Truncating unsigned numbers

• Assume that $x^{\prime }$ is the result of truncating $x$
• If $x=B2U_w(\vec{x})$ and $x^{\prime }=B2U_k(\overrightarrow{x^{\prime }})$, then $x^{\prime }=x\mathrm{mod}{2}^k$.
  • This is equivalent to dropping the extra higher bits since the $(k+1{)}^{th}$ bit has the weight of $2^k$.

Sources

• Computer Systems - A Programmer’s Perspective

Related Notes

• Representing different data types in binary