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Home/ Questions/Q 139685
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Asked: May 11, 20262026-05-11T07:28:46+00:00 2026-05-11T07:28:46+00:00

I know strings in Erlang can be costly to use. So how do I

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I know strings in Erlang can be costly to use. So how do I convert '5'to 5?

Is there anything like io:format('~p',[5]) that would return a formatted string instead of printing to a stream?

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  1. 2026-05-11T07:28:47+00:00Added an answer on May 11, 2026 at 7:28 am

    The following is probably not the neatest way, but it works:

    1> lists:flatten(io_lib:format('~p', [35365])). '35365' 

    EDIT: I’ve found that the following function comes in useful:

    %% string_format/2 %% Like io:format except it returns the evaluated string rather than write %% it to standard output. %% Parameters: %%   1. format string similar to that used by io:format. %%   2. list of values to supply to format string. %% Returns: %%   Formatted string. string_format(Pattern, Values) ->     lists:flatten(io_lib:format(Pattern, Values)). 

    EDIT 2 (in response to comments): the above function came from a small program I wrote a while back to learn Erlang. I was looking for a string-formatting function and found the behaviour of io_lib:format/2 within erl counter-intuitive, for example:

    1> io_lib:format('2 + 2 = ~p', [2+2]). [50,32,43,32,50,32,61,32,'4'] 

    At the time, I was unaware of the ‘auto-flattening’ behaviour of output devices mentioned by @archaelus and so concluded that the above behaviour wasn’t what I wanted.

    This evening, I went back to this program and replaced calls to the string_format function above with io_lib:format. The only problems this caused were a few EUnit tests that failed because they were expecting a flattened string. These were easily fixed.

    I agree with @gleber and @womble that using this function is overkill for converting an integer to a string. If that’s all you need, use integer_to_list/1. KISS!

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