问题描述
我正在尝试改变一个位,例如:
Double x: -1.500912597
这是:
二进制: 10111111 11111000 00000011 10111100 11101101 01100100 01001111 10010011
更改底层二进制代码中的一位(例如,位16),以便:
二进制: 10111111 11111001 00000011 10111100 11101101 01100100 01001111 10010011
Double x: -1.563412596999999903
是否有一些C ++代码可以用于此?
方式是使用 memcpy
(是的,我知道你在想什么,而不是没有效率)。
请注意,此解决方案不考虑字节顺序。您也需要适应这种情况,以免严格便携。
#include< cstdlib>
#include< cstring>
#include< utility>
#include< iostream>
//只编译此函数,如果Integer的大小与double
模板相同大小< class Integer,std :: enable_if_t< sizeof(Integer)== sizeof(double)> * = nullptr>
double or_bits(double input,Integer bits)
{
整数拷贝;
//将double转换为整数
std :: memcpy(& copy& input,sizeof(input))的一个位表示;
//执行按位操作
copy | = bits;
//将这些位转换回一个double
std :: memcpy(& input,& copy,sizeof(copy));
//返回double
返回输入;
}
int main()
{
double d = 1.0;
d = or_bits(d,0x10ul);
std :: cout<<< d<的std :: ENDL;
}
gcc5.3上的汇编输出:
double or_bits< unsigned long,(void *)0>(double,unsigned long):
pre>
movq%xmm0,%rax
orq% rdi,%rax
movq%rax,-8(%rsp)
movsd -8(%rsp),%xmm0
ret
I am trying to change one bit in a double so that for example:
Double x:
-1.500912597
which is:Binary:
10111111 11111000 00000011 10111100 11101101 01100100 01001111 10010011
Change one bit in the underlying binary code (for example, bit 16) so that:
Binary:
10111111 11111001 00000011 10111100 11101101 01100100 01001111 10010011
Double x:
-1.563412596999999903
Is there some C++ code I can use for this?
解决方案The only portable way is to use
memcpy
(yes, I know what you're thinking, and no it's not inefficient).Note that this solution does not take into account byte-ordering. You'd need to cater for that too to be strictly portable.
#include <cstdlib> #include <cstring> #include <utility> #include <iostream> // only compile this function if Integer is the same size as a double template<class Integer, std::enable_if_t<sizeof(Integer) == sizeof(double)>* = nullptr> double or_bits(double input, Integer bits) { Integer copy; // convert the double to a bit representation in the integer std::memcpy(©, &input, sizeof(input)); // perform the bitwise op copy |= bits; // convert the bits back to a double std::memcpy(&input, ©, sizeof(copy)); // return the double return input; } int main() { double d = 1.0; d = or_bits(d, 0x10ul); std::cout << d << std::endl; }
assembly output on gcc5.3:
double or_bits<unsigned long, (void*)0>(double, unsigned long): movq %xmm0, %rax orq %rdi, %rax movq %rax, -8(%rsp) movsd -8(%rsp), %xmm0 ret
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