I was just wondering if anyone can explain to me why this piece of simple code isn't working.Code:#include <stdio.h> main() { long nc; nc = 0; while (getchar() != EOF) ++nc; printf("%1d\n", nc); }
Im using GNU/GCC compiler(although i don't think it has anything to do with it...).
Thanks,Filster7.
1. Are you counting the '\n'?
2. What are you using to emulate EOF?
3. Have you tried with simple file passed to stdio of the program?
4. Do you mean to have %ld instead of %1d format?
All problems in computer science can be solved by another level of indirection,
except for the problem of too many layers of indirection.
– David J. Wheeler
The program was designed to count characters...not the \n.
I did try the %ld format,but it didn't work....
If you're on Linux, you need to press ctrl-d to signal EOF to the program.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Im using Windows OS-64bit.Originally Posted by Salem
In that case, try ctrl-z after you have pressed enter.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
How embarassing,i already did that,but didn't notice anything.Thanks for solving this.Sorry to bother you...
What isn't working?
The following works for me:
Testing it with the source code itself as input:Code:#include <stdio.h> main() { long nc; nc = 0; while (getchar() != EOF) ++nc; printf("%ld\n", nc); }
Code:Test$ ls -lB | grep getcharTest.c -rw-rw-r-- 1 guraknugen guraknugen 137 jul 13 20:17 getcharTest.c Test$ ./a.out < getcharTest.c 137 Test$
