# Common pitfalls¶

## xarray initialization¶

```
xt::xarray<double> a({1, 3, 4, 2});
```

does not initialize a 4D-array, but a 1D-array containing the values `1`

, `3`

,
`4`

, and `2`

.
It is strictly equivalent to

```
xt::xarray<double> a = {1, 3, 4, 2};
```

To initialize a 4D-array with the given shape, use the `from_shape`

static method:

```
auto a = xt::xarray<double>::from_shape({1, 3, 4, 2});
```

The confusion often comes from the way `xtensor`

can be initialized:

```
xt::xtensor<double, 4> a = {1, 3, 4, 2};
```

In this case, a 4D-tensor with shape `(1, 3, 4, 2)`

is initialized.

## Intermediate result¶

Consider the following function:

```
template <class C>
auto func(const C& c)
{
return (1 - func_tmp(c)) / (1 + func_tmp(c));
}
```

where `func_tmp`

is another unary function accepting an xtensor expression. You may
be tempted to simplify it a bit:

```
template <class C>
auto func(const C& c)
{
auto tmp = func_tmp(c);
return (1 - tmp) / (1 + tmp);
}
```

Unfortunately, you introduced a bug; indeed, expressions in `xtensor`

are not evaluated
immediately, they capture their arguments by reference or copy depending on their nature,
for future evaluation. Since `tmp`

is an lvalue, it is captured by reference in the last
statement; when the function returns, `tmp`

is destroyed, leading to a dangling reference
in the returned expression.

Replacing `auto tmp`

with `xt::xarray<double> tmp`

does not change anything, `tmp`

is still an lvalue and thus captured by reference.

## Random numbers not consistent¶

Using a random number function from xtensor actually returns a lazy generator. That means, accessing the same element of a random number generator does not give the same random number if called twice.

```
auto gen = xt::random::rand<double>({10, 10});
auto a0 = gen(0, 0);
auto a1 = gen(0, 0);
// a0 != a1 !!!
```

You need to explicitly assign or eval a random number generator, like so:

```
xt::xarray<double> xr = xt::random::rand<double>({10, 10});
auto xr2 = eval(xt::random::rand<double>({10, 10}));
// now xr(0, 0) == xr(0, 0) is true.
```

## variance arguments¶

When `variance`

is passed an expression and an integer parameter, this latter
is the axis along which the variance must be computed, but the degree of freedom:

```
xt::xtensor<double, 2> a = {{1., 2., 3.}, {4., 5., 6.}};
std::cout << xt::variance(a, 1) << std::endl;
// Outputs 3.5
```

If you want to specify an axis, you need to pass an initializer list:

```
xt::xtensor<double, 2> a = {{1., 2., 3.}, {4., 5., 6.}};
std::cout << xt::variance(a, {1}) << std::endl;
.. Outputs { 0.666667, 0.666667 }
```