Sudoku input program skipping a prompt - c++

Question

I wrote a program for my computer science class that validates and solves sudoku puzzles from .txt files, but I wanted to take it one step further and write a program that made it easy to input and sudoku game. I'm sure you can figure out the format of the files based on this code. My only problem is that the last cin gets skipped, and that option is important to me. Any insight will be appreciated!!
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
struct s {
s();
~s() {/*zzzz*/}
void show_grid();
void set (int &r, int &c, int &v) {g[r][c] = v;}
private:
int g[9][9];
};
//************************************************************************
void s::show_grid() {
//print game out to check it
cout << " | ------------------------------- |" << endl;
for (int k=0; k<81; k++) {
if (k%3 == 0)
cout << " |";
cout << " " << g[k/9][k%9];
if (k%9 == 8) {
cout << " |" << endl;
if ((k/9)%3 == 2)
cout << " | ------------------------------- |" << endl;
}
}
cout << endl;
}
//************************************************************************
s::s() {
//initialize all elements to zero
for (int i=0; i<9; i++) {
for (int j=0; j<9; j++) {
g[i][j] = 0;
}
}
}
//************************************************************************
void create_name (string &name) {
//append .txt extension LIKE IT OR NOT
string ext = name;
ext.erase(ext.begin(), ext.end() - 4);
if (ext.compare(".txt")!=0)
name.append(".txt");
}
//************************************************************************
int main () {
s g;
string name;
string yon("");
int count = 0;
int row, col, val, rcv;
ofstream os;
cout << "Enter game file name: ";
cin >> name;
create_name(name);
//open and do typical checks
os.open(name.c_str());
if (os.fail()) {
cerr << "Could not create " << name << ". Waaaah waaaaaaaaaah...\n\n";
return 0;
}
//useful output (hopefully)
cout << "Enter grid coordinates and value as a 3-digit number,\n"
<< "from left to right, row by row.\n"
<< "(e.g. 2 in first box would be 112)\n";
//take input as one int, to be user friendly
while (cin >> rcv && count < 81) {
row = (rcv / 100) - 1;
col = ((rcv / 10) % 10) - 1;
val = rcv % 10;
os << row << " " << col << " " << val << endl;
g.set (row, col, val);
count++;
}
os.close();
//From here down is broken, but it still compiles, runs, and works
cout << "Show grid input(y/n)?\n";
cin >> yon;
if (yon.compare("y")==0)
g.show_grid();
else if (yon.compare("n")==0)
cout << "Peace!\n";
return 0;
}

Answer 1

The problem is here:
while (cin >> rcv && count < 81)
Consider what happens when count==81: First, rcv will be input from cin, and only then the condition count < 81 will be evaluated to false. The loop will stop, and the value of rcv will be ignored. So effectively you read one input too many.
You should change the order of evaluation, so that count is checked first:
while (count < 81 && cin >> rcv)
Edit:
According to your comment above you are actually expecting to read less than 81 values. In that case, I recommend having the user input a special value (for example, 0) to terminate the loop. You'd just need to add if (rcv==0) break;. If you just input an invalid value as you are apparently doing, the cin stream will be put in a failed state and further input will not succeed.

Answer 2

Try something like:
//useful output (hopefully)
cout << "Enter grid coordinates and value as a 3-digit number,\n"
<< "from left to right, row by row.\n"
<< "(e.g. 2 in first box would be 112)\n"
<< "or Z to end the loop\n"; // 1
//take input as one int, to be user friendly
while (count < 81 && cin >> rcv ) { // 2
row = (rcv / 100) - 1;
col = ((rcv / 10) % 10) - 1;
val = rcv % 10;
os << row << " " << col << " " << val << endl;
g.set (row, col, val);
count++;
}
if(!std::cin) { // 3
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
1) Let the user know that he can enter in invalid char. It doesn't have to be Z, actually any non-numeric char will work.
2) Fix off-by-one error in the order of the &&.
3) If std::cin is in error state, clear the error and ignore the Z.

Answer 3

cin >> yon
still actually reads in a variable, it just reads in the variable that the while loop found to be false. When the while loop condition returns false rcv is ignored, so the number remains in the input stream waiting for the next cin statement. When yon is called that number meant for rcv is read into yon, giving you some strange errors.
it would be better to use interjay's method:
while (count < 81 && cin >> rcv)