Can a string be assigned to a char pointer, without it becoming a literal?

Question

The question sounds a tad dumb, allow me to demonstrate what I mean.

I know that if I were to do something along the lines of:

(const) char *ptr = "I'm text!";

It'd be a literal I can't modify by any means later on. However, I figured, as there is a way to set up a pointer to work just like an array (on the heap), wouldn't it work to set up a string that way too? If yes, what'd be the easy way?

I tried the following, but it seems rather redundant, compared to just making an array and then assigning a pointer to it.

char *ptr = malloc(sizeof(char)*256);

ptr[0]='S';
ptr[1]='u';
ptr[2]='p';
ptr[3]='e';
ptr[4]='r';
ptr[5]='\0';

printf("%s\n", ptr);  

free(ptr);

Answer 1

You can do

char str[] = "eureka, this works";

Now you can modify the char s in it, using str , because it is essentially a char array. This means that certain operation like incrementing str++ will not work.

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However, if you strictly want to work with a pointer, then you can add another line to the above code.

char str[] = "eureka, this works";
char* ptr = str;

Now you can use ptr , operations like incrementing and all will work, since it is a pointer.

Answer 2

After allocating space to char * ( as you talk about it in example ), instead of doing character by character , you can use strcpy -

char *ptr = malloc((sizeof *ptr)*256);
if(ptr!=NULL) {                     // check return 
     strcpy(ptr,"super");
     //do something 
} 
free(ptr); 

Answer 3

There is difference between character array initialization and char pointer initialization.

Whenever you initialize a char pointer to point at a string literal, the literal will be stored in the code section. You can not modify code section memory. If you are trying to modify unauthorised memory then you will get a segmentation fault.

But if you initialize a char array, then it will be stored in the data or stack section, depending on at where you declared the array. So you can then modify the data.