Thread: string arrays

"When a string constant is assigned to a pointer, the C compiler will allocate memory space to hold the string constant,store the starting address of the string in the pointer.It is important to note that the target array must have enough space to hold the string.

The declaration

char target[40];

is used instead of

char *target;

It is because the second declaration does not have space allocated to hold the string."

Let's say a target string constant is assigned to a pointer which is much longer than the string to be copied. Why is it that strncpy() still wont work?

For example(the following code is compilable),

[code]

#include <stdio.h>
#include <string.h>


main(void)
{
int n;
char *target = "Target string.(additional space)";
char *source = "Source string.";

printf("Input n:");/* n is no of characters to copy */
scanf("%d", &n);
printf("Before:\n");
/* why does the compiler recognise this as an array when it is a pointer?*/
printf("Target: %s\n",&target[0]);
printf("Source: ");
puts(source);
/* the program doesnt work even though *target have enough space to store source string */
strncpy(target,source,n);
if (target[n] != '\0')
target[n] = '\0';

printf("After:\n");
printf("Target: %s\n",target);
printf("Source: ");
puts(source);
return 0;
}

[\code]


This second example uses an array to store the target string and then assign the array to a pointer. This time, the pointer can be passed to the strncpy() and operate the copy function.

[code]

#include <stdio.h>
#include <string.h>



main(void)
{
int n;
char target[40] = "Target string.";
char *source = "Source string.";
/* pointer to point to target[40] */
char *tar = target;

printf("Input n:");
scanf("%d", &n);
printf("Before:\n");
printf("Target: %s\n",tar);
printf("Source: ");
puts(source);
/* passing tar to strncpy() */
strncpy(tar,source,n);
if (target[n] != '\0')
target[n] = '\0';

printf("After:\n");
printf("Target: %s\n",target);
printf("Source: ");
puts(source);
return 0;
}


[\code]

Sorry for the lengthy post. Hope someone could give me some guidance. Thanks![CODE]

Code:

a

can you explain more on what is string literals?

Great!...I finally understand on string literals. People like Quzah make this a lovely site.

As the world of knowledge is limitless, i still have more doubts.

Code:

i declared this:
char *target = "Target string.";


 /* output is "Target string."  Why does the compiler can recognise this as an array when it is a pointer?*/
printf("Target: %s\n",&target[0]);

You should take a look at assembly if you really want to understand *why* it is so. As quzah said, it is what you must know in C and it will work well with it but it won't tell you why it is so. Just take it as it is or learn assembly.

And anyway, try to do something like the code below to see what happens.

Code:

int **pt;
int ar[10][10];

pt = ar;

Indeed, the compiler will emit a warning. It simply means that multi-dimensionnal arrays cannot be used as pointers to pointer and vice versa.

If you want to know a little more, to the CPU, an array is actually a constant whereas a pointer is a variable.

Does it means that it takes a longer process to fetch a certain elements in a string using pointers than arrays?

Does it means that it takes a longer process to fetch a certain elements in a string using pointers than arrays?

Actually, yes and not really. It does when you have to load the pointer and it does not really when it is still in a register after some use.

Depending on how you implement it, using a pointer is faster than array subscripting. For example, to access an array like so:
myarray[i][j] = 1;
the compiler needs to perform some maths to work out the memory location to store the 1. However, using pointers in the right way, you could have ended up achieving the same result with
*p = 1;
Here p is a pointer that gets dereferenced, and the 1 jumps straight in, no maths needed.

Well, this is the case of a loop, and, yes, in this case, using pointers is appropriate. However, the use of pointers can be sometimes tricky.

Can someone explain how do i intepret this function prototype?

Code:

char *strcat(char *s1, const char *s2);

I know this function appends a copy of the string pointed by s2 to end of string pointed by s1.
But what does the "*" before the "strcat" mean? What does it serve to place a "*" before a function name. I am only taught to use "*" for pointers.
Furthermore, what does "const" before char *s2 mean? Am i right to say it means what ever string s2 is pointed to cannot be change. It must remain constant.

As an aside,
what is the difference between static and global variables? How do i use them? I believe both can be used in any function once declared and they retain any value assigned to it.
Thanks!