Reputation: 55
Assume you're coding in golang, you can do something like:
str := fmt.Sprintf("%d is bigger than %d", 6, 4)
How about Erlang?
Upvotes: 3
Views: 2558
Reputation: 21013
The Erlang equivalent would be
Str = io_lib:format("~p is bigger than ~p", [6, 4])
Note that, even if the result may be not technically a string, normally there is no need to convert it to the string by calling lists:flatten
. The result of the format function usually is a special case of iolist. Virtually all Erlang functions expecting a string accept iolists as arguments as well.
"Usually" above means "if Unicode modifier is not used in the format string". In most cases there is no need to use Unicode modifiers, and the result of format
can be used directly as described above.
Upvotes: 3
Reputation: 48659
To use io_lib:format/2
with unicode characters:
50> X = io_lib:format("~s is greater than ~s", [[8364], [36]]).
** exception error: bad argument
in function io_lib:format/2
called as io_lib:format("~s is greater than ~s",[[8364],"$"])
51> X = io_lib:format("~ts is greater than ~s", [[8364], [36]]).
[[8364],
32,105,115,32,103,114,101,97,116,101,114,32,116,104,97,110,
32,"$"]
52> io:format("~s~n", [X]).
** exception error: bad argument
in function io:format/2
called as io:format("~s~n",
[[[8364],
32,105,115,32,103,114,101,97,116,101,114,32,116,
104,97,110,32,"$"]])
*** argument 1: failed to format string
53> io:format("~ts~n", [X]).
€ is greater than $
ok
Upvotes: -1
Reputation: 7169
There is io_lib:format/2
, that does the job, but note that it returns a possibly nested list of chars, not a string. For a proper string, you have to flatten/1
it afterwards:
lists:flatten(io_lib:format("~p is bigger than ~p", [6, 4]))
Upvotes: 2