$$a=f\left( {{\log }_{e}}\frac{1}{ \pi } \right)=-{{\log }_{e}} \pi \in (-2,-1)$$，$$b=f({{\log }_{ \pi }}\frac{1}{e})=-{{\log }_{ \pi }}e\in (-1,0)$$，$$c=f\left( {{\log }_{\frac{1}{e}}}\frac{1}{{{ \pi }^{2}}} \right)=-2{{\log }_{2}} \pi \in (-2,-4)$$，由此能求出结果．  【解答】解：∵$$f(x)=-\textbar x\textbar$$，  ∴$$a=f(lo{{g}_{e}}\frac{1}{ \pi })=-\left\textbar{} {{\log }_{e}}\frac{1}{ \pi } \right\textbar=-{{\log }_{e}} \pi \in (-2,-1)$$，  $$b=f\left( {{\log }_{ \pi }}\frac{1}{e} \right)=-\left\textbar{} {{\log }_{ \pi }}\frac{1}{e} \right\textbar=-{{\log }_{ \pi }}e\in (-1,0)$$，  $$c=f\left( {{\log }_{\frac{1}{e}}}\frac{1}{{{ \pi }^{2}}} \right)=-\left\textbar{} {{\log }_{\frac{1}{e}}}\frac{1}{{{ \pi }^{2}}} \right\textbar=-{{\log }_{e}}{{ \pi }^{2}}=-2{{\log }_{e}} \pi \in (-2,-4)$$，  ∴$$b\textgreater a\textgreater c$$．  故选$$\text{D}$$． 
设$$f(x)=-\textbar x\textbar$$，$$a=f\left( {{\log }_{e}}\frac{1}{ \pi } \right)$$，$$b=f\left( {{\log }_{ \pi }}\frac{1}{e} \right)$$，$$c=f\left( {{\log }_{\frac{1}{e}}}\frac{1}{{{ \pi }^{2}}} \right)$$，则下述关系式正确的是（~ ）．
$$a\textgreater b\textgreater c$$ 
$$b\textgreater c\textgreater a$$ 
$$c\textgreater a\textgreater b$$ 
$$b\textgreater a\textgreater c$$ 