Pointer incompatible type assignment to character.

[gk245]

I have something like this:

#include <stdio.h>

main ()
{
struct line
{
char write[20];
char read[20];

struct line *next;
};

struct line n1;

n1.write= "concepts";

}

However, if i try to compile it, i get a compiler error saying that
"incompatible types in assignment". Whats strange is that if i set
write as an integer, i don't get such a error and it compiles. Does
something special need to be done with character strings and pointers?

Thanks.

 

[Ian Collins]

I have something like this:

#include <stdio.h>

main ()
{
struct line
{
char write[20];
char read[20];

struct line *next;
};

struct line n1;

n1.write= "concepts";

}

However, if i try to compile it, i get a compiler error saying that
"incompatible types in assignment". Whats strange is that if i set
write as an integer, i don't get such a error and it compiles. Does
something special need to be done with character strings and pointers?

They are different. You have declared write as an array of 20 char and
you are attempting to assign a pointer to const char to it.

I think you are confused regarding accessing an array through a pointer
and assigning to an array. You have to copy the string literal into the
array.

 

[Giorgos Keramidas]

I have something like this:

#include <stdio.h>

main ()
{
struct line
{
char write[20];
char read[20];

struct line *next;
};

struct line n1;

n1.write= "concepts";

}

This is probably the usual confusion around arrays and pointers. You
cannot treat an array "as if" it was a pointer in the left part of an
assignment. Array names "decay" to pointers (to their first element)
when they are in the right part of an assignment, but the reverse is not
true.

You will have to use strncpy() or strlcpy() to copy the data from your
constant string into the array member of the structure, i.e. with:

size_t len;

len = sizeof(n1.write);
strncpy(n1.write, "concepts", len - 1);
n1.write[len - 1] = '\0';

or

strlcpy(n1.write, "concepts", sizeof(n1.write));

 

[Keith Thompson]

I have something like this:

#include <stdio.h>

main ()
{
struct line
{
char write[20];
char read[20];

struct line *next;
};

struct line n1;

n1.write= "concepts";

}

However, if i try to compile it, i get a compiler error saying that
"incompatible types in assignment". Whats strange is that if i set
write as an integer, i don't get such a error and it compiles. Does
something special need to be done with character strings and pointers?

<http://www.c-faq.com/>. Read all of section 6, "Arrays and Pointers".

 

[gk245]

Keith Thompson formulated on Wednesday :

I have something like this:

#include <stdio.h>

main ()
{
struct line
{
char write[20];
char read[20];

struct line *next;
};

struct line n1;

n1.write= "concepts";

}

However, if i try to compile it, i get a compiler error saying that
"incompatible types in assignment". Whats strange is that if i set
write as an integer, i don't get such a error and it compiles. Does
something special need to be done with character strings and pointers?

<http://www.c-faq.com/>. Read all of section 6, "Arrays and Pointers".

Thx for the link and the help guys. ^^

 

[Chad]

This is probably the usual confusion around arrays and pointers. You

cannot treat an array "as if" it was a pointer in the left part of an
assignment. Array names "decay" to pointers (to their first element)
when they are in the right part of an assignment, but the reverse is not
true.

Okay, I'm confused. Then how come something like this works.

include <stdio.h>
#define BUF 6

int main(void) {
char arr[BUF] = "la";

printf("the value is: %s\n", arr);

return 0;
}


Chad

$gcc -Wall arr.c -o arr
$./arr
the value is: la


But something like this:

include <stdio.h>
#define BUF 6

int main(void) {
int arr[BUF] = "la";

printf("the value is: %s\n", arr);

return 0;
}


produces
$gcc -Wall arr.c -o arr
arr.c: In function `main':
arr.c:5: error: invalid initializer
arr.c:7: warning: char format, different type arg (arg 2)

 

[Chad]

This is probably the usual confusion around arrays and pointers. You

cannot treat an array "as if" it was a pointer in the left part of an
assignment. Array names "decay" to pointers (to their first element)
when they are in the right part of an assignment, but the reverse is not
true.

