Mixing
TripleDES encrypting and decrypting gives strange results
I have a working implementation of TripleDESCng (tested against some test vectors), but the following happens:
When I encrypt plain text This is a sample message (24 bytes, thus for this it would be 3 blocks) (hex for it is 5468697320697320612073616D706C65206D657373616765) with an example key, I get E81F113DD7C5D965E082F3D42EC1E2CA39BCDBCCBC0A2BD9. However, when I decrypt this with the same example key, I get 5468697320697320612073616D706C650000000000000000, which, when converted back to ASCII, is:
This is a sample.
Any reason other than my code why this would behave this way? To encrypt and decrypt, I use 24 byte keys (ECB mode).
EDIT:
using (var tripleDES = new TripleDESCryptoServiceProvider())
{
byte data = ASCIIEncoding.ASCII.GetBytes("This is a sample message");
Console.WriteLine(BitConverter.ToString(data));
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var encryptor = tripleDES.CreateEncryptor();
byte result = new byte[data.Length];
encryptor.TransformBlock(data, 0, data.Length, result, 0);
var decryptor = tripleDES.CreateDecryptor();
byte result2 = new byte[result.Length];
decryptor.TransformBlock(result, 0, result.Length, result2, 0);
Console.WriteLine(BitConverter.ToString(result2));
}
Console.ReadLine();
c# .net cryptography
asked Jul 31 '18 at 12:04
karolyzzkarolyzz
13815
add a comment |
I have a working implementation of TripleDESCng (tested against some test vectors), but the following happens:
When I encrypt plain text This is a sample message (24 bytes, thus for this it would be 3 blocks) (hex for it is 5468697320697320612073616D706C65206D657373616765) with an example key, I get E81F113DD7C5D965E082F3D42EC1E2CA39BCDBCCBC0A2BD9. However, when I decrypt this with the same example key, I get 5468697320697320612073616D706C650000000000000000, which, when converted back to ASCII, is:
This is a sample.
Any reason other than my code why this would behave this way? To encrypt and decrypt, I use 24 byte keys (ECB mode).
EDIT:
using (var tripleDES = new TripleDESCryptoServiceProvider())
{
byte data = ASCIIEncoding.ASCII.GetBytes("This is a sample message");
Console.WriteLine(BitConverter.ToString(data));
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var encryptor = tripleDES.CreateEncryptor();
byte result = new byte[data.Length];
encryptor.TransformBlock(data, 0, data.Length, result, 0);
var decryptor = tripleDES.CreateDecryptor();
byte result2 = new byte[result.Length];
decryptor.TransformBlock(result, 0, result.Length, result2, 0);
Console.WriteLine(BitConverter.ToString(result2));
}
Console.ReadLine();
c# .net cryptography
asked Jul 31 '18 at 12:04
karolyzzkarolyzz
13815
The default mode is CBC, not EBC. Since you don't explicitly set theMode, why do you think this is EBC? Plus it's 2018. Outside of historic interest, why are you writing new code that uses outdated ciphers and outdated modes?
– Damien_The_Unbeliever
Jul 31 '18 at 12:24
@Damien_The_Unbeliever the mode is not a problem, IV is set to 0s either way
– karolyzz
Jul 31 '18 at 12:28
add a comment |
I have a working implementation of TripleDESCng (tested against some test vectors), but the following happens:
When I encrypt plain text This is a sample message (24 bytes, thus for this it would be 3 blocks) (hex for it is 5468697320697320612073616D706C65206D657373616765) with an example key, I get E81F113DD7C5D965E082F3D42EC1E2CA39BCDBCCBC0A2BD9. However, when I decrypt this with the same example key, I get 5468697320697320612073616D706C650000000000000000, which, when converted back to ASCII, is:
This is a sample.
Any reason other than my code why this would behave this way? To encrypt and decrypt, I use 24 byte keys (ECB mode).
