Consider the following C Program. #include int main() int m = 10; int n, n1; n = ++m; n1 = m++; n--; --n1; n -= n1; printf("%d", n); return 0; The output of the program is …

Official answer: 0. Explanation: Let us trace the execution of the program step by step: 1. `int m = 10;` - Variable m is initialized to 10. 2. `n = ++m;` - Pre-increment operator: m is incremented first, so m = 11. - Then n is assigned the value of m, so n = 11. 3. `n1 = m++;` - Post-increment operator: n1 receives the current value of m, so n1 = 11. - Then m is incremented, so m = 12. 4. `n--;` - Post-decrement: n is decremented by 1, so n = 10. 5. `--n1;` - Pre-decrement: n1 is decremented by 1, so n1 = 10. 6. `n -= n1;` - Compound assignment: n = n - n1 = 10 - 10 = 0. 7. `printf("%d", n);` prints the value of n, which is 0.

Programming and Data Structures · Programming in C

Official IIT answer key · IIT Roorkee · Audited Aug 2026

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GATE CSE 2017 Set 2 Q54 · NAT · 2 marks

Consider the following C Program. `#include<stdio.h> int main() { int m = 10; int n, n1; n = ++m; n1 = m++; n--; --n1; n -= n1; printf("%d", n); return 0; }` The output of the program is _________.

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Programming in C

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How GATE tests Programming in C

In GATE CS, Programming in C is tested extensively through rigorous code tracing, pointer mechanics, and fundamental semantics rather than high-level syntax trivia. Questions emphasize pointer arithmetic and multidimensional array decay, storage classes (especially static local variables), evaluation order, side effects of increment/decrement operators, parameter passing mechanisms (pass-by-value vs pass-by-reference), and recursive program state tracking. The topic features a nearly balanced split between NAT and MCQ items across 1-mark and 2-mark categories, demanding precise, bug-free execution traces without a compiler.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.

Programming and Data Structures · Programming in C

Official IIT answer key · IIT Roorkee · Audited Aug 2026

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GATE CSE 2017 Set 2 Q54 · NAT · 2 marks

Consider the following C Program. `#include<stdio.h> int main() { int m = 10; int n, n1; n = ++m; n1 = m++; n--; --n1; n -= n1; printf("%d", n); return 0; }` The output of the program is _________.

Numerical answer

Enter a value to check

Key concept

Programming in C

Step-by-step verified solution

Examiner trap & misconception

Topic notes

How GATE tests Programming in C

Full Programming in C guide →

More on Programming in C

All 74 questions on Programming in C →

Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.

Consider the following C Program. `#include<stdio.h> int main() { int m = 10; int n, n1; n = ++m; n1 = m++; n--; --n1; n -= n1; printf("%d", n); return 0; }` The output of the program is _________.

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Consider the following C Program. `#include<stdio.h> int main() { int m = 10; int n, n1; n = ++m; n1 = m++; n--; --n1; n -= n1; printf("%d", n); return 0; }` The output of the program is _________.

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Check you can still solve Programming in C under a clock

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More on Programming in C

Consider the following C Program. #include<stdio.h> int main() { int m = 10; int n, n1; n = ++m; n1 = m++; n--; --n1; n -= n1; printf("%d", n); return 0; } The output of the program is _________.

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Check you can still solve Programming in C under a clock

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Consider the following C Program. #include<stdio.h> int main() { int m = 10; int n, n1; n = ++m; n1 = m++; n--; --n1; n -= n1; printf("%d", n); return 0; } The output of the program is _________.

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Check you can still solve Programming in C under a clock

More from GATE CSE 2017 Set 2

More on Programming in C

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Consider the following C Program. `#include<stdio.h> int main() { int m = 10; int n, n1; n = ++m; n1 = m++; n--; --n1; n -= n1; printf("%d", n); return 0; }` The output of the program is _________.

Consider the following C Program. `#include<stdio.h> int main() { int m = 10; int n, n1; n = ++m; n1 = m++; n--; --n1; n -= n1; printf("%d", n); return 0; }` The output of the program is _________.