how can i compute the length of an integer?

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

1 Stimme

int=-123456789
max(ceil(log10(abs(int))),1)

1 Kommentar

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

1 Stimme

Edit
nnz(num2str(int) - '-')

4 Kommentare

Friedrich
Friedrich am 4 Jan. 2013
Bearbeitet: Friedrich am 4 Jan. 2013
Why this cast to an double array? Shouldn't
numel(num2str(int))
be enough?
Sean de Wolski
Sean de Wolski am 4 Jan. 2013
Friedrich, neither work actually, consider -12345.
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

0 Stimmen

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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