For a sheaf $\mathcal{F}$, denote by $\mathcal{F}[n]$ ($n\in \mathbb{ Z}$) the complex of sheaves given by $$\mathcal{F}[n]_i=\begin{cases} \mathcal{F}&\text{if }i= n\\ 0&\text{otherwise}. \end{cases} $$
Let $C$ be an object in the category of derived constructible sheaves on $\mathbb{C}$, that I denote by $D_c(\mathbb C)$. It seems there is (maybe under some mild assumptions) an exact triangle $$H^{-1}(C)[1]\rightarrow C\rightarrow H^0(C)[0]$$ in $D_c(\mathbb C)$, where $H^i(C)$ denotes the sheaf obtained taking the $i^{\text{th}}$-homology of $C$.
I know little about derived sheaves. For instance I know how we construct the categories $D(X)$ and $D_c(X)$, but I'm not used to manipulate them. Is the above true? What is the map $C\rightarrow H^0(C)[0]$, how do I describe it? Do I have a similar exact triangle for an object $C$ in $D_c(X)$ for an arbitrary topological spaces/scheme $X$?
To have a triangle $$H^{-1}(C)[1]\to C\to H^0(C)[0]\overset{+1}\to$$ it is necessary and sufficient that $C$ is concentrated in (cohomological) degree $-1$ and $0$. Indeed, $C$ is an extension of a complexe concentrated in degree $0$ by a complex concentrated in degree $-1$ so this is clearly necessary. It also sufficient, for in that case, $C$ is quasi-isomorphic to a complex of the form $0\to C^{-1}\overset{f}{\to} C^0\to 0$. In that case, the map $C\to H^0(C)[0]$ is induced by the map of complexes : $$\require{AMScd} \begin{CD} C^{-1}@>f>> C^0\\ @.@VVV\\ @.\operatorname{coker} f \end{CD} $$ (note that $H^0(C)=\operatorname{coker}f$). And similarly, the map $H^{-1}(C)[1]\to C$ is induced by the map $$\begin{CD} \ker f\\ @VVV\\ C^{-1}@>f>> C^0 \end{CD} $$ (since $H^{-1}(C)=\ker f$).
This kind of complexes is appears for example as the cone of a morphism of (constructible) sheaves. (In fact, this triangle holds in any derived category, not just the category of sheaves).