Okay, I'm confused. Then how come something like this works.


#include <stdio.h>
#define BUF 6

int main(void) {
char arr[BUF] = "la";

printf("the value is: %s\n", arr);

return 0;

}


$gcc -Wall arr.c -o arr
$./arr
the value is: la

But something like this:

#include <stdio.h>
#define BUF 6

int main(void) {
int arr[BUF] = "la";

printf("the value is: %s\n", arr);

return 0;

}

produces
$gcc -Wall arr.c -o arr
arr.c: In function `main':
arr.c:5: error: invalid initializer
arr.c:7: warning: char format, different type arg (arg 2)

Chad

 

[sweety.rathore]

Integer arrays cannot be assigned strings, you can assign one character
at a time to the array--

like arr[0]='l';
arr[1]='a';

Because when characters are assigned to integers, theri ascii value
gets transferred, but the ascii value of strings cannot be calculated.

 

[Old Wolf]

Array names "decay" to pointers (to their first element) when they
are in the right part of an assignment, but the reverse is not true.

Okay, I'm confused. Then how come something like this works.

char arr[BUF] = "la";

This is not an assignment; it is an initialization. Some
programming languages use a different symbol for initialization
than they do for assignment. But C uses the equals sign for both.

In the initialization case, it means that "la" is an initializer for
arr. The C standard defines specifically that arrays of char can
be initialized from string literals.

int arr[BUF] = "la";

Other arrays can only be initialized by an initializer list, eg:
int arr[BUF] = { 1, 2 };

The case of initializing from a string literal is only for arrays of
char.

 

[Peter Shaggy Haywood]

Groovy hepcat Ian Collins was jivin' on Thu, 20 Apr 2006 11:06:02
+1200 in comp.lang.c.
Re: Pointer incompatible type assignment to character.'s a cool scene!
Dig it!

struct line
{
char write[20];
char read[20];
struct line *next;
};

struct line n1;

n1.write= "concepts";

[Snipage.]

They are different. You have declared write as an array of 20 char and
you are attempting to assign a pointer to const char to it.

No, he's trying to assign a pointer to char to it. There is no const
qualification on a string literal. It's not modifiable, but not const
qualified either.

--

Dig the even newer still, yet more improved, sig!

http://alphalink.com.au/~phaywood/
"Ain't I'm a dog?" - Ronny Self, Ain't I'm a Dog, written by G. Sherry & W. Walker.
I know it's not "technically correct" English; but since when was rock & roll "technically correct"?

 

[Peter Shaggy Haywood]

Groovy hepcat (e-mail address removed) was jivin' on 19 Apr 2006
20:54:38 -0700 in comp.lang.c.
Re: Pointer incompatible type assignment to character.'s a cool scene!
Dig it!

Integer arrays cannot be assigned strings, you can assign one character
at a time to the array--

like arr[0]='l';
arr[1]='a';

Because when characters are assigned to integers, theri ascii value
gets transferred, but the ascii value of strings cannot be calculated.

What on Earth are you talking about? Without seeing any quoted text,
it's hard to know what you're trying to say; especially when what
you've written is unclear in itself.
Now, as to what you said, it's bunk. An array of char is an "integer
array", since char is an integer type. And you certainly can
initialise an array of char with a string literal. And EBCDIC based
implementations don't use ASCII. Nor do implementations that use other
non-ASCII character sets. And just what is "the ascii value of
strings" supposed to mean? Do you mean the values of the characters in
a string? Why would you want to calculate their values? You can,
though, should the need arise.

--

Dig the even newer still, yet more improved, sig!

http://alphalink.com.au/~phaywood/
"Ain't I'm a dog?" - Ronny Self, Ain't I'm a Dog, written by G. Sherry & W. Walker.
I know it's not "technically correct" English; but since when was rock & roll "technically correct"?