Thread: argument testing

Why not just use a flag?

Code:

struct person {
  /* Blah blah */
  int npc; /* 1=nonplayer character, 0=player */
};

That way you don't have to work with strings.

>if(gunner=="player") {do stuff};
You can't compare strings like this. Well, you can...but it doesn't work like you think it does.

>How does it really work?
When used in an expression, the name of an array is treated as a pointer to the first item in the array. A string literal is really a pointer to char. What does the following print?

Code:

#include <stdio.h>

int main ( void )
{
  char a[] = "Test";

  printf ( "%p\n", (void *)a );
  printf ( "%p\n", (void *)"Test" );

  return 0;
}

Not two equal addresses if that's what you were going to say. The chances of that are slim and none, and I'd lean more toward none if I were a betting girl.

Anyway, using the above knowledge, what will the following print (feel free to ignore the warning about logical comparison with a string literal if you get one):

Code:

#include <stdio.h>

int main ( void )
{
  char a[] = "Test";

  if ( a == "Test" )
    printf ( "Equal?!\n" );

  return 0;
}

What you would expect is that the contents of a is compared lexographically with the contents of "Test". Of course, if you hang around C programmers, this tendency goes away with time. What really happens is that two addresses that are highly unlikely to be the same are compared.

Now, since we C programmers are kewler () than everyone else, we use a sneaky method of weeding out the pretenders by making string comparison illogical unless you already know C or something similar:

Code:

#include <stdio.h>
#include <string.h>

int main ( void )
{
  char a[] = "Test";

  if ( strcmp ( a, "Test" ) == 0 )
    printf ( "Equal!\n" );

  return 0;
}

> Not two equal addresses if that's what you were going to say. The chances of that are slim and none
Always none - they're two different objects, so they really need to be in two different places


But

Code:

#include <stdio.h>

int main ( void )
{
  char *a = "Test";

  if ( a == "Test" )
    printf ( "Equal?!\n" );

  return 0;
}

could well produce equality for a different reason than you might expect

> might that be because a only points to "T" because of only being a single char pointer,
No, you've just described
if ( *a == *"Test" )

Which would also be true if you wrote
if ( *a == *"Trouble" )


But
if ( a == "Trouble" )
would always be false

It depends on the compiler. A smart one might see that the same literal was used in two places and instead create a single one to store in the initialized area of the executable. But I would think that most compilers won't go to so much trouble. The point is, you'd be comparing pointers - not array contents.

But back to the original question. You should use something like Prelude suggested - it's faster than string comarison anyway.

>> Is it because they are both string constants that it will cause equality?

No.

>> ...and give it the same pointer value as *a?

No, what Salem was trying to point out is that given:

const char * a = "Test", * b = "Trouble";

*a == *b because dereferencing either yields the letter 'T'.