Character, Line and Word Counter in C [closed] - c

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I am sorry to ask so many questions but I am clueless what to do when my Program gives me errors which I don't understand how to fix. I use Code Blocks. The book name is The C Programming Language 2nd Edition.
The Code:
#include <stdio.h>
#include <stdlib.h>
#define IN 1
#define OUT 0
int main()
{
int c, nl, nw, nc, state;
state = OUT;
while((c = getchar()) != EOF)
{
++nc;
if (c == '\n')
if (c == '' || c == '\n' || c = '\t')
state = OUT;
else if (state == OUT) {
state = IN;
++nw;
}
}
printf("%d %d %d \n", nl, nw, nc);
}
So when I build it it gives me error on line 14 saying:
||=== Build: Debug in Line Counter v2 (compiler: GNU GCC Compiler) ===|
C:\Users\Uddhava\Desktop\Uddhava\Learning\Programming\C C+\Line Counter v2\
main.c|14|error: empty character constant|
C:\Users\Uddhava\Desktop\Uddhava\Learning\Programming\C C+\Line Counter v2\
main.c|14|error: lvalue required as left operand of assignment|
||=== Build failed: 2 error(s), 2 warning(s) (0 minute(s), 0 second(s)) ===|

Answer 1

if (c == '' || c == '\n' || c = '\t')
The first thing wrong with this line is that it should be:
if (c == ' ' || c == '\n' || c == '\t')

Answer 2

if (c == '' || c == '\n' || c = '\t')
should be
if (c == ' ' || c == '\n' || c == '\t')
if (c == '\n')
if (c == '' || c == '\n' || c = '\t')
In your code, the first if statement makes the second meaningless.

Answer 3

You are testing the value of c against an empty character constant '', which is not allowed. I think you want to test it against a space ' '.

Answer 4

If you want to test the value of your variable c with an empty space, you should compare it with ' ' instead of ''. Furthermore, comparison operator is ==, not = (assignment), which you used to compare the tab character '\t' in the last condition of your if statement, thus your if statement should be:
if (c == ' ' || c == '\n' || c == '\t')

Answer 5

this code block:
while((c = getchar()) != EOF)
{
++nc;
if (c == '\n')
if (c == '' || c == '\n' || c = '\t')
state = OUT;
else if (state == OUT) {
state = IN;
++nw;
}
}
does not indicate (via the indentation) exactly which 'if' clause
is related to the 'else clause.
this is a VERY good reason to include ALL the '{' and '}' braces
this is what is actually written in the code:
while((c = getchar()) != EOF)
{
++nc;
if (c == '\n')
{
if (c == '' || c == '\n' || c = '\t') // two errors on this line
{
state = OUT;
}
else if (state == OUT)
{
state = IN;
++nw;
}
}
}
the question becomes: is this what you really wanted?
we already know that the code is not correct
because the second 'if' will ALWAYS fail due to 'c' being
already confirmed as containing a '\n'
and the comparison of 'c' to ''
and the assignment of '\t' to 'c'
therefore, I suspect it should be:
while((c = getchar()) != EOF)
{
++nc;
if (('\n' == c) || (' ' == c) || ('\n' == c) || ('\t' == c) )
{
state = OUT;
}
else if (state == OUT)
{
state = IN;
++nw;
}
}
which would have also caught the assignment of '\t' to c at compile time
rather than you having to spend many an hour debugging the code.
I.E. put the literal on the left side of the comparison operator '=='

Answer 6

Well this worked, I used ideas from Amadeus, user3629249 and Min Fu. Thanks everyone for your help.
Working Code:
#include <stdio.h>
#include <stdlib.h>
#define IN 1
#define OUT 0
int main()
{
int c, nl, nw, nc, state;
state = OUT;
nc = 0;
nl = 0;
nw = 0;
while((c = getchar()) != EOF)
{
++nc;
if (('\n' == c) || (' ' == c) || ('\n' == c) || ('\t' == c) )
{
state = OUT;
++nl;
}
else if (state == OUT)
{
state = IN;
++nw;
}
}
printf("%d %d %d \n", nl, nw, nc);
return 0;
}