Char arrays and functions

[cnewbie1]

I know I should read more, and I am, but sometimes it helps to get clear example answers to my more or less clear example questions. So, here goes:

This works:

char *my_strcpy (char *);

int main (int argc, char *argv[]) {
	char str[64];
	
	my_strcpy(str);
        puts(str);

	return 0;
}

char *my_strcpy(char *dest) {
	char *p = dest;			
	
	// Do something...
	
	return dest;
}

However, I'm at a loss if my array is an array of strings - a two-dimensional char array.

This does not work (naturally):

char *my_strcpy (char *);

int main (int argc, char *argv[]) {
	char str[5][64];
	
	my_strcpy(str);

	return 0;
}

char *my_strcpy(char *dest) {
	char *p = dest;			
	
	// Do something...
	
	return dest;
}

And neither does this:

char *my_strcpy (char dest[5][64]);

int main (int argc, char *argv[]) {
	char str[5][64];
	
	my_strcpy(str);
 	
	return 0;
}

char *my_strcpy(char dest[5][64]) {
	char *p = dest;  (*)	
	
	// Do something...
	
	return dest;  (**)
}

- "warning: initialization from incompatible pointer type" on line (*)
- "warning: return from incompatible pointer type" on line (**)

I'd like to use pointer notation instead of array notation, but at this point I cannot get either to work.

[claudiu]

p is type char *. dest is char *[64] (1)
my_strcpy returns char*. you are returning dest which is not a char*

[cnewbie1]

Hm...do I need to redefine the my_strcpy() function? If I remove everything inside the function:

char *my_strcpy (char *);

int main (int argc, char *argv[]) {
	char str[5][64];
	
	my_strcpy(str);
 
	return 0;
}

char *my_strcpy(char *dest) {
}

I get:
temp.c:9:2: warning: passing argument 1 of ‘my_strcpy’ from incompatible pointer type
temp.c:4:7: note: expected ‘char *’ but argument is of type ‘char (*)[64]’

[whiteflags]

I find it weird that your strcpy knock-off takes only one argument. Standard strcpy takes two, so I'm wondering where the copy part is.

The type is kinda a non-issue.

my_strcpy(str[0]);
my_strcpy(str[1]);
...
my_strcpy(str[4]);

These are all fine.

[cnewbie1]

The strcpy name is only because I forgot to change the name from the previous example I worked on. The thing I'm trying to find out is how to do the same as this with a 2-d array:

char *populate (char *);

int main (int argc, char *argv[]) {
	char str[64];
	
	my_strcpy(str);

	return 0;
}

char *populate (char *dest) {
	char *p = dest;			
	
	// Do something...
	
	return dest;
}

That is, pass the pointer to a 2-d array to a function which then populates the 2-d array and returns the pointer.

[whiteflags]

Oh, well the compiler gave you the correct type in the error message. You would just need to name your variable, like char (*foo)[64].

[cnewbie1]

Ah, thanks - think I got it now:

char *populate (char (*)[64]);

int main (int argc, char *argv[]) {
	char str[5][64];
	
	populate(str);
    	puts(str[0]);
	puts(str[1]);	

	return 0;
}

char *populate(char (*dest)[64]) {
	char (*p)[64] = dest;			

	strcpy(p[0], "one");
	strcpy(p[1], "two");
	
	return *dest;
}

Thanks a heap :-)

[ssharish2005]

char *populate(char (*dest)[64]) {
	char (*p)[64] = dest;			

	strcpy(p[0], "one");
	strcpy(p[1], "two");
	
	return *dest;
}

You dont have to return anything from this function. As your not changing the base pointer to anything.

ssharish

[King Mir]

Yeah, it doesn't make sense to return that. You could return dest, though the syntax for that is tricky.

[cnewbie1]

I thought so too. Also, why should it be necessary to return anything from that function? The array is created in the calling function, and the populate() function populates the array using the pointer.

I got this from here: A Tutorial on Pointers and Arrays in C
The author claims that this is "the practice used in the standard routine of returning a pointer to the destination":

int main(void) {
    my_strcpy(strB, strA);
    puts(strB);
} 

char *my_strcpy(char *destination, char *source) {
   char *p = destination;
   while (*source != '\0') {
      *p++ = *source++;
   }
   *p = '\0';
   return destination;
}

In that example, I don't see a reason for returning the destination pointer? The contents of the array is set by the function, by dereferencing the pointer, aren't they?

[tabstop]

Presumably destination is returned because the "real" strcpy also returns the destination pointer.

[ssharish2005]

i couldn't work out how to return 2D array. I tried this all the following

char (*)[64] populate( char (*des)[64] );
char [][64] populate( char (*des)[64] );
char ** populate( char (*des)[64] );

all gives me an error :-/ Any ideas?

ssharish

[cnewbie1]

ssharish2005: You're getting and error using the code I pasted? Exactly what are you doing?

[ssharish2005]

ssharish2005: You're getting and error using the code I pasted? Exactly what are you doing?

I'm trying to return a 2D array from a function. And i was trying to form a function declaration. Thats wasn't from your code.
ssharish

[CommonTater]

I'm trying to return a 2D array from a function. And i was trying to form a function declaration. Thats wasn't from your code.
ssharish

C cannot return arrays from functions... it doesn't know how. Inside any function except the one where the array is declared, C does not know how big it is, all size information is lost and without that, it cannot copy the array out to the calling function.

That is why you mostly see the array's pointer and size passed in the top of the function as parameters.

Run this small program for an example of why you should never try to return an array from a function...

#include <stdio.h>

int* MyFunction(int a, int b, int c)
  {  static int array[3];
     array[0] = a;
     array[1] = b;
     array[2] = c;
     return array;  } // return a pointer.


int main (void)
  { int *a1, *a2;  // int pointers

    printf("calling a1 = MyFunction(10,20,30);\t");
    a1 = MyFunction(10,20,30);
    printf("a1 has %d %d %d\n",a1[0],a1[1],a1[2]);

    printf("calling a2 = MyFunction(100,200,300);\t");
    a2 = MyFunction(100,200,300);
    printf("a2 has %d %d %d\n",a2[0],a2[1],a2[2]);

    printf("\nLooks good, except...\t"); 
    printf("a1 now has %d %d %d\n",a1[0],a1[1],a1[2]);

    getchar();
    return 0; }