#include <iostream>
using namespace std;
template<class T>
void toBinary(T num)
{
char * numi = reinterpret_cast<char*>(&num);
for (int i = 1; i <= sizeof(T); i++)
{
for( int j = 1 ; j <= 8; ++j )
{
char byte = numi[i];
cout << ( byte & j ? 1 : 0);
}
}
cout << endl << endl;
}
int main()
{
toBinary(1);
std::cin.get();
}
输出是0000000000000 ...
你能告诉我我的错误在哪里吗?
编辑:
#include <iostream>
#include <bitset>
#include <iomanip>
#include <boost/format.hpp>
using namespace std;
template<class T> bitset<sizeof(T)*CHAR_BIT> toBinary(const T num)
{
bitset<sizeof(T)*CHAR_BIT> mybits;
const char * const p = reinterpret_cast<const char*>(&num);
for (int i = sizeof(T)*CHAR_BIT-1 ; i >= 0 ; --i)
mybits.set(i, (*(p)&(1<<i)));
return mybits;
}
template<class T> void printBinary(T num, ostream& stream = cout)
{
stream << boost::format("%-35s %-8s %-32s\n") % typeid(T).name() % num % toBinary(num).to_string();
}
struct Foo{void bar(){}};
int main()
{
printBinary(-8);
printBinary(8u);
printBinary('a');
printBinary(8.2f);
printBinary("Overflow");
printBinary(main);
printBinary(&Foo::bar);
printBinary(8.2);
std::cin.get();
}
最佳答案
我想,如果我真的想按原样修复此代码,我会这样做:
#include <iostream>
#include <string>
using namespace std;
template<class T>
void toBinary(const T& num)
{
const char *const asbytes = reinterpret_cast<const char* const>(&num);
for (const char* byte=asbytes + sizeof(T) - 1; byte>=asbytes; byte--)
{
for ( int bitnr = 7; bitnr>=0; bitnr-- )
{
cout << ( (*byte & (1<<bitnr)) ? 1 : 0);
}
}
cout << endl << endl;
}
int main()
{
toBinary(1);
std::cin.get();
}
关于c++ - 模板类型转换成二进制表示形式,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/5653958/