在 C++ 中读取格式化输入的最简单方法?

The easiest way to read formatted input in C++?(在 C++ 中读取格式化输入的最简单方法?)

问题描述

有没有办法读取这样的格式化字符串，例如:48754+7812=Abcs.

Is there any way to read a formatted string like this, for example :48754+7812=Abcs.

假设我有三个字符串 X、Y 和 Z，我想要

Let's say I have three stringz X,Y and Z, and I want

X = 48754 
Y = 7812
Z = Abcs

两个数字的大小和字符串的长度可能会有所不同，所以我不想使用 substring() 或类似的东西.

The size of the two numbers and the length of the string may vary, so I dont want to use substring() or anything like that.

是否可以给C++这样的参数

Is it possible to give C++ a parameter like this

":#####..+####..=SSS.."

所以它直接知道发生了什么?

so it knows directly what's going on?

推荐答案

一种可能性是 boost::split()，它允许指定多个分隔符并且不需要输入大小的先验知识:

A possibility is boost::split(), which allows the specification of multiple delimiters and does not require prior knowledge of the size of the input:

#include <iostream>
#include <vector>
#include <string>

#include <boost/algorithm/string.hpp>
#include <boost/algorithm/string/split.hpp>

int main()
{
    std::vector<std::string> tokens;
    std::string s(":48754+7812=Abcs");
    boost::split(tokens, s, boost::is_any_of(":+="));

    // "48754" == tokens[0]
    // "7812"  == tokens[1]
    // "Abcs"  == tokens[2]

    return 0;
}

或者，使用sscanf():

#include <iostream>
#include <cstdio>

int main()
{
    const char* s = ":48754+7812=Abcs";
    int X, Y;
    char Z[100];

    if (3 == std::sscanf(s, ":%d+%d=%99s", &X, &Y, Z))
    {
        std::cout << "X=" << X << "
";
        std::cout << "Y=" << Y << "
";
        std::cout << "Z=" << Z << "
";
    }

    return 0;
}

然而，这里的限制是字符串的最大长度 (Z) 必须在解析输入之前确定.

However, the limitiation here is that the maximum length of the string (Z) must be decided before parsing the input.

这篇关于在 C++ 中读取格式化输入的最简单方法?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！