EDIT:
using (var tripleDES = new TripleDESCryptoServiceProvider())
{
byte data = ASCIIEncoding.ASCII.GetBytes("This is a sample message");
Console.WriteLine(BitConverter.ToString(data));
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var encryptor = tripleDES.CreateEncryptor();
byte result = new byte[data.Length];
encryptor.TransformBlock(data, 0, data.Length, result, 0);
var decryptor = tripleDES.CreateDecryptor();
byte result2 = new byte[result.Length];
decryptor.TransformBlock(result, 0, result.Length, result2, 0);
Console.WriteLine(BitConverter.ToString(result2));
}
Console.ReadLine();
c# .net cryptography
asked Jul 31 '18 at 12:04
karolyzzkarolyzz
13815
I have a working implementation of TripleDESCng (tested against some test vectors), but the following happens:
When I encrypt plain text This is a sample message (24 bytes, thus for this it would be 3 blocks) (hex for it is 5468697320697320612073616D706C65206D657373616765) with an example key, I get E81F113DD7C5D965E082F3D42EC1E2CA39BCDBCCBC0A2BD9. However, when I decrypt this with the same example key, I get 5468697320697320612073616D706C650000000000000000, which, when converted back to ASCII, is:
This is a sample.
Any reason other than my code why this would behave this way? To encrypt and decrypt, I use 24 byte keys (ECB mode).
EDIT:
using (var tripleDES = new TripleDESCryptoServiceProvider())
{
byte data = ASCIIEncoding.ASCII.GetBytes("This is a sample message");
Console.WriteLine(BitConverter.ToString(data));
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var encryptor = tripleDES.CreateEncryptor();
byte result = new byte[data.Length];
encryptor.TransformBlock(data, 0, data.Length, result, 0);
var decryptor = tripleDES.CreateDecryptor();
byte result2 = new byte[result.Length];
decryptor.TransformBlock(result, 0, result.Length, result2, 0);
Console.WriteLine(BitConverter.ToString(result2));
}
Console.ReadLine();
c# .net cryptography
c# .net cryptography
asked Jul 31 '18 at 12:04
karolyzzkarolyzz
13815
asked Jul 31 '18 at 12:04
karolyzzkarolyzz
13815
edited Jul 31 '18 at 12:23
karolyzz
asked Jul 31 '18 at 12:04
karolyzzkarolyzz
13815
asked Jul 31 '18 at 12:04
karolyzzkarolyzz
13815
asked Jul 31 '18 at 12:04
karolyzzkarolyzz
13815
13815
The default mode is CBC, not EBC. Since you don't explicitly set theMode, why do you think this is EBC? Plus it's 2018. Outside of historic interest, why are you writing new code that uses outdated ciphers and outdated modes?
– Damien_The_Unbeliever
Jul 31 '18 at 12:24
@Damien_The_Unbeliever the mode is not a problem, IV is set to 0s either way
– karolyzz
Jul 31 '18 at 12:28
add a comment |
The default mode is CBC, not EBC. Since you don't explicitly set theMode, why do you think this is EBC? Plus it's 2018. Outside of historic interest, why are you writing new code that uses outdated ciphers and outdated modes?
– Damien_The_Unbeliever
Jul 31 '18 at 12:24
@Damien_The_Unbeliever the mode is not a problem, IV is set to 0s either way
– karolyzz
Jul 31 '18 at 12:28
The default mode is CBC, not EBC. Since you don't explicitly set the
Mode, why do you think this is EBC? Plus it's 2018. Outside of historic interest, why are you writing new code that uses outdated ciphers and outdated modes?– Damien_The_Unbeliever
Jul 31 '18 at 12:24
The default mode is CBC, not EBC. Since you don't explicitly set the
Mode, why do you think this is EBC? Plus it's 2018. Outside of historic interest, why are you writing new code that uses outdated ciphers and outdated modes?– Damien_The_Unbeliever
Jul 31 '18 at 12:24
@Damien_The_Unbeliever the mode is not a problem, IV is set to 0s either way
– karolyzz
Jul 31 '18 at 12:28
@Damien_The_Unbeliever the mode is not a problem, IV is set to 0s either way
– karolyzz
Jul 31 '18 at 12:28
add a comment |
2 Answers
2
active
oldest
votes
With almost all modes1, you should make sure that the final part of your data is pushed through TransformFinalBlock rather than TransformBlock2, to make sure it knows that no more data is coming and to ensure final blocks are flushed/written.
