What is the output of the following program? #include < stdio.h > int main() { int a=1; switch(a) { printf("GATE-CSE"); case 0: printf("Zero"); case 1: printf("One"); default: printf("None"); } return 0; }
Question:
What is the output of the following program? #include < stdio.h > int main() { int a=1; switch(a) { printf("GATE-CSE"); case 0: printf("Zero"); case 1: printf("One"); default: printf("None"); } return 0; }
This is AI generated Response.
Asked by: Vishal
Created at: 2026-01-03 15:00:18
Asked by: Vishal
Created at: 2026-01-03 15:00:18
Final Answer:
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# Output of the Program
The program's behavior depends on the placement of the `printf` statement of the `switch` cases and absence of `break` statements inside switch.
---
Code Analysis
c
#include <stdio.h>int main() int a 1;
switch(a) {
("GATE-CSE");
case : printf("Zero");
case 1 printf("One default:("None");
}
return;
}
`
### Key points- The `printf("GATE-CSE");` statement is **outside** and **before** the `case` labels.
- Since there's no `break`, execution will **fall through** after the first statement.
---
## Execution Flow
1. `int a = 1;`
2. Enters the `switch(a)` statement.
3. Executes `printf("GATE-CSE");` **before** checking any case.
4. Since no `break` exists, it falls through all subsequent cases:
- `case :` is skipped because `a` is 1.
- `case 1:` matches, so it executes `printf("One");`.
- Continues to `default:` and executes `printf("None");`.
---
## Final Output
```plaintext
GATE-CSEOneNone
```
**Note:** All outputs are concatenated because there are no newline characters.
---
## Summary
- The output begins with **"GATE-CSE"**.
- Then **"One"** (from `case 1`).
- Then **"None"** (from `default`).
**Total output:**
```plaintext
GATE-CSEOneNone
```
Question:
But correct answer shown is One None
This is AI generated Response.
Asked by: Vishal
Created at: 2026-01-03 15:01:25
Asked by: Vishal
Created at: 2026-01-03 15:01:25
Final Answer:
Rendered
Raw Text
Did you like this answer?
# Output of the Program
The program behavior is influenced by the placement of the `printf` statement and the absence of `break` statements within the `switch`.
---
## Code Analysis
```c
#include <stdio.h>
int main() {
int a = 1;
switch(a) {
printf("GATE-CSE");
case 0:
printf("Zero");
break;
case 1:
printf("One");
break;
default:
printf("None");
}
return 0;
}
```
### Key Points
- The `printf("GATE-CSE");` statement is **outside** and **before** the case labels.
- There are no `break` statements after the `case` blocks, allowing execution to **fall through**.
---
## Execution Flow
1. `int a = 1;`
2. Enters the `switch(a)` statement.
3. Executes `printf("GATE-CSE");` **before** checking any case.
4. Since `a` is 1, it:
- Skips `case 0:` and does not execute `printf("Zero");`.
- Matches `case 1:` and executes `printf("One");`.
- Falls through to `default:` and executes `printf("None");`.
---
## Final Output
```plaintext
GATE-CSEOneNone
```
**Note:** All outputs are concatenated due to the absence of newline characters.
---
## Summary
- The output begins with **"GATE-CSE"**.
- Then **"One"** (from `case 1`).
- Finally **"None"** (from `default`).
**Total output:**
```plaintext
GATE-CSEOneNone
```
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