Does anybody know?

A line of code i could use for my function so when i insert a vector such as [1 2 3 4], the function i am making see's it as the value 1234(one thousand, two hundred and thirty four) instead? Thanks :)

[Andrei Bobrov]

c = [1 2 3 4]

out = c*10.^(numel(c)-1:-1:0).'

Answer 1

[Walter Roberson]

str2double(sprintf('%d',V))

or

str2double(char(V+'0'))

which should be more efficient.

Warning: if your vectors might have 15 to 19 elements, then there would be additional work involved in order to get the correct integer. From 19 elements up, only an approximation of the integer can be produced.

[Ben Davis]

So far the code for my function is this:

function [q,r]=Divide(x,y)

q=x/y;

q=fix(q);

r=x-(q*y);

end

So if i just insert either str2double(sprintf('%d',V)) or str2double(char(V+'0')) somewhere into there. It will read vectors for x and y that i submit into the command window of the function,as integers? and also produce the ans as a vector also?

[Walter Roberson]

Well you would use x or y instead of V.

x = str2double(char(x+'0'));

y = str2double(char(y+'0'));

It would not create your answer as a vector unless you added more code at the end:

q = sprintf('%d',q) - '0';

r = sprintf('%d',r) - '0';

[Ben Davis]

This is good thanks, but only seems to work for when my x and y vector consist of 4 digits :(.

The function i am trying to build is one which will dividie the x value by the y value. But both x and y must be inserted as a vector (but actually a whole number) and come out as a vector (but the vector represents a whole number).

function [q,r]=Divide(x,y)

q=x/y;

q=fix(q);

r=x-(q*y);

x = str2double(char(x+'0'));

y = str2double(char(y+'0'));

q = sprintf('%d',q) - '0';

r = sprintf('%d',r) - '0';

end

thats my code so far, could you give me some tips please? i really cant get my head around whats going wrong !

Thanks!

[Walter Roberson]

The x= and y= have to go as the first things after the "function" line.

[Ben Davis]

Thanks :)

[Ben Davis]

Sorry i only just noticed your warning about the 19 elements up needing additional work, is there much more i would need to put into my program to get it to work for over 19 elements etc?

function [q,r]=Divide(x,y)

for x = str2double(char(x+'0'));

y = str2double(char(y+'0'));

end

assert(0<y<10, 'Y can only only be between the values of 1...9')

q=x/y;

q=floor(q);

r=x-(q*y);

q = sprintf('%d',q) - '0';

r = sprintf('%d',r) - '0';

end

[Ben Davis]

Or am i going to have to basically make a whole new function?

[Walter Roberson]

If you need to handle more than 19 elements, then you are going to have to write a whole new function based on very different computation principles. I think it likely that it was intended that you use those computation principles right from the beginning.

Your algorithm will need to implement "long division" or the equivalent.

[Ben Davis]

hoooray -_-

Answer 2

[the cyclist]

x = [1 2 3 4];
str2num(num2str(x')')

[Ben Davis]

Sorry i deleted my previous reply, what i meant to say is that this only works for when x and y are the same size vector (x=[2 2 2 2], y=[1 1 1 1]. Sorry for being a pain, but do you know what i could do to my code so that it will work even if x and y do not contain the same amount of digits inside of them. My functions plans on dividing x by y.