在C ++中读取格式化输入的最简单方法是什么? [英] The easiest way to read formatted input in C++?
问题描述
有任何方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
.
让我们假设我有三个stringz 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因此我不想使用 substring()
或任何类似的东西。
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.
这样的参数
":#####..+####..=SSS.."
,因此它直接知道发生了什么事。
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()
:
Or, using 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 << "\n";
std::cout << "Y=" << Y << "\n";
std::cout << "Z=" << Z << "\n";
}
return 0;
}
但是,这里的限制是字符串的最大长度c $ c> Z )必须在解析输入之前决定。
However, the limitiation here is that the maximum length of the string (Z
) must be decided before parsing the input.
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