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What will be the output of the following C code?
#include <stdio.h>
    int main()
    {
        j = 10;
        printf("%d\n", j++);
        return 0;
    }
  • a)
    10
  • b)
    11
  • c)
    Compile time error
  • d)
    0
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
What will be the output of the following C code?#include <stdio.h&g...
Variable j is not defined.
Output:
$ cc pgm3.c
pgm3.c: In function ‘main’:
pgm3.c:4: error: ‘j’ undeclared (first use in this function)
pgm3.c:4: error: (Each undeclared identifier is reported only once
pgm3.c:4: error: for each function it appears in.)
Free Test
Community Answer
What will be the output of the following C code?#include <stdio.h&g...


Explanation:

The code provided will result in a compile-time error. This is because the variable 'j' is being used without declaration.

Reasoning:
- In the code, the variable 'j' is used without being declared. This will result in a compile-time error as the compiler does not know the data type of 'j'.

Fix:
To fix this issue, you can declare 'j' as an integer before using it.
Example:
c
#include
int main()
{
int j = 10;
printf("%d\n", j++);
return 0;
}


By declaring 'j' as an integer, the code will no longer have a compile-time error and will output the value of 'j' (which is 10) before incrementing it due to the postfix increment operator (++).
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What will be the output of the following C code?#include <stdio.h> int main() { j = 10; printf("%d\n", j++); return 0; }a)10b)11c)Compile time errord)0Correct answer is option 'C'. Can you explain this answer?
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