8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
\begin{array}{l}
{\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\cdots+\left[\log _{2} 2012\right]} \\
=
\end{array}
$$
8. 18084.

When $2^{k} \leqslant x<2^{k+1}$, then $\left[\log _{2} x\right]=k$.
Given $1024=2^{10}<2012<2^{11}=2048$, we know
$$
\begin{array}{l}
{\left[\log _{2} 1024\right]+\left[\log _{2} 1025\right]+\cdots+\left[\log _{2} 2012\right]} \\
=10 \times(2012-1023)=9890 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence }\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\cdots+\left[\log _{2} 1023\right] \\
=1 \times 2+2 \times 2^{2}+\cdots+9 \times 2^{9} .
\end{array}
$$

Let $S=1 \times 2+2 \times 2^{2}+\cdots+9 \times 2^{9}$.
Then $2 S=1 \times 2^{2}+2 \times 2^{3}+\cdots+9 \times 2^{10}$.
Subtracting the above two equations, we get
$$
\begin{array}{l}
S=-2-2^{2}-\cdots-2^{9}+9 \times 2^{10} \\
=-2^{10}+2+9 \times 2^{10}=8194
\end{array}
$$

Therefore, the original expression $=8194+9890=18084$.