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.

Practice

1. What does varargin do in a MATLAB function?

easy

A. Returns multiple output arguments as a cell array

B. Collects all input arguments into a cell array

C. Limits the number of input arguments to one

D. Specifies the data type of input arguments

Solution

1. Step 1: Understand the role of varargin

   varargin is used to collect any number of input arguments passed to a function into a single cell array.

2. Step 2: Differentiate from varargout

   varargout is used for output arguments, not input. So it does not collect inputs.

3. Final Answer:

   Collects all input arguments into a cell array -> Option B

4. Quick Check:

   varargin = input collector [OK]

Hint: Remember: 'varargin' collects inputs, 'varargout' returns outputs [OK]

Common Mistakes:

• Confusing varargin with varargout
• Thinking varargin limits inputs
• Assuming varargin specifies data types

2. Which of the following is the correct way to define a MATLAB function that accepts a fixed input followed by any number of additional inputs using varargin?

easy

A. function output = myFunc(input1, varargin)

B. function output = myFunc(varargout)

C. function output = myFunc(varargin)

D. function output = myFunc(varargin, input1)

Solution

1. Step 1: Check correct varargin placement

   varargin must be the last input argument to collect all remaining inputs.

2. Step 2: Evaluate options

   function output = myFunc(input1, varargin): function output = myFunc(input1, varargin) correctly places varargin after the fixed input input1. function output = myFunc(varargout) incorrectly uses varargout as an input. function output = myFunc(varargin) has no fixed input. function output = myFunc(varargin, input1) places varargin before a fixed input, which is invalid.

3. Final Answer:

   function output = myFunc(input1, varargin) -> Option A

4. Quick Check:

   varargin must be last input [OK]

Hint: Put varargin last in input list to accept extra arguments [OK]

Common Mistakes:

• Placing varargin before fixed inputs
• Using varargout as input
• Omitting parentheses in function definition

3. What will be the output of the following MATLAB code?

function varargout = exampleFunc(varargin)
  n = length(varargin);
  for k = 1:n
    varargout{k} = varargin{k} * 2;
  end
end

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

medium

A. Error: varargout not assigned properly

B. [3 5]

C. [6 10]

D. [9 25]

Solution

1. Step 1: Understand varargin and varargout usage

   varargin collects inputs {3, 5}. The loop doubles each input and assigns to varargout.

2. Step 2: Calculate output values

   3 * 2 = 6, 5 * 2 = 10. So outputs are 6 and 10.

3. Final Answer:

   [6 10] -> Option C

4. Quick Check:

   Each input doubled = output [OK]

Hint: varargin inputs, varargout outputs; loop doubles inputs here [OK]

Common Mistakes:

• Confusing input and output values
• Forgetting to assign varargout elements
• Assuming varargout is a numeric array, not cell

4. Identify the error in this MATLAB function using varargin:

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

medium

A. Incorrect indexing of varargin with parentheses

B. Missing output argument varargout

C. No error, function works correctly

D. Using varargin as a number in the for loop

Solution

1. Step 1: Check for loop range

   The for loop uses 1:varargin, but varargin is a cell array, not a number.

2. Step 2: Correct way to get number of inputs

   Use length(varargin) or nargin to get number of inputs for the loop.

3. Final Answer:

   Using varargin as a number in the for loop -> Option D

4. Quick Check:

   Loop needs numeric length, not cell array [OK]

Hint: Use length(varargin) to loop over inputs, not varargin itself [OK]

Common Mistakes:

• Using varargin directly as a number
• Forgetting to use curly braces for cell access
• Assuming varargout is required for outputs

5. You want to write a MATLAB function that accepts any number of numeric inputs and returns their squares as multiple outputs. Which function definition and body correctly uses varargin and varargout to achieve this?

hard

A. function varargout = squareAll(varargin)
for k = 1:length(varargin)
varargout{k} = varargin{k}^2;
end
end

B. function varargout = squareAll(varargin)
for k = 1:nargin
varargout{k} = varargin(k)^2;
end
end

C. function varargout = squareAll(varargin)
for k = 1:length(varargin)
varargout(k) = varargin{k}^2;
end
end

D. function varargout = squareAll(varargin)
for k = 1:nargout
varargout{k} = varargin{k}^2;
end
end

Solution

1. Step 1: Check input count method

   Using length(varargin) correctly gets the number of inputs. nargin inside the function counts all inputs including fixed ones, but here only varargin inputs matter.

2. Step 2: Check cell indexing and assignment

   Accessing inputs with varargin{k} and assigning outputs with varargout{k} using curly braces is correct. Parentheses would cause errors.

3. Final Answer:

   function varargout = squareAll(varargin) for k = 1:length(varargin) varargout{k} = varargin{k}^2; end end -> Option A

4. Quick Check:

   Use length(varargin) and curly braces for cell access [OK]

Hint: Use length(varargin) and curly braces {} for cell arrays [OK]

Common Mistakes:

• Using parentheses instead of curly braces for cell access
• Using nargin instead of length(varargin)
• Assigning outputs with parentheses instead of curly braces