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Compiling errors with braces!

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Join Date: Jan 2009

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Compiling errors with braces!

Jan 28th, 2009

As some of you would already know, I made a program which accepts login and password from users and shows "*" instead of actual password characters.
I know the program is grossly incorrect but I still tried to apply whatever I picked up using books and example programs.

*****DISCLAIMER- I am not an IT wizkid and calling me a beginner would be an extrapolated statement aswell. I am here to learn and discuss and not to be admonished at***************

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
.
.
 
#include<stdio.h>
#include<string.h> 
#include<stdlib.h> 
#include<ctype.h> 
#include<math.h> 
 
void
fnlogincheck();
int main(int argc,char**argv);
{
system("cls");
fnlogincheck();
return 0;
}
void
fnlogincheck();
{
char *pcloginid[3]={"user1","user2","user3"};
char *pcpwd[3]={"Potato","Tomato","Turnip"};
int l,p,i,c;
char login[5];
char password[6];
printf("\n\n\n\n\t\tEmployee Information System\n"); 
printf("\t\t===========================\n\n"); 
printf("\t\tLogin Screen\n"); 
printf("\t\t============\n"); 
printf("\n\t\tEnter User Id\t: ");
scanf("%s",&login);
printf("\n\t\tEnter Password\t: ");
scanf("%s",&password);
}
 
{
l=strcmp(&loginid[i], login);
p=!strcmp(&pwd[i],password);
for(i=o,i<=3,i++)
}
{
password[c]= password[6]
printf("*");
for(c=0,c<=6,c++)
}
{
if( l=0,p=0)
    printf("LOGIN ACCEPTED");
else 
    printf("LOGIN DENIED");
 
exit
} 
.
.
.
.

#include<stdio.h>
#include<string.h> 
#include<stdlib.h> 
#include<ctype.h> 
#include<math.h> 

void
fnlogincheck();
int main(int argc,char**argv);
{
system("cls");
fnlogincheck();
return 0;
}
void
fnlogincheck();
{
char *pcloginid[3]={"user1","user2","user3"};
char *pcpwd[3]={"Potato","Tomato","Turnip"};
int l,p,i,c;
char login[5];
char password[6];
printf("\n\n\n\n\t\tEmployee Information System\n"); 
printf("\t\t===========================\n\n"); 
printf("\t\tLogin Screen\n"); 
printf("\t\t============\n"); 
printf("\n\t\tEnter User Id\t: ");
scanf("%s",&login);
printf("\n\t\tEnter Password\t: ");
scanf("%s",&password);
}

{
l=strcmp(&loginid[i], login);
p=!strcmp(&pwd[i],password);
for(i=o,i<=3,i++)
}
{
password[c]= password[6]
printf("*");
for(c=0,c<=6,c++)
}
{
if( l=0,p=0)
    printf("LOGIN ACCEPTED");
else 
    printf("LOGIN DENIED");

exit
} 
.
.

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Join Date: Jan 2009

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Re: Compiling errors with braces!

Jan 28th, 2009

compiling errors->

compiling...
practice1.c
- c:\program files\microsoft visual studio\myprojects\practice\practice1.c(11) : error C2449: found '{' at file scope (missing function header?)
- c:\program files\microsoft visual studio\myprojects\practice\practice1.c(15) : error C2059: syntax error : '}'
- c:\program files\microsoft visual studio\myprojects\practice\practice1.c(18) : error C2449: found '{' at file scope (missing function header?)
- c:\program files\microsoft visual studio\myprojects\practice\practice1.c(32) : error C2059: syntax error : '}'
- c:\program files\microsoft visual studio\myprojects\practice\practice1.c(39) : error C2449: found '{' at file scope (missing function header?)
- c:\program files\microsoft visual studio\myprojects\practice\practice1.c(43) : error C2059: syntax error : '}'
Error executing cl.exe.

practice1.obj - 6 error(s), 0 warning(s)

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Join Date: May 2008

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riahc3

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Re: Compiling errors with braces!

Jan 28th, 2009

WOW....

