BinaryReader vs byte[]+shifts
0
I must be misunderstanding what BinaryReader is doing. Why are these outputs different?
var data = File.ReadAllBytes(testFile);
var pos = 0;
var read8 = new Func<uint>(() => data[pos++]);
var read32 = new Func<uint>(() => (read8() << 24)
using (var reader = new BinaryReader(File.Open(testFile, FileMode.Open)))
Console.WriteLine(reader.ReadUInt32());
c# binaryreader
edited Nov 15 '18 at 18:58
GSerg
60.1k15106231
asked Nov 15 '18 at 18:56
Brandon PrudentBrandon Prudent
1057
add a comment |
0
I must be misunderstanding what BinaryReader is doing. Why are these outputs different?
var data = File.ReadAllBytes(testFile);
var pos = 0;
var read8 = new Func<uint>(() => data[pos++]);
var read32 = new Func<uint>(() => (read8() << 24)
using (var reader = new BinaryReader(File.Open(testFile, FileMode.Open)))
Console.WriteLine(reader.ReadUInt32());
c# binaryreader
edited Nov 15 '18 at 18:58
GSerg
60.1k15106231
asked Nov 15 '18 at 18:56
Brandon PrudentBrandon Prudent
1057
add a comment |
0
0
0
I must be misunderstanding what BinaryReader is doing. Why are these outputs different?
var data = File.ReadAllBytes(testFile);
var pos = 0;
var read8 = new Func<uint>(() => data[pos++]);
var read32 = new Func<uint>(() => (read8() << 24)
using (var reader = new BinaryReader(File.Open(testFile, FileMode.Open)))
Console.WriteLine(reader.ReadUInt32());
c# binaryreader
edited Nov 15 '18 at 18:58
GSerg
60.1k15106231
asked Nov 15 '18 at 18:56
Brandon PrudentBrandon Prudent
1057
I must be misunderstanding what BinaryReader is doing. