从文件中逐个字符输入，用 C++

发布于 2024-09-26 13:53:55 字数 384 浏览 5 评论 0原文

有没有办法一次从文件中获取一个数字的输入？例如，我想将以下整数存储在整数向量中，因为它太长了，甚至不能用 long long int 来保存。

12345678901234567900

那么我如何从文件中读取这个数字，以便我可以：

vector<int> numbers;
number.push_back(/*>>number goes here<<*/)

我知道上面的代码并不完整，但我希望它能解释我正在尝试做的事情。我也尝试过谷歌，到目前为止它被证明是无效的，因为只有 C 的教程出现了，这并没有真正帮助我太多。

感谢提前， 丹·谢瓦利埃

原文

Is there any way to get input from a file one number at a time?
For example I want to store the following integer in an vector of integers since it is so long and can't be held by even a long long int.

12345678901234567900

So how can I read this number from a file so that I can:

vector<int> numbers;
number.push_back(/*>>number goes here<<*/)

I know that the above code isn't really complete but I hope that it explains what I am trying to do.
Also I've tried google and so far it has proved innefective because only tutorials for C are coming up which aren't really helping me all too much.

Thank is advance,
Dan Chevalier

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沧笙踏歌 2024-10-03 13:53:55

这可以通过多种方式完成，所有这些方式都可以归结为将每个字符“0”..“9”转换为相应的整数 0..9。以下是如何通过单个函数调用来完成此操作：

#include <string>
#include <iostream>
#include <vector>
#include <iterator>
#include <functional>
#include <algorithm>
int main()
{
        std::string s = "12345678901234567900"; 
        std::vector<int> numbers;
        transform(s.begin(), s.end(), back_inserter(numbers),
                  std::bind2nd(std::minus<char>(), '0'));
        // output
        copy(numbers.begin(), numbers.end(),
             std::ostream_iterator<int>(std::cout, " "));
        std::cout << '\n';
}

从文件读取时，您可以读取字符串和transform()，甚至可以直接从istream迭代器读取transform()，如果该文件中除了您的数字之外没有其他内容：

    std::ifstream f("test.txt");
    std::vector<int> numbers;
    transform(std::istream_iterator<char>(f),
              std::istream_iterator<char>(),
              back_inserter(numbers),
              std::bind2nd(std::minus<char>(), '0'));

This could be done in a variety of ways, all of them boiling down to converting each char '0'..'9' to the corresponding integer 0..9. Here's how it can be done with a single function call:

#include <string>
#include <iostream>
#include <vector>
#include <iterator>
#include <functional>
#include <algorithm>
int main()
{
        std::string s = "12345678901234567900"; 
        std::vector<int> numbers;
        transform(s.begin(), s.end(), back_inserter(numbers),
                  std::bind2nd(std::minus<char>(), '0'));
        // output
        copy(numbers.begin(), numbers.end(),
             std::ostream_iterator<int>(std::cout, " "));
        std::cout << '\n';
}

When reading from a file, you could read the string and transform(), or even transform() directly from istream iterators, if there is nothing else in that file besides your number:

    std::ifstream f("test.txt");
    std::vector<int> numbers;
    transform(std::istream_iterator<char>(f),
              std::istream_iterator<char>(),
              back_inserter(numbers),
              std::bind2nd(std::minus<char>(), '0'));

回复收藏 0 原文

零時差 2024-10-03 13:53:55

在我的脑海中，这应该填充一个字符数组，然后您可以对其进行迭代。我意识到这并不完全是您想要的，但这是我的首选方法。

void readfile(char *string)
{
    ifstream NumberFile; 
    NumberFile.open("./Number"); //For a unix file-system
    NumberFile >> string;
    NumberFile.close();
}

另外，要对实际数字执行操作，您可以使用：

CharacterArray[ElementNumber] - '0'

并在数字足够小以适合数据类型时获取数字，您将数组的每个元素乘以 10 的索引次方相加。

Off the top of my head this should fill up a character array which you can then iterate through. I realize it's not exactly what you were after but it's my preferred method.

void readfile(char *string)
{
    ifstream NumberFile; 
    NumberFile.open("./Number"); //For a unix file-system
    NumberFile >> string;
    NumberFile.close();
}

Also, to perform operations on the actual numbers you can use: