How to assign a value to a char* using hex notation?

I usually use pointers in the following manner

    char *ptr = malloc( sizeof(char) * 100 );
    memset( ptr, 0, 100 ) ;
    strncpy( ptr, "cat" , 100 - 1 );

But this time instead of using "cat", I want to use it ASCII equivalent in hex.

cat = 0x63, 0x61, 0x74, 0x00

I tried

    strncpy( ptr, "0x630x61" , 100 - 1 );

But it fails as expected.

What is the correct syntax?

Do I need to put a 0x00 too? For a moment lets forget about memset, now do I need to put a 0x00? Because in "cat" notation, a null is automatically placed.

Regards