It's bad form, in general, to assume the output size is going to match the input size.
the mode is not a problem, IV is set to 0s either way
Yes, that'll mean that the first block was not affected by your choice of Mode. But all subsequent blocks will be, because they will use the chaining mode and the previous block, not the IV. So if you want ECB (you shouldn't3) you need to explicitly set that mode.
1Your code is using CBC, not EBC as you claim in your narrative. CBC is the default mode for .NET encryption classes.
2And when using this second method, pay attention to it's return value, as mjwills commented.
3You've picked an outdated crypto algorithm, paired it with an outdated mode of operation, and your words I've quoted above mean that you don't understand modes. Added together, I would suggest that you're not well placed to be writing code that uses crypto currently. The .NET classes can make it seem easy to write crypto code but you still have to understand how to make good choices in using them. Better to spend more time on researching these things before writing code.
answered Jul 31 '18 at 12:30
Damien_The_UnbelieverDamien_The_Unbeliever
194k17248335
I was well aware the default mode is CBC. I'm sorry, but your suggestion seems to be based on vague assumptions about my knowledge and the choice of algorithm, which, by the way, I purposefully ignored, since I have my own reasons (don't worry, this code is not supposed to reach production)
– karolyzz
Jul 31 '18 at 12:56
the actual problem was with the final block indeed, I agree I should've studied it a bit more carefully, thanks
– karolyzz
Jul 31 '18 at 12:59
add a comment |
I think that your problem is in the method of the encryptor / decryptor that you are using: the TransformBlock method is conceived to transform a block when you will be encrypting multiple blocks.
That is not the case in your code, where you want to transform a single block, and therefore you should be using the TransformFinalBlock method instead. BTW I took the liberty of making your sample buildable.
using System;
using System.Text;
namespace Tests
{
class Program
{
static void Main(string args)
{
System.Security.Cryptography.TripleDESCryptoServiceProvider tripleDES = new System.Security.Cryptography.TripleDESCryptoServiceProvider();
byte data = Encoding.UTF8.GetBytes("This is a sample message");
byte key = Encoding.UTF8.GetBytes("NOSTROMOHASSOMEGODPOWERS");
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var encryptor = tripleDES.CreateEncryptor();
byte result = new byte[data.Length];
result = encryptor.TransformFinalBlock(data, 0, data.Length);
string res = BitConverter.ToString(result).Replace("-","");
Console.WriteLine(BitConverter.ToString(result).Replace("-",""));
byte data2 = result;
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var decryptor = tripleDES.CreateDecryptor();
byte result2 = new byte[data2.Length];
result2 = decryptor.TransformFinalBlock(data2, 0, data2.Length);
Console.WriteLine(Encoding.UTF8.GetString(result2));
}
}
}
answered Jul 31 '18 at 12:49
xtoikxtoik
65738
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
With almost all modes1, you should make sure that the final part of your data is pushed through TransformFinalBlock rather than TransformBlock2, to make sure it knows that no more data is coming and to ensure final blocks are flushed/written.
It's bad form, in general, to assume the output size is going to match the input size.
the mode is not a problem, IV is set to 0s either way
Yes, that'll mean that the first block was not affected by your choice of Mode. But all subsequent blocks will be, because they will use the chaining mode and the previous block, not the IV. So if you want ECB (you shouldn't3) you need to explicitly set that mode.
1Your code is using CBC, not EBC as you claim in your narrative. CBC is the default mode for .NET encryption classes.
2And when using this second method, pay attention to it's return value, as mjwills commented.
3You've picked an outdated crypto algorithm, paired it with an outdated mode of operation, and your words I've quoted above mean that you don't understand modes. Added together, I would suggest that you're not well placed to be writing code that uses crypto currently. The .NET classes can make it seem easy to write crypto code but you still have to understand how to make good choices in using them. Better to spend more time on researching these things before writing code.
answered Jul 31 '18 at 12:30
Damien_The_UnbelieverDamien_The_Unbeliever
194k17248335
I was well aware the default mode is CBC. I'm sorry, but your suggestion seems to be based on vague assumptions about my knowledge and the choice of algorithm, which, by the way, I purposefully ignored, since I have my own reasons (don't worry, this code is not supposed to reach production)
– karolyzz
Jul 31 '18 at 12:56
the actual problem was with the final block indeed, I agree I should've studied it a bit more carefully, thanks
– karolyzz
Jul 31 '18 at 12:59
add a comment |
With almost all modes1, you should make sure that the final part of your data is pushed through TransformFinalBlock rather than TransformBlock2, to make sure it knows that no more data is coming and to ensure final blocks are flushed/written.
