[英]How does c++ interpret 6e(or any number with e) as an input?
當運行此代碼並輸入 6e 或任何其他帶有 e 的數字時,程序無法識別它並以 else 狀態結束。 我在這里做錯了什么?
#include "../std_lib_facilities.h"
int main() {
cout << "Please enter amount of money followed by the currency name:"
+ "y for yen, p for pound or e for euro, like 3.40e. \n"
<< "This app will convert it to dollars. \n";
double amount;
char currency = ' ';
double dollars;
cin >> amount >> currency;
cout<<"amount: "<< amount <<"\n";
cout<<"currency :" << currency<<"\n";
if (currency == 'y'){
cout <<amount<<"Yuan = "<< amount * 0.15<<" dollars";
}
else if (currency == 'e') {
cout << amount << "Euro = " << amount * 1.18 << " dollars";
}
else if (currency == 'p') {
cout << amount << "Pounds = " << amount * 1.29<< " dollars";
}
else {
dollars = 0;
cout << "Unknown currency\n";
}
}
operator >>
讀取由空格分隔的元素(或直到無法解析的字符)。
6e
不能被解析為double
6e
值(但e
是科學記數法的有效字符,因此它被消耗),並且cin
進入失敗狀態並且每個后續>>
調用都失敗。
對於cin >> amount >> currency;
工作輸入應該看起來像6 e
(或6e0e
)。
如果您希望6e
也用作輸入,則需要將其作為字符串讀取(或使用getline
讀取整行)並getline
解析(例如提取和后綴字符,然后將剩余數字解析為double
)。
另請參閱這個類似的問題: 從沒有“E”的輸入流中讀取浮點數。
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