how can i compute the length of an integer?
显示 更早的评论
if i have
int = 12345;
length_int = 5;
???
采纳的回答
Azzi Abdelmalek
2013-1-4
编辑：Azzi Abdelmalek
2013-1-4
int=-123456789
max(ceil(log10(abs(int))),1)
1 个评论
Christian Ziegler
2020-11-14
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)
更多回答（2 个）
Sean de Wolski
2013-1-4
编辑：Sean de Wolski
2013-1-4
Edit
nnz(num2str(int) - '-')
4 个评论
Why this cast to an double array? Shouldn't
numel(num2str(int))
be enough?
Sean de Wolski
2013-1-4
Friedrich, neither work actually, consider -12345.
Friedrich
2013-1-4
Good point Sean. But then
numel(num2str(abs(int)))
should do ;)
Sean de Wolski
2013-1-4
arghh, you win!
Davide Ferraro
2013-1-4
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 个评论
why, it works:
>> int = 12345678901234567890123
numel(num2str(int))
int =
1.2346e+22
ans =
23
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