why i dong get accsesing null memory error stuff..

if at some point ,my string is "aa"
and my pointer points to the first "a"
then i go 2 places fowrward
so it goes "out of band"(java term)
why i dont get an error in a compiler

If your string is "aa", then it has a null terminator, so dereferencing a pointer to the char at index 2 is perfectly fine. Also, having a pointer that points one past the last element of an array is also perfectly fine, and is in fact useful for iterating over an array (to keep track of the end point).

EDIT:
Incidentally, "out of band" sounds more like a networking or telecommunications term than a Java term. You are probably thinking about "out of bounds", but that term is not specific to Java.

no \0 char
i defined it as
char* st="aa";

why i dont get out of bounds
??

Quote:

---

Originally Posted by transgalactic2

no \0 char
i defined it as
char* st="aa";

---

Then it has a null terminator. Note that you should write it as:

Code:

---

const char* st = "aa";

---

why const??
so if i write it like this then it adds \0 in the end??

>> (java term)
>> why i dont get an error in a compiler

Because C is not Java. when you program in C you must practice "safe driving" and debug with separate tools like valgrind.

Quote:

---

Originally Posted by transgalactic2

why const??

---

Because the pointer points to the first char of a string literal, and string literals may not be modified.

Quote:

---

Originally Posted by transgalactic2

so if i write it like this then it adds \0 in the end??

---

Yes.

there is no such thing as string in C we are talking about
array of chars and the last char is \0

so what you are saying is if i have

Code:

---

char *st={'a','c'};

---

(which could be an array of chars)
then we cant do
*(st+1)='h';
because its a const??

Quote:

---

Originally Posted by transgalactic2

there is no such thing as string in C

---

You are mistaken. In C, a string is "a contiguous sequence of characters terminated by and including the first null character".

Quote:

---

Originally Posted by transgalactic2

so what you are saying is if i have

---

That is not a string. That is a pointer to the first element of an array of two chars.

Quote:

---

Originally Posted by transgalactic2

then we cant do
*(st+1)='h';
because its a const??

---

Yes.

EDIT:
Actually, that syntax is wrong. It seems that what it really does is convert 'a' to pointer to char, or something like that.

i ran this code
and it worked fine
i didnt get an error

Code:

---

#include <stdio.h>
#include <string.h>
int main()
{

        char str1[2]={'a','b'};
       
        char* str=str1;
        *(str+1)='h';
                return 0;
}

---

Quote:

---

Originally Posted by transgalactic2

i ran this code
and it worked fine
i didnt get an error

---

Why did you think you would get an error? Note that your pointer now points to the first element of an array that is modifiable.

because you said
"pointer points to the first char of a string literal, and string literals may not be modified."

thats what i did
i changed the first char too
and it works fine

Code:

---

#include <stdio.h>
#include <string.h>
int main()
{

        char str1[2]={'a','b'};
       
        char* str=str1;
        *(str)='h';
                return 0;
}

---

Let me be painstakingly clear.

If you have an array of char like

Code:

---

str = {'a', 'a'};

---

and you do something like, *(str+1) = 'b'; there will be no error because you are in bounds. Accessing anything greater than the first index is out of bounds and is undefined behavior hence.

Likewise when I said that C has no string type in another thread, I only said that out of frustration because you wouldn't allow people to explain the difference between pointers and arrays. There are certainly types that hold strings, but those types are not called "string". Try not to be too literal though. I'm sorry if I caused confusion.

laser light said:
"pointer points to the first char of a string literal, and string literals may not be modified."
so this code works because its not ending with \0
??

Code:

---

#include <stdio.h>
#include <string.h>
int main()
{

        char str1[2]={'a','b'};
       
        char* str=str1;
        *(str)='h';
                return 0;
}

---

?

But in this case

Code:

---

{'a', 'b'}

---

are not part of a string literal.

Also note that not ALL compilers generate code that fail when accessing string literals - the standard only says that string literals mustn't be modified, not what actually happens - that is part of the "undefined land", which as I said in another thread "Makes Alice in Wonderland seem quite normal". It may crash, it probably won't give you a nice error message, and it may appear to work just fine (until you modify the right bit in the right way, e.g. concatenating a string, and then some other string that got overwritten is modified as a result).

--
Mats