Initialization of Char Pointers?

I don't understand how this works:

Code:

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char *set = "abcdefghijklmnopqrstuvwxyz";

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Does the c compiler magically convert that into:

Code:

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char *set = malloc(28 * sizeof(*set));
strcpy(set, "abcdefghijklmnopqrstuvwxyz"

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??

No.

It's more like

Code:

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const char anonymous_name_generated_by_compiler[] = "abcdefghijklmnopqrstuvwxyz";
char *set = anonymous_name_generated_by_compiler;

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The "characters" are used to fill in an array of chars, and then the assignment makes a pointer to the first element of that array.

Code:

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const char temp[] = "abcdefghijklmnopqrstuvwxyz";
char *set = temp;

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Is almost the same thing (except you don't have to create a named location for the string)

The main difference is that with malloc()'d memory, you can modify the string later on at some point. But the memory that is pointed to using char *set = "some string" can't be modified later on (i.e. you couldn't do set[0] = 'r').

itsme86 : you should take a looong vacation and give us a chance to reply to some threads :p

posted by kleid-0

Code:

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Does the c compiler magically convert that into:

Code:
char *set = malloc(28 * sizeof(*set));
strcpy(set, "abcdefghijklmnopqrstuvwxyz");

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Code:

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                      yes !!!                 

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Um, compilers don't use magic. His first code is a string literal. His second is using a block of memory allocated to it.

The first code does not magically act like the second. The behavior's are different, and respond differently.

Want to make sure? Use the function free() on your second code after you malloc() it. Did it crash? Probably not. Try it on your first code. Did it crash? Yes. Because a string literal isn't your memory, so you can't modify it nor free it.

Big difference.


- Stack Overflow

Quote:

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Originally Posted by Stack Overflow

Um, compilers don't use magic.

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I beg to differ: http://jargon.watson-net.com/jargon.asp?w=automagically

;)

Once again, enjoy shows how little he knows...

But at least they used code tags! Oh, wait...

Quzah.

/b] sorry [/b]

Code:

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but i'm a student not a professonal programmer ... !

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Quote:

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Originally Posted by enjoy

/b] sorry [/b]

Code:

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but i'm a student not a professonal programmer ... !

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I know Quzah's humor is very subtle, but YOU REALLY DON"T NEED TO PUT CODE TAGS AROUND YOUR REPLIES.

Use code tags plz, when using code.
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Quote:

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Originally Posted by Salem

Code:

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const char temp[] = "abcdefghijklmnopqrstuvwxyz";
char *set = temp;

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I see now, thank you very much Salem :)

Quote:

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Originally Posted by Kleid-0

Who cares if they're freakin' code tags? It's not that big of a deal.

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We care. The creator of the forum cares. Everyone who has to read unindented piles of steaming (I am sillyI am sillyI am sillyI am silly) care. Use them or ... well find out fast that no one will help you ever again.

Quzah.

Ok I understand, I edited my last post. I need to start being more positive here.

Quote:

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Originally Posted by itsme86

The main difference is that with malloc()'d memory, you can modify the string later on at some point. But the memory that is pointed to using char *set = "some string" can't be modified later on (i.e. you couldn't do set[0] = 'r').

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You should point out (no pun intended) that char *set is immutable but it can still point to another area in memory. Not a big thing but helpful

Code:

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char *set = "some string"
set = "now i'm changed"

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