It's bad form, in general, to assume the output size is going to match the input size.
the mode is not a problem, IV is set to 0s either way
Yes, that'll mean that the first block was not affected by your choice of Mode. But all subsequent blocks will be, because they will use the chaining mode and the previous block, not the IV. So if you want ECB (you shouldn't3) you need to explicitly set that mode.
1Your code is using CBC, not EBC as you claim in your narrative. CBC is the default mode for .NET encryption classes.
2And when using this second method, pay attention to it's return value, as mjwills commented.
3You've picked an outdated crypto algorithm, paired it with an outdated mode of operation, and your words I've quoted above mean that you don't understand modes. Added together, I would suggest that you're not well placed to be writing code that uses crypto currently. The .NET classes can make it seem easy to write crypto code but you still have to understand how to make good choices in using them. Better to spend more time on researching these things before writing code.
answered Jul 31 '18 at 12:30
Damien_The_UnbelieverDamien_The_Unbeliever
194k17248335
I was well aware the default mode is CBC. I'm sorry, but your suggestion seems to be based on vague assumptions about my knowledge and the choice of algorithm, which, by the way, I purposefully ignored, since I have my own reasons (don't worry, this code is not supposed to reach production)
– karolyzz
Jul 31 '18 at 12:56
the actual problem was with the final block indeed, I agree I should've studied it a bit more carefully, thanks
– karolyzz
Jul 31 '18 at 12:59
add a comment |
With almost all modes1, you should make sure that the final part of your data is pushed through TransformFinalBlock rather than TransformBlock2, to make sure it knows that no more data is coming and to ensure final blocks are flushed/written.
It's bad form, in general, to assume the output size is going to match the input size.
the mode is not a problem, IV is set to 0s either way
Yes, that'll mean that the first block was not affected by your choice of Mode. But all subsequent blocks will be, because they will use the chaining mode and the previous block, not the IV. So if you want ECB (you shouldn't3) you need to explicitly set that mode.
1Your code is using CBC, not EBC as you claim in your narrative. CBC is the default mode for .NET encryption classes.
2And when using this second method, pay attention to it's return value, as mjwills commented.
3You've picked an outdated crypto algorithm, paired it with an outdated mode of operation, and your words I've quoted above mean that you don't understand modes. Added together, I would suggest that you're not well placed to be writing code that uses crypto currently. The .NET classes can make it seem easy to write crypto code but you still have to understand how to make good choices in using them. Better to spend more time on researching these things before writing code.
answered Jul 31 '18 at 12:30
Damien_The_UnbelieverDamien_The_Unbeliever
194k17248335
With almost all modes1, you should make sure that the final part of your data is pushed through TransformFinalBlock rather than TransformBlock2, to make sure it knows that no more data is coming and to ensure final blocks are flushed/written.
It's bad form, in general, to assume the output size is going to match the input size.
the mode is not a problem, IV is set to 0s either way
Yes, that'll mean that the first block was not affected by your choice of Mode. But all subsequent blocks will be, because they will use the chaining mode and the previous block, not the IV. So if you want ECB (you shouldn't3) you need to explicitly set that mode.
1Your code is using CBC, not EBC as you claim in your narrative. CBC is the default mode for .NET encryption classes.
2And when using this second method, pay attention to it's return value, as mjwills commented.
