Question

假设我有

int a; cin >> a;   
int b; cin >> b;
int c; cin >> c;

我想知道如何从两个输入中得到相同的结果：

1.2.3和1 2 3

如果我输入1.2.3

，则a为1，b为2，c为3

Answer 1

您可以使用std::scanf：

执行此操作

if (std::scanf("%d%*1[ .]%d%*1[ .]%d", &a, &b, &c) == 3) {
    ... // Input was successful
}

棘手的部分是mat说明符：%*1[ .]它扫描并忽略单个空格或单个点。

Demo.

Answer 2

这篇文章被标记为C ++。

这是一些需要考虑的C ++：

#include <iostream>
#include <iomanip>
#include <sstream>
#include <string>
#include <cassert>

// forward 

// stream input, extract and discard junk char
void trial1(std::string s);     

// stream input, ignore() junk char,
//   with error check for each integer input
void trial2(std::string s); 

// sscanf of string
void trial3(std::string s); 



int main(int , char** )
{
   std::cout << "\n";  // stream input, extract and discard junk char
   trial1("1.2.3");
   trial1("1,2,3");
   trial1("1 2 3");    // fail


   std::cout << "\n";  // stream input, ignore junk char, error checks
   trial2("1.2.3");
   trial2("1,2,3");
   trial2("1 2 3");
   trial2("11.12.13");
   trial2("111,222,333");
   trial2("1234 1234 1234");
   trial2("1234 -12345678 1234");
   trial2("1x2y3");
   trial2("1xx2yy3");    // fail - ignore skips 1 char, not 2


   std::cout << "\n"; // sscanf of std::string
   trial3("1.2.3");
   trial3("1 2 3");
   trial3("1,2,3");
   trial3("111,222,333");
   trial3("1234 -12345678 1234");
   trial3("1234,1234,1234");
   trial3("1xx2yy3");    // fail 


   std::cout << "\n";  // requirements?
   trial1("foo bar");
   trial2("foo bar");
   trial3("foo bar");
}


void trial1(std::stringstream& sin)
{
   char junk;
   int a = -1;  sin >> a >> junk;
   int b = -1;  sin >> b >> junk;
   int c = -1;  sin >> c >> junk;

   std::cout << " -->  " << a << " " << b << " " << c;
}
void trial1(std::string s)
{
   std::cout << '\n' << __FUNCTION__ << "  '" << s << "'";
   std::stringstream sin (s);
   trial1(sin);
}


void trial2(std::stringstream& sin)
{
   int a = -1;
   int b = -1;
   int c = -1;

   do{
      sin >> a;
      if(!sin.good() && !sin.eof()) // check for error
         std::cerr << "  error on a " << std::flush;

      sin.ignore();   // ignore 1 char

      sin >> b;
      if(!sin.good() && !sin.eof()) // check for error
         std::cerr << "  error on b " << std::flush;

      sin.ignore();

      sin >> c;
      if(!sin.good() && !sin.eof()) // check for error
         std::cerr << "  error on c " << std::flush;

   }while(0);

   std::cout << " -->  " << a << " " << b << " " << c;
}
void trial2(std::string s)
{
   std::cout << '\n' << __FUNCTION__ << "  '" << s << "'";
   std::stringstream sin (s);
   trial2(sin);
}



void trial3(std::string s)
{
   std::cout << '\n' << __FUNCTION__ << "  '" << s << "'";
   int a = -1;
   int b = -1;
   int c = -1;    // note added ','----v---------v
   if (std::sscanf(s.c_str(), "%d%*1[ .,]%d%*1[ .,]%d", &a, &b, &c) == 3)
      std::cout << " -->  " << a << " " << b << " " << c;
   else
      std::cout << " -->  " << a << " " << b << " " << c << "         FAILED." << std::flush;
}

带输出：

trial1  '1.2.3' -->  1 2 3
trial1  '1,2,3' -->  1 2 3
trial1  '1 2 3' -->  1 3 -1

trial2  '1.2.3' -->  1 2 3
trial2  '1,2,3' -->  1 2 3
trial2  '1 2 3' -->  1 2 3
trial2  '11.12.13' -->  11 12 13
trial2  '111,222,333' -->  111 222 333
trial2  '1234 1234 1234' -->  1234 1234 1234
trial2  '1234 -12345678 1234' -->  1234 -12345678 1234
trial2  '1x2y3' -->  1 2 3
trial2  '1xx2yy3'  error on b   error on c  -->  1 0 -1

trial3  '1.2.3' -->  1 2 3
trial3  '1 2 3' -->  1 2 3
trial3  '1,2,3' -->  1 2 3
trial3  '111,222,333' -->  111 222 333
trial3  '1234 -12345678 1234' -->  1234 -12345678 1234
trial3  '1234,1234,1234' -->  1234 1234 1234
trial3  '1xx2yy3' -->  1 -1 -1         FAILED.

trial1  'foo bar' -->  0 -1 -1
trial2  'foo bar'  error on a   error on b   error on c  -->  0 -1 -1
trial3  'foo bar' -->  -1 -1 -1         FAILED.

Answer 3

您可以使用getline（）获取整个输入，使用strtok（）来标记字符串。您可能需要验证输入等。以下是一个示例。

#include <iostream>
#include <string.h>
using namespace std;

void DoSomething(char* inputChar){
    printf("%s\n", inputChar);
}

int main() {
    char myInput[256];
    char* pch;
    char* delimiters = " .";

    //Loop Through Input
    while (cin.getline(myInput,256)) {
        cout << "-" << myInput << endl;
        pch = strtok(myInput, delimiters);
        while (pch != NULL){
            //Individual input chars here
            DoSomething(pch);
            //Set NULL to find next delimited set of char
            pch = strtok(NULL, delimiters);
        }
    }
    return 0;
}

Try it now!

您可以修改分隔符列表以及是否使用输入存储或执行其他操作。如果你只需要使用一次，显然getline（）可以从外部while（）中取出。 atoi（）也可以尝试将char *转换为int。

我使用了输入

1 2 3
1.2.3
1

Answer 4

您可以将文字作为字符串阅读，替换＆＃39;。＆＃39;使用空格，然后使用std::istringstream转换为数字：

std::string text_read;
std::getline(cin, text_read);
std::string::size_type position = text_read.find('.');
while (position != std::string::npos)
{
  text_read[position] = ' ';
  position = text_read.find('.');
}
int a;
int b;
int c;
std::istringstream text_stream(text_read);
text_stream >> a;
text_stream >> b;
text_stream >> c;

这是一种蛮力技术。您可以使用std::transform进行优化。

C ++点作为空格？

4 个答案: