What is the output of the following C code?#include #include void main(){int index;for(index=1; index{printf('%d',index);if(i == 3)continue;}}a)2b)1c)0d)3Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Understanding the Code
The given C code snippet is incomplete and contains a few errors, but let's analyze what it seems to intend:
c
#include
void main() {
int index;
for(index = 1; index < 5;='' index++)='' />
printf('%d', index);
if(index == 3) continue;
}
}
Key Components of the Code
- Loop Initialization: The loop starts with `index` initialized to 1 and runs while `index` is less than 5.
- Printing Values: It prints the value of `index` during each iteration.
- Condition Check: If `index` equals 3, it uses `continue`, which skips the rest of the loop's body and proceeds to the next iteration.
Flow of Execution
- First Iteration: `index = 1`, prints `1`.
- Second Iteration: `index = 2`, prints `2`.
- Third Iteration: `index = 3`, prints `3`, but the `continue` statement is triggered, skipping the next steps.
- Fourth Iteration: `index = 4`, prints `4`.
Final Output Analysis
The output of the program will be `1234`. However, let's clarify the confusion regarding the options provided.
- The correct output is not among the options stated (0, 1, 2, 3).
- If the question asked for the number of printed integers, it would be 4.
Conclusion
Given the choices, if 'C' is marked as the correct answer, it may be due to misunderstanding the output context. The actual output of the program is `1234`, and the logic flow ensures that all numbers from 1 to 4 (except the skipped logic for `3`) are printed.
This code's execution results in the printing of four integers, not zero.

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