No. In general, 64 bits are needed.
MIPS contains two 32-bit registers called hi and lo. These are not general purpose registers. When two 32-bit operands are multiplied, hi and lo hold the 64 bits of the result. Bits 32 through 63 are in hi and bits 0 through 31 are in lo.
Here are the instructions that do this. The operands are contained in general-purpose registers.
mult s,t # hilo ← $s * $t (two's comp operands) multu s,t # hilo ← $s * $t (unsigned operands)
There is a multiply instruction for unsigned operands, and a multiply instruction for signed operands (two's complement operands). Integer multiplication never causes a trap.
Confusion Altert! add and addu,
both perform the same operation.
The "u" means "don't trap overflow".
With mult and multu,
different operations are carried out.
Neither instruction ever causes a trap.
If the operands are too big, overflow will happen. But with these instructions the overflow will not cause a trap.
Two small integers are multiplied. Where is the result?