3You've picked an outdated crypto algorithm, paired it with an outdated mode of operation, and your words I've quoted above mean that you don't understand modes. Added together, I would suggest that you're not well placed to be writing code that uses crypto currently. The .NET classes can make it seem easy to write crypto code but you still have to understand how to make good choices in using them. Better to spend more time on researching these things before writing code.
answered Jul 31 '18 at 12:30
Damien_The_UnbelieverDamien_The_Unbeliever
194k17248335
edited Jul 31 '18 at 12:40
answered Jul 31 '18 at 12:30
Damien_The_UnbelieverDamien_The_Unbeliever
194k17248335
answered Jul 31 '18 at 12:30
Damien_The_UnbelieverDamien_The_Unbeliever
194k17248335
answered Jul 31 '18 at 12:30
Damien_The_UnbelieverDamien_The_Unbeliever
194k17248335
194k17248335
I was well aware the default mode is CBC. I'm sorry, but your suggestion seems to be based on vague assumptions about my knowledge and the choice of algorithm, which, by the way, I purposefully ignored, since I have my own reasons (don't worry, this code is not supposed to reach production)
– karolyzz
Jul 31 '18 at 12:56
the actual problem was with the final block indeed, I agree I should've studied it a bit more carefully, thanks
– karolyzz
Jul 31 '18 at 12:59
add a comment |
I was well aware the default mode is CBC. I'm sorry, but your suggestion seems to be based on vague assumptions about my knowledge and the choice of algorithm, which, by the way, I purposefully ignored, since I have my own reasons (don't worry, this code is not supposed to reach production)
– karolyzz
Jul 31 '18 at 12:56
the actual problem was with the final block indeed, I agree I should've studied it a bit more carefully, thanks
– karolyzz
Jul 31 '18 at 12:59
I was well aware the default mode is CBC. I'm sorry, but your suggestion seems to be based on vague assumptions about my knowledge and the choice of algorithm, which, by the way, I purposefully ignored, since I have my own reasons (don't worry, this code is not supposed to reach production)
– karolyzz
Jul 31 '18 at 12:56
I was well aware the default mode is CBC. I'm sorry, but your suggestion seems to be based on vague assumptions about my knowledge and the choice of algorithm, which, by the way, I purposefully ignored, since I have my own reasons (don't worry, this code is not supposed to reach production)
– karolyzz
Jul 31 '18 at 12:56
the actual problem was with the final block indeed, I agree I should've studied it a bit more carefully, thanks
– karolyzz
Jul 31 '18 at 12:59
the actual problem was with the final block indeed, I agree I should've studied it a bit more carefully, thanks
– karolyzz
Jul 31 '18 at 12:59
add a comment |
I think that your problem is in the method of the encryptor / decryptor that you are using: the TransformBlock method is conceived to transform a block when you will be encrypting multiple blocks.
That is not the case in your code, where you want to transform a single block, and therefore you should be using the TransformFinalBlock method instead. BTW I took the liberty of making your sample buildable.
using System;
using System.Text;
namespace Tests
{
class Program
{
static void Main(string args)
{
System.Security.Cryptography.TripleDESCryptoServiceProvider tripleDES = new System.Security.Cryptography.TripleDESCryptoServiceProvider();
byte data = Encoding.UTF8.GetBytes("This is a sample message");
byte key = Encoding.UTF8.GetBytes("NOSTROMOHASSOMEGODPOWERS");
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var encryptor = tripleDES.CreateEncryptor();
byte result = new byte[data.Length];
result = encryptor.TransformFinalBlock(data, 0, data.Length);
string res = BitConverter.ToString(result).Replace("-","");
Console.WriteLine(BitConverter.ToString(result).Replace("-",""));
byte data2 = result;
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var decryptor = tripleDES.CreateDecryptor();
byte result2 = new byte[data2.Length];
result2 = decryptor.TransformFinalBlock(data2, 0, data2.Length);
Console.WriteLine(Encoding.UTF8.GetString(result2));
}
}
}
answered Jul 31 '18 at 12:49
xtoikxtoik
65738
add a comment |
I think that your problem is in the method of the encryptor / decryptor that you are using: the TransformBlock method is conceived to transform a block when you will be encrypting multiple blocks.
