Tutorial: char, pointer in C

g and h in printf with %d have different interpretation since g is pointer to char,
so *(g+1) can locate the second char of ("bc"), Right ?

I search it in goole, the example link is the best to
do explode() in C.
Those ***arr or  arr[] or pointer address arithemtic or pointer of potiner made me
to lose track to the code.

So now I try to get it detail to printf out every step as debugger.
C is much faster than PHP in speed as ASM lanauage  (if no doing DOM access) and can
handle a lot of memory allocation and data structure .

But C is  not easy  to be remembered that I used it before in long long time ago
and C  sytnax is easy to forget

it is good

I can let the g char pointe into g[0] or g[1], sometimes, I don't know why it can
go into array.
Anyway let me try it before closing this thread.


I will open new thread for the example link, please look into it
http://www.it.uu.se/katalog/larme597/explode
Duncan

I did look at the example. It is taking a string and converting it into an array of strings. Or in this case, a pointer to character and converting it to an array of pointers to characters. Is something in that example not working for you?

I can run the example sucessfully but I don't understand how it works for every code step
so I will made new thread for it instead of this thread

Anyway, I will do the array first as your suggestion and close this thread

Duncan

I tried char *g="bc";

printf("%c\n", g[0]) will echo out "b";

so it is not need for declaration as  array such as

char *g[]="bc"; Why  no need ?



Duncan

Thanks for you reply

And you answer me the final questioon

and check my new thread of explode.c

The reason you do not have to declare g as an array before referencing g[0] is because in C (or at least, in most implementations of C I've ever used), C automatically knows how to dereference char * variables as arrays of character, its kind of a built in feature.

So char* g = "bc"

printf("%c %c %c %c", *g, *(g+1), g[0], g[1]);

prints: b c b c

a g[1] translates to *(g+1) when dealing with pointer to character (pointers to bytes, essentially).

and g[x] translates to *(g+x) where x is an offset in bytes from the original starter position of the character pointer.

Note that if you wrote this:

char *g = "ab"

printf("%c", g[2]);
or
printf("%c", *(g+2);

Depending on the compiler you may get an error or may get a run time error because you are attempting to reference memory outside the boundary for  char *g.

And if it does not error, it could produce unpredictable results. Whatever happens to be stored in the memory location 2 bytes after *g will be converted to char and printed.

On most OS if this causes you to attempt to read a part of memory not allocated to your program, you will get a memory access violation error.