I'm a bit confused about Example 22.26.6 from Stacks project on Hom complexes. Let $R$ a ring and $A^{\bullet}, B^{\bullet}$ two complexes of $R$-modules. Then we can define the Hom complex $\mathop{\mathrm{Hom}}\nolimits ^ {\bullet}(A^\bullet , B^\bullet ) $ as
$$\mathop{\mathrm{Hom}}\nolimits ^ n(A^\bullet , B^\bullet ) = \prod \nolimits _{p + q = n} \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(A^{-q}, B^ p)$$
end endow it with a differential as follows. Let $f \in \mathop{\mathrm{Hom}}\nolimits ^ n(A^\bullet , B^\bullet )$. Then
$$\text{d}(f) = \text{d}_ B \circ f - (-1)^ n f \circ \text{d}_ A$$
In terms of components, if $f=(f_{p,q})$ with $f_{p, q} : A^{-q} \to B^ p$ we have
$$\begin{equation} \label{dga-equation-differential-hom-complex} \text{d}(f_{p, q}) = \text{d}_ B \circ f_{p, q} - (-1)^{p + q} f_{p, q} \circ \text{d}_ A \end{equation}$$
Note that the first term of this expression is in $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(A^{-q}, B^{p + 1})$ and the second term is in $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(A^{-q - 1}, B^ p)$.
Then it is claimed in (2) that
an element $f$ in $\mathop{\mathrm{Hom}}\nolimits ^ n(A^\bullet , B^\bullet )$ has $d(f)=0$ if and only if the morphism $f : A^\bullet \to B^\bullet [n]$ of graded objects in the category of complexes over $A$ is actually a map of complexes.
I'm not sure why this is true. Let $f=(f_{p,q}):A^\bullet \to B^\bullet [n]$. Then $f_{p,q}: A^{-q} \to B^{-q}[n]= B^{n-q}$ with $p=n-q$ and since it is a map of complexes we have $d_B \circ f_{p-1, q+1} = f_{p,q} \circ d_A$.
Assume $n$ is odd. Then $-(-1)^n=1$ and $-(-1)^{(p) + (q)} f_{p,q} \circ d_A$ cannot "neutralize" $d_B \circ f_{p-1, q+1}$. Thus $d(f) \neq 0$, I think. This contradicts the claim (2) or do I misunderstand the issue?