Compiler warning question

After reading Salem's post about using -Wall with the gcc compiler I decided to give it a try. I'm getting a warning that has me a little confused.

warning is

Quote:

---

adt.c:104: warning: char format, different type arg (arg 2)
adt.c:106: warning: char format, different type arg (arg 2)

---

The code for those two lines are:

Code:

---

printf("  Last Name:  ");
scanf("%s", &tln);
printf("  First Name: ");
scanf("%s", &tfn);

---

The two variables are:

Code:

---

char tln[name_len], tfn[name_len];

---

and name_len is defined by:

Code:

---

#define name_len        20

---

Anyone want to venture a guess on what this means? Thanks

> scanf("%s", &tln);
Should be
scanf("%s", tln);

For an array, ltn and &ltn have the same value, but different types (which is why it works).


If on the other hand you had pointers

char *ltn = malloc( 20 );
scanf("%s", &tln);
That would be very wrong (both the type and the value would be wrong).

char *ltn = malloc( 20 );
scanf("%s", tln);
is correct (as far as the types are concerned)

I think you are saying that for arrays you do not need to provide the address when using scanf? What about when you want to pass by reference to another routine?

I did remove the & from the calls and that did get rid of the warning. I understand what you are saying about the values and types. However I thought that using & passed the address of the variable and not it's value.

Man I'm glad I'm taking a class next semester. I see my self taught C have left a lot of holes.

Thanks Salem.

For an array, ltn and &ltn have the same value.

Now this is something new, I thought &ltn == &&ltn[0], it turned out that &ltn == &ltn[0]
weird, why does work so? Isn't ltn supposed to be a pointer to the first element of the array?

When you pass an array to a function, you get (as you know) a pointer to the first element of the array

So arr and &arr[0] are the same thing (by value and by type).

A less frequent (and occasionally damn useful) feature is a pointer to the whole array, which you get by saying &arr
This has the same value (like what you would see if you were to print each of the following using %p with printf), but the type is completely different. And yes, the () are necessary - char *r[10] is a completely different animal to char (*r)[10]

char arr[10];
char *p = arr;
char *q = &arr[0];
char (*r)[10] = &arr;

Now r is useful because you can also allocate a char matrix[x][10] in one easy step using
r = malloc ( x * sizeof *r );

If you look carefully, you'll see that sizeof *p and sizeof *q are 1, but sizeof *r is 10.

Thanks Salem, I understand that now.

Mmm, thanks Salem for the array pointer thing :)