在C ++中打印浮点数的二进制表示 [英] Print binary representation of a float number in C++

问题描述

可能重复：

在C ++中浮动到二进制

在C ++中的浮点数的二进制表示。不太实用，只是出于好奇。

I want to print out the binary representation of a float number in C++. Not very practical, just out of curiosity.

以下程序不编译。 reinterpret_cast失败。我可以使用什么样的演员，以便我可以做&（1<

The following program doesn't compile though. The reinterpret_cast fails. What kind of cast can I use so that I can do the " &(1 << i) " part?

#include <iostream>
using namespace std;


void toBinary(float num) {
  int numi = reinterpret_cast<int>(num);
  cout << num << " " << numi << endl;
  for (int i = 0; i < 8 * sizeof(num); i++){
    if (numi & (1<<i)) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
  cout << endl << endl;
}

int main() {
  float a;
  cout << sizeof(int) << " " << sizeof(float) << endl;
  a = 13.5;
  toBinary(a);
  toBinary(13.9);
  toBinary(2 * a);
  toBinary(-a);
}

推荐答案

取指向浮点的指针，并将其重新指向char的指针。现在循环通过sizeof（float）并将每个字符转换为8个二进制数字。此方法也适用于双打。

There's a much easier way. Take a pointer to the float, and reinterpret_cast it to a pointer to char. Now loop through sizeof(float) and convert each char to 8 binary digits. This method works for doubles too.

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