问题
After read this FAQ, i choose to use istringstream to convert my input string to numerical value.
My code is look like this:
<template T>
T Operand<T>::getValue(const std::string &s)
{
T _value;
std::istringstream v_ss(s);
v_ss >> _value;
return _value;
}
When T is int, short, long or float, no problem i get correct value. But when T is int8_t, this code doesn't work.
By exemple, if my input string is "10", getValue return me a int8_t with value equals 49.
With 49 == '1' in ASCII table, i guess the >> operator just read the first char in input string and stop.
Is there a trick or something i don't understand in the FAQ ?
回答1:
The problem is int8_t
is implemented as char
.
The implementation of the input stream is working like this:
char x;
std::string inputString = "abc";
std::istringstream is(inputString);
is >> x;
std::cout << x;
The result is 'a', because for char
the input stream is read char for char.
To solve the problem, provide a specialised implementation for your template method. and reading into a int
, then check the bounds and convert the value into a int8_t
.
来源:https://stackoverflow.com/questions/22127538/istringstream-to-int8-t-produce-unexpected-result