Question

我想知道如何从C ++中的字符串中获取数字。我从输入中获取字符串并将获得多行，但我已经有了读取行的工作。注意：这些行总是具有偶数个整数

以下是我希望代码看起来像：

std::getline(std::cin, line);// line looks something like "10 3 40 45 8 12"
int a, b;
while(!line.empty() /*line still has ints to extract*/) {
    a = someMethod(line);//gets first int.  So, 10 first loop, 40 second, 8 third
    b = someMethod(line);//gets second int. So, 3 first loop, 45 second, 12 third
    myMethod(a,b);//a method from elsewhere in my code.  It's here so you know that I need both a and b
}

任何类似的东西都会有所帮助。非常感谢你！

Answer 1

这是一个完整的例子。

#include <sstream>
#include <string>
#include <iostream>


int main(){
    std::string line = "2 4 56 6";
    std::stringstream stream(line);
    int i;
    while (stream >> i) {
        std::cout << i << std::endl;
    }
}

以下也可以正常工作，因此阅读多行不应该是个问题。

#include <sstream>
#include <string>
#include <iostream>


int main(){
    std::string line = "2 4 56 6";
    std::stringstream stream(line);
    int i;
    while (stream >> i) {
        std::cout << i << std::endl;
    }

    line =  "32 62 44 6 22 58 34 60 71 86";
    stream.clear();
    stream.str(line);
    int a,b;
    while(stream >> a && stream >> b){ 
        std::cout << a << " " << b << "\n";
    }
}

Answer 2

你可以这样做：

string line;
getline(std::cin, line);// line looks something like "10 3 40 45 8 12"
int a,  b;

vector<string> tokens;
istringstream iss(line);

copy(istream_iterator<string>(iss), 
     istream_iterator<string>(),
     back_inserter<vector<string> >(tokens));

stringstream s;
for (int i=0;i<tokens.size()/2;i++)
{   
  s<< tokens[i];
  s>>a;
  s.clear();
  s<<tokens[i+2];
  s>>b;
  s.clear();
  myMethod(a,b);
}

Answer 3

从字符串行获取标记并根据需要使用它们。

#include <iostream>
#include <string>
#include <boost/tokenizer.hpp>

using namespace std;

typedef boost::tokenizer<boost::char_separator<char> >
    tokenizer;


void myMethod(int a, int b)
{
    cout<<a<<" "<<b<<endl;
}

void getNumber(string line)
{            
    boost::char_separator<char> sep(" ");
    tokenizer tokens(line, sep);    
    string evenStr, oddStr;

    for(tokenizer::iterator iterToken=tokens.begin();
          iterToken != tokens.end(); ++iterToken)
    {
        evenStr = *iterToken;                       
        ++iterToken;
        if(iterToken != tokens.end())
        {
            oddStr = *iterToken;                                   
        myMethod(atoi(evenStr.c_str()), atoi(oddStr.c_str()));
        }        
    }    
}

int main()
{
   string line("10 3 40 45 8 12");
   getNumber(line);

   return 0;
}

Answer 4

有多种方法可以实现这一目标。我更喜欢使用boost。

示例：

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

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

int main()
{
    std::string line = "10 3 40 45 8 12 9"; //std::getline(std::cin, line);// line looks something like "10 3 40 45 8 12"

    std::vector<std::string> intNumbers;
    std::string s;
    boost::split(intNumbers, line, boost::is_any_of(" "), boost::token_compress_on);

    unsigned int i=0;
    while(i < intNumbers.size())
    {
        try{
            int a = boost::lexical_cast<int>(intNumbers[i++]);

            if(i >= intNumbers.size())
            {
                std::cout << "invlaid values" << std::endl;
                break;
            }
            int b = boost::lexical_cast<int>(intNumbers[i++]);

            std::cout << "value a: " << a << std::endl;
            std::cout << "value b: " << b << std::endl;

            std::cout << "my method (multiply) a*b: " << (a*b) << std::endl;
        }
        catch(boost::bad_lexical_cast &e)
        {
            std::cout << "invlaid values" << std::endl;
        }
    }
}

从字符串C ++中取出整数

4 个答案: