Handling argc==0 in the kernel

> If argc is 0, argv[1] uses an index outside of the bounds of the array. Thus even writing argv[1] would be UB.

Another comment by areilly hinted at this already, but evaluating argv[1] for an argv array of size 1 (where argc == 0) is not undefined behavior. You are allowed to construct (but not dereference) a pointer to the element immediately after the end of an array. (Taking the address of a dereference or array indexing expression does not count as actually dereferencing the pointer: even though argv[1] in an expression by itself would be UB, &argv[1] and &*(argv + 1) are both semantically identical to the pointer arithmetic operation argv + 1. The address-of operator "cancels out" the dereference.) So there is no undefined behavior in the example, even if argc == 0, as &argv[1] and &argv[0] are both valid pointers.