我正在从小端到大端转换回小检查 是对还是请帮帮我
#include <stdio.h>
#include <stdlib.h>
#include <netinet/in.h>
double dx = 122992001003000;
//A method little to big endian
double bEndian(double d) {
int *p0 = (int *) &d;
int *p1 = p0 + 1;
printf("1: %x %x\n", *p0, *p1);
int tmp = *p1;
*p1 = htonl(*p0);
*p0 = htonl(tmp);
printf("2: %x %x\n", *p0, *p1);
return *(double *)p0;
}
//A method big to little endian
double lEndian(double d) {
int *p0 = (int *) &d;
int *p1 = p0 + 1;
printf("3: %x %x\n", *p0, *p1);
int tmp = *p1;
*p1 = ntohl(*p0);
*p0 = ntohl(tmp);
printf("4: %x %x\n", *p0, *p1);
return *(double *)p0;
}
int main (int argc, char *argv[])
{
double d1 = bEndian(dx);
double d2 = lEndian(d1);
printf("5: %.0f\n", d2);
if ((*(double*) &d2) == dx)
printf("6: %.0lf\n", *(double*) &d2);
else
printf("7:%d\n", d2);
return 0;
}
有一段时间它给我的结果122992001003129为什么? 还使用ntohl或htonl没有区别?它只适用于惯例 即网络主机和主机网络
答案 0 :(得分:4)
这是一个会改变字节序的函数。
#include <algorithm>
#include <cstdio>
template <class T>
T change_endian(T in)
{
char* const p = reinterpret_cast<char*>(&in);
for (size_t i = 0; i < sizeof(T) / 2; ++i)
std::swap(p[i], p[sizeof(T) - i - 1]);
return in;
}
int main()
{
double d = 122992001003000;
printf("%f\n", d);
d = change_endian(d);
printf("%f\n", d);
d = change_endian(d);
printf("%f\n", d);
}
输出:
122992001003000.000000
0.000000
122992001003000.000000
答案 1 :(得分:1)
简单:
#include <endian.h>
#include <stdint.h>
double src_num = YOUR_VALUE;
int64_t tmp_num = htobe64(le64toh(*(int64_t*)&src_num));
double dst_num = *(double*)&tmp_num;
但是,有些怪癖:https://docs.microsoft.com/en-us/azure/aks/virtual-nodes-portal#known-limitations
答案 2 :(得分:0)
typedef union
{
double d;
unsigned char s[8];
} Union_t;
// reverse little to big endian or vice versa as per requirement
double reverse_endian(double in)
{
int i, j;
unsigned char t;
Union_t val;
val.d = in;
// swap MSB with LSB etc.
for(i=0, j=7; i < j; i++, j--)
{
t = val.s[i];
val.s[i] = val.s[j];
val.s[j] = t;
}
return val.d;
}
答案 3 :(得分:-2)
四个字节的简单反转应该有效
double ReverseDouble( const double indouble )
{
double retVal;
char *doubleToConvert = ( char* ) & indouble;
char *returndouble = ( char* ) & retVal;
// swap the bytes into a temporary buffer
returndouble[0] = doubleToConvert[3];
returndouble[1] = doubleToConvert[2];
returndouble[2] = doubleToConvert[1];
returndouble[3] = doubleToConvert[0];
return retVal;
}