I have no idea what you are trying to do but you are WAY over your head.
1: Learn to use "tab"
2: password[c]= password[6] with no ; at the end? (Thatll problably solve 99% of the complying errors
3:
What the **** is this:

 Help with Code Tags 
 C Syntax (Toggle Plain Text) 
{
l=strcmp(&loginid[i], login);
p=!strcmp(&pwd[i],password);
for(i=o,i<=3,i++)
}
 
{
{
l=strcmp(&loginid[i], login);
p=!strcmp(&pwd[i],password);
for(i=o,i<=3,i++)
}

{

Personally I think you come from another programming language that should either stick to it or reread from C manuals.

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Join Date: May 2008

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riahc3

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Re: Compiling errors with braces!

Jan 28th, 2009

This complys:

 Help with Code Tags 
 C Syntax (Toggle Plain Text) 
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
#include<math.h>
 
void fnlogincheck();
int main(int argc,char**argv)
{
    system("cls");
    fnlogincheck();
    return 0;
}
 
void fnlogincheck()
{
char *pcloginid[3]={"user1","user2","user3"};
char *pcpwd[3]={"Potato","Tomato","Turnip"};
int l,p,i,c;
char login[5];
char password[6];
printf("\n\n\n\n\t\tEmployee Information System\n");
printf("\t\t===========================\n\n");
printf("\t\tLogin Screen\n");
printf("\t\t============\n");
printf("\n\t\tEnter User Id\t: ");
scanf("%s",&login);
printf("\n\t\tEnter Password\t: ");
scanf("%s",&password);
 
 
 
//l=strcmp(&loginid[i], login);
//p=!strcmp(&pwd[i],password);
 
password[c]= password[6];
printf("*");
for(c=0;c<=6;c++)
 
{
if( l=p)
    printf("LOGIN ACCEPTED");
else
    printf("LOGIN DENIED");
 
exit;
}
}
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
#include<math.h>

void fnlogincheck();
int main(int argc,char**argv)
{
    system("cls");
    fnlogincheck();
    return 0;
}

void fnlogincheck()
{
char *pcloginid[3]={"user1","user2","user3"};
char *pcpwd[3]={"Potato","Tomato","Turnip"};
int l,p,i,c;
char login[5];
char password[6];
printf("\n\n\n\n\t\tEmployee Information System\n");
printf("\t\t===========================\n\n");
printf("\t\tLogin Screen\n");
printf("\t\t============\n");
printf("\n\t\tEnter User Id\t: ");
scanf("%s",&login);
printf("\n\t\tEnter Password\t: ");
scanf("%s",&password);



//l=strcmp(&loginid[i], login);
//p=!strcmp(&pwd[i],password);

password[c]= password[6];
printf("*");
for(c=0;c<=6;c++)

{
if( l=p)
    printf("LOGIN ACCEPTED");
else
    printf("LOGIN DENIED");

exit;
}
}

But again, What the **** are you trying to do with this code???

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Join Date: Jan 2009

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Re: Compiling errors with braces!

Jan 28th, 2009

•

Originally Posted by riahc3

This complys:
Help with Code Tags

C Syntax (Toggle Plain Text)
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
#include<math.h>
 
void fnlogincheck();
int main(int argc,char**argv)
{
    system("cls");
    fnlogincheck();
    return 0;
}
 
void fnlogincheck()
{
char *pcloginid[3]={"user1","user2","user3"};
char *pcpwd[3]={"Potato","Tomato","Turnip"};
int l,p,i,c;
char login[5];
char password[6];
printf("\n\n\n\n\t\tEmployee Information System\n");
printf("\t\t===========================\n\n");
printf("\t\tLogin Screen\n");
printf("\t\t============\n");
printf("\n\t\tEnter User Id\t: ");
scanf("%s",&login);
printf("\n\t\tEnter Password\t: ");
scanf("%s",&password);
 
 
 
//l=strcmp(&loginid[i], login);
//p=!strcmp(&pwd[i],password);
 
password[c]= password[6];
printf("*");
for(c=0;c<=6;c++)
 
