IO in C standard - c

Question

The following program is from a homework assignment.
I need to write a program where it:
asks a Seed and a Range,
generates and displays a sequence of random numbers based on the seed and range input,
converts the above into a sequence of "A"s, "B"s, or "C"s depending on the generated sequence: turn 1 into A, 2 into B, 3 into C;
asks whether to continue; if user puts "n", stop; else starts over again.
#include <stdlib.h>
#include <stdio.h>
int main() {
int seed;
int Range;
char ch;
while (ch != "n") {
printf("Enter a seed: ");
scanf("%d", &seed);
printf("Enter a Range: ");
scanf("%d", &Range);
srand(seed);
for (int i = 0; i < 10; i++) {
scheduler(rand() % Range + 1);
}
printf("Continue?\n");
scanf("%c", &ch);
}
return 0;
}
I put two arbitrary numbers and it generates a random sequence as expected. But then it just prints "Continue?" without taking an input from me and then asks me to enter a seed. What's wrong with my code? What's the proper way to do this?

Answer 1

Change this:
scanf("%c", &ch);
to this:
scanf(" %c", &ch);
in order to eat the leftover character (the newline character you pressed when you entered the numbers (seed andRange`) before).
I had wrote more about that issue here.
Moreover, your code invokes Undefined Behavior (UB), since you check ch before it gets initialized here:
char ch;
while (ch != "n") {
...
}
Use a do-while loop structure instead, like this:
do {
...
} while (ch != 'n');
Furthermore, you should get a compiler warning like this:
warning: comparison between pointer and integer
while (ch != "n") {
^~
warning: comparison with string literal results in unspecified behavior [-Waddress]
Change "n" to 'n', since ch is of type char, not char*.
Putting everything together, your code should like this:
#include <stdlib.h>
#include <stdio.h>
int main() {
int seed;
int Range;
char ch;
do {
printf("Enter a seed: ");
scanf("%d", &seed);
printf("Enter a Range: ");
scanf("%d", &Range);
srand(seed);
for (int i = 0; i < 10; i++) {
scheduler(rand() % Range + 1);
}
printf("Continue?\n");
scanf(" %c", &ch);
} while (ch != 'n');
return 0;
}

Answer 2

#include <stdlib.h>
#include <stdio.h>
int main()
{
int seed;
int Range;
char ch = '\0'; /* initialize ch to something other than 'n' */
/* if ch not initialized, u take your chances on being able to enter while loop */
while (ch != 'n') / single quote characters, double quote for strings imply null character \0 after last character in string */
{
printf("Enter a seed: ");
fflush( stdout ); /* because of no \n in printf */
scanf("%d", &seed);
printf("Enter a Range: ");
fflush( stdout ); /* because of no \n in printf */
scanf("%d", &Range);
srand(seed);
for (int i = 0; i < 10; i++)
{
scheduler(rand() % Range + 1);
}
printf("Continue?\n"); /* bet u this works, because of \n */
scanf("%c", &ch);
}
return 0;
}