$8[x]$ represents the greatest integer not exceeding the real number $x$, then $\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+$ $\left[\log _{2} 3\right]+\cdots+\left[\log _{2} 2012\right]=$ $\qquad$ .
818084 Prompt: When $2^{k} \leqslant x<2^{k+1}$, $\left[\log _{2} x\right]=k$, because
$$
1024=2^{10}<2012<2^{11}=2048,
$$

so
$$
\begin{array}{l}
{\left[\log _{2} 1024\right]+\left[\log _{2} 1025\right]+\cdots+\left[\log _{2} 2012\right]} \\
=10 \times(2012-1023)=9890 \text {. } \\
{\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+\cdots+\left[\log _{2} 1024\right]} \\
=1 \times 2+2 \times 2^{2}+3 \times 2^{3}+\cdots+9 \times 2^{9} \text {. } \\
\end{array}
$$

Let
$$
S=1 \times 2+2 \times 2^{2}+3 \times 2^{3}+\cdots+9 \times 2^{9},
$$

then
$$
\begin{aligned}
2 S=1 & \times 2^{2}+2 \times 2^{3}+3 \times 2^{4}+\cdots+9 \times 2^{10}, \\
-S & =2+2^{2}+2^{3}+\cdots+2^{9}-9 \times 2^{10} \\
& =2^{10}-2-9 \times 2^{10} \\
& =-8 \times 2^{10}-2 \\
& =-8194,
\end{aligned}
$$

so $S=8194$, therefore,
the original expression $=8194+9890=18084$.