C beginner question: Array is not printing - c

Question

I'm trying to produce a simple program which will read a string and print it along with the number of characters in it.
The character count seems fine, but it won't print the full string. For whatever reason, it will print only the second character. I have been reviewing my code and I still cannot figure out why this is happening.
If I input: abcdef
It will print out: 1
b 2
3
4
5
6
Instead of the intended: a 1 b 2 c 3 d 4 e 5 f 6
#include <stdio.h>
int main()
{
char c;
char str[0] = {};
int i;
int charcount;
charcount = 0;
for (i = 0; (c = getchar()) != '#' && c != '\n'; i++) {
c = str[i];
charcount++;
printf("%c %d \n", str[i], charcount);
//if(charcount > 80)
// printf("%d", z[i]);
}
return 0;
}
Any help is appreciated. Thanks.

Answer 1

You have a variable str that is of zero length, and you try to access various elements in it (but it has none, so this is undefined behavior)
Just get rid of it and use c directly:
#include <stdio.h>
int main()
{
char c;
int i;
int charcount;
charcount = 0;
for (i = 0; (c = getchar()) != '#' && c != '\n'; i++) {
charcount++;
printf("%c %d \n", c, charcount);
}
return 0;
}

Answer 2

For starters according to the C Standard the function main without parameters shall be declared like
int main( void )
This declaration
char str[0] = {};
is invalid with two respects. The array size shall be greater than zero. And the the braces that initialize the array shall not be empty.
In fact the array is redundant because what you are trying to do is just output entered characters.
This assignment
c = str[i];
does not make sense even if the array was be declared correctly because the entered character is overwritten by the (non-existent) element of the array.
The variable c itself should have the type int.
If you just need to perform this
I'm trying to produce a simple program which will read a string and
print it along with the number of characters in it.
then the program can look for example the following way
#include <stdio.h>
int main(void)
{
size_t i = 0;
for ( int c = getchar(); c != EOF && c != '\n' && c != '#'; c = getchar() )
{
++i;
printf( "%c %zu ", c, i );
}
putchar( '\n' );
return 0;
}
If to enter
abcdef
then the output will be
a 1 b 2 c 3 d 4 e 5 f 6
If you want to use a character array and store entered characters in the array then the program can look like
#include <stdio.h>
int main(void)
{
enum { N = 100 };
char s[N];
int c;
for ( size_t i = 0; i < N && ( c = getchar( ) ) != EOF && c != '\n' && c != '#'; i++ )
{
s[i] = c;
printf( "%c %zu ", s[i], i + 1 );
}
putchar( '\n' );
return 0;
}
The program output will be the same as above if to enter for example abcdef.

Answer 3

I took your question to mean you were looking to store the characters that are entered into an array and print them out as you go. You want to give the array an initial size (I chose 99) and then add characters to the array as you go. You can keep track of the length so you know where the "end" is. There are better ways to manage string data in C, but here you go.
#include <stdio.h>
int main()
{
char c;
char str[99] = {};
int i;
int charcount;
charcount = 0;
for (i = 0; (c = getchar()) != '#' && c != '\n'; i++) {
// c = str[i];
str[i] = c;
charcount++;
printf("%c %d \n", str[i], charcount);
//if(charcount > 80)
// printf("%d", z[i]);
}
return 0;
}
Some additional thoughts:
You probably want to look at something like string.h
If you are going to keep the char array, you need to protect against going over the size of the array
Adding a little blurb about what you expect your program to do, what the requirements are, example input, output, etc would have helped get better answers from the community