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# 1  
Old 10-07-2005
array

#include<stdio.h>
int main()
{
char r[10];
char let;
int i;
for(i = 0; i < 10; i++)
{
printf("Enter a letter:");
scanf("%c",&let);
r[i] = let;
}
puts(r);
}

In the above code, on execution we r prompted to enter only five times.. Can anyone please tl me y it is so... Only using character array has this problem.

abey
# 2  
Old 10-07-2005
Hi,

I am not a C guru but what I feel in the code portion of yours is that the newline ('\n') is also getting accepted as typed character. So if you see the output each digit is one below the other.
1
2
3
4
5

Try modifying the code by putting scanf("%c\n", &let);

Output 1234567890

This will work..

Regards,
Parasa Kiran.
# 3  
Old 10-07-2005
You are reading char type. I'm preety sure you press enter after you input the character. scanf("%c", &let); read a char and only a char. If you input a char and enter then you have char enter in stream buffer and the next time scanf will be called will read the enter key from stream buffer.
# 4  
Old 10-07-2005
Hi,

True..

ChMar, thanks for the help.. Smilie

abey
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