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Tricky Questions or Puzzles in C ( Updated for 2026)


Updated for 2026
This article was originally written when C/C++ puzzles were commonly asked in interviews. While such language-specific puzzles are less frequent today, the problem-solving and logical reasoning skills tested here remain highly relevant for modern Software Engineering, Data Engineering, SQL, and system design interviews
.

Why These Puzzles Still Matter in 2026

Although most Software & Data Engineering interviews today focus on Programming, SQL, data pipelines, cloud platforms, and system design, interviewers still care deeply about how you think.

These puzzles test:

  • Logical reasoning

  • Edge-case handling

  • Understanding of execution flow

  • Ability to reason under pressure

The language may change, but the thinking patterns do not.


How These Skills Apply to Data Engineering Interviews

The same skills tested by C/C++ puzzles appear in modern interviews as:

  • SQL edge cases and NULL handling

  • Data pipeline failure scenarios

  • Incremental vs full refresh logic

  • Debugging incorrect data outputs

If you can reason through these puzzles, you’ll find SQL and pipeline logic much easier.

Check how these problem solving skills is useful in Data Engineering Debugging (with real world examples)

Real-World Data Engineering Debugging Scenarios


1. Output of Nested printf

Question

printf("%d", printf("%d", printf("%d", printf("%s","ILOVECPROGRAM"))));

Output

ILOVECPROGRAM1321

Explanation

  • printf("%s","ILOVECPROGRAM") prints 13 characters

  • That value is printed by the outer printf13

  • 13 has 2 digits, so next printf prints 2

  • 2 has 1 digit, so the final printf prints 1

➡️ Final output: 1321

Interview takeaway:
Understanding return values and execution order is critical—similar to how functions behave in pipelines.


2. Output of Nested Conditional Operator

Question

a = 0 ? (3>2 ? 23 : (2>5 ? (7<6 ? 34 : 48) : 64)) : 1; printf("%d", a);

Output

1

Explanation

  • The first condition is 0 (false)

  • The entire nested expression is ignored

  • Value after : is returned

Interview takeaway:
Always evaluate the outermost condition first—don’t overthink nested logic.


3. Printing "I Love C Language" Without Directly Printing It

Question

if (condition) printf("I love"); else printf("C Language");

Correct Condition

if (!printf("I Love"))

Explanation

  • printf("I Love") prints 6 characters

  • !6 evaluates to 0

  • else block executes

Interview takeaway:
Side effects of functions matter—similar to how SQL functions or transformations behave.


4. Even or Odd Without Arithmetic or Conditional Operators

(Asked in Microsoft written test)

Solution

(no & 1) ? printf("odd") : printf("even");

Explanation

  • Bitwise AND with 1

  • Even numbers → last bit 0

  • Odd numbers → last bit 1

Interview takeaway:
Bitwise logic tests low-level understanding, similar to partitioning and hashing logic.


5. Sum of Digits in a Single Statement

int sum(int x) { int s; for (s = 0; x > 0; s += x % 10, x /= 10); return s; }

Interview takeaway:
Compact logic tests clarity of thought—important for SQL transformations.


6. Print Numbers 1–100 Without Loop, Recursion, or Goto

(Macro-based expansion approach)

#define STEP1 step(); #define STEP2 STEP1 STEP1 #define STEP4 STEP2 STEP2 #define STEP8 STEP4 STEP4 #define STEP16 STEP8 STEP8 #define STEP32 STEP16 STEP16 #define STEP64 STEP32 STEP32 #define STEP128 STEP64 STEP64 int n = 0; int step() { if (++n <= 100) printf("%d\n", n); } int main() { STEP128; }

Interview takeaway:
Creative constraint solving—similar to designing pipelines under system limitations.


7. Deallocate Memory Without free()

realloc(ptr, 0);

Explanation

  • realloc(ptr, 0) behaves like free(ptr)

Interview takeaway:
Understanding system behavior matters more than memorizing APIs.


🎯 Want to Become a Data Engineer? (2026 Path)

If you’re preparing for modern Data Engineering interviews, focus more on:

  • SQL and data modeling

  • Data pipelines and transformations

  • Cloud platforms and analytics engineering

  • AI-ready data foundations

Recommended Reads:

Comments

  1. Very good question... really helpful!!!

    ReplyDelete
  2. Thanks for your appreciation..vl post some more C puzzles :-)

    ReplyDelete
  3. really good work....thanx a lot frnd

    ReplyDelete
  4. really gud. microsoft puzzle was nice,.......

    ReplyDelete
  5. Awesome question Thank you for it

    ReplyDelete
  6. thnxx for such questions. Please post more such questions looking for more articles from you. and I too have collected some of the tricky programming questions from my personal experience and from my friends. Please have a review on this and i hope that it will really help you to improve your logics. :) http://csetips.catchupdates.com/tricky-programming-questions-interview/

    ReplyDelete
  7. For 3rd one
    Condition could be if(fork())

    ReplyDelete
  8. Thanks...Will post some more Tricky C C++ questions and answers soon :-)

    ReplyDelete
  9. Thanku very much

    ReplyDelete
  10. nice questions...,

    ReplyDelete
  11. int a=0.7;
    if(a>0.7)
    {
    printf("hello");
    }
    else
    {
    printf("bye");
    }

    and same do with 0.8 my question is what will be the output ??

    ReplyDelete
  12. Write a C program which takes one integer number from user and then checks whether number is Even or Odd without using Modulus (%) and Division (/) operator.

    void main()
    {
    int i,n =2567;
    for(i=n;i>0;)
    {
    i=i-2;
    if(i==1)
    {
    printf("Odd");
    break;
    }
    else if(i==0)
    printf("Even");
    }
    }

    ReplyDelete
  13. very very helpful

    ReplyDelete
  14. Awesome..
    its very useful for me. thank u

    ReplyDelete
  15. Brilliant tricky question its so many helpful for us.

    ReplyDelete
  16. I am trying to find out all factors of a number like for 20

    1x20
    2x10
    4x5
    2x2x5

    please help

    ReplyDelete
  17. I am trying to find out all factors of a number like for 20

    1x20
    2x10
    4x5
    2x2x5

    please help

    ReplyDelete

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