why i dong get accsesing null memory error stuff..

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:)