I recently asked this question [Unbounded sectional curvature implies infinite diameter?](https://mathoverflow.net/questions/348751/unbounded-sectional-curvature-implies-infinite-diameter).
I would like now to ask something similar, but in another context. 


Suppose you have a complete metric space $(M,d)$. Assume that the curvature of $M$ nas no upper bound. Can one concludes that the diameter of $M$ is infinite?


If it helps, one can assume that $(M,d)$ is the limit of a sequence of compact manifolds.


**EDIT**


Sorry for being vague on the first time. I am assuming I have a length space $(M,d)$ which is actually complete. It is obtained as the Gromov-Hausdorff limit of a sequence of compact Riemannian manifolds. So my question is: if the Alexandrov curvature of $(M,d)$ is not bounded from above, is the diameter of $(M,d)$ infinite?



Unbounded curvature implies infinite diameter on complete metric spaces
Here is counter example. Consider the surface of the unit cube in $\mathbb{R}^3$. This is compact and has infinite curvature at the corners. One can also construct a sequence of smooth compact manifolds that converge to this surface, some care needs to be taken of what kind of convergence you want.


**Edit:** Thanks to the helpful discussion in the comments. As a result we can say that the surface of the unit cube does have infinite curvature in the sense that it is not a $CAT(k)$ space for any finite $k$.


