c++ - C++ 中变量的默认值使用 cin >>

Question

我在 CodeBlocks IDE 中用 C++ 编写了这段代码，但是当我运行它时，如果它没有读取数字，它不会给我 -1，它会给我 0。代码有问题吗？

#include "iostream"

using namespace std;

int main()
{
    cout<<"Please enter your first name and age:\n";
    string first_name="???"; //string variable
                            //("???" means "don't know the name")
    int age=-1; //integer variable (-1 means "don't know the age")
    cin>>first_name>>age; //read a string followed by an integer
    cout<<"Hello, " <<first_name<<" (age "<<age<<")\n";

    return 0;
}

Answer 1

std::basic_istream::operator>> 的行为已从 C++11 更改。从 C++11 开始，

If extraction fails, zero is written to value and failbit is set. If extraction results in the value too large or too small to fit in value, std::numeric_limits::max() or std::numeric_limits::min() is written and failbit flag is set.

请注意，直到 C++11，

If extraction fails (e.g. if a letter was entered where a digit is expected), value is left unmodified and failbit is set.

可以通过std::basic_ios::fail查看结果或 std::basic_ios::operator!并自行设置默认值。比如，

string first_name;
if (!(cin>>first_name)) {
    first_name = "???";
    cin.clear(); //Reset stream state after failure
}

int age;
if (!(cin>>age)) {
    age = -1;
    cin.clear(); //Reset stream state after failure
}

cout<<"Hello, " <<first_name<<" (age "<<age<<")\n";

另请参阅:Resetting the State of a Stream