Shifting a value wider than the shifter
A single shift instruction reaches across 32 bits, so it can't carry bits over the seam between the two halves of a u64. That's why a variable-distance 64-bit shift hands off to a runtime helper, the same shape division uses, with a prologue, a bl, and an epilogue:
stwu r1, -16(r1)
mflr r0
mr r5, r6 # marshal the shift amount into r5
stw r0, 20(r1)
bl __shl2i # 64-bit shift-left helper
lwz r0, 20(r1)
mtlr r0
addi r1, r1, 16
blr
Here bl __shl2i is the left-shift helper. Right shifts split by signedness: __shr2u for a logical (unsigned) shift and __shr2i for an arithmetic (signed) one. The mr r5, r6 just ahead of the call is the compiler lining up the arguments the helper wants, dropping the shift distance into place, much as a divide marshals its operands.
That rounds out the short list of 64-bit operations the hardware can't do inline. Divide, modulo, and shift are the three, and each one announces itself with a bl __…2i or __…2u. Add, subtract, multiply, and the bitwise ops coming up next all stay inline.
Your task
Write shl_64 to match the target.