Arrays initialization and declaration

why is this wrong to declare arrays and then initialize it
For instance

int x;
x =10;

the above code is valid.
but the below code raises an error
char searchstring[] = {"am not a fool"};
char tracks[10];
tracks ={"am not ready"};

char searchstring[] = {"am not a fool"};
This is initialisation, which you can do for an array.

> char tracks[10];
> tracks ={"am not ready"};
This is assignment, which you can't do for arrays.

Use instead
strcpy(tracks,"am not ready"); //!! needs 13 chars, not 10

Making sure of course that the string will fit.

Thanks alot

I see, by experiment, that what you say is true. Isn't this a hole in the c language definition?

"abd" is a pointer.
{ } initialization is an array.
{"abc"} should be an array of pointers.

How c handles this depends on the declaration:

Not if it's been defined that way. The 'holes' in C are called undefined behavior.

char*a[] = {"am not a fool"};
a points to a pointer containing the string.
a is also an array of pointers.

char b[] = {"am not a fool"};
b points to the string.

printf("a %%p %p\n",a);
Display the pointer a points to.

printf("a[0] %%c %c\n",a[0]);
Display the 1st byte of the 1st pointer as a character.

printf("b[0] %%c %c\n",b[0]);
Display the first byte of the string as a character.

So what's the problem?

Clearly the behavior is defined, and I wasn't included on the c standards committee. My misunderstanding was on the right hand side, not on the left hand side.

In the initializer
{"am not a fool"}
The curly braces indicate an array. That's one level of indirection.

The "string" represents a pointer. That's a second level of indirection.

thus, while
char b[] = {"am not a fool"};

might be unambiguous and supported, it looks wrong.

Furthermore, as you say,

b is an array.
a string is a pointer.

I wouldn't write code like that, intentionally.

They do? I thought they simply indicated a group. What's inside might be an array... Braces do not indicate an array.



Yep. And nope. It's still just a string.


It's the same as
char b[] = "am not a fool";

What's the difference between:




One has curly braces. It must be different according to your logic.


Fine. Then don't. You need to write code to your own style. Just understand that others may have a different style and you need to be able to distinguish correct and incorrect syntax in either style.

For example, my style for an IF is:


I hate this format:


and despise this format:


All are acceptable conventions and I cannot tell the other programmers they are wrong. I just have to live with it. Under my breath, though...

Mr.waltp AM a newbie to this language and i just started experimenting.

But still this point is not clear

par with my instance

i.e
int x;
x= 20;
/*if above code is correct */
then

char search[40] ;
search = {"am sid"};

/*why the above code is raising compilation erros*/

Please do explain me in terms of memory also.....

[QUOTE=salem]char searchstring[] = {"am not a fool"};
This is initialisation, which you can do for an array.

> char tracks[10];
> tracks ={"am not ready"};
This is assignment, which you can't do for arrays.

sir if we declare an array and take input from stdin then how itz able to assign our value to array .if its possible in that case then why itz not possible in the above code.

I already showed you - use strcpy()

To WaltP:
I recommend you run a pretty printer on c code with your favorite settings. Then the format of the code doesn't matter.

For instance, bcpp
which I originally learned as
cb

Dave.

Why?

With a pretty printer you'd transform your hate into love.

Because in the 1st code x is an int.
It can be assigned to, among other things.

On the other hand, in the 2nd code search is an array.
Arrays, in C, cannot be assigned to (they can be initialized).

Why can't arrays be assign to, you ask.

... Well ... arrays are, in C, second class citizens. They only exist as arrays in a few places (definition, arguments to sizeof, and arguments to unary & operator). In all other places, they get automatically converted to a pointer to their first element. So, the assignment in the 2nd code is (would be if it were legal) converted to

and, obviously, the address of the first letter in the string cannot be changed.

As I said, I can live with it.

And if I couldn't, why use someone else's program? I'd just write my own that does exactly what I want...