是否保证标准提取运算符>>在失败的情况下不会更改参数？
Is it guaranteed that standard extraction operator>> does not change argument in case of failure?
如果调用input_stream >> i;之类的，其中i是算术类型，抛出异常或设置badbit等，是否保证i没有改变？
参考 std::basic_istream::operator>> std::num_get::get, std::num_get::do_get 的 cppreference 文档：
1-4) Behaves as a FormattedInputFunction. After constructing and checking the sentry object, which may skip leading whitespace, extracts an integer value by calling std::num_get::get()
然后
Stage 3: conversion and storage:
[...]
- If the conversion function fails to convert the entire field, the value 0 is stored in v
在C++11之前，该值保持原样，[reference]:
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)
但在 C++11 之后，没有。如果提取失败，则设置为 0（相同参考）：
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<T>::max() or std::numeric_limits<T>::min() is written and failbit flag is set. (since C++11)
如果调用input_stream >> i;之类的，其中i是算术类型，抛出异常或设置badbit等，是否保证i没有改变？
参考 std::basic_istream::operator>> std::num_get::get, std::num_get::do_get 的 cppreference 文档：
1-4) Behaves as a FormattedInputFunction. After constructing and checking the sentry object, which may skip leading whitespace, extracts an integer value by calling std::num_get::get()
然后
Stage 3: conversion and storage:
[...]
- If the conversion function fails to convert the entire field, the value 0 is stored in v
在C++11之前，该值保持原样，[reference]:
If extraction fails (e.g. if a letter was entered where a digit is expected),
valueis left unmodified andfailbitis set. (until C++11)
但在 C++11 之后，没有。如果提取失败，则设置为 0（相同参考）：
If extraction fails, zero is written to
valueandfailbitis set. If extraction results in the value too large or too small to fit in value,std::numeric_limits<T>::max()orstd::numeric_limits<T>::min()is written andfailbitflag is set. (since C++11)