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Can't be done. That's what it means by "decaying to a pointer" -- there is no way to get an array into a function -- either you pass a specific element of the array to the function (which can be...

If you want to modify some chars, you need to pass a pointer-to-char. An array name already is a pointer-to-char, so you're done.

char *foo;
char bar[32];

In the above code, then:
foo is of type pointer-to-char
bar is of type pointer-to-char (array names by themselves decay to pointers to their element)
&foo is of type...