error: incompatible types in assignment

[Richard Bos]

I have the following source code. It seems wierd to me why gca's value
cannot be reassigned. It's afterall a pointer and has a pointer value.

char gca[7];

That's not a pointer.

No, it's not even a "constant pointer".

RTFFAQFFS: <http://c-faq.com/aryptr/index.html>.

Richard

 

[fei.liu]

I have the following source code. It seems wierd to me why gca's value
cannot be reassigned. It's afterall a pointer and has a pointer value.
I am aware that the standard says it's not allowed. But I cannot
convince myself why this is prohibited.

static int ia[4] = {64, 64, 64, 64};
static char ca[4] = {'a', 'b', 'c', 'd'};
static int iaa[] = {64, 64, 64, 64};
static char caa[] = {'a', 'b', 'c', 'd'};
static char cab[] = "abcd";
static char * cac = "abcd";

char gca[7];
char gcaa[] = "hello!";

int x[4];
int * px;
int main(void){
char sgca[7];
char sgcaa[] = "hello!";
//cab = "xyz";
cab[0] = 'x';
cac = "xyz";

gca = gcaa;
//sgca = sgcaa;
// gcaa = gca;

px = &main;
}


gcc -o a ary.c
ary.c: In function `main':
ary.c:20: error: incompatible types in assignment
ary.c:24: warning: assignment from incompatible pointer type

 

[Eric Sosman]

I have the following source code. It seems wierd to me why gca's value
cannot be reassigned. It's afterall a pointer and has a pointer value.
I am aware that the standard says it's not allowed. But I cannot
convince myself why this is prohibited.
[snipped attempt to assign `array1 = array2']

An array is not a pointer. Please see Section 6 of
the comp.lang.c Frequently Asked Questions (FAQ) list

http://www.c-faq.com/

.... with special attention to Questions 6.5 and 6.7
through 6.10.

 

[fei.liu]

I have the following source code. It seems wierd to me why gca's value
cannot be reassigned. It's afterall a pointer and has a pointer value.

char gca[7];

That's not a pointer.

No, it's not even a "constant pointer".

RTFFAQFFS: <http://c-faq.com/aryptr/index.html>.

Richard

Arrays are ``second-class citizens'' in C; one upshot of this prejudice
is that you cannot assign to them.

Well, I guess it's just reiterating my confusion. Why should arrays be
second class citizens? gca does have a memory location. The value of
gca has to be stored in memory so gca[0] etc can be accessed.

 

[Simon Biber]

I have the following source code. It seems weird to me why gca's value
cannot be reassigned. It's after all a pointer and has a pointer value.
I am aware that the standard says it's not allowed. But I cannot
convince myself why this is prohibited.

gca is not a pointer. It is an array. An array is an area of memory. The
area of memory stores 7 char objects. No pointer to char is stored in
memory associated with gca.

In most cases when you use gca as an identifier in your program, the
compiler substitutes the name of the array for a pointer to the first
element. It doesn't generate code to read that pointer out of memory. It
simply writes the pointer value into the generated code, since it knows
where the array will be placed in memory.

The C standard says that the name of an array is an lvalue, but not a
modifiable lvalue. That is, you are not allowed to modify gca. You
cannot use it on the left hand side of an assignment operator.

static int ia[4] = {64, 64, 64, 64};
static char ca[4] = {'a', 'b', 'c', 'd'};
static int iaa[] = {64, 64, 64, 64};
static char caa[] = {'a', 'b', 'c', 'd'};
static char cab[] = "abcd";
static char * cac = "abcd";

char gca[7];
char gcaa[] = "hello!";

int x[4];
int * px;
int main(void){
char sgca[7];
char sgcaa[] = "hello!";
//cab = "xyz";
cab[0] = 'x';
cac = "xyz";

gca = gcaa;
//sgca = sgcaa;
// gcaa = gca;

px = &main;

px is a pointer to int.
main is a function (void) returning int.
&main is a pointer to function (void) returning int.
&main and px have incompatible types.
The assignment is a constraint violation.

