kalindkaria
Map.update inside for loop
Hi, I’ve just started learning Elixir.
For one of the applications I’m developing, I have the following code:
use Bitwise
sensor_data = %{sensor1: 0}
for n <- 0..9, do: sensor_data = Map.update!(sensor_data, :sensor1, &(&1 ||| (1 <<< n)))
This produces following output:
[
%{sensor1: 1},
%{sensor1: 2},
%{sensor1: 4},
%{sensor1: 8},
%{sensor1: 16},
%{sensor1: 32},
%{sensor1: 64},
%{sensor1: 128},
%{sensor1: 256},
%{sensor1: 512}
]
But the expected output should be:
[
%{sensor1: 1},
%{sensor1: 3},
%{sensor1: 7},
%{sensor1: 15},
%{sensor1: 31},
%{sensor1: 63},
%{sensor1: 127},
%{sensor1: 255},
%{sensor1: 511},
%{sensor1: 1023}
]
If I run the following statement in iex session the sensor_data map gets updated like this:
sensor_data = Map.update!(sensor_data, :sensor1, &(&1 ||| (1 <<< 0)))
%{sensor1: 1}
sensor_data = Map.update!(sensor_data, :sensor1, &(&1 ||| (1 <<< 1)))
%{sensor1: 3}
sensor_data = Map.update!(sensor_data, :sensor1, &(&1 ||| (1 <<< 2)))
%{sensor1: 5}
Can you help me understand why does the Map.update!/3 doesn’t work inside a for loop?
I also tried with Map.update/4 Map.replace!/3 but no luck.
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stefanchrobot
The “assignment” inside the for loop has no effect, it creates a new variable on each loop iteration. I think you might be looking for Enum.reduce:
Enum.reduce(0..9, %{sensor1: 0}, fn n, sensor_data ->
Map.update!(sensor_data, :sensor1, &(&1 ||| (1 <<< n)))
end)
LostKobrakai
If you really want to use for you can. Since a few versions back it also supports reductions.
use Bitwise
sensor_data = %{sensor1: 0}
new_sensor_data =
for n <- 0..9, reduce: sensor_data do
acc -> Map.update!(acc, :sensor1, &(&1 ||| (1 <<< n)))
end
stefanchrobot
See also: https://elixir-lang.org/getting-started/case-cond-and-if.html#if-and-unless
This is also a good opportunity to talk about variable scoping in Elixir. If any variable is declared or changed inside
if,case, and similar constructs, the declaration and change will only be visible inside the construct. For example:
iex> x = 1
1
iex> if true do
...> x = x + 1
...> end
2
iex> x
1
In said cases, if you want to change a value, you must return the value from the
if:
iex> x = 1
1
iex> x = if true do
...> x + 1
...> else
...> x
...> end
2
Last Post!
kalindkaria
Thanks @LostKobrakai and @kokolegorille !
I missed out this :reduce option in Comprehensions. I am able to recall now.
Documentation about :reduce for anyone to refer in future:
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