Is it guaranteed that standard extraction operator>> does not change argument in case of failure?

问题

If calling something like input_stream >> i; where i is of arithmetic type, throws exception or sets badbit etc., is it guaranteed that i has not changed?

回答1:

Before C++11, the value was left as it was, [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)

But after C++11, no. It is set to 0 if extraction fails (same reference):

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)

回答2:

Referring to the cppreference documentation for std::basic_istream::operator>> std::num_get::get, std::num_get::do_get :

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()

And then

Stage 3: conversion and storage:
[...]
- If the conversion function fails to convert the entire field, the value ​0​ is stored in v

来源：https://stackoverflow.com/questions/40828481/is-it-guaranteed-that-standard-extraction-operator-does-not-change-argument-in