Queries on declaring & initialising string and also string memory region

Question

My question comes in three part.

char str[] = "hello!";
    char *str1 = "heello!";
    puts(str);
    str1[1] = 1
    puts(str1);
    printf("%s\n", str1);
    printf("%p\n", str1);

Ouput of code:

hello!
h1ello!
heello!
0x10735ef9b

1) I understand that when you're declaring a char * type, you're declaring a pointer that points to char type. And usually for pointer, since it contains the address of the variable its pointing to, when you print the pointer itself, you'll get the address stored in the pointer and only when you dereference the pointer, then you'll get the content stored in the variable the pointer is pointing to. However for a char * type, why is it that it doesn't function like a regular pointer? You can print the content of the variable which the 'char *' pointer is pointing to without dereferencing it and also use it like a string variable(rather than a pointer) by changing the value of the str1[1] as shown above.

2) I was taught that you shouldn't declare a pointer variable as such in the topic of pointer:

char *str1 = "heello!";

because the pointer variable would be pointing to whatever address happens to be in the memory at that time, thus running a risk of changing the value in a memory location that you don't mean to and resulting in segfault. Hence you should do the following:

char *str1;
char c;
str1 = &c;
*addr = "hello!"

By doing the above, you'll make sure that your pointer is pointing to the right location before dereferencing and writing to the location.

However in the notes on string, it says to declare and initialise as such:

char *str1 = "heello!";

I'm quite confused as to what I should follow now when declaring a char * pointer. Please help.

3) It says in my notes that the location of string is stored in a read only region of the memory called the text region. Hence doing this:

    char *str1 = "hello!";
    str[1] = '.';

Will crash your program.

But when you do this:

char str2[7] = "hello!";
str2[1] = '.';

This is okay and will successfully change the second element in the string from 'e' to '.'

The explanation to this was that str1 points to a read only region in the memory, while str2 contains a copy of the string on the stack. I do not understand how the above 2 different ways of declaring the string variable will make such a big difference when executing the code. Please help.

Thank you!

score 4 · Accepted Answer · answered Mar 20 '19 at 11:45

String literals like "heello!" are really read-only arrays of characters (including the string null-terminator).

With

char *str1 = "heello!";

you make str1 point to the first character of such an array.

And when you do str1[1] = 1 you attempt to modify the read-only array, which is forbidden and leads to undefined behavior (which can but doesn't have to lead to crashes).

That's why you either should use your own arrays for strings, or use const char * to point to literal strings.

Queries on declaring & initialising string and also string memory region

1 Answers1