j]))
uqideni[j]=1;
}
}
}
for(i=1;i<opi+1;i++) //removing duplicate operators
{
if(uqopi[i-1]==0)
{
strcpy(uqop[uqoperi++],oper[i-1]);
for(j=i;j<opi;j++)
{
if(!strcmp(oper[i-1],oper[j]))
uqopi[j]=1;
}
}
}
}
void final()
{
int i=0;
idi=0;
for(i=0;i<uqidi;i++)
{
if(Iskey(uqiden[i])) //identifying keywords
strcpy(keyword[kdi++],uqiden[i]);
else
if(isdigit(uqiden[i][0])) //identifying constants
strcpy(constant[ci++],uqiden[i]);
else
strcpy(iden[idi++],uqiden[i]);
}
// printing the outputs
printf(" Delemeter are : ");
for(i=0;i<uqdi;i++)
printf(" %c ",uqdel[i]);
printf(" Operators are : ");
for(i=0;i<uqoperi;i++)
{
printf(" ");
puts(uqop[i]);
}
printf(" Identifiers are : ");
for(i=0;i<idi;i++)
{
printf(" ");
puts(iden[i]);
}
printf(" Keywords are : ");
for(i=0;i<kdi;i++)
{
printf(" ");
puts(keyword[i]);
}
printf(" Constants are : ");
for(i=0;i<ci;i++)
{
printf(" ");
puts(constant[i]);
}
printf(" Literals are : ");
for(i=0;i<liti;i++)
{
printf(" ");
puts(litral[i]);
}
}
int main()
{
char str[50];
FILE *f1;
if((f1=fopen("a.txt","r"))!=NULL){
lexanalysis(str);
}
else
{
printf("Error");
}
fclose(f1);
}
Explanation / Answer
Update main function with below given main function code:
int main()
{
char str[50];
FILE *f1;
f1 = fopen ( "a.txt" , "r" );
if (f1==NULL)
{
fputs ("Error",stderr);
exit (1);
}
lexanalysis(str);
fclose(f1);
}
-------------------------------------------------------------------------------------------------------------------------------------------------------------------
1. Check whether the file specified is present in the same location where this file resides.
2. Code tries to open the file specified and if any error occurs, reports the user by printing a message on console.
3. If error doesn't occurs, opens the file and performs the remaining operations.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.