Delay output till EOF rather than newline - c

Question

I am writing a program that copies input to output character by character on Linux terminal. The code is as follows (from Dennis Ritchie's C book)
#include <stdio.h>
/* copy input to output; 2nd version*/
main()
{
int c;
while ((c = getchar()) != EOF)
putchar(c);
}
The program and its execution works fine. But I want a slight modification.
The output appears on terminal for each new line character (when I press enter). I want to delay the output till I signal the end of file by pressing Ctrl + D. What modifications I have to do for the program in-order to delay my output on terminal.
Sample output I am getting is as follows:
abcd (enter)
abcd
llefn;elnf(enter)
llefn;elnf
(ctrl+d)
Sample output I wants to get is as follows:
abcd(enter)
llefn;elnf(ctrl+d)
abcd
llefn;elnf

Answer 1

You need to store these character's to buffer, control indexes where are you writing and when you receive EOF just print that buffer via printf.
If you cant solve it you can inspire here
#include <stdio.h>
#define BUFFER_SIZE 1024
int main()
{
int c, i = 0;
char buffer[BUFFER_SIZE];
while ((c = getchar()) != EOF)
{
if (i < BUFFER_SIZE - 1)
{
buffer[i] = c;
i++;
}
else
{
buffer[BUFFER_SIZE - 1] = '\0';
printf("%s", buffer);
i = 0;
}
}
buffer[i] = '\0';
printf("%s", buffer);
return 0;
}

Answer 2

An easy, albeit sloppy and partial solution is to configure stdout to fully buffered with a large buffer:
#include <stdio.h>
int main(void) {
int c;
setvbuf(stdout, NULL, _IOFBF, 32767);
while ((c = getchar()) != EOF) {
putchar(c);
}
return 0;
}
Notes:
On a 32- or 64-bit system, you could make the buffer ridiculously large.
Passing NULL to setvbuf lets it allocate the buffer with malloc().

Answer 3

As already mentioned in the comments: put the characters into a buffer and display the whole buffer at the end:
#include <stdio.h>
#define BUF_SIZE 1024
int main()
{
char str[BUF_SIZE]; // the buffer
int c, i = 0;
while (i < BUF_SIZE-1 && (c = getchar()) != EOF)
str[i++] = (char) c;
str[i] = '\0';
printf("%s\n", str); // display the contents of the buffer
}

Answer 4

You could set buffer to stdout with setvbuf(3).
#include <stdio.h>
main()
{
int c;
char buf[BUFSIZ];
setvbuf(stdout,buf,_IOFBF,BUFSIZ);
while ((c=getchar())!=EOF)
putchar(c);
}
The key here is the fully buffered mode specified with the _IOFBF constant.
The size of the buffer is set to BUFSIZ which usually equals 8192.
As was correctly pointed in the comments by Jonathan Leffler the limited size of the buffer may cause the program to suddenly spit out its contents garbling the terminal. To avoid that one could track the utilization of the buffer and extend its size when it gets filled.