What is the output for following?
main()
{
printf(“%%%%”);
}

1. %%%%
2. %%
3. Error
4. Garbage Value

• The answer is %% 2 times only. As % is exit delimiter so to print % itself we have to mention it twice for 1-time printing.
• 6 years agoHelpfull: Yes(16) No(2)
• output is 2. i.e. %% it will print two times
• 6 years agoHelpfull: Yes(6) No(0)
• ans 3.Error
• 6 years agoHelpfull: Yes(2) No(4)
• Output is %%%%
• 6 years agoHelpfull: Yes(1) No(5)
• to print % once we have to give it two times to print one so the o/p will be option (2)%%
• 5 years agoHelpfull: Yes(1) No(0)
• %%
  %d%d...replace d with %
• 5 years agoHelpfull: Yes(1) No(0)
• garbage value
• 5 years agoHelpfull: Yes(0) No(1)
• Option 1 because printf prints the information in between the double qoutes
• 5 years agoHelpfull: Yes(0) No(3)
• Answer is option 2 because of delimeters both of %% it will take one %.so four %%%% it takes two %%.
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• it gives you an error
• 5 years agoHelpfull: Yes(0) No(1)
• The output is %%
• 5 years agoHelpfull: Yes(0) No(0)
• %% will be output this question because we are executing this programm.
• 5 years agoHelpfull: Yes(0) No(0)
• printf will print the characters within " ".So ans is option 1
• 5 years agoHelpfull: Yes(0) No(0)
• Ans : %%
  Because : Delimeters both of %% it will take one %.so four %%%% it takes two %%.
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• 2 times %%
• 4 years agoHelpfull: Yes(0) No(0)
• If the statement given with apaspere with with print statement it will print the same statement
  Output:%%%%
• 2 years agoHelpfull: Yes(0) No(0)