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What will be the output of the following C code?
    #include <stdio.h>
    int main()
    {
        const int p;
        p = 4;
        printf("p is %d", p);
        return 0;
    }
  • a)
    p is 4
  • b)
    Compile time error
  • c)
    Run time error
  • d)
    p is followed by a garbage value
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
What will be the output of the following C code? #include <stdio.h...
Explanation:
The code will result in a compile time error because the variable `p` is declared as a constant using the `const` keyword but is then assigned a value of 4. This is not allowed because a constant variable cannot be modified after initialization.

Reason:
- The variable `p` is declared as a constant using the `const` keyword, making it read-only.
- The assignment `p = 4;` tries to modify the value of `p`, which is not allowed for constant variables.
- This violation of the constant variable rule will result in a compile time error.
Therefore, the output of the code will be a compile time error.
Free Test
Community Answer
What will be the output of the following C code? #include <stdio.h...
Since the constant variable has to be declared and defined at the same time, not doing it results in an error.
Output:
$ cc pgm10.c
pgm10.c: In function ‘main’:
pgm10.c:5: error: assignment of read-only variable ‘p’
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What will be the output of the following C code? #include <stdio.h> int main() { const int p; p = 4; printf("p is %d", p); return 0; }a)p is 4b)Compile time errorc)Run time errord)p is followed by a garbage valueCorrect answer is option 'B'. Can you explain this answer?
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