[Solved] Examine the following nested if code snippet. What is the logical error?int num = 7; if (num > 0) { if (num % 2 == 0) printf("Even\n"); else printf("Odd\n"); } else printf("Non-positive\n"); | C

C - onditional Statements

Examine the following nested if code snippet. What is the logical error?

int num = 7;
if (num > 0) {
  if (num % 2 == 0)
    printf("Even\n");
  else
    printf("Odd\n");
} else
  printf("Non-positive\n");

AThe else statement is incorrectly paired with the outer if instead of the inner if.

BThere is no error; the code correctly distinguishes even and odd positive numbers.

CThe code does not handle zero correctly as a positive number.

DThe modulus operator (%) cannot be used inside nested if statements.

Step-by-Step Solution

Solution:

Step 1: Analyze the code logic
The outer if checks if num is positive. The inner if checks if num is even or odd.
Step 2: Check else pairing
The else after the inner if correctly pairs with the inner if to print "Odd" if num is not even.
Step 3: Consider zero handling
Zero is not greater than zero, so it goes to the outer else and prints "Non-positive" which is correct.
Final Answer:
The code correctly distinguishes even and odd positive numbers without logical errors. -> Option B
Quick Check:
Inner else pairs with inner if; outer else handles non-positive. [OK]

Quick Trick: Check else pairs with correct if block [OK]

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