Why this code does not print value of s2

Question

#include <stdio.h>

int main()
{
char s1[100],s2[100];
int c;

fgets(s1,100,stdin);

scanf("%d",&c);

fgets(s2,100,stdin);

puts(s1);
printf("%d",c);
puts(s2);

return 0;
}

Answer 1

Solution: getchar() after scanf().

After you type the number for c, you press "enter". "enter" means '\n', which is left after scanf(). And the second fgets() immediately stop reading the input after reading the '\n'. So the second input will never be read.

Answer 2

How your giving the inputs?
For ex:
If you're giving like
Ssss
7
Aaaa

Then s1=Ssss, c=7, s2="" \\\nevery thing after 7 until \n
If you're giving like:

Ssss
7 aaaa

Now
s1=Ssss, C=7, S2=aaaa

Answer 3

Yes.. Ssss you can follow as CarrieForle said in first reply.. And I said the same..
Just you need to consume \n after Integer.
So alternatively instead of last fgets you can use scanf without worry about enter pressing or not..
Like scanf("%s", s2);

Answer 4

If we use gets instead of scanf then also problem sustains

Answer 5

Then suggest solution

Answer 6

Can you explain more?
What else solution you need?
What is your program requirements?
If you don't need c value to be taken then remove, give 2 lines of input..

Else
Take input like:
Ssss
7 Aaaa

Or else put getchar() after scanf, then you can enter 3 lines of input but press enter after inputting value to C, immediately.....

Answer 7

I want to take three inputs in different lines and then print

Answer 8

That's what my solution does.