Thread: A question on pointers

Hi all.

I have two questions.

1) Please take a look at the following code:

Code:

char test[] = "hi all";
char *pt = test;

This bit of code makes perfect sence, since test == &test[0]. Now look at

Code:

char *test = "hi all";

Now this bit I cannot wrap my head around. Where is the logic in this statement? Is the string "hi all" an address?


2) Why is it not possible in C to use

Code:

int *test = {1,2,3};

Thank you very much in advance.

Best regards,
Niles.

So in this case "hi all" is a string terminated by a '\0' character. Remember strings are nothing but array of characters in c. So here "hi all\0" is stored in contiguous memory locations. So hi all is not an address in itself but test has the pointer to the beginning of the string and knows that it extends till the '\0' character.

Also in the second case the way you are doing the assignment is illegal.

You are trying to have an array of integers so you need to do the declaration like this

int test[] = {1,2,3};

Originally Posted by roaan

So hi all is not an address in itself

If that is the case, then why is

char *pt = "hi all"

logical? The address of the pointer is set the be the string.

Originally Posted by roaan

Also in the second case the way you are doing the assignment is illegal.

Why is this? Is it decided that the assignment is legal, or is there a logical explanation of why it is illegal?

Thank you.

Originally Posted by Niels_M

If that is the case, then why is

char *pt = "hi all"

logical? The address of the pointer is set the be the string.






Why is this? Is it decided that the assignment is legal, or is there a logical explanation of why it is illegal?

Thank you.

I didn't get your first question. What are you trying to ask ?

Also for the second assignment the way arrays are declared in C that is the correct syntax for them is

[insert]

Code:

int test[] = {1,2,3};

or for your case if you are trying to get a pointer to the array of integers then you should do like this

[insert]

Code:

int *p;
int test[] = {1,2,3};
p = test; // now p has the address to the first element of test

My first question is the following:

Please take a look at the following code

Code:

char test[] = "hi all";
char *pt = test;

This bit of code makes perfect sence, since test == &test[0].

Now look at the following code

Code:

char *test = "hi all";

Now this piece of code I do not understand. When declaring a pointer, we should say where it points to, i.e. we have to specify the address. But it is not equalled to an address, but to a string. Why is it we are allowed to do that?

Okay so thats good that you know that we need to specify the address when assigning something to a pointer as in

char *pt = test;

But as i mentioned before that strings are special in C. They are treated as an array of characters. So when i say

char *test = "hi all";

what this implies is that the address of the first array element(just consider this for the sake of understanding) which is 'h' in this case is assigned to test. So this is how strings are treated in c.

If you were to do the following on the test

printf("\n %c", test); // this would give you 'h'
so now you see that test is having the address of the first element of the array "hi test". Hope this makes it a bit clearer. So in this case test which is a pointer to a char is pointing to the first character 'h' and you can iterate through the array till you hit the terminating character '\0'.

[insert]

Code:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
	
	char *str = "Hello Sir";
	int *p;
	
	int test[] = {1,2,3};
	p = test;

	printf("\n %c", *str);
	while(*str != '\0')
	{
		printf("\n %c", *str);
		++str;
	}

	getch();
	return 0;
}

I understand everything in your post. My question is a lot more fundemental: Did K&R decide that when we have

pointer = character array,

then pointer gets the address of the first element of the array? I.e., is it just a "coincidence" or is there a logic behind the notation, just like the following piece of code is logical

Code:

char test[] = "hi all";
char *pt = test;

?

Originally Posted by Elysia

For the second question, an initializer list is basically a syntax that allows you to put data into an array (or struct) when you first define it.

In that case, a string is also an initalizer list. And since that is the case, how can we initialize a pointer with it?

Originally Posted by Elysia

For the first question, yes, the string is copied into the array, so it's modifiable.

Ok, so when I have char test[] = "hello", "hello" is NOT a string literal?

Thank you for being patient with me.

I think I get it now. So there is actually a difference between the two following assignments, even though my book says they are equivalent:

Code:

char str[] = "test"

and

Code:

char str1[] = {'t', 'e', 's', 't', '\0'}

The first is a string literal, but the content is also copied into the array str[], and thusly is modifiable. The second is an initializer, and thus I cannot replace str1[] with *str.

Thanks to all for helping.