如果输入类型与目标类型不同，"cin" 是否会将变量重置为某个默认值？

Question

我对 "cin" 的行为有疑问（我不明白）。我的 IDE 是 Windows OS 下的 Netbeans（使用 Cygwin）。

这是一个代码示例：

int main()
{
    int temp = -1;
    std::cin >> temp;    // here user enters string of characters (string) or a single character

    if (temp == 0)
        std::cout << "temp = " << temp << ".\n";
    if (temp == -1)
        std::cout << "temp = " << temp << ".\n";

    return 0;
}

如果我输入某种 character/string 字符，此代码将显示消息 temp = 0。这就像 char 到 int 的转换，并且转换总是以值 0.

结束

谢谢你能解释一下这个行为。

Answer 1

如果读取失败 operator>> 会将值设置为零 (cppreference):

If extraction fails, zero is written to value and failbit is set.

Answer 2

这是 std::basic_istream::operator>> 的预期行为；从C++11开始，如果提取失败，变量将被设置为0。在C++11之前，变量不会被修改，它的原始值仍然存在。

If extraction fails (e.g. if a letter was entered where a digit is expected), value is left unmodified and failbit is set. (until C++11)

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. (since C++11)

如果输入类型与目标类型不同，"cin" 是否会将变量重置为某个默认值？

Does "cin" reset variable to some default value if input type differs from destination type?

c++

cin