 

[Simon Biber]

I have the following source code. It seems wierd to me why gca's value
cannot be reassigned. It's afterall a pointer and has a pointer value.
char gca[7];

That's not a pointer.

No, it's not even a "constant pointer".

RTFFAQFFS: <http://c-faq.com/aryptr/index.html>.

Richard

Arrays are ``second-class citizens'' in C; one upshot of this prejudice
is that you cannot assign to them.
Indeed.

Well, I guess it's just reiterating my confusion. Why should arrays be
second class citizens? gca does have a memory location. The value of
gca has to be stored in memory so gca[0] etc can be accessed.

Arrays are second class citizens because that's the way they are defined.

gca does have a memory location, in the same way that an int variable
has a memory location. The int variable's memory location is known to
the compiler, but it is not stored in memory.

You are attempting to do the equivalent of this:

int foo;
int bar;

&foo = &bar; /* constraint violation */

Why can't you assign to the address of foo? Foo has a (fixed) memory
location, but &foo does not have its own memory location. It's just a
value that the compiler calculated for you.

Similarly, the array gca has a memory location but its location is not
stored in memory; it is fixed. You cannot change its address.

 

[fei.liu]

I have the following source code. It seems wierd to me why gca's value
cannot be reassigned. It's afterall a pointer and has a pointer value.
char gca[7];
That's not a pointer.

No, it's not even a "constant pointer".

RTFFAQFFS: <http://c-faq.com/aryptr/index.html>.

Richard

Arrays are ``second-class citizens'' in C; one upshot of this prejudice
is that you cannot assign to them.
Indeed.

Well, I guess it's just reiterating my confusion. Why should arrays be
second class citizens? gca does have a memory location. The value of
gca has to be stored in memory so gca[0] etc can be accessed.

Arrays are second class citizens because that's the way they are defined.

gca does have a memory location, in the same way that an int variable
has a memory location. The int variable's memory location is known to
the compiler, but it is not stored in memory.

You are attempting to do the equivalent of this:

int foo;
int bar;

&foo = &bar; /* constraint violation */

Why can't you assign to the address of foo? Foo has a (fixed) memory
location, but &foo does not have its own memory location. It's just a
value that the compiler calculated for you.

Similarly, the array gca has a memory location but its location is not
stored in memory; it is fixed. You cannot change its address.

Thanks, you've made it very clear now. The key is the value of gca as a
pointer is a compiler calculated value that has no memory storage.
That's why it cannot be reassigned.

Fei

 

[Keith Thompson]

Richard Bos wrote:

I have the following source code. It seems wierd to me why gca's value
cannot be reassigned. It's afterall a pointer and has a pointer value.
char gca[7];
That's not a pointer.

[...]
Thanks, you've made it very clear now. The key is the value of gca as a
pointer is a compiler calculated value that has no memory storage.
That's why it cannot be reassigned.

Right.

Just to pound on the point a bit harder, gca is an array object, not a
pointer. Arrays are not pointers, and pointers are not arrays.

What causes the confusion is that, in most contexts, an expression of
array type is implicitly converted to a pointer to its first element.
The resulting pointer is a value, not an object. It might be clearer
to say that the array expression is converted to *the address of* its
first element. (This conversion does not occur if the array
expression is the operand of a unary "&" or "sizeof" operator, or if
it's a string literal used in an initializer for an array.)

And, as you know by now, this is well explained in section 6 of the
FAQ, <http://www.c-faq.com/>.

 

[Mark McIntyre]

I have the following source code. It seems wierd to me why gca's value
cannot be reassigned. It's afterall a pointer and has a pointer value.

NO. Pointers are not arrays, and arrays are not pointers.
Arrays are blocks of memory.
Pointers are addresses of memory.

A house is not an address, or vice versa.

--
Mark McIntyre

"Debugging is twice as hard as writing the code in the first place.
Therefore, if you write the code as cleverly as possible, you are,
by definition, not smart enough to debug it."
--Brian Kernighan