{
if( l=p)
    printf("LOGIN ACCEPTED");
else
    printf("LOGIN DENIED");
 
exit;
}
}
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<ctype.h> #include<math.h> void fnlogincheck(); int main(int argc,char**argv) { system("cls"); fnlogincheck(); return 0; } void fnlogincheck() { char *pcloginid[3]={"user1","user2","user3"}; char *pcpwd[3]={"Potato","Tomato","Turnip"}; int l,p,i,c; char login[5]; char password[6]; printf("\n\n\n\n\t\tEmployee Information System\n"); printf("\t\t===========================\n\n"); printf("\t\tLogin Screen\n"); printf("\t\t============\n"); printf("\n\t\tEnter User Id\t: "); scanf("%s",&login); printf("\n\t\tEnter Password\t: "); scanf("%s",&password); //l=strcmp(&loginid[i], login); //p=!strcmp(&pwd[i],password); password[c]= password[6]; printf("*"); for(c=0;c<=6;c++) { if( l=p) printf("LOGIN ACCEPTED"); else printf("LOGIN DENIED"); exit; } }
But again, What the **** are you trying to do with this code???

ok...well..we were given this assignment to do a case senitive comaprison for teh password and a case insensitive comparison for the login..and show a "*" when we accept passwords form users..typical like gmail/orkut or for that matter daniweb...

thats what i did..and In my 17 years of studying, I have never ever undergone soft-skills training...so I guess that would account for the logic and syntax deprived code..!!

ofcourse there were examples of programs using logical operands like || but somehow I wanted to make it this way..

one more thins, since strcmp returns ansi c value and for everymatch, it returns 0, shouldnt l and p be equal to o..?

thanks

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Join Date: Jan 2009

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Re: Compiling errors with braces!

Jan 28th, 2009

ahh!!

again, I feel there is a bug in Visual C++ 6.0 because my compiler seems hung-up on the braces!!!!

errors after compiling the above mentioned code-
practice1.c
- C:\Program Files\Microsoft Visual Studio\MyProjects\practice\practice1.c(46) : warning C4550: expression evaluates to a function which is missing an argument list
- C:\Program Files\Microsoft Visual Studio\MyProjects\practice\practice1.c(19) : warning C4101: 'i' : unreferenced local variable
- C:\Program Files\Microsoft Visual Studio\MyProjects\practice\practice1.c(57) : error C2084: function 'int __cdecl main(int ,char ** )' already has a body
- C:\Program Files\Microsoft Visual Studio\MyProjects\practice\practice1.c(64) : error C2084: function 'void __cdecl fnlogincheck()' already has a body
- C:\Program Files\Microsoft Visual Studio\MyProjects\practice\practice1.c(94) : warning C4550: expression evaluates to a function which is missing an argument list
Error executing cl.exe.

practice1.obj - 2 error(s), 3 warning(s)

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riahc3

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Re: Compiling errors with braces!

Jan 28th, 2009

How can you put p=o?

P is a integer. Do you even know what a integer is?

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riahc3

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Re: Compiling errors with braces!

Jan 28th, 2009

BTW use wvDev-C++; Much better IDE.

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Re: Compiling errors with braces!

Jan 28th, 2009

•

Originally Posted by riahc3

How can you put p=o?

P is a integer. Do you even know what a integer is?

Even if I (or for that matter anyonw who has reached here) would not be a programmer, I would still know what an integer is!!

The question here is not whether we can allot value like zero to it, but to the best of my knowledge, on making a case insensitive/sensitive comparison for two strings, a differnce is returned if they are not found to be equal..(diffrnce of ansi c values)

Now for both the cases we are comparing strings with unequal lengths, login has a strlen of 5 characters and pwd has 6 characters..

how can we equate l to p?

the whole point of me making l =0 and p=0 was to make sure that l and p accept values which are already defined in the arrays!

still, I might be wrong, feel free to correct me.

Modesty is the true virtue of the educated. Conceit always comes as a bonus to the hollow and haughty..!

Know thyself..