我正在创建一个函数来返回十进制和整数位数,并使用sstreams将插入的typename号转换为字符串.

但是,转换为字符串时的数字用科学记数法表示,这对于计算正常数字中的数字位数没有用.如何在下面的函数中阻止这种情况发生？

enum { DECIMALS = 10, WHOLE_NUMBS = 20, ALL = 30 };

template < typename T > int Numbs_Digits(T numb, int scope)
{
    stringstream ss(stringstream::in | stringstream::out);
    stringstream ss2(stringstream::in | stringstream::out);
    unsigned long int length = 0;
    unsigned long int numb_wholes;

    ss2 << (int) numb;
    numb_wholes = ss2.str().length();
    ss2.flush();
    bool all = false;

    switch (scope) {
    case ALL:
        all = true;

    case DECIMALS:
        ss << numb;
        length += ss.str().length() - (numb_wholes + 1);  // +1 for the "."
        if (all != true)
            break;

    case WHOLE_NUMBS:
        length += numb_wholes;
        if (all != true)
            break;

    default:
        break;
    }
    return length;
}

解决方法:

使用std::fixed流操纵器：

ss << fixed << numb;

–

例,

#include <iostream>
using namespace std;

int main () {
  double a,b,c;
  a = 3.1415926534;
  b = 2006.0;
  c = 1.0e-10;
  cout.precision(5);
  cout       <<         a << '\t' << b << '\t' << c << endl;
  cout <<   fixed    << a << '\t' << b << '\t' << c << endl;
  cout << scientific << a << '\t' << b << '\t' << c << endl;
  return 0;
}

输出：

3.1416          2006            1e-010
3.14159         2006.00000      0.00000
3.14159e+000    2.00600e+003    1.00000e-010

示例取自here.

并且您可以使用std :: stringstream而不是cout,但结果将是相同的.在这里试验：

http://www.ideone.com/HUrRw