How to initialize a char array using a char pointer in C

Question

Let's say I have a char pointer called string1 that points to the first character in the word "hahahaha". I want to create a char[] that contains the same string that string1 points to.

How come this does not work?

char string2[] = string1;

Answer 1

"How come this does not work?"

Because that's not how the C language was defined.

You can create a copy using strdup() [Note that strdup() is not ANSI C]

Refs:

• C string handling
• strdup() - what does it do in C?

Answer 2

1) pointer string2 == pointer string1

change in value of either will change the other

From poster poida

char string1[] = "hahahahaha";
char* string2 = string1;

2) Make a Copy

char string1[] = "hahahahaha";
char string2[11]; /* allocate sufficient memory plus null character */
strcpy(string2, string1);

change in value of one of them will not change the other

Answer 3

In C you have to reserve memory to hold a string.
This is done automatically when you define a constant string, and then assign to a char[].

On the other hand, when you write string2 = string1 ,
what you are actually doing is assigning the memory addresses of pointer-to-char objects. If string2 is declares as char* (pointer-to-char), then it is valid the assignment:

char* string2 = "Hello.";

The variable string2 now holds the address of the first character of the constanta array of char "Hello.".

It is fine, also, to write string2 = string1 when string2 is a char* and string1 is a char[] .

However, it is supposed that a char[] has constant address in memory. Is not modifiable.
So, it is not allowed to write sentences like that:

 char string2[];
 string2 = (something...);

However, you are able to modify the individual characters of string2, because is an array of characters:

 string2[0] = 'x'; /* That's ok! */

Answer 4

What you write like this:

char str[] =  "hello";

... actually becomes this:

char str[] =  {'h', 'e', 'l', 'l', 'o'};

Here we are implicitly invoking something called the initializer .
Initializer is responsible for making the character array, in the above scenario.
Initializer does this, behind the scene:

char str[5];

str[0] = 'h';
str[1] = 'e';
str[2] = 'l';
str[3] = 'l';
str[4] = 'o';

---

C is a very low level language. Your statement:

char str[] = another_str;

doesn't make sense to C. It is not possible to assign an entire array, to another in C. You have to copy letter by letter, either manually or using the strcpy() function. In the above statement, the initializer does not know the length of the another_str array variable. If you hard code the string instead of putting another_str , then it will work.

Some other languages might allow to do such things... but you can't expect a manual car to switch gears automatically. You are in charge of it.

Answer 5

You can assign the address of string1 to string2 using a pointer like this:

#include <stdio.h>

int main(int argc, char** argv) {
    char* string1 = "hahahahaha";
    char* string2 = string1;

    printf("%s\n", string2);
}