That is not the case in your code, where you want to transform a single block, and therefore you should be using the TransformFinalBlock method instead. BTW I took the liberty of making your sample buildable.
using System;
using System.Text;
namespace Tests
{
class Program
{
static void Main(string args)
{
System.Security.Cryptography.TripleDESCryptoServiceProvider tripleDES = new System.Security.Cryptography.TripleDESCryptoServiceProvider();
byte data = Encoding.UTF8.GetBytes("This is a sample message");
byte key = Encoding.UTF8.GetBytes("NOSTROMOHASSOMEGODPOWERS");
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var encryptor = tripleDES.CreateEncryptor();
byte result = new byte[data.Length];
result = encryptor.TransformFinalBlock(data, 0, data.Length);
string res = BitConverter.ToString(result).Replace("-","");
Console.WriteLine(BitConverter.ToString(result).Replace("-",""));
byte data2 = result;
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var decryptor = tripleDES.CreateDecryptor();
byte result2 = new byte[data2.Length];
result2 = decryptor.TransformFinalBlock(data2, 0, data2.Length);
Console.WriteLine(Encoding.UTF8.GetString(result2));
}
}
}
answered Jul 31 '18 at 12:49
xtoikxtoik
65738
add a comment |
I think that your problem is in the method of the encryptor / decryptor that you are using: the TransformBlock method is conceived to transform a block when you will be encrypting multiple blocks.
That is not the case in your code, where you want to transform a single block, and therefore you should be using the TransformFinalBlock method instead. BTW I took the liberty of making your sample buildable.
using System;
using System.Text;
namespace Tests
{
class Program
{
static void Main(string args)
{
System.Security.Cryptography.TripleDESCryptoServiceProvider tripleDES = new System.Security.Cryptography.TripleDESCryptoServiceProvider();
byte data = Encoding.UTF8.GetBytes("This is a sample message");
byte key = Encoding.UTF8.GetBytes("NOSTROMOHASSOMEGODPOWERS");
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var encryptor = tripleDES.CreateEncryptor();
byte result = new byte[data.Length];
result = encryptor.TransformFinalBlock(data, 0, data.Length);
string res = BitConverter.ToString(result).Replace("-","");
Console.WriteLine(BitConverter.ToString(result).Replace("-",""));
byte data2 = result;
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var decryptor = tripleDES.CreateDecryptor();
byte result2 = new byte[data2.Length];
result2 = decryptor.TransformFinalBlock(data2, 0, data2.Length);
Console.WriteLine(Encoding.UTF8.GetString(result2));
}
}
}
answered Jul 31 '18 at 12:49
xtoikxtoik
65738
I think that your problem is in the method of the encryptor / decryptor that you are using: the TransformBlock method is conceived to transform a block when you will be encrypting multiple blocks.
That is not the case in your code, where you want to transform a single block, and therefore you should be using the TransformFinalBlock method instead. BTW I took the liberty of making your sample buildable.
using System;
using System.Text;
namespace Tests
{
class Program
{
static void Main(string args)
{
System.Security.Cryptography.TripleDESCryptoServiceProvider tripleDES = new System.Security.Cryptography.TripleDESCryptoServiceProvider();
byte data = Encoding.UTF8.GetBytes("This is a sample message");
byte key = Encoding.UTF8.GetBytes("NOSTROMOHASSOMEGODPOWERS");
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var encryptor = tripleDES.CreateEncryptor();
byte result = new byte[data.Length];
result = encryptor.TransformFinalBlock(data, 0, data.Length);
string res = BitConverter.ToString(result).Replace("-","");
Console.WriteLine(BitConverter.ToString(result).Replace("-",""));
byte data2 = result;
tripleDES.Key = key;
tripleDES.IV = new byte[tripleDES.BlockSize / 8];
var decryptor = tripleDES.CreateDecryptor();
byte result2 = new byte[data2.Length];
result2 = decryptor.TransformFinalBlock(data2, 0, data2.Length);
Console.WriteLine(Encoding.UTF8.GetString(result2));
}
}
}
answered Jul 31 '18 at 12:49
xtoikxtoik
65738
answered Jul 31 '18 at 12:49
xtoikxtoik
65738
answered Jul 31 '18 at 12:49
xtoikxtoik
65738
answered Jul 31 '18 at 12:49
xtoikxtoik
65738
65738
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The default mode is CBC, not EBC. Since you don't explicitly set the
Mode, why do you think this is EBC? Plus it's 2018. Outside of historic interest, why are you writing new code that uses outdated ciphers and outdated modes?– Damien_The_Unbeliever
Jul 31 '18 at 12:24
@Damien_The_Unbeliever the mode is not a problem, IV is set to 0s either way
– karolyzz
Jul 31 '18 at 12:28