我有一个解密base64的方法，我需要将其转换为c＃，我尝试使用C＃的Convert.FromBase64String以及我在这里看到的其他一些技巧，但最终的字符串不匹配在c ++上的结果。
有人可以帮忙理解C ++代码的这种和平吗？
示例输入：lWXtYpNpwBBXoLmcuktDqg==
预期输出：AC0105E92ED496ACD373D7496B76C1C8
使用Convert.FromBase64的输出：95-65-ED-62-93-69-C0-10-57-A0-B9-9C-BA-4B-43-AA
static BYTE BackFrom64[]={
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0x3e,0xff,0xff,0xff,0x3f,
0x34,0x35,0x36,0x37,0x38,0x39,0x3a,0x3b,
0x3c,0x3d,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0x00,0x01,0x02,0x03,0x04,0x05,0x06,
0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,
0x0f,0x10,0x11,0x12,0x13,0x14,0x15,0x16,
0x17,0x18,0x19,0xff,0xff,0xff,0xff,0xff,
0xff,0x1a,0x1b,0x1c,0x1d,0x1e,0x1f,0x20,
0x21,0x22,0x23,0x24,0x25,0x26,0x27,0x28,
0x29,0x2a,0x2b,0x2c,0x2d,0x2e,0x2f,0x30,
0x31,0x32,0x33,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff
};
static BYTE * TabBase64=(BYTE *)"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
int FromBase64(BYTE *Output,BYTE *Input,int BufSize,int MaxSize) {
int xut,txut;
int i,t,k,shift;
BYTE c;
dWORD dr,dr1;
xut=txut=t=0;
dr=0L;
for(i=0; i < BufSize; ++i) {
c=*Input++;
if(c == '=') {
break;
}
if((c=BackFrom64[c]) == 0xff) {
continue;
}
dr1=(dWORD)c;
shift=(32-(6*(t+1)));
dr1<<=shift;
dr|=dr1;
t++;
if(t == 4) {
t=0;
for(k=0; k < 3; ++k) {
dr1=dr;
dr1>>=(32-(8*(k+1)));
dr1&=0xff;
*Output++=(BYTE) dr1;
if(++txut >= MaxSize) {
return(-1);
}
}
dr=0;
}
}
if(t > 0) {
for(k=0; k < (t-1); ++k) {
dr1=dr;
dr1>>=(32-(8*(k+1)));
dr1&=0xff;
*Output++=(BYTE) dr1;
if(++txut >= MaxSize) {
return(-1);
}
}
}
return(txut);
}
答案 0 :(得分：1)
只需很少的工作即可完成非常幼稚的转换。
唯一的其他更改是将未提供的BackFrom64数组计算为TabBase64数组的逆数组，并为提供的输入和输出添加单元测试。如果它执行的不是标准算法，则我会整理计数器，其中一些是多余的（可能i和iInput都可以用foreach(var c in input)
using System;
using Microsoft.VisualStudio.TestTools.UnitTesting;
namespace SO
{
[TestClass]
public class UnitTest1
{
private const string TabBase64 = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
private static readonly byte[] BackFrom64 = CreateBackFrom64();
private static byte[] CreateBackFrom64()
{
var backFrom64 = new byte[128];
for (var i = 0; i < backFrom64.Length; ++i)
backFrom64[i] = 0xff;
for (byte i = 0; i < TabBase64.Length; i++)
{
var c = TabBase64[i];
backFrom64[c] = i;
}
return backFrom64;
}
int FromBase64(byte[] Output, string Input, int BufSize, int MaxSize)
{
int xut, txut;
int i, t, k, shift;
byte c;
int dr, dr1;
xut = txut = t = 0;
dr = 0;
var iInput = 0;
var iOutput = 0;
for (i = 0; i < BufSize; ++i)
{
c = (byte)Input[iInput++];
if (c == '=')
{
break;
}
if ((c = BackFrom64[c]) == 0xff)
{
continue;
}
dr1 = c;
shift = (32 - (6 * (t + 1)));
dr1 <<= shift;
dr |= dr1;
t++;
if (t == 4)
{
t = 0;
for (k = 0; k < 3; ++k)
{
dr1 = dr;
dr1 >>= (32 - (8 * (k + 1)));
dr1 &= 0xff;
Output[iOutput++] = (byte)dr1;
if (++txut >= MaxSize)
{
return (-1);
}
}
dr = 0;
}
}
if (t > 0)
{
for (k = 0; k < (t - 1); ++k)
{
dr1 = dr;
dr1 >>= (32 - (8 * (k + 1)));
dr1 &= 0xff;
Output[iOutput++] = (byte)dr1;
if (++txut >= MaxSize)
{
return (-1);
}
}
}
return (txut);
}
[TestMethod]
public void TestMethod1()
{
var input = @"lWXtYpNpwBBXoLmcuktDqg==";
var expectedOutput = "AC0105E92ED496ACD373D7496B76C1C8";
var converted = new byte[(input.Length * 2) / 3];
var result = FromBase64(converted, input, input.Length, int.MaxValue);
Assert.AreNotEqual(-1, result, "result is valid");
Assert.AreEqual(
BitConverter.ToString(converted).Replace("-",""),
BitConverter.ToString(Convert.FromBase64String(input)).Replace("-",""),
"Supplied algorithm matches standard algorithm" );
Assert.AreEqual(
expectedOutput,
BitConverter.ToString(converted).Replace("-",""),
"Supplied algorithm matches expected output" );
}
}
}
现在，第一个断言通过了，因为移植的代码返回的结果与Convert.FromBase64String完全相同，但是第二个断言失败了，因为这不是您说的从C ++获得的输出。
由此我可以得出结论，造成差异的最可能原因是在调用C ++版本时，BackFrom64的内容尚未正确初始化，并且包含垃圾值-{{1}的值}在编辑中提供给OP的结果与计算得出的结果相同，因此预期结果会发生其他变化。编译C ++版本并使用给定的输入运行：
BackFrom64生成输出
int main()
{
const char* input = "lWXtYpNpwBBXoLmcuktDqg==";
BYTE Input[100];
BYTE Output[100];
int length = strlen(input);
memcpy(Input, input, length);
int result = FromBase64(Output, Input, 100, length);
for (int i = 0; i < result; ++i)
std::cout << std::hex << (int)Output[i] << " ";
std::cout << "\n";
}
再次表明它与标准算法相同。