如何在 C++ 中正确关闭我的程序?
How do I close my program correctly in C++?
我正在编写我的第一个程序,它从文件中读取并允许您玩游戏,我被告知 exit 函数不是一个好主意。
我正在尝试回调 main 以正确关闭程序,但出现以下错误:
C3861 'main': 未找到标识符。
我哪里出错了或者我如何正确调用主函数有什么想法吗?
代码如下:
#include <fstream>
#include <iostream>
#include <string>
using namespace std;
void extra() {
int lives = 3;
int UI, intAnswer;
int opt = 0;
string IN, NoQ, q, c1, c2, c3, Answer;
fstream quiz;
cout << "Welcome to the guessing game!" << endl;
quiz.open("QuizQuestions.txt");
getline(quiz, IN);
cout << "There are " << IN << " Questions" << endl;
while (quiz.good() && opt !=2) {
getline(quiz, q);
cout << "Question " << q << endl;
getline(quiz, c1);
cout << c1 << endl;
getline(quiz, c2);
cout << c2 << endl;
getline(quiz, c3);
cout << c3 << endl;
getline(quiz, Answer);
intAnswer = stoi(Answer);
cout << "What answer do you think it is? ";
cin >> UI;
if (UI == intAnswer) {
lives++;
cout << "You got it right! You now have " << lives << " lives left " << endl << endl;
//i = 0;
}
else {
cout << "You got the answer wrong sorry, the correct answer is " << Answer << endl;
lives--;
cout << "You now have " << lives << " lives" << endl;
//i = 0;
if (lives < 1) {
cout << "You lose, would you like to play again? 1 for yes, 2 for no? ";
cin >> opt;
if (opt = 1) {
cout << endl;
extra();
}
else if (opt = 2) {
quiz.close();
return;
}
}
}
}
quiz.close();
}
int main() {
int UI;
cout << "Would you like to do the quiz? 1 - yes other - no ";
cin >> UI;
if (UI = 1) {
extra();
}
return 0;
}
您可以从 extra
函数简单地 return 而不是调用 main。然后程序从您调用 extra
.
的地方继续执行
只是 return 到 main
。
else {
quiz.close();
;
}
您不能给自己打电话 main
。
当你调用一个函数并且它结束时,函数 pointer/flow 将 return 到调用代码。
让我们考虑一下您的代码的一般结构:
void extra() {
for (int i = 0; i = 1; i++) {
//^---I suspect you don't mean this, maybe i<1, or 3, or...
// recall == and -= are different
//snipped some details
if (UI == intAnswer) {
lives++;
cout << "You got it right! You now have " << lives << " lives left " << endl << endl;
i = 0;
}
else {
cout << "You got the answer wrong sorry, the correct answer is " << Answer << endl;
lives--;
cout << "You now have " << lives << " lives" << endl;
i = 0;
if (lives < 1) {
cout << "You lose, would you like to play again? 1 for yes, 2 for no? ";
cin >> UI;
if (UI = 1) {
cout << endl;
extra();
//^--- I suspect you don't need this recursive call
}
else {
quiz.close();
return;
// ^---- return back to where we started
}
}
}
}
}
quiz.close();
system("pause");
}
int main() {
int UI;
cout << "Would you like to do the quiz? 1 - yes other - no ";
cin >> UI;
if (UI = 1) {
extra();//we come back here after the function stops
}
return 0;
}
请注意,我只是将 return
放在您想要结束 function/program 的位置。
我正在编写我的第一个程序,它从文件中读取并允许您玩游戏,我被告知 exit 函数不是一个好主意。
我正在尝试回调 main 以正确关闭程序,但出现以下错误:
C3861 'main': 未找到标识符。
我哪里出错了或者我如何正确调用主函数有什么想法吗?
代码如下:
#include <fstream>
#include <iostream>
#include <string>
using namespace std;
void extra() {
int lives = 3;
int UI, intAnswer;
int opt = 0;
string IN, NoQ, q, c1, c2, c3, Answer;
fstream quiz;
cout << "Welcome to the guessing game!" << endl;
quiz.open("QuizQuestions.txt");
getline(quiz, IN);
cout << "There are " << IN << " Questions" << endl;
while (quiz.good() && opt !=2) {
getline(quiz, q);
cout << "Question " << q << endl;
getline(quiz, c1);
cout << c1 << endl;
getline(quiz, c2);
cout << c2 << endl;
getline(quiz, c3);
cout << c3 << endl;
getline(quiz, Answer);
intAnswer = stoi(Answer);
cout << "What answer do you think it is? ";
cin >> UI;
if (UI == intAnswer) {
lives++;
cout << "You got it right! You now have " << lives << " lives left " << endl << endl;
//i = 0;
}
else {
cout << "You got the answer wrong sorry, the correct answer is " << Answer << endl;
lives--;
cout << "You now have " << lives << " lives" << endl;
//i = 0;
if (lives < 1) {
cout << "You lose, would you like to play again? 1 for yes, 2 for no? ";
cin >> opt;
if (opt = 1) {
cout << endl;
extra();
}
else if (opt = 2) {
quiz.close();
return;
}
}
}
}
quiz.close();
}
int main() {
int UI;
cout << "Would you like to do the quiz? 1 - yes other - no ";
cin >> UI;
if (UI = 1) {
extra();
}
return 0;
}
您可以从 extra
函数简单地 return 而不是调用 main。然后程序从您调用 extra
.
只是 return 到 main
。
else {
quiz.close();
;
}
您不能给自己打电话 main
。
当你调用一个函数并且它结束时,函数 pointer/flow 将 return 到调用代码。
让我们考虑一下您的代码的一般结构:
void extra() {
for (int i = 0; i = 1; i++) {
//^---I suspect you don't mean this, maybe i<1, or 3, or...
// recall == and -= are different
//snipped some details
if (UI == intAnswer) {
lives++;
cout << "You got it right! You now have " << lives << " lives left " << endl << endl;
i = 0;
}
else {
cout << "You got the answer wrong sorry, the correct answer is " << Answer << endl;
lives--;
cout << "You now have " << lives << " lives" << endl;
i = 0;
if (lives < 1) {
cout << "You lose, would you like to play again? 1 for yes, 2 for no? ";
cin >> UI;
if (UI = 1) {
cout << endl;
extra();
//^--- I suspect you don't need this recursive call
}
else {
quiz.close();
return;
// ^---- return back to where we started
}
}
}
}
}
quiz.close();
system("pause");
}
int main() {
int UI;
cout << "Would you like to do the quiz? 1 - yes other - no ";
cin >> UI;
if (UI = 1) {
extra();//we come back here after the function stops
}
return 0;
}
请注意,我只是将 return
放在您想要结束 function/program 的位置。