Variables & Scope — Test 1

Correct Answer: A. UnboundLocalError is raised
Explanation: Because x is assigned somewhere in f, Python treats x as local throughout the function. print(x) runs before that local is assigned, which raises UnboundLocalError. ALSO KNOW (related points): • What is the output of this C++ code?: int f(){ static int c = 0; return ++c; }; int main(){ f(); f(); cout << f(); } → 3 • What are the scope and lifetime of a static local variable declared inside a C++ function? → Block scope, but a lifetime equal to the whole program run • What does this Python code print?: a = [1, 2]; b = a; b.append(3); print(a) → [1, 2, 3]

Correct Answer: C. 3
Explanation: A static local variable is initialized once and keeps its value between calls, so the calls return 1, 2, and then 3.

Correct Answer: A. Indeterminate (garbage); reading it is undefined behavior
Explanation: Only static and global variables are zero-initialized. Automatic locals hold whatever was previously in that memory.

Correct Answer: C. 0
Explanation: Objects with static storage duration (globals and statics) are zero-initialized.

Correct Answer: B. Block scope, but a lifetime equal to the whole program run
Explanation: Scope (where the name is visible) and lifetime (how long the object exists) are separate properties. A static local is visible only in its block but lives until the program ends.

Correct Answer: C. [2, 2, 2]
Explanation: Closures bind late: each lambda looks up i when it is called, and by then i is 2. Writing lambda i=i: i would capture the value at creation instead. ALSO KNOW (related points): • What is the value of an uninitialized local (automatic) int variable in C++? → Indeterminate (garbage); reading it is undefined behavior • Which keyword lets a Python function assign a new value to a module-level variable? → global • What happens if a const int variable is declared in C++ without an initializer (const int k;)? → Compilation error

Correct Answer: C. global
Explanation: nonlocal refers to the nearest enclosing function scope, not the module. Python has no static or extern keywords.

Correct Answer: C. 1
Explanation: The inner x shadows the outer one only inside the block. The outer x is unchanged.

Correct Answer: D. [1, 2, 3]
Explanation: Assignment binds another name to the same list object. It does not copy the list.

Correct Answer: B. [1, 2]
Explanation: A default argument is evaluated once, when the function is defined, so the same list is shared across calls. A common fix is to use L=None and create the list inside the function. ALSO KNOW (related points): • What is the value of a global int variable declared without an initializer in C++? → 0 • What does this C++ code print?: int x = 1;; { int x = 2; x++; }; cout << x; → 1 • What does this Python code print?: def outer():; n = 0; def inner():; nonlocal n; n += 1; inner(); inner(); return n; print(outer()) → 2