Pointer error.

Hello,

I just have a quick question to ask...

I am getting an "assignment from incompatible pointer type" in my code and I can't understand why.

I have actually got this program to compile and run without any errors now, but I would like to know the reason for the problem that was occuring, if anyone could help me out?

The program is very simple. It asks the user to input a string and uses a function for the inputting. I am still learning pointers so this is only for my own learning purposes.

Here is the bugged code.

Code:

---

#include <stdio.h>

main()
{
        char input[50];
        char *ptrInput;
       
        ptrInput = &input;
       
        puts("Enter a string: ");
       
        my_scanf(ptrInput);
       
        puts(input);
       
        fflush(stdin);
        getchar();
}

my_scanf(char *ptrInput)
{
        gets(ptrInput);
}

---

Here is the working code. Can anyone tell me why a cast is required when I'm using the same type?

Code:

---

#include <stdio.h>

main()
{
        char input[50];
        char *ptrInput;
       
        ptrInput = (char *)&input;
       
        printf("Enter a string: ");
       
        my_scanf(ptrInput);
       
        puts(input);
       
        fflush(stdin);
        getchar();
}

my_scanf(char *ptrInput)
{
        gets(ptrInput);
}

---

Thanks.

input is a char[50]. &input, then, is a pointer to an array. A char* is a pointer to a char. Since an array is not a char, you have an assignment from an incompatible pointer type.

The correct solution is to use the fact that an array is converted to a pointer to its first element when passed as an argument. I would expect something like this:

Code:

---

#include <stdio.h>

void my_scanf(char *ptrInput);

int main()
{
        char input[50];

        puts("Enter a string: ");

        my_scanf(input);

        puts(input);

        getchar();

        return 0;
}

void my_scanf(char *ptrInput)
{
        gets(ptrInput);
}

---

Note that I have removed the fflush(stdin) as it is undefined. I have also given main() and my_scanf() their correct return types (though you may or may not want to return something from my_scanf()), and declared a prototype for my_scanf().

Also, note that you should not be using gets() as will allow a buffer overrun. I presume that you will changing the implementation of my_scanf() to something more sane. You probably also need to change my_scanf() to take the maximum length of the string.

Thanks for clearing that up :)

Also, I've just been using fflush(); getchar(); because that's what I was shows to do from the beginning.

I've been following the C programming tutorials by Mark Virtue.

actualyl &input isnt an array its a pointer to a pointer to an array, so you have incompatible levels of indirection. You shoudl use the line -

Code:

---

  ptrInput = (char*)input;
 
or
 
  ptrInput = (char*)&input[0];

---

Quote:

---

Originally Posted by abachler

actualyl &input isnt an array its a pointer to a pointer to an array, so you have incompatible levels of indirection. You shoudl use the line -

Code:

---

  ptrInput = (char*)input;
 
or
 
  ptrInput = (char*)&input[0];

---

---

The cast it not at all necessary. The cast was being used in the original code to silence a warning (this is usually not the right way to fix such a warning), but since input has type char* (effectively—I know it's really an array), the cast is superfluous.

Also, given

Code:

---

char input[50];

---

it's correct to say that &input is not an array, but neither is it a pointer to a pointer to an array. Laserlight was correct in saying it's a pointer to an array. So you can have something like:

Code:

---

char input[50];
char (*p)[50]; /* p is a pointer to an array of 50 char */
p = &input; /* thus this is compatible */

---

In C it is superfluous, but 99.99% of the code I write is C++ so I always cast, which is not incorrect in C. It also makes the code more portable if you decide to port it to C++.

Quote:

---

Originally Posted by george_1988

I've been following the C programming tutorials by Mark Virtue.

---

Perhaps it's time to switch?
Also, here's more information on gets and avoiding it: http://cpwiki.sourceforge.net/Gets

Quote:

---

Originally Posted by abachler

In C it is superfluous, but 99.99% of the code I write is C++ so I always cast, which is not incorrect in C. It also makes the code more portable if you decide to port it to C++.

---

But the point was that it's perfectly valid in both C and C++ without a cast because you don't use & on the array, it simply decays to a char* pointer to which you assign it.

As mentioned already, just not sure on your intent to allocate a 50 char array to simply pass it to a pointer...a nono. I would advise using malloc for char *ptrinput to allocate room. Then use fgets() over scanf to get your data from the user.

> C programming tutorials by Mark Virtue

"taught by an expert C programmer, Mark Virtue, who has been using C for over 15 years, and has been teaching C programming for over 5 years. "
If he's pumping out garbage with gets() and fflush(stdin), that's a hell of a lot of brain-rot.

This one will be much better
http://www.eskimo.com/~scs/cclass/cclass.html

Quote:

---

Originally Posted by abachler

actualyl &input isnt an array its a pointer to a pointer to an array.

---

No it is not.

input is an array of 50 char. &input is the address of an array of 50 char (of type pointer to array of 50 char).

Quote:

---

Originally Posted by abachler

Code:

---

  ptrInput = (char*)input;
 
or
 
  ptrInput = (char*)&input[0];

---

---

These two lines are equivalent. In the first line, the name of the array is implicitly converted to a pointer to the first character. The second line does that explicitly.

In both cases, the conversion to char * is not required.