Echoing Standard input to Standard Output - c

Question

Write a program that will read standard input and echo each line to standard output with a line number and tab preceding it. When you run this program and enter lines from the terminal, lines of input will be interspersed with lines of output. If your system has output redirection and you redirect output to a file, the file will look like the input with the lines numbered.
Here is an example of how the script should work.
User input in bold.
Enter your text:
This is line 1.
1 This is line 1.
This is line 2.
2 This is line 2.
This is the last line of input.
3 This is the last line of input.
The last line will end the process.
Here is what I have so far:
#include <stdio.h>
#include <stdlib.h>
int main()
{
printf("Enter your text: \n");
while (1)
{
int ch = getc(stdin);
fflush(stdout);
if(ch == EOF) break;
putc(ch, stdout);
}
return 0;
}
I have been attempting at this for a few hours now to no luck, any help is appreciated. I basically cant get my script to display the example output. I can get my script to display the stdin as an stdout and thats all. This is a C assignment.

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
int main(void){
size_t no = 0;
bool line_top = true;
printf("Enter your text: \n");
while (1) {
int ch = fgetc(stdin);
if(ch == EOF) break;
if(line_top)
printf("%zu ", ++no);
line_top = (ch == '\n');
fputc(ch, stdout);
}
fflush(stdout);
return 0;
}

Answer 2

To properly complete the assignment, you need to understand what is the line. The code you have in place reads and prints data symbol-by-symbol. This is not going to print the numbers requested!
There are two solutions here. First is to switch from reading symbol by symbol to line by line, and than print an ever incrementing counter every time you print a line. Second is leave reading symbol-by-symbol, but print an incremented number every time you've read a new line symbol from the input. A slight challenge here is that you would need a state - that is, once you've seen the new line character, you should remember that fact, but not print the number right away - this might be the last line in the input. Instead, print a number whenever your saved state tells you so, and reset the flag after this.
A separate question is what to do with the empty lines - do they participate in counter increment or not - but this is probably beyond the assignment.

Answer 3

Ok you should only do what the assignment requires. Here's the most straight-forward solution I can think of:
#include <stdio.h>
int main(void)
{
char buf[1024];
int num = 0;
/* read a line as long as we can read (no error / EOF) */
while (fgets(buf, 1024, stdin))
{
/* print line number */
printf("%d\t", ++num);
/* print what we actually read */
fputs(buf, stdout);
}
return 0;
}
Test run:
> ./numlines < numlines.c
1 #include <stdio.h>
2
3 int main(void)
4 {
5 char buf[1024];
6 int num = 0;
7
8 /* read a line as long as we can read (no error / EOF) */
9 while (fgets(buf, 1024, stdin))
10 {
11 /* print line number */
12 printf("%d\t", ++num);
13
14 /* print what we actually read */
15 fputs(buf, stdout);
16 }
17
18 return 0;
19 }
20

Answer 4

Code needs to keep track of the beginning of a line. This can be done by reading a line using fgets() or by looking at individual char.
fgets() is likely faster, but it limits line length to some constant.
fgetc() gets 1 unsigned char at a time (or EOF).
The below follows OP's lead, yet adds the inner while loop.
Not much code needed.
#include <stdio.h>
#include <stdlib.h>
int main(void) {
unsigned long long line_number = 0;
int ch;
while ((ch = fgetc(stdin)) != EOF) {
printf("%llu\t", ++line_number);
while (ch != EOF && fputc(ch, stdout) != '\n') {
ch = fgetc(stdin);
}
}
return 0;
}

Answer 5

const int buflen = 1024;
char buffer[buflen];
int count = 0;
while (1)
{
char* line = fgets(buffer, buflen, stdin);
if (line == null)
break;
fprintf(stdout, "%d: %s\n", ++count, line);
}

Answer 6

Here's a c++ solution using getline and cout.
#include <string>
#include <iostream>
using namespace std;
int main()
{
unsigned int i=0;
string s;
cout<<"Enter your text: "<<endl;
while (std::getline(cin,s))
{
cout<<++i<<'\t'<<s<<endl;
}
return 0;
}
Here a c solution:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int exit=0, i=0, newline=1;
printf("Enter your text: \n");
while (!exit)
{
int ch = getc(stdin);
if(newline){
printf("\n%d\t",++i);
newline=0;
}
switch(ch){
case '\n':
newline=1;
printf("\n");
break;
case EOF:
printf("\n");
exit=1;
break;
default:
printf("%c",(char)ch);
break;
}
}
return 0;
}

Answer 7

#include<stdio.h>
int main()
{
char line[1000];
int lineno=0;
while((fgets(line,1000, stdin)) != NULL)
{
lineno++;
printf("%d ",lineno);
fputs(line, stdout);
}
}
It will work as you are expected. It is a simple method to do this. You don't have to put the while(1) and if condition and break the loop.