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What will be the output of the following code snippet?
#include <stdio.h>
void solve() {
    bool ok = false;
    printf(ok ? "YES" : "NO");
}
int main() {
    solve();
    return 0;
}
  • a)
    Yes
  • b)
    No
  • c)
    Compilation Error
  • d)
    None of the above
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
What will be the output of the following code snippet?#include <std...
The code snippet will result in undefined behavior because the variable ok is uninitialized. In C, variables that are not explicitly initialized can have unpredictable values.
The printf statement uses the ternary operator (ok ? "YES" : "NO") to determine the string to be printed based on the value of ok. However, since ok is uninitialized, its value is indeterminate.
Therefore, the output of the printf statement cannot be determined reliably. It could print "YES" or "NO" or any other unexpected result depending on the uninitialized value of ok.
As a result, the correct answer is C. None of the above.
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Community Answer
What will be the output of the following code snippet?#include <std...
The code snippet will result in undefined behavior because the variable ok is uninitialized. In C, variables that are not explicitly initialized can have unpredictable values.
The printf statement uses the ternary operator (ok ? "YES" : "NO") to determine the string to be printed based on the value of ok. However, since ok is uninitialized, its value is indeterminate.
Therefore, the output of the printf statement cannot be determined reliably. It could print "YES" or "NO" or any other unexpected result depending on the uninitialized value of ok.
As a result, the correct answer is C. None of the above.
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What will be the output of the following code snippet?#include <stdio.h>void solve() { bool ok = false; printf(ok ? "YES" : "NO");}int main() { solve(); return 0;}a)Yesb)Noc)Compilation Errord)None of the aboveCorrect answer is option 'C'. Can you explain this answer?
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