adding vectors: positive single digit values and returning digits of sum as elements of vector

I am trying to write a function that takes two vectors of positive single digit integer values as its input and return the digits of the sum as elements of a vector.
I have two vectors but am unsure how to start a function that will allow me to perform what is listed in the previous statement. If someonme could please help with me a starting po0int, it would be appreciated.

5 Comments

so if I have vector [x] and vector [y] both of which are single digit integer values and I add the sum together so sum=[x}+[y]. This should calculate the he sum of two scalars composed from the digits in the vector and then return the sum of those digits in the form of a vector.
what happens if i have two vectors that are of different sizes...say for example one vector has four numbers while the other vector has five values
Is it the following that you want to do?
x = [1,2,3] ;
y = [7,9] ;
and then, taking these as sets of digits, add the two numbers that they represent the way we do it by hand? I.e. adding 3 and 9, storing 2 and holding 1, etc.. ?

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 Accepted Answer

Please give an example to show what you want. It could either be
theSum = vector1 + vector2;
like Matt J suggested, or
theSum = sum(vector1) + sum(vector2);
which will work for vectors of different lengths. Or maybe you want to interpolate them to make them equal sizes before summing - we have no idea what you really want. Don't use sum as the name of a variable like you did because sum() is the name of a built in function and you'll disable it if you use the same name for your own functions or variables.

1 Comment

My guess is that she wants something like an addition the way we do it by hand, with numbers given by digits stored in vectors.
PS: thank you for your comment yesterday. I replied but quite late!

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More Answers (2)

Hints:
>> x=str2double(char([7 2 4]+'0'))
x =
724
>> y=num2str(x+1)
y =
725
>> whos x y
Name Size Bytes Class Attributes
x 1x1 8 double
y 1x3 6 char

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Asked:

on 29 Oct 2013

Edited:

on 30 Oct 2013

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