Thread: Third newbie question

This method takes two strings, then first compares the length and then uses memcmp to compare them if the length is identical.

Code:

int strcomp(char, char);

int strcomp(char s1, char s2) {
        int l1 = strlen(s1);
        int l2 = strlen(s2);
        if (l1 == l2) {
                return memcmp(s1, s2, l1);
        } else {
                return 0;
        }
}

When compiling, it gives the following warning:

test.c:25: warning: passing argument 1 of ‘strlen’ makes pointer from integer without a cast
test.c:26: warning: passing argument 1 of ‘strlen’ makes pointer from integer without a cast
test.c:28: warning: passing argument 1 of ‘memcmp’ makes pointer from integer without a cast
test.c:28: warning: passing argument 2 of ‘memcmp’ makes pointer from integer without a cast

I've barely started looking at what pointers are yet, but I didn't think I "made" any pointers in this code.

You didn't 'make' any pointer but you're using a function that takes a pointer as an argument:

Code:

size_t strlen ( const char * str );

str is a pointer to char.

Pointer are variables that hold addresses, instead of int or char values.

We don't print the actual addresses the pointers hold, too often. (You can do it, but it's rare that you need to). You can also use the pointer to print out what the pointer has the address of. THAT gets used all the time.

Could you give me a hint on how to write the obove method correctly, then? I kick-in-the-butt to get to understand pointers? :-)

A hint, but you should try to implement it yourself:

Code:

int strcomp(char *, char *);

or even better:

Code:

int strcomp(const char *, const char * );

because the function is not supposed to change the strings.

Try to write the implementation yourself and post it here.

Another important fact to know is that if

Code:

char a[6] = "hello";
int i = 0;

then these two statements are the same: a[i] and *(a + i)

This means that an array name actually is a pointer! This is very important to know. The following explains this:

If you declare :

Code:

char a[6] = "hello";
char * p;

then p = a is the same as p =&a[0]

Originally Posted by laserlight

No, an array is converted to a pointer to its first element in that context.

Should I have said: "An array name is an address?" Is this correct? In any case you can do pointer arithmetics with an array name (like *(a+1) in my example). This let me make the conclusion that an array name is equivalent with a pointer.

(but you look much more experienced then me. I'm a beginner myself)

Ok, I will try. But first - why does this compile without warning?

Code:

        char str[10];
        printf("%s is %d characters long\n", str, strlen(str));

There's no pointer passed to strlen() here...?

Originally Posted by cnewbie1

Ok, I will try. But first - why does this compile without warning?

Code:

        char str[10];
        printf("%s is %d characters long\n", str, strlen(str));

There's no pointer passed to strlen() here...?

There is a pointer passed: in my post above I said that array names are equivalent with pointers. If you take str, it is seen as the address of the first element of the array.

Laserlight, you are right:

Code:

#include <stdio.h>

int main(void)
{
        char a[6] = "hello";
	char * p = a;
	printf("%d\n", sizeof(a));
	printf("%d\n", sizeof(p));
	return 0;
}

gives 6 and 4. I'm a bit confused now. I'll look it up in the documentation. Thanks for the hint!