# Complexity of zero-one loss vs logistic loss functions

In Machine Learning: The Basics, Alexander Jung, Spinger, the author states that using the zero-one loss function defined by:$$\mathit{L}((\mathbf{x}, y), h)= \begin{cases} 1, y \ne \hat{y}\\ 0\ otherwise, \end{cases}\\ (with\ \hat{y}=1\ for\ h(\mathbf{x}) \ge 0, and\ \hat{y} = −1 \ for \ h(\mathbf{x})\lt0).$$

results in computationally challenging problems and the logistic loss is a computationally attractive alternative.

However, consider(ing) that the logistic function defined as follows:

$$L((\mathbf{x}, y), h) := log(1 + exp(−yh(\mathbf{x}))),$$

I don't see any "disadvantages" of the zero-one loss in terms of computational efficiency. To my understanding, for each training example (or instance), the zero-one loss function needs only 2 operations (1 comparison and 1 assignment) and it's done. Meanwhile with the logistic loss, it's a bunch of operations (1 multiplication and exponentiate with base e and then 1 addition, and taking log). And not only the number of operations is larger the complexity of those operations is also higher.

Can someone help to point out what I'm missing here?