Question about C Pointers (just learning)

Question

I am curious as to why this is an error and what the error message means. Here is some code:

int *x[] = {"foo", "bar", "baz"};
int *y[] = {"foo", "bar", "baz"};

x = y;

I try to compile and I get this:

error: incompatible types when assigning to type ‘char *[3]’ from type ‘char **’

Question #1 why is this an error? and Question #2 why are the types different?

Answer 1

As written, there are quite a few problems. First, the array literals are of type char*[], while x and y are of type int*[]. Second, you can't assign arrays directly, as they are in effect constant pointers.

Answer 2

There is an incompatible types error because you are assigning arrays of strings (type char * in C) to arrays of pointers to ints (such as int *x[]). The error message given by the compiler is a little confusing because C does a lot behind the scenes to try to convert variables from one type to another.

As chars are represented internally as numbers (letters correspond to their ASCII values), C is able to convert characters to ints, so it attempts to treat variables x and y as arrays of pointers to chars instead of ints, hence the char *[3]. It sees {"foo", "bar", "baz"} as type char ** because strings are type char * and arrays essentially stored as pointers in C, so it is a pointer to char *, or char **.

While this doesn't completely relate to your question, I also wonder what you're trying to do with x = y; As it is written, that will make x point to the same array as y, leaving the array that x used to point to inaccessible. To check whether two variables in C are equal, you would use the == operator. Testing equality isn't as simple for arrays or strings, but that's completely outside the scope of this question.

Answer 3

I'd recommend starting with a good tutorial on pointers and arrays. It looks like you might be coming from an interpreted language like JavaScript or PHP. Things are quite different in C, and you need to get a handle on the notions of stack memory, heap memory, pointers, etc. Once you do this will all make sense and you will be loving C.

Answer 4

It is an error, because an array is a non-modifiable lvalue - meaning you can't assign directly to it. You can only modify the array's members individually.

The types are different because if an array is used in a context where an lvalue is not required, it is evaluated to a pointer to its first element (somearray becomes effectively &somearray[0]).

Since the array y is used in such a context, it is evaluated as the address of its first element, which has type int **. Since the array x is in an lvalue context, it is evaluated as an array (int *[3]). (Your error message doesn't actually match your code - I suspect your arrrays are actually arrays of char *).

("lvalue context" means, pretty much, one of: The left hand side of an assignment operator (which is where the name comes from); the subject of the sizeof operator; or the subject of the & operator).

Answer 5

Hi guys, for some reason I can't use the same account which I published the question with.

The "int" thing was a typo on my part I really meant it to be "char". Sorry, for the confusion.

However, I think caf answered my question.

"It is an error, because an array is a non-modifiable lvalue - meaning you can't assign directly to it. You can only modify the array's members individually."

I'm am just trying to understand how pointers work in c that's why I am trying these random things you probably normally wouldn't do. I thought this might work since I thought arrays = pointers so I could just make x point to y. I guess not?

Thanks for all your response guys.