Undefined behaviour of scanf() in a do-while loop - c

Question

I'm currently learning C by a book "C Programming a modern approach" and encountered this code. When I tried to run it, after typing consecutive characters like 'abc' and hitting Enter (new line), nothing was printed. Please explain what is going on here.
char ch;
do {
scanf("%c" , &ch);
} while (ch != '\n');
printf("%c", ch);

Answer 1

You're asking the user to input a character using scanf. This is happening in a loop until the user inputs a '\n' or newline character (the same as pressing the enter key), which is when the loop will break.
Your print statement will then print the character in the variable ch, which at that point will be '\n' (since this variable just stores one character, the last one you typed).
This newline character will probably be invisible when you run your program so you may not be seeing it. You can add another print statement after the loop and if that print statement starts at a newline, you know that the '\n' was printed on the previous line.
Something like:
#include <stdio.h>
int main()
{
char ch;
do
{
scanf("%c" , &ch);
} while (ch != '\n');
printf("%c", ch);
printf("I should show up on a newline");
return 0;
}

Answer 2

The code you provided reads characters from the input using the scanf() function and stores them in the variable ch until a newline character (\n) is encountered. After that, the program prints the last character that was read, which is the newline character.
The reason you are not seeing any output when you enter characters followed by a newline character is because the printf() statement is only executed after the loop has finished running. So, the program is waiting for you to enter a newline character to terminate the loop and print the last character that was read.
If you want to see the characters you enter, you can add a printf() statement inside the loop, like this:
char ch;
do {
scanf("%c" , &ch);
printf("%c", ch);
} while (ch != '\n');
This will print out each character as it is read from the input, so you can see what you're typing. Happy coding :)

Answer 3

When I tried to run it, after typing consecutive characters like abc and hitting Enter (new line), nothing was printed.
Well with the posted code, if the loop even finishes, the last byte read by scanf("%c", &ch) and stored into ch is the newline character. Hence printf("%c", ch) outputs this newline and it seems nothing is printed but something is, the newline which is invisible on the terminal but does move the cursor to the next line.
You can make this more explicit by changing the printf call to this:
printf("last value: '%c'\n", ch);
Note however that the posted code is not a recommended way to read the contents of the input stream:
scanf("%c", &ch) may fail to read a byte if the stream is at end of file. Failure to test this condition leads to undefined behavior (ch is unmodified, hence stays uninitialized if the input stream is an empty file) or to an infinite loop as ch may never receive a newline.
this code is a typical example of a do / while with a classic bug. It would be much better to write the code using getchar() and a while loop.
Here is a modified version:
#include <stdio.h>
int main(void) {
int c; // must use int to distinguish EOF from all valid byte values
int count = 0; // to tell whether a byte was read at all
char ch = 0; // the last byte read
// read all bytes from the input stream until end of file or a newline
while ((c = getchar()) != EOF && c != '\n') {
ch = (char)c;
count++;
}
if (count == 0) {
printf("no characters entered: ");
if (c == EOF) {
printf("end of file or read error\n");
} else {
printf("empty line\n");
}
} else {
printf("last character on line is '%c'\n", ch);
if (c == EOF) {
printf("end of file or input error encountered\n");
}
}
return 0;
}