英文:
Why is my user-defined function not being read in my main function?
问题
我遇到了一个问题，我的用户定义函数无法在主函数中读取，有什么建议或提示可以导致这种情况吗？
我尝试了不同的参数，将函数从主函数下面移到上面，但它仍然完全忽略了这段代码
#include <iostream>
using namespace std;
int Fahrenheit, Celsius ;
char y, Y ;
using namespace std;
char Answer ;
int FtoCTemp (int argc, char** argv){
cout << "What Fahrenheit Value Do You Want in Celsius" << endl;
cin >> Fahrenheit ;
Celsius = (Fahrenheit -32); // 使用两个方程式以获得正确的输出
Celsius = (Celsius * 5/9);
cout << Celsius << " Celsius" << endl; // 在方程式后打印摄氏度
cout << "Would you like to enter another Fahrenheit" << endl;
cin >> Answer; // 捕获答案
}
int main(int argc, char** argv) {
do{
int FtoCTemp (int argc, char** argv);
cout << "Would you like to enter another Fahrenheit" << endl;
cin >> Answer; // 捕获答案
} while (Answer == 'y' || Answer == 'Y');
cout << "Thank you!";
return 0;
}
英文:
Im having an issue with getting my user-defined function to read in my main function, any advice or tips on what can cause this?
i tried different parameters, moving the function from below the main to above and it still ignores the code all together
#include <iostream>
using namespace std;
int Fahrenheit, Celsius ;
char y, Y ;
/* run this program using the console pauser or add your own
getch, system("pause") or input loop */
using namespace std;
char Answer ;
int FtoCTemp (int argc, char** argv){
cout << "What Fahrenheit Value Do You Want in Celsius" << endl;
cin >> Fahrenheit ;
Celsius = (Fahrenheit -32); //2 equations used to get correct ouput
Celsius = (Celsius * 5/9);
cout << Celsius << " Celsius" << endl; // prints Celsius after equation
cout << "Would you like to enter another Fahrenheit" << endl;
cin >> Answer; // Captures answer
}
int main(int argc, char** argv) {
do{
int FtoCTemp (int argc, char** argv);
cout << "Would you like to enter another Fahrenheit" << endl;
cin >> Answer; // Captures answer
} while (Answer == 'y' || Answer == 'Y');
cout << "Thank you!";
return 0;
}
答案1
得分: 2
试试只写 FtoCTemp (argc, argv);，而不是在循环中重新声明函数，而不是 int FtoCTemp (int argc, char** argv);。
英文:
it seems like you're redeclaring the function in the loop, instead of
int FtoCTemp (int argc, char** argv); , try just writing FtoCTemp (argc, argv);
答案2
得分: 0
以下是您要翻译的代码部分：
主要问题如前一个答案中所指出的那样。还有一些其他问题，包括两次要求“...输入另一个...”和FtoCTemp()的返回值。我在下面放了一个合适的版本并附上一些解释。
#include <iostream>
using namespace std;
// 声明在所有函数体外部的变量是全局变量，可以被任何函数访问。为了代码清晰起见，尽量不要多次使用它们。
// int Fahrenheit, Celsius; // 由于这些变量仅在FtoCTemp()中使用，因此将它们移到FtoCTemp()体内。
// char y, Y; // 这些变量没有被使用。
// using namespace std; // 重复。
// char Answer; // 由于这个变量仅在main()中使用，因此将它移到main()体内。
// 将此函数的返回类型从“int”更改为“void”，因为它没有返回值。
void FtoCTemp(int argc, char **argv)
{
int Fahrenheit, Celsius;
cout << "要将华氏温度转换为摄氏温度，请输入华氏温度：" << endl;
cin >> Fahrenheit;
// 这两行代码可以合并成一行。
// Celsius = (Fahrenheit -32);
// Celsius = (Celsius * 5/9);
Celsius = (Fahrenheit - 32) * 5 / 9;
cout << Celsius << " 摄氏度" << endl;
// 这里的询问“...输入另一个...”已被删除，因为在main()中已经询问过了。
// cout << "您想输入另一个华氏温度吗？" << endl;
// cin >> Answer; // 捕获答案
return; // 为了更好的编码规范，始终在最后一行使用“return”。
}
int main(int argc, char **argv)
{
char Answer;
do
{
FtoCTemp(argc, argv);
cout << "您想输入另一个华氏温度吗？" << endl;
cin >> Answer; // 捕获答案
} while (Answer == 'y' || Answer == 'Y');
cout << "谢谢！";
return 0;
}
英文:
The main problem is just as pointed out in the previous answer. There are also a few other problems, including double asking "...enter another..." and return value of FtoCTemp(). I put a proper version below with some explaination.
#include <iostream>
using namespace std;
// Variables declared outside all function bodies are global variables, which can be accessed by any function. For code clarity, try not to use them many.
//int Fahrenheit, Celsius ; // Since these variables are only used in FtoCTemp(), so move them into FtoCTemp() body.
//char y, Y ; // These variables are not used.
//using namespace std; // Duplicated.
//char Answer ; // Since this variable is only used in main(), so move it into main() body.
// Change return type of this function from "int" to "void" since it has no return value.
void FtoCTemp (int argc, char** argv){
int Fahrenheit, Celsius;
cout << "What Fahrenheit Value Do You Want in Celsius" << endl;
cin >> Fahrenheit ;
// These two line code can be combined into one.
//Celsius = (Fahrenheit -32);
//Celsius = (Celsius * 5/9);
Celsius = (Fahrenheit - 32) * 5 / 9;
cout << Celsius << " Celsius" << endl;
// The asking "...enter another..." here is removed because it has been asked in main().
//cout << "Would you like to enter another Fahrenheit" << endl;
//cin >> Answer; // Captures answer
return; // For better coding custom, always "return" at the last line.
}
int main(int argc, char** argv) {
char Answer;
do{
FtoCTemp (argc, argv);
cout << "Would you like to enter another Fahrenheit" << endl;
cin >> Answer; // Captures answer
} while (Answer == 'y' || Answer == 'Y');
cout << "Thank you!";
return 0;
}
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