Variable arguments (varargin, varargout) in MATLAB - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a function using varargin

What is the output of the following MATLAB function call?

function result = sumAll(varargin)
    result = 0;
    for k = 1:length(varargin)
        result = result + varargin{k};
    end
end

output = sumAll(2, 4, 6); disp(output);

MATLAB

function result = sumAll(varargin)
    result = 0;
    for k = 1:length(varargin)
        result = result + varargin{k};
    end
end

output = sumAll(2, 4, 6); disp(output);

A 12

B 246

C Error: Undefined function or variable 'varargin'

D 0

💡 Hint

Remember varargin is a cell array containing all input arguments.

The function sums all input arguments passed via varargin. Each varargin{k} is a number, so the sum is 2 + 4 + 6 = 12.

❓ Predict Output

intermediate

Output when using varargout to return multiple outputs

What will be the output of the following MATLAB code?

function varargout = splitNumber(x)
    varargout{1} = floor(x);
    varargout{2} = x - varargout{1};
end

[a, b] = splitNumber(5.75);
disp([a, b]);

MATLAB

function varargout = splitNumber(x)
    varargout{1} = floor(x);
    varargout{2} = x - varargout{1};
end

[a, b] = splitNumber(5.75);
disp([a, b]);

A [5 0.75]

B [6 0.25]

C Error: Too many output arguments.

D [5.75 0]

💡 Hint

varargout allows returning multiple outputs as a cell array.

The function returns the integer part and fractional part of the input number. floor(5.75) is 5, fractional part is 0.75.

🧠 Conceptual

advanced

Understanding varargin behavior with no inputs

Consider the function below:

function count = countInputs(varargin)
    count = length(varargin);
end

result = countInputs(); disp(result);

What is the output of this code?

MATLAB

function count = countInputs(varargin)
    count = length(varargin);
end

result = countInputs(); disp(result);

A Error: varargin is undefined

B 1

C 0

D NaN

💡 Hint

Think about what varargin contains when no inputs are given.

When no inputs are passed, varargin is an empty cell array, so length(varargin) is 0.

❓ Predict Output

advanced

Output of function with mixed varargin and varargout

What is the output of the following MATLAB code?

function varargout = processData(varargin)
    n = length(varargin);
    varargout{1} = n;
    if n > 0
        varargout{2} = varargin{1} * 2;
    else
        varargout{2} = 0;
    end
end

[a, b] = processData(3, 5, 7);
disp([a, b]);

MATLAB

function varargout = processData(varargin)
    n = length(varargin);
    varargout{1} = n;
    if n > 0
        varargout{2} = varargin{1} * 2;
    else
        varargout{2} = 0;
    end
end

[a, b] = processData(3, 5, 7);
disp([a, b]);

A [3 10]

B Error: Index exceeds array bounds.

C [7 14]

D [3 6]

💡 Hint

varargin contains all inputs; varargout returns outputs based on inputs.

The function returns the number of inputs (3) and twice the first input (3*2=6).

🔧 Debug

expert

Identify the error in using varargin and varargout

What error will the following MATLAB function produce when called as myFunc(1,2)?

function varargout = myFunc(varargin)
    varargout{1} = varargin{1} + varargin{2};
    varargout{2} = varargin{3};
end

MATLAB

function varargout = myFunc(varargin)
    varargout{1} = varargin{1} + varargin{2};
    varargout{2} = varargin{3};
end

myFunc(1,2);

A No error, returns [3, []]

B Index exceeds array bounds because varargin{3} does not exist

C Syntax error: Missing semicolon

D Error: varargout must have at least one output argument

💡 Hint

Check how many inputs are passed and what varargin contains.

The function tries to access varargin{3} but only two inputs are given, causing an index error.