Thread: copying the arrays

Generally speaking, copy no more than needed, and ensure you don't overrun either of the arrays. Overrunning an array you're copying from is just as bad an idea as overrunning an array you are copying to.

There are plenty of methods. Look up the memcpy() function, for one.

Also, the word "static" does not mean what you think it does. Your code has no static arrays.

Are you sure it works?

Code:

$ gcc -Wall foo.c
foo.c: In function ‘main’:
foo.c:10:6: warning: implicit declaration of function ‘copy_array’ [-Wimplicit-function-declaration]
foo.c:17:9: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat]
foo.c: At top level:
foo.c:24:2: warning: return type defaults to ‘int’ [-Wreturn-type]
foo.c: In function ‘copy_array’:
foo.c:29:17: warning: assignment makes integer from pointer without a cast [enabled by default]
foo.c:37:9: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat]
foo.c:41:1: warning: control reaches end of non-void function [-Wreturn-type]
foo.c: In function ‘main’:
foo.c:23:1: warning: control reaches end of non-void function [-Wreturn-type]

Things like this
array[i]= array1;
are not copying the array.
It's just making every element 'point' to the start of another array. Except it's a pointer coerced into an integer.

You need
array[i]= array1[i];

> sorry its array[i]
Where?

Why do I get the feeling all of a sudden that the code you posted is not your real code, but something you half remember because you're not "at your computer" at the moment.