i am clearing my concepts on programming and figured out this concept of eof using getchar()
This program never prints the number of characters inputted in getchar .
please help
where n how it gets eof
and why is it not printing tota number of counts
Code:#include<stdio.h> main(){ int c ,count; while ((c=getchar())!=EOF) count++; printf("%d characters /n",count); }
First of all, review your post before pressing submit. All your code has been wrapped onto a single line.
It's only saving grace is that the program is short enough to be still readable (just).
> printf("%d characters /n",count);
Try a \n instead of a /n
\n is an actual newline.
> while ((c=getchar())!=EOF) count++;
What is the initial value of count?
> where n how it gets eof
If you're typing stuff in on your terminal, you normally press either ctrl-d (Unix/Linux) or ctrl-z (windows) at the start of a line.
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.
Originally Posted by Salem
ok i will surely keep your suggestion in mind
And now please help me with this code
why this code is not displaying the number of characters in count variableCode:#include<stdio.h> main(){ int c,count=0; while ((c = getchar()) != EOF){ count++; } printf("%d\n",count); }
please explain this
Your loop never ends because you never reach EOF.
You are not reading from a file.
Using Windows 10 with Code Blocks and MingW.
Did you read the post by Salem? Look at the bottom part about signalling EOF from the keyboard.
Is this going to turn into another one of those "why is my console window disappearing" questions?
We have a FAQ for that.
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.
this is the code
In what case will it print end .Code:#include<stdio.h> main(){ int c,count=0; if((c=getchar())=!EOF) printf("process"); printf("end"); }
if we press ctrl+d/ ctrl+c ,it should first print "end"
Please explain me in which case my control will be in else condition.
You don't have an else condition in that code. Also, not equals is written `!='. What you have won't work.
If you press CTRL+C nothing will happen because the program will be terminated. Pressing CTRL+D (Linux, MacOS, etc.) or CTRL+Z (Windows) and enter will cause end to be printed.Code:int main(void) { int c; if((c=getchar())!=EOF) printf("processed character %c\n", c); else printf("end\n"); return 0; }
Are you just typing in code for the fun of it just to see what we say?
Your latest missive won't even compile, so the question "does it print...." is meaningless.Code:$ gcc -Wall foo.c foo.c:2:1: warning: return type defaults to ‘int’ [-Wreturn-type] foo.c: In function ‘main’: foo.c:4:17: error: lvalue required as left operand of assignment foo.c:3:7: warning: unused variable ‘count’ [-Wunused-variable] foo.c:3:5: warning: variable ‘c’ set but not used [-Wunused-but-set-variable] foo.c:9:1: warning: control reaches end of non-void function [-Wreturn-type]
> if((c=getchar())=!EOF)
Notice you've written =! and not !=
Also, if you want to be reasonably sure of seeing what printf prints, then always put a \n at the end of each line.
> Please explain me in which case my control will be in else condition.
What else - there is no else in your code.
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.
