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  • error in finding out the lexems and no of lines of a text file in C

    - by mekasperasky
    #include<stdio.h> #include<ctype.h> #include<string.h> int main() { int i=0,j,k,lines_count[2]={1,1},operand_count[2]={0},operator_count[2]={0},uoperator_count[2]={0},control_count[2]={0,0},cl[13]={0},variable_dec[2]={0,0},l,p[2]={0},ct,variable_used[2]={0,0},constant_count[2],s[2]={0},t[2]={0}; char a,b[100],c[100]; char d[100]={0}; j=30; FILE *fp1[2],*fp2; fp1[0]=fopen("program1.txt","r"); fp1[1]=fopen("program2.txt","r"); //the source file is opened in read only mode which will passed through the lexer fp2=fopen("ccv1ouput.txt","wb"); //now lets remove all the white spaces and store the rest of the words in a file if(fp1[0]==NULL) { perror("failed to open program1.txt"); //return EXIT_FAILURE; } if(fp1[1]==NULL) { perror("failed to open program2.txt"); //return EXIT_FAILURE; } i=0; k=0; ct=0; while(ct!=2) { while(!feof(fp1[ct])) { a=fgetc(fp1[ct]); if(a!=' '&&a!='\n') { if (!isalpha(a) && !isdigit(a)) { switch(a) { case '+':{ i=0; cl[0]=1; operator_count[ct]=operator_count[ct]+1;break;} case '-':{ cl[1]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '*':{ cl[2]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '/':{ cl[3]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '=':{a=fgetc(fp1[ct]); if (a=='='){cl[4]=1; operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1;} else { cl[5]=1; operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1; ungetc(1,fp1[ct]); } break;} case '%':{ cl[6]=1; operator_count[ct]=operator_count[ct]+1;i=0;break;} case '<':{ a=fgetc(fp1[ct]); if (a=='=') {cl[7]=1; operator_count[ct]=operator_count[ct]+1;} else { cl[8]=1; operator_count[ct]=operator_count[ct]+1; ungetc(1,fp1[ct]); } break; } case '>':{ ; a=fgetc(fp1[ct]); if (a=='='){cl[9]=1; operator_count[ct]=operator_count[ct]+1;} else { cl[10]=1; operator_count[ct]=operator_count[ct]+1; ungetc(1,fp1[ct]); } break;} case '&':{ cl[11]=1; a=fgetc(fp1[ct]); operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1; variable_used[ct]=variable_used[ct]-1; break; } case '|':{ cl[12]=1; a=fgetc(fp1[ct]); operator_count[ct]=operator_count[ct]+1; operand_count[ct]=operand_count[ct]+1; variable_used[ct]=variable_used[ct]-1; break; } case '#':{ while(a!='\n') { a=fgetc(fp1[ct]); } } } } else { d[i]=a; i=i+1; k=k+1; } } else { //printf("%s \n",d); if((strcmp(d,"if")==0)){ memset ( d, 0, 100 ); i=0; control_count[ct]=control_count[ct]+1; } else if(strcmp(d,"then")==0){ i=0;memset ( d, 0, 100 );control_count[ct]=control_count[ct]+1;} else if(strcmp(d,"else")==0){ i=0;memset ( d, 0, 100 );control_count[ct]=control_count[ct]+1;} else if(strcmp(d,"while")==0){ i=0;memset ( d, 0, 100 );control_count[ct]=control_count[ct]+1;} else if(strcmp(d,"int")==0){ while(a != '\n') { a=fgetc(fp1[ct]); if (isalpha(a) ) variable_dec[ct]=variable_dec[ct]+1; } memset ( d, 0, 100 ); lines_count[ct]=lines_count[ct]+1; } else if(strcmp(d,"char")==0){while(a != '\n') { a=fgetc(fp1[ct]); if (isalpha(a) ) variable_dec[ct]=variable_dec[ct]+1; } memset ( d, 0, 100 ); lines_count[ct]=lines_count[ct]+1; } else if(strcmp(d,"float")==0){while(a != '\n') { a=fgetc(fp1[ct]); if (isalpha(a) ) variable_dec[ct]=variable_dec[ct]+1; } memset ( d, 0, 100 ); lines_count[ct]=lines_count[ct]+1; } else if(strcmp(d,"printf")==0){while(a!='\n') a=fgetc(fp1[ct]); memset(d,0,100); } else if(strcmp(d,"scanf")==0){while(a!