从std :: string解析整数，但是如果float则失败 [英] Parse integer from std::string, but fail if float

问题描述

在C ++和C中，有多种方法可以将字符串转换为整数，但是我还没有找到一种转换方法无法解析浮点数.

In C++ and C there are multiple methods to convert a string to integer, but I haven't found a conversion method that fails on parsing a floating point number.

const float fnum = std::stof("1.5");
std::cout << fnum << std::endl; // prints "1.5", all okay

const int inum = std::stoi("1.5");
std::cout << inum << std::endl; // prints "1", but wrong!

我需要它来分析CSV文件的列类型.如果一列中的所有字段都是整数，则将该列存储为std :: vector<int>，如果为float，则为std :: vector<float>，否则将其存储为字符串.

I need this to analyse a CSV file for column type. If all fields from one column are integers, then store the column as std::vector< int>, if float, then std::vector< float>, else store it as strings.

唯一有希望的方法是:

std::string num = "1.5";
char *end = nullptr;

const long lnum = strtol(num.data(), &end, 10);
if (end != &*num.end()) {
    std::cout << "Float? " << l << " / " << num << std::endl;
} else {
    std::cout << "Integer! " << l << " / " << num << std::endl;
}

这有效，但是非常难看.有解决这个问题的C ++方法吗?

This works, but is quite ugly. Is there a C++-way to solve this?

推荐答案

您可以使用boost lexical_cast.如果强制转换失败，它将引发异常

You could use boost lexical_cast. It throws an exception if the cast fails

try
{
    number = boost::lexical_cast<int>(your_string);
}
catch (const boost::bad_lexical_cast& exec)
{
    // do something on fail
}

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