C Programming - Input / Output - Discussion
Discussion Forum : Input / Output - Find Output of Program (Q.No. 10)
10.
What will be the output of the program ?
#include<stdio.h>
int main()
{
printf("%%%%\n");
return 0;
}
Discussion:
36 comments Page 2 of 4.
K Jagannadham said:
1 decade ago
We can print % in C.
But another way to print % we need %% in C.
But to print %% we need %%% or %%%% in C.
And next to print %%% we need %%%%% or %%%%%% in C.
And so on.....
But another way to print % we need %% in C.
But to print %% we need %%% or %%%% in C.
And next to print %%% we need %%%%% or %%%%%% in C.
And so on.....
Bhavesh jain said:
1 decade ago
printf("%\n")
It will give output % . and to print % in c we have to add \n or \t or \r ..so it will terminate the % which starts finding for the d,s,c,f char to print something. so if we give \t ,\n then it will stop finding and simply print %.
It will give output % . and to print % in c we have to add \n or \t or \r ..so it will terminate the % which starts finding for the d,s,c,f char to print something. so if we give \t ,\n then it will stop finding and simply print %.
Narayan said:
1 decade ago
I don't get anyone answer. Can anyone please explain about :
1. printf("%%%\n") (what will be o/p).
2. printf("%%\n") (what will be o/p).
3. printf("%\n") (what will be o/p).
Give ans respectively with proper explanation.
1. printf("%%%\n") (what will be o/p).
2. printf("%%\n") (what will be o/p).
3. printf("%\n") (what will be o/p).
Give ans respectively with proper explanation.
Pavani said:
1 decade ago
#include<stdio.h>
#include<conio.h>
main()
{
clrscr();
printf("%\n");//only one % is used
getch();
}
o/p:
%
Q:why it happened?
#include<conio.h>
main()
{
clrscr();
printf("%\n");//only one % is used
getch();
}
o/p:
%
Q:why it happened?
Ranjitha said:
1 decade ago
@josna:
If we print %%% the result is % and not %%.
If we print %%% the result is % and not %%.
Dheeraj said:
1 decade ago
Why printf("%??%%%%%\n"); giving output of %??%%% ?
Sneha said:
1 decade ago
Yes I got it.
Baadal said:
1 decade ago
Each % read the next single character hence it is printing only two %.
Ganesh said:
1 decade ago
My suggestion is, In C, we need a memory reference variable after the % symbol. In the above code, it considers % as memory reference variable and it prints. Try with %%%%%. It will print %%%. Its because, %% %% %.Split it into pair and then try it.
Sumedh said:
1 decade ago
I think it uses the first % as formate string and remaining two are printed and used \n to move cursor to next line.
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