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

i turned into string and it still working??

Code:

---

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

        char str1[3]={'a','b','\0'};
       
        char* str=str1;
        *(str)='h';
                return 0;
}

---

Quote:

---

Originally Posted by transgalactic2

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

---

An array of characters is not a string literal.

Neither is this:

Code:

---

char str[] = "foo";

---

In that case, "foo" itself is a literal, but it will be COPIED into the str[] array. So it would be writable.

i made it a string literal

Code:

---

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

        char str1[3]="ab";
       
        char* str=str1;
        *(str)='h';
                return 0;
}

---

its working
??

Quote:

---

Originally Posted by transgalactic2

so this code works because its not ending with \0

---

No, it works because you are dealing with an array whose contents can be modified, not a string literal whose contents cannot be modified.

This is correct:

Code:

---

#include <stdio.h>

int main(void)
{
    char str[] = "ab";
    char *p = str;
    *p = 'b';
    printf("%s\n", str);
    return 0;
}

---

This is wrong:

Code:

---

#include <stdio.h>

int main(void)
{
    char *str = "ab";
    char *p = str;
    *p = 'b';
    printf("%s\n", str);
    return 0;
}

---

Spot the difference.

Quote:

---

Originally Posted by transgalactic2

i made it a string literal

---

You didn't. You declared an array, not a "char *". As I already said, doing it this way causes the string to be copied into the array. The literal is the "ab" in the code, not the array. Had you declared a char *, you would be pointing at this literal, and could not modify it, but that's not what you did.

hooooray i got my error

Code:

---

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

       
       
        char* str="ab";
        *(str)='h';
                return 0;
}

---

it weird because by the definitions that i know

Code:

---

char* str="ab";

---

equals

Code:

---

char str1[3]={'a','b','\0'};
char* str=str1;

---

so i cant see why one is working and the other doesnt
??

so
this is a literal

Code:

---

char* str="ab";

---

and we cant change any of it cells
thanks
the other stuff that i showed are arrays of chars
correct??

Quote:

---

Originally Posted by transgalactic2

by the definitions that i know

---

start learning correct definition in this case

Quote:

---

Originally Posted by transgalactic2

this is a literal

---

This is a string literal: "ab"
It is an array of three chars, including the null terminator.

Quote:

---

Originally Posted by transgalactic2

the other stuff that i showed are arrays of chars

---

Arrays of chars are always involved where strings are involved.

you say that they both are arrays of chars.

so what is the difference between

Code:

---

char str[] = "ab";
and
char* str = "ab";

---

??
why the first is allowed and the last not
except the fact that the first is not a literal ??
what happens in the compiler that allowes it?

Quote:

---

Originally Posted by transgalactic2

so what is the difference between

---

The difference is that in the former, str is an array of chars that is a copy of the string literal "ab", but in the latter, str is a pointer to the first char of the string literal.

This difference is critical when you attempt to change the contents of str, since with the former you safely change the elements of an array under your control, but for the latter you attempt to change a string literal that is stored in an area not under your control.

Quote:

---

Originally Posted by transgalactic2

you say that they both are arrays of chars.

so what is the difference between

Code:

---

char str[] = "ab";
and
char* str = "ab";

---

??
why the first is allowed and the last not
except the fact that the first is not a literal ??
what happens in the compiler that allowes it?

---

Do you know difference of the array and pointer?

first is array of 3 chars that is initialized with 'a','b',0

and because it is array - you can modify each element of it

the second is just a pointer that points to some external memory, which could be read-only (it is compiler dependent)

you can modify your pointer but you have no right tomodify contents of the external memory

aaahh the memory is Read only
so in this case char* str = "ab";
it changes the memory

thanks:)