how can i compute the length of an integer?

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if i have
int = 12345;
length_int = 5;
???

Risposta accettata

Azzi Abdelmalek
Azzi Abdelmalek il 4 Gen 2013
Modificato: Azzi Abdelmalek il 4 Gen 2013
int=-123456789
max(ceil(log10(abs(int))),1)
  1 Commento
Christian Ziegler
Christian Ziegler il 14 Nov 2020
this doesn't work for 10, 100, 1000...
for example max(ceil(log10(abs(10))),1) equals 1 but it should be 2
simple fix:
int = 10;
max(ceil(log10(abs(int)+1)),1)

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Più risposte (2)

Sean de Wolski
Sean de Wolski il 4 Gen 2013
Modificato: Sean de Wolski il 4 Gen 2013
Edit
nnz(num2str(int) - '-')
  4 Commenti
Friedrich
Friedrich il 4 Gen 2013
Good point Sean. But then
numel(num2str(abs(int)))
should do ;)

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Davide Ferraro
Davide Ferraro il 4 Gen 2013
Casting the variable into a string may be risky because you may get to "unexpected" cases such as:
int = 12345678901234567890123
numel(num2str(int))
ans =
12
You may consider a numeric approach using LOG10: floor(log10(int))+1 all numbers between 10 and 100 will have a LOG10 between 1 and 2 so you can use FLOOR to get the lower value (1 in this case) and then you need to add the value 1 cause you are trying to compute the number of digits and not the power of ten.
  1 Commento
G A
G A il 4 Gen 2013
Modificato: G A il 4 Gen 2013
why, it works:
>> int = 12345678901234567890123
numel(num2str(int))
int =
1.2346e+22
ans =
23

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