istringstream到int8_t产生意外的结果 [英] istringstream to int8_t produce unexpected result
问题描述
阅读此常见问题,我选择使用istringstream将我的输入字符串转换为数值。
After read this FAQ, i choose to use istringstream to convert my input string to numerical value.
我的代码看起来像这样:
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;
}
当T是int,short,long或float时,值。
但是当T是int8_t时,这个代码不工作。
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.
例如,如果我的输入字符串是10,getValue返回一个int8_t值等于49。
By exemple, if my input string is "10", getValue return me a int8_t with value equals 49.
在ASCII表中使用49 =='1',我想>>运算符只读取输入字符串中的第一个字符并停止。
With 49 == '1' in ASCII table, i guess the >> operator just read the first char in input string and stop.
推荐答案
执行输入流的工作方式如下:
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;
结果是'a',因为 char
输入流读取char的char。
The result is 'a', because for char
the input stream is read char for char.
要解决这个问题,请为您的模板方法提供一个专门的实现。并读入 int
,然后检查边界并将该值转换为 int8_t
。
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
.
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