='\n') a=fgetc(fp1[ct]); memset(d,0,100);} else if (isdigit(d[i-1])) { memset ( d, 0, 100 ); i=0; constant_count[ct]=constant_count[ct]+1; operand_count[ct]=operand_count[ct]+1; } else if (isalpha(d[i-1]) && strcmp(d,"int")!=0 && strcmp(d,"char")!=0 && strcmp(d,"float")!=0 && (strcmp(d,"if")!=0) && strcmp(d,"then")!=0 && strcmp(d,"else")!=0 && strcmp(d,"while")!=0 && strcmp(d,"printf")!=0 && strcmp(d,"scanf")!=0) { memset ( d, 0, 100 ); i=0; operand_count[ct]=operand_count[ct]+1; } else if(a=='\n') { lines_count[ct]=lines_count[ct]+1; memset ( d, 0, 100 ); } } } fclose(fp1[ct]); operand_count[ct]=operand_count[ct]-5; variable_used[0]=operand_count[0]-constant_count[0]; variable_used[1]=operand_count[1]-constant_count[1]; for(j=0;j<12;j++) uoperator_count[ct]=uoperator_count[ct]+cl[j]; fprintf(fp2,"\n statistics of program %d",ct+1); fprintf(fp2,"\n the no of lines ---> %d",lines_count[ct]); fprintf(fp2,"\n the no of operands --->%d",operand_count[ct]); fprintf(fp2,"\n the no of operator --->%d",operator_count[ct]); fprintf(fp2,"\n the no of control statments --->%d",control_count[ct]); fprintf(fp2,"\n the no of unique operators --->%d",uoperator_count[ct]); fprintf(fp2,"\n the no of variables declared--->%d",variable_dec[ct]); fprintf(fp2,"\n the no of variables used--->%d",variable_used[ct]); fprintf(fp2,"\n ---------------------------------"); fprintf(fp2,"\n \t \t \t"); ct=ct+1; } t[0]=lines_count[0]+control_count[0]+uoperator_count[0]; t[1]=lines_count[1]+control_count[1]+uoperator_count[1]; s[0]=operator_count[0]+operand_count[0]+variable_dec[0]+variable_used[0]; s[1]=operator_count[1]+operand_count[1]+variable_dec[1]+variable_used[1]; fprintf(fp2,"\n the time complexity of program 1 is %d",t[0]); fprintf(fp2,"\n the time complexity of program 2 is %d",t[1]); fprintf(fp2,"\n the space complexity of program 1 is %d",s[0]); fprintf(fp2,"\n the space complexity of program 2 is %d",s[1]); if((t[0]>t[1]) && (s[0] >s[1])) fprintf(fp2,"\n the efficiency of program 2 is greater than program 1"); else if(t[0]<t[1] && s[0] < s[1]) fprintf(fp2,"\n the efficiency of program 1 is greater than program 2 " ); else if (t[0]+s[0] > t[1]+s[1]) fprintf(fp2,"\n the efficiency of program 1 is greater than program 2"); else if (t[0]+s[0] < t[1]+s[1]) fprintf(fp2,"\n the efficiency of program 2 is greater than program 1"); else if (t[0]+s[0] == t[1]+s[1]) fprintf(fp2,"\n the efficiency of program 1 is equal to that of program 2"); fclose(fp2); return 0; } this code basically compares two c codes and finds out the no. of variables declared , used , no. of control statements , no. of lines and no. of unique operators , and operands , so as to find out the time complexity and space complexity of of the two programs given in the text file program1.txt and program2.txt ... Lets say program1.txt is this #include<stdio.h> #include<math.h> int main () { FILE *fp; fp=fopen("output.txt","w"); long double t,y=0,x=0,e=5,f=1,w=1; for (t=0;t<10;t=t+0.01) { //if (isnan(y) || isinf(y)) //break; fprintf(fp,"%ld\t%ld\n",y,x); y = y + ((e*(1 - (x*x))*y) - x + f*cos(w*0.1))*0.1; x = x + y*0.1; } fclose(fp); return (0); } i havent indented it as its just a text file . But my output is totally faulty . Its not able to find the any of the ouput that i need . Where is the bug in this ? I am not able to figure out as the algorithm looks fine .

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