Did this program initialize its pointer? - c

Question

This is a simple program to copy a string to a new string from a C programming book (Programming in C, S. Kochan). It runs fine, but there is a pointer that I do not see initialized. I'm struggling with pointers here, but I have imbedded in my coding the C golden commandment: "THOU SHALT NOT USE THY POINTERS THAT ARE NOT INITIALIZED". Here is the code:
#include <stdio.h>
void copyString(char *to, char *from) //I do not see where the 'from' pointer is initilized?
{
for( ; *from != '\0'; ++from, ++to)
{
*to = *from;
}
*to = '\0';
}
int main(void)
{
void copyString (char *to, char *from);
char string1[] = "A string to be copied.";
char string2[50];
copyString(string2, string1);
printf("%s\n", string2);
copyString(string2, "So is this.");
printf("%s\n", string2);
return 0;
}
I was under the impression that all pointers must be initialized in the manner:
*ptr = &variable;
or risk something important in your system being overridden. But I am seeing many programs in my books that are not initializing pointers explicitly, and this makes me very uncomfortable. Please give me some tips about prudent pointer use so I don't destroy my machine, especially anything pertaining to strings. Thanks in advance all!

Answer 1

This confuses you - void copyString (char *to, char *from);. This is just a declaration in main. char *to and char *from in main are not in use so do not worry. The above code is as good as:
#include <stdio.h>
void copyString(char *to, char *from) //I do not see where the 'from' pointer is initilized?
{
for( ; *from != '\0'; ++from, ++to)
{
*to = *from;
}
*to = '\0';
}
int main(void)
{
void copyString (char *, char *);
char string1[] = "A string to be copied.";
char string2[50];
copyString(string2, string1);
printf("%s\n", string2);
copyString(string2, "So is this.");
printf("%s\n", string2);
return 0;
}
//I do not see where the 'from' pointer is initilized?
When you pass arguments like this copyString(string2, string1); - They will be copied into the args on function call of copyString(char *to, char *from). So in the first call:
copyString(string2, string1); - to = string2, and, from = string1
"A string to be copied."
from--^
And in the second call:
to = string2, and, from = "So is this."
"So is this."
from--^
string2 does not cause any problems (even though it was not initialized) since your overwriting its values (which had garbage values before).

Answer 2

to and from are function parameters, when the function is called, they are assigned the value of the arguments:
copyString(string2, string1);
to is assigned the value of stirng2 while from is assigned the value of string1. Note that even though string2 and string1 has type of char arrays, but they "decayed" into a pointer to the first element when passed as function arguments.

Answer 3

You are correct - all pointer must be initialized as you said.
When you're declaring the strings, as an array of characters as below:
char string1[] = "A string to be copied.";
char string2[50];
string1 and string2 are pointer to arrays of strings. So, they are already pointers.
When you used them in the copyString function, you are already using pointers, and the assignent is done implicitly. So:
to = string2 = &string2[0];

Answer 4

Here is a commented version of your code - the "from" variable has been renamed to source, and the printf's for the pointer values should illustrate the source. But suffice to say that the arguments to the copyString function are provided by the two calls to the function in your main program. The compiler generates code which takes the arguments you provide in the calls to the function with instructions to save these arguments onto the function call stack frame. Then when the function is called, these values are referenced in the function.
#include <stdio.h>
//the char pointer dest is provided as an argument
//the char pointer source is provided as an argument
char*
copyString(char *dest, char *source)
{
if(!dest) return dest;
if(!source) return dest;
printf("dest %x, source %x\n", dest, source);
//for each character in source, until *source == '\0'
for( ; *source; )
{
//assign the char at *dest into *source
//then post-increment source, dest
*dest++ = *source++;
}
//ensure the dest string is terminated
*dest = '\0';
return dest;
}
int main(void)
{
char* copyString (char *to, char *source);
char string1[] = "A string to be copied.";
char string2[50];
printf("string2 %x, string1 %x\n", string2, string1);
copyString(string2, string1);
printf("%s\n", string2);
copyString(string2, "So is this.");
printf("%s\n", string2);
return 0;
}
When you run the program, you will see the locations of the source and destination pointers provided,
$ ./copystr
string2 bff0dbe7, string1 bff0dc19
dest bff0dbe7, source bff0dc19
A string to be copied.
dest bff0dbe7, source 80485f0
So is this.