Two conditions for DO WHILE loop in C - c

Question

Could you please explain me why this condition works with || and does not with &&.
I fixed this on my own. I just do not understand why this is OR and not AND.
This program must quit the loop when user inputs Q0.
#include <stdio.h>
int main() {
char character;
int integer;
printf("Loop started\n\n");
do {
printf("Enter Q0 to quit the loop: ");
scanf(" %c%d", &character, &integer);
if (character != 'Q' || integer != 0) {
printf("Invalid value(s)!\n\n");
}
else {
printf("Correct value(s)!\n");
}
} while (character != 'Q' || integer != 0);
printf("\nLoop ended\n");
return 0;
}
Code link

Answer 1

The code is correct and the loop will be exited if and only if the user enters a Q followed by a integer with a value of 0. The test (character != 'Q' || integer != 0) is equivalent to !(character == 'Q' && integer == 0). In plain English, the latter is a more direct translation of continue unless the user typed a Q and the number 0.
Note however these remarks:
using a do / while loop makes you repeat the test inside the loop at as the exit condition. It would be simpler to user a for (;;) loop, aka for ever loop, and use a break statement to exit the loop explicitly when the user types the appropriate sequence, using a positive test.
you do not check the scanf() return value, which must be 2 for proper operation. If the user types a sequence that cannot be parsed with " %c%d", the behavior may be incorrect and possibly undefined. For example an input of 10QA will be misinterpreted as a correct value. It is much more reliable to read the input as a string with fgets(), parse it with sscanf and check the return values.
%d will convert sequences such as 00, +0 and -0 as the number 0 which may or may not be expected. If the input sequence must match Q0 exactly, the test must be performed some other way.
Here is a modified version:
#include <stdio.h>
int main() {
printf("Loop started\n\n");
for (;;) {
char buf[128];
char character, c2;
int integer;
printf("Enter Q0 to quit the loop: ");
if (!fgets(buf, sizeof buf, sizeof buf)) {
printf("Unexpected end of file!\n\n");
break;
}
if (sscanf(" %c%d %c", &character, &integer, &c2) != 2) {
printf("Invalid format: must have a character and an integer!\n\n");
continue;
}
if (character == 'Q' && integer == 0) {
printf("Correct values!\n");
break;
}
printf("Invalid values: %c%d!\n\n", character, integer);
printf("Try again.\n");
}
printf("\nLoop ended\n");
return 0;
}

Answer 2

if (character != 'Q' || integer != 0) {
printf("Invalid value(s)!\n\n");
}
else {
printf("Correct value(s)!\n");
}
These statements translate to:
"If character is not equal to Q OR if integer is not equal to 0, print invalid value.
Else, print correct value."
So, if character is Q or integer is 0, the expression will evaluate to true and correct value would be printed.
Could you please explain me why this condition works with || and does
not with &&.
Because if the first operand of the logical AND operator evaluates to false, the second operator is not evaluated. So if character is not equal to Q, then the second expression is never evaluated (even if integer is 0).
If the above code was written as:
if (character != 'Q' && integer != 0) {
printf("Invalid value(s)!\n\n");
} else {
printf("Correct value(s)!\n");
}
which translates to:
"If character is not equal to Q AND integer is not equal to 0, print incorrect value. Else, print correct value."
It'll only print correct value when the input was Q0 and both the expressions evaluated to true.