Thread: Confused

#include <stdio.h>

int main(void) {
char a="a";
return 0;
}

what's wrong of that ?
compile error : Cannot convert 'char *' to 'char' in function main()

after replacing with this, it has no error:
char a=(char)"a";

I am really confused.......

string a='abcdef';
string b="abcdef";

they are also different because b contain a null terminator and a does not, am I right?

int StrCmp(
LPCTSTR lpStr1,
LPCTSTR lpStr2
);

that means strcmp requires two pointer parameters.
but why strcmp(&a, "/?"); can compile successfully without any error and warning? "/?" is not a point ....
I am totally confused
Please help...

Code:

#include <stdio.h>

int main(void) {
  printf("%d , %d",sizeof("/?"), sizeof('/?'));
  return 0;
}

the output is 3, 4 why ?

Originally posted by Hammer
>am I right?
No.

What's that "string" type? There's no such thing in C (but there is one in C++).

You cannot define character constants this way either:
>a='abcdef';

If you want a char array without a NULL terminator, you can do so this way:
>char a[] = {'H','a','m','m','e','r'};
For example:

Code:

#include<stdio.h>
int main()
{  
	int i;  
	char a[] = {'H','a','m','m','e','r'}; 
	
	for (i = 0; i < sizeof (a); i++)
		putchar(a[i]);
		
   	return 0;
}

I see you point!
thanks^^

Originally posted by Hammer
>"/?" is not a point(er) ....
Yes, it is. Well, kind of. When you hardcode a string like this
>"My Name"
you are creating a char array. The compiler will see this as a pointer to an array of chars containing 8 bytes of data (7 chars + 1 null).

does that mean if the parameter is array, then need not to pass a pointer even other type of variables (e.g. int, long)?

Originally posted by Hammer

I'm not sure I understand your question, but I'll have a go anyway:

If the function takes an array as a parameter, you must pass an array name or a pointer to that function. Note that array names are really pointer is disguise.

Eg:
>void myfunc(char myarray[]);
To call this you can:
>char a[10];
>myfunc(a);

Code:

  char a[]="abc";
  char b='a';
  strcmp(&a,&b);
//can not compile successfully

Code:

  char a[]="abc";
  char b='a';
  strcmp(a,&b);
//compile successfully

Code:

cout <<&a;  //no error when compile, print out a address

then just one question: why I can't pass &a into strcmp() ?

Originally posted by Hammer
>char a[]="abc";
>char b='a';
>strcmp(&a,&b);
>then just one question: why I can't pass &a into strcmp() ?

Because:

Also, the b variable is incorrect, you should not pass this to strcmp(), because it is not a string, it's only a single char. The compiler won't complain, because you are passing &b, which is the address of b (and hence is a pointer). But your program will crash, or at best, produce unexpected results.

This is the correct method:

Code:

char a[] = "Hammer";
char b[] = "Kelvin";
strcmp(a, b);

Note, this code doesn't do anything with the return code from strcmp(), and therefore does nothing. To actually use it properly, you embed it in a if statement.

Also, cout is C++. This is a C board, where we only talk about C.

got it! sorry><