How has this array been changed?

// A UDF to copy a String

#include <stdio.h>
#include <conio.h>
void stringcopy(char[],char[]);

void main()
{
clrscr();

char x[20];
char y[20] = "Aquib Mir";

stringcopy(x,y);

printf("X is %s",x);

getch();
}

void stringcopy(char s[20],char t[20])
{
int i = 0;
while(t[i] != '\0')
{
s[i] = t[i];
i++;
}
s[i] = '\0';
}

How has 'y' been copied to 'x'? Because I performed the operation in my own finction, not in main.?// A UDF to copy a String #include <stdio.h>

Your function copies the second parameter to the first parameter, not the other way around. When you pass the array names to the function, it actually gets pointers to the address of the first element in the arrays. So the string copy function is actually acting on the arrays you passed, not copies of them.

I don't fully understand what you are asking, perhaps you could reword it?

p.s. - main returns an int

-Prelude

Quote:

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How has 'y' been copied to 'x'? Because I performed the operation in my own finction

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because array names are pointers, they store the address of the first element of the array.

Quote:

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stringcopy(x,y);

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So what your doing here is telling your function to use the address's stored in x, y. So s and t are actually using the arrays x and y.

[edit]
Grrrrr beaten again
[/edit]

Thanks to both of you.

Actually, I was trying to ask how the arrays in main were copied, when I performed the operation in my own function.

Anyway, its clear now. I had the notion that the whole array is passed to the UDF. But, I now understand that the variables in the UDF are actually working as pointers. They are not receiving any array. The whole operation is actually performed in main.

Thanks a lot.

>The whole operation is actually performed in main
Actually, the operation is performed in your function on pointers that point to objects local to main. :)

-Prelude