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What will be the output of this program on an implementation where int occupies 2 bytes?
#include <stdio.h>
void main()
{
      int i = 3;
      int j;
      j = sizeof(++i + ++i);
      printf("i=%d j=%d", i, j);
}
  • a)
    i=4 j=2
  • b)
    i=3 j=2
  • c)
    i=5 j=2
  • d)
    the behavior is undefined
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
What will be the output of this program on an implementation where int...
Explanation:
- The `sizeof` operator does not actually evaluate the expression, it only determines the size in bytes of the result.
- In the expression `++i + ++i`, the behavior is undefined in C language because it modifies `i` more than once between two sequence points.
- Sequence points in C are points in the execution of a program at which all side effects from previous evaluations are guaranteed to have been performed.
- In this case, the result of `sizeof(++i + ++i)` is 2 because `++i` is evaluated twice, but the value of `i` remains the same as it was before the expression was evaluated.
- Therefore, the value of `i` remains 3 and the value of `j` is 2, so the output will be `i=3 j=2`.
So, the correct answer is option b) i=3 j=2.
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Community Answer
What will be the output of this program on an implementation where int...
Evaluating ++i + ++i would produce undefined behavior, but the operand of sizeof is not evaluated, so i remains 3 throughout the program. The type of the expression (int) is reduced at compile time, and the size of this type (2) is assigned to j.
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What will be the output of this program on an implementation where int occupies 2 bytes?#include <stdio.h>void main(){ int i = 3; int j; j = sizeof(++i + ++i); printf("i=%d j=%d", i, j);}a)i=4 j=2b)i=3 j=2c)i=5 j=2d)the behavior is undefinedCorrect answer is option 'B'. Can you explain this answer?
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