Am I assigning a value directly to a memory location with a pointer variable in C?

Question

I've decided to complicate my life by learning C. Currently I am trying to fully understand pointers.

So far this is what I understand:

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

    char var = 'A';
    // Iniialize a pointer variable to the ADDRESS of variable "var"
    char *ptr = &var;
    printf("%c", var);    // Output: 'A'

    // Change the VALUE at the memory location of variable "var"
    *ptr = 'B';
    printf("%c", var);    // Output: 'B'

    return (0);
}

But then I was looking into re-assigning a string to a variable, and came across this post, where a pointer variable is assigned a value directly.

This is what I am trying to understand:

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

    // Declare a pointer variable
    char *ptr;

    // Assign a VALUE directly to a memory location???
    ptr = "String 1";
    printf("%s", ptr);    // Output: "String 1"

    return (0);
}

Also, if I assign a new string value to ptr, such as ptr = "String 2";, and also output the memory location after each assignment, I get a different memory location:

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

    // Declare a pointer variable
    char *ptr;

    // Assign a VALUE directly to a memory location???
    ptr = "String 1";
    printf("%s\n", ptr);    // Output: "String 1"
    printf("%p\n", ptr);    // Output: 0000000000404003 <- one memory location

    // Assign a new VALUE to the pointer variable
    ptr = "String 2";
    printf("%s\n", ptr);    // Output: "String 2"
    printf("%p", ptr);    // Output: 000000000040400F <- a new memory location

    return (0);
}

Am I really assigning a VALUE to the memory location the pointer variable is referencing?

Why does the memory location change when I assign a new value to the pointer variable?

`"String 1"` and `"String 2"` are constant strings necessarily at different memory locations. They aren't even the same string value. However, you indicate in the code comments that, `ptr = "String 2"; printf("%s\n", ptr); ` prints `String 1`. That's not making sense. Check that again and make sure. — lurker, Dec 25 '21 at 17:36
When you assign to dereferenced pointer, i.e. `*ptr = ...`, you are changing *the value at the location* where the pointer is pointing to. When you assign to pointer, i.e., `ptr = ...`, you are changing *where* the pointer is pointing to. — danadam, Dec 25 '21 at 17:43
@lurker My mistake. I wrote the wrong output. I corrected it. — howdoicode, Dec 25 '21 at 17:43
@howdoicode thanks, that makes sense. The rest of my comment explains the difference in pointer values. — lurker, Dec 25 '21 at 17:59

0___________ · Answer 1 · 2021-12-25T17:52:26.550

pointer hold reference to (address of) the object.

char *ptr = "string 1";

is the same as

char *ptr;
ptr = "string 1";

it assigns the pointer ptr with the address of the string literal "string 1". String literals are null character terminated const char arrays.

When you :

ptr = "string 2";

you assign the pointer ptr with the address of another string literal. And as I wrote above is simply a const char array. Addresses of both string literals are different.

ptr = "String 2";
printf("%s\n", ptr);    // Output: "String 1"

It is not possible, and you post here different code than you run on your computer. It will print String 2 for sure.

// Assign a VALUE directly to a memory location???
ptr = "String 1";

No, this assignment only assigns pointer ptr with the address of string literal which resides somewhere in memory. It does not allocate the space or copy this string anywhere.

The magic *:

When you use the dereference operator * you access the object the pointer is referencing.

char str[] = "String 1";
char *ptr;

ptr = str;
*ptr = 'X';

printf("str = `%s` And its address is %p\n", str, (void *)str);
printf("ptr = `%s` it is holging %p address. Its address is %p\n", ptr, (void *)ptr, (void *)&ptr);

Output:

str = `Xtring 1` And its address is 0x7ffdc9047827
ptr = `Xtring 1` it is holging 0x7ffdc9047827 address. Its address is 0x7ffdc9047818

As you see it changes the first character of the str as pointer ptr holds the address of the first character of this array.

https://godbolt.org/z/x8PsW9vsP

For `ptr = "String 2";` is mistakenly wrote the wrong output. I corrected it in my original post. — howdoicode, Dec 25 '21 at 17:51
@howdoicode be more careful when you ask questions. Such mistakes waste time of many people — 0___________, Dec 25 '21 at 17:52
When you are assigning `ptr = str;`, what is happening here? How come you don't use the `&` "address-of" operator, such as `ptr = &str;`? — howdoicode, Dec 25 '21 at 19:42

score 2 · Answer 2 · answered Dec 25 '21 at 17:44

To assign a value to an object using a pointer that points to the object you need to dereference the pointer as you are doing in the first your program.

*ptr = 'B';

As for this code snippet

char *ptr;

ptr = "String 1";

then there is declared a pointer that then is assigned with the address of the first character of the string literal "String 1".

You can reassign the pointer like

ptr = "String 2";

but this reassigns the pointer itself not the object previously pointed to by the pointer. That is now the pointer points to the first character of the string literal "String 2".

Pay attention to that you may not change the string literal using the pointer like for example

*ptr = 's';

Any attempt to change a string literal results in undefined behavior.

But you can change a character array initialized by a string literal like

char s[] = "String 1";
char *ptr = s;
*ptr = 's';

Am I assigning a value directly to a memory location with a pointer variable in C?

2 Answers2