在数字中的特定数字处查找数字

Finding a number at a particular digit in a number

我想获取特定索引处的号码。

假设号码是“123456789”,我需要return左边第3位的号码。

我是否应该将此数字转换为字符串并将特定索引处的字符重新转换为整数以获得数字值?

在 C++ 中是否有任何内置函数可以这样做?

这应该是return数字的第三位!

cout << "Enter an integer";
int number;
cin >> number;
int n = number / 100 % 10

或(对于所有数字):

 int number = 12345;
 int numSize = 5;
for (int i=numSize-1; i>=0; i--) {
  int y = pow(10, i);
  int z = number/y;
  int x2 = number / (y * 10);
  printf("%d-",z - x2*10 );
}

要获取数字中任意位置的数字,您可以使用简单的字符串转换:

int foo = 123456789;
int pos = 3;
// convert to string and use the [] operator
std::cout << std::to_string(foo)[pos - 1];//need to subtract 1 as C++ is 0 index based

使用std::stringstream

std::string showMSD3(int foo)
{
   std::stringstream ssOut;

   std::string sign;
   std::stringstream ss1;
   if(foo < 0) {
      ss1 << (-1 * foo);
      sign = '-';
   }
   else
      ss1 << foo;

   std::string s = ss1.str();

   ssOut << "               foo string: '"
         << sign << s << "'" << std::endl;

   if(s.size() > 2)
   {
      ssOut << " Most Significant Digit 3: "
            << s[2] // 1st digit at offsset 0
            << "\n            has hex value: 0x"
            << std::setw(2) << std::setfill('0') << std::hex
            << (s[2] - '0')
            << std::endl;
   }
   else
      ssOut << "                      Err: foo has only "
            << s.size() << " digits" << std::endl;

   return (ssOut.str());
}

以及靠近主干道的某个地方(可能在主干道中):

   // test 1
   {
      std::cout << showMSD3(123456789) << std::endl;
   }

   // test 2
   {
      std::cout << showMSD3(12) << std::endl;
   }

   // test 3 - handle negative integer
   {
      std::cout << showMSD3(-123) << std::endl;
   }

有输出

               foo string: '123456789'
 Most Significant Digit 3: 3
            has hex value: 0x03

               foo string: '12'
                      Err: foo has only 2 digits

               foo string: '-123'
 Most Significant Digit 3: 3
            has hex value: 0x03