what is wrong =(

[Unregistered]

i keep tring and trying but i cant seem to find out what is wrong with this damn program.. this is the error that i get..

error C2143: syntax error : missing ';' before 'type'
Error executing cl.exe.

sec.obj - 1 error(s), 0 warning(s)

and here is the code...

#include <stdio.h>

    char name(void);

int main()
{

	char nam;

	 printf("What is your name?\n");
	 scanf("%c",&nam);

	 if (nam != name())
	 {
		 printf("No your name is not %c\n",nam);
		 printf("Your name is %c \n",name());
	 }
	 else
 {
		 printf("Hmm i dont recognize this name!\n");
		 return 0;
 }

	 char name(void)
{
		 char nick[0] = ****y;
		 return 0;
	 }

any help... ?

[Sebastiani]

A char holds ONE character. An array of chars (a string) holds more (as many as you specify). Strings, and indeed ALL arrays cannot be compared like:
if(a == b) ...etc.
The strcmp() function returns 0 if two strings match, non-zero otherwise...

#include <stdio.h>

    char* name(void);
    int Identical(char a[], b[] ) { return strcmp(a, b) ? 0 : 1 } 
    
int main()
{

	char nam[100];

	 printf("What is your name?\n");
	 scanf("%s",&nam);

 if ( !Identical(nam, name() ) )
 {
    printf("No your name is not %s\n",nam);
    printf("Your name is %s \n",name());
  }
 else
  {
    printf("I know you, %s!\n", name() );
  }

system("PAUSE"); //..edit: you may need this too :)

return 0;
}






char *name(void)
{
 char nick[] = "****y";
 return nick;
 }

[shaik786]

1.) Try to indent your code, looks neat and helps in organising your code
2.) Close the } in main(), you have missed it
3.) int main() should return a value
4.) char nick[0] = ****y;
     Cannot have 0 there
     What is this 'y'?
5.) Can you have a name with just one character? Make it a string

[master5001]

I'm looking at your name() function and seeing some new problems that may arise.

Here is what you should do:

char *name(void) {
    char *nick = (char *)malloc(5);
    strcpy(nick, "****y");
    return nick;
}

The reason why you should do it this way is because before you were returning the address of a local variable. A local variable is temporary and will be deleted once the function is done

[Hammer]

Originally posted by master5001
I'm looking at your name() function and seeing some new problems that may arise.

Here is what you should do:

char *name(void) {
    char *nick = (char *)malloc(5);
    strcpy(nick, "****y");
    return nick;
}

The reason why you should do it this way is because before you were returning the address of a local variable. A local variable is temporary and will be deleted once the function is done

Almost right The string ****y is 5 in length, therefore you need to malloc 6 bytes to hold it. Also, you should really check the return code from malloc to ensure it's not NULL. To be honest, though, the use of malloc() is not really required for storing hardcoded value. If you are going to use it though, don't forget to free() the memory later.