c++ - 数独输入程序跳过提示

Question

我为我的计算机科学类(class)编写了一个程序，用于验证和解决来自 .txt 文件的数独谜题，但我想更进一步，编写一个程序来简化输入和数独游戏。我相信你可以根据这段代码找出文件的格式。我唯一的问题是最后一个 cin 被跳过了，这个选项对我来说很重要。任何见解将不胜感激!!

#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

问题出在这里:

while (cin >> rcv && count < 81)

考虑当 count==81 时会发生什么: 首先，rcv将从cin输入，然后才出现条件 count < 81将被评估为假。循环将停止，并且 rcv 的值将被忽略。如此有效，您阅读一个输入太多了。

你应该改变评估的顺序，这样count首先检查:

while (count < 81 && cin >> rcv)

---

编辑:

根据您上面的评论，您实际上期望读取的值少于 81 个。在这种情况下，我建议让用户输入一个特殊值(例如 0)来终止循环。您只需要添加 if (rcv==0) break; .如果你只是输入了一个无效的值，就像你显然在做的那样，cin流将处于失败状态，进一步的输入将不会成功。