how can i compute the length of an integer?

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Zaza
Zaza am 4 Jan. 2013
if i have
int = 12345;
length_int = 5;
???

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Azzi Abdelmalek
Azzi Abdelmalek am 4 Jan. 2013
Bearbeitet: Azzi Abdelmalek am 4 Jan. 2013
int=-123456789
max(ceil(log10(abs(int))),1)
  1 Kommentar
Christian Ziegler
Christian Ziegler am 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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Weitere Antworten (2)

Sean de Wolski
Sean de Wolski am 4 Jan. 2013
Bearbeitet: Sean de Wolski am 4 Jan. 2013
Edit
nnz(num2str(int) - '-')
  4 Kommentare
Friedrich
Friedrich am 4 Jan. 2013
Good point Sean. But then
numel(num2str(abs(int)))
should do ;)
Sean de Wolski
Sean de Wolski am 4 Jan. 2013
arghh, you win!

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Davide Ferraro
Davide Ferraro am 4 Jan. 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 Kommentar
G A
G A am 4 Jan. 2013
Bearbeitet: G A am 4 Jan. 2013
why, it works:
>> int = 12345678901234567890123
numel(num2str(int))
int =
1.2346e+22
ans =
23

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