--- In [email protected], Rick <[EMAIL PROTECTED]> wrote:
>
> I have a program I'm trying to write and can't figure out
> how to do it the way I want.
<snip>
> Can someone show me the proper way to reference the char arrays
> within a function? Do I actually need to de-reference them?
>
> Here is what I've written which is, obviously, wrong.
>
> #include <stdio.h>
> #include <stdlib.h>
>
> extern void loadcArray(char [][], int);
> extern void printcArray(char [][], int);
>
> #define MAXSTRLN 16
>
> int main(int argc, char *argv[])
> {
> int result;
> char array2[5][MAXSTRLN] = {"string1", "String2",
> "String3", "string4", "string5"};
> int a2size = (sizeof(array2) / MAXSTRLN);
It's better to change this to
int a2size = sizeof( array2) / sizeof( array2 [0]);
This way you can easily get the allocated size of any array, no matter
what its elements consist of (be it structs, unions, or pointers).
> loadcArray(array2, a2size);
> printcArray(array2, a2size);
>
> system("PAUSE");
>
> return 0;
> }
>
>
> void loadcArray(char cArray[][], int alen)
> {
> int ix;
> char val[MAXSTRLN];
>
> for (ix = 0; ix < alen; ix++)
> {
> printf("Enter a string value for array[%d]: ", ix);
> fscanf(stdin, "%s", val);
> strncpy(cArray[ix][0], val, MAXSTRLN-1);
> }
Why do you use "strncpy()" here? You have NOT terminated the copied
string explicitly. While this might not impose trouble if you're
dealing with global or static variables, it's still a good chance to
introduce unnecessary errors. I would use "strcpy()" here.
> printf("Finished loading %d values\n", ix);
>
> return;
> }
>
>
> void printcArray(char cArray[][], int alen)
> {
> int ix;
>
> printf("\n");
> for (ix = 0; ix < alen; ix++)
> {
> printf("Array[%d] = %s\n", ix, cArray[ix][0]);
> }
Leave out the "[0]" after "cArray"; "cArray [ix]" denotes a char[]
whereas your code denotes one single char although the printf() format
string requires a char* here. This is your main error.
> return;
> }
Please let us know whether there are any errors which I have overlooked.
Regards,
Nico
