Explain why gives output %%d

Question

#include<iostream>
using namespace std;
int main(){
int a=10;
printf("%%%%%%d"+1);
return 0;
}

Answer 1

Abhishek Bairwa

because first two %% output as %,
then next two %% output as %, third one similary, and finally d...

% is the character used to escape variable placeholder, so to output % you need to escape it ;)

Answer 2

Are you sure it shows %%d? Running in the code playground shows me %%{random large number}, which is an output I will expect, so I will instead explain the output I get.

"%%%%%%d"+1 becomes "%%%%%d". The number of % becomes 5 because the +1.
%% outputs % in literal
So %%%% outputs %%.

And the final %d, because you didn't give any parameter for it. It outputs random number which I think is a garbage number, but I can't ensure about that.