Q.State True or False: While defining a function in Python, the positional parameters in the function header must always be written after the default parameters.

When a Python function has no explicit return, it returns None. This question tests the difference between what a function prints and what it returns, and how print handles None and its keyword arguments.

The function f1(a, b=1) computes a+b and prints it with end='-' instead of a newline. There is no return, so calling f1(1,2) prints the sum and evaluates to None, which is then assigned to c.

Trace of the execution:

1. Call f1(1, 2) — inside, a=1 and b=2, so a+b is 3.

2. print(a+b, end='-') — outputs 3 and, instead of a newline, ends with -. So far the console shows:

   3-
   

3. Function returns None — with no return statement, f1 returns None, so c = f1(1,2) stores None in c. …

The print() function in Python is incredibly versatile, but it has specific rules about how you pass arguments to it. Understanding these rules, especially the distinction between positional and keyword arguments, is crucial for writing correct Python code.

When you call a function in Python, you can pass arguments in two main ways:

1. Positional Arguments: These are arguments whose meaning is determined by their position in the function call. For example, in print("Hello", "World"), "Hello" is the first item to print, and "World" is the second.
2. Keyword Arguments: These are arguments identified by a keyword (their parameter name) followed by an equals sign and a value. They are used to specify optional parameters like sep (separator) and end (what to print at the end). For example, print("A", "B", sep="-") explicitly tells print to use a hyphen as a separator.

The fundamental rule for mixing these two types of arguments is that all positional arguments must come before any keyword arguments. If you place a positional argument after a keyword argument, Python will raise a SyntaxError.

Let's examine each option:

1. Option (A): print("A",10,end="*")

   • Here, "A" and 10 are positional arguments.
   • end="*" is a keyword argument.
   • The keyword argument end="*" correctly follows all positional arguments.
   • This statement is valid. It will print A 10* (note the space between A and 10, which is the default separator).

2. Option (B): print("A",sep="*",10)

   • Here, "A" is a positional argument.
   • sep="*" is a keyword argument.
   • 10 is another positional argument.
   • This statement attempts to place a positional argument (10) after a keyword argument (sep="*"). This violates Python's syntax rules.
   • This statement is invalid. Python would raise a SyntaxError: positional argument follows keyword argument. …

Concept

When a function is called, Python matches arguments to parameters from left to right. A parameter with a default value is optional, while a plain positional parameter is mandatory. If a mandatory parameter came after an optional one, Python could not tell whether a single supplied value was meant for the optional parameter or the mandatory one, so the language forbids that ordering.

Correct order

def calculate_area(length, width=10):   # valid: positional first, default after
    return length * width

Calling calculate_area(5) gives length=5, width=10; calculate_area(5, 20) gives length=5, width=20.

The order the statement claims …

This question tests your understanding of how Python handles variable scope — specifically, the difference between global and local variables. The code defines a global variable a = 20, then defines a function convert(a) that takes a parameter also named a. Inside the function, a new local variable b is set to 20, and then a (the local parameter) is increased by b. The function is called with the argument 10, so inside the function, the local a starts at 10, becomes 30 after adding b, and then the function ends. The crucial point: this change affects only the local a inside the function, not the global a outside.

When the function call finishes, the program moves to the next line: print(a). At this point, a refers to the global variable, which was never modified — it remains 20. The local a that was changed to 30 is gone, as if it never existed outside the function.

Why scope matters in Python

When you call a function with a parameter, Python creates a local variable inside that function's namespace. Any modifications to that variable stay confined to the function—they never reach back out to change variables in the outer (global) scope, even if they share the same name.

This is fundamentally about namespaces: the X inside Update and the X in the global scope are two distinct entities that happen to have the same identifier. The function parameter X shadows the global X for the duration of the function call.

Tracing the execution

Let's walk through what Python does, line by line:

1. Function definition

   def Update(X=10):
       X += 15
       print('X = ', X)
   

   This defines a function Update with a default parameter value of 10 for X. The parameter X is local to this function.

2. Global assignment

   X = 20
   

   A global variable X is created and set to 20. This lives in the module's global namespace.

3. Function call

   Update()
   

   Since no argument is passed, the default value kicks in: the local X starts at 10.

   Then X += 15 increments the local X to 10 + 15 = 25.

   The print statement outputs:

   X =  25
   

   When the function returns, the local X is destroyed.

4. Global print

   print('X = ', X)
   ``` …