Failing Tictactoe in c++ - c++

Question

I'm a beginner in c++, i'm trying to make a tictactoe. My program fails at the part acfter i enter input. When I enter an input, there is no next action from the program like i expected it to ("unvalid", check if win or loose, ask for input again). It only shows blank. like below: after I enter "1", nothing happens, and I can keep enter more input.
terminal example photo
I know it is a simple activity but I just ca't figure it out ToT. Thank you for helping!
//tictactoe
#include <iostream>
#include <vector>
using namespace std;
//declared variables
vector<char> out = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'};
int in = 2;
char player_out;
bool loose = false;
char x;
bool filled = false;
bool end(){ //won yet?
bool loose = false;
//horizontal
if (out[1] == out[2] && out[3] == out[2]){
loose = true;
}
else if (out[4] == out[5] && out[6] == out[5]){
loose = true;
}
else if (out[7] == out[8] && out[9] == out[8]){
loose = true;
}
//vertical
else if (out[1] == out[4] && out[7] == out[1]){
loose = true;
}
else if (out[2] == out[5] && out[8] == out[2]){
loose = true;
}
else if (out[3] == out[6] && out[9] == out[3]){
loose = true;
}
else if (out[1] == out[5] && out[9] == out[5]){
loose = true;
}
else if (out[3] == out[5] && out[7] == out[5]){
loose = true;
}
else{
loose = false;
}
return loose;
}
void game_start_display(){ //display the board
cout << "TIC TAC TOE\n";
cout << " | | \n";
cout << " " << out[1] << " | " << out[2] << " | " << out[3] << " \n";
cout << "______|______|______\n";
cout << " | | \n";
cout << " " << out[4] << " | " << out[5] << " | " << out[6] << " \n";
cout << "______|______|______\n";
cout << " | | \n";
cout << " " << out[7] << " | " << out[8] << " | " << out[9] << " \n";
cout << " | | \n\n";
}
int change_player(){ //take turn 1st and 2nd player
if (in == 1){
in++;
}
else{
in--;
}
return in;
}
bool filled_f() { //check if the spot is filled
if (out[x] != 'X' and out[x] != 'O'){
filled = true;
out[x] = player_out; //fill the input into the spot
}
else if (out[x] == 'X' or out[x] == 'O')
cout << "The square has already been used!\n";
filled = false;
return filled;
}
char player_out_f(){ //change output sign for each players (X, O)
if (in == 1){
player_out = 'X';
}
else if (in == 2){
player_out = 'O';
}
return player_out;
}
void c_player_display(){ //tell players to enter a number
cout << "Player " << in << "'s turn, please enter a number:\n";
}
int main(){
//intro
int loose = false;
game_start_display();
while(loose == false){ //when the game is still happening
change_player(); //change player (player start is set 2 so this comes first and change it to 1)
player_out_f(); //change player output sign (X, O)
c_player_display(); //print the line to ask for input
while(filled == false){ //when the there is no input yet (the spot is not filled)
cin >> x; // input
if (x > 0 && x < 10){ //check if input is in range 1-9
filled_f(); //check if the spot is occupied
}
else if(x < 0 && x > 10) { //if input is out of range
cout << "Invalid! Enter again!\n";
filled = false; //repeat the asking input circle (the while)
}
}
game_start_display(); //output the board again with new char (X or O)
end(); //check if anyone's won yet, if no, repeat the circle
}
cout << "Player " << in << " won! GG";
}

Answer 1

You have infinite loop at while (filled == false) { ... }, because filled_f always sets filled to false (and the else if branch of the condition inside this loop as well does so). It's because you most likely missed figure brackets when writing else if block in filled_f. Your indentation hints that you wanted 2 statements to be in that block, but as of now, only the first is executed conditionally, and the second (filled = false;) is executed after the branch. In other words, with intuitive indentation this function looks like this:
bool filled_f() { //check if the spot is filled
if (out[x] != 'X' and out[x] != 'O') {
filled = true;
out[x] = player_out; //fill the input into the spot
}
else if (out[x] == 'X' or out[x] == 'O')
cout << "The square has already been used!\n";
filled = false;
return filled;
}
It sets filled = false; in any case, since if/else execute (depending on condition) the statement immediately following one of them (see here), and indentation is ignored (unlike in, e.g., Python, where indentation alone determines boundaries of conditions, loops, function etc), so only cout << ... is conditionally executed. To do what you want put figure brackets around appropriate statements the same way you already did for the first if branch to create compound statement (block) from them, which is a statement itself and does what you want - groups several other statements within it, executed in sequence:
bool filled_f() { //check if the spot is filled
if (out[x] != 'X' and out[x] != 'O') {
filled = true;
out[x] = player_out; //fill the input into the spot
}
else if (out[x] == 'X' or out[x] == 'O') {
cout << "The square has already been used!\n";
filled = false;
}
return filled;
}
Additional remarks
Note that logically it's not needed to have if condition in else since if the first if condition is dissatisfied, else if condition is definitely satisfied (look De Morgan's Laws) and you can just write else:
// ...
else {
cout << "The square has already been used!\n";
filled = false;
}
// ...
Also in your main loop, you use this:
if (x > 0 && x < 10){ //check if input is in range 1-9
filled_f(); //check if the spot is occupied
}
else if(x < 0 && x > 10) { //if input is out of range
cout << "Invalid! Enter again!\n";
filled = false; //repeat the asking input circle (the while)
}
to test whether x is within range, but your condition in else if is wrong (should be x <= 0 || x >= 10 instead of x < 0 && x > 10) and can be omitted altogether (again see De Morgan's Laws) by using just else.