https://ide.codingblocks.com/s/189205
Why it isnt showed any output!
Subsequences show no ouput
@Kinjal
hello kinjal,
out[j]=inp[i]; should be after subsequences(inp,out,i+1,j); otherwise it will get overwrite
something like 
it will get overwrite by NULL char but I didnt understand that why it was overwritten in every subsequences by NULL!
Like in “ab” - there is no output
it was intentional because I did dry run. What I have found out in output [" " a a ab] That was my output in pen and paper. Can you tell me where I did wrong, Bhaiya!
@Kinjal
after making first call (highlighted part) output[j]=’\0’ right?
and then in next call output[j] will remain same i,e output[j]=’\0’. same procedure will repeat for j+1,j+2,j+3 …so on therefore evrytime evry generated subsequence is empty string .
Bhaiya, i’m thinking like a tree. So, in the first call, it will go to the end of left sub tree and satisfied the base case and out[j]=’\0’ and return nothing. After that it will go to the right child of that branch node (2nd node from the end of left sub tree) where only ‘a’ is there so base case meets and print ‘a’ and return nothing and it goes on like this.
@Kinjal
in u rightchild node when u will do output[j]=‘a’ and call subsequences(inp,out,i+1,j) and if u hit base case then u will again assign out[j]=’\0’ thats why u will again get empty string and not ‘a’ in output
but bhaiya, rightchild node come to the picture by calling subsequences(inp,out,i+1,j+1) so it means, after out[j]=a, the j pointer increment by 1 place. Now when it again calls subsequences(inp,out,i+1,j) like you said then it will assign out[j]=’\0’. So, it’s like, [a|\0].
@Kinjal
when ur control will come back from highlighted part, b will be overwritten with null (becuase we are making chnages in original string out) and then when it go back to a same it will be null then in next brach same thing will happen
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thank you so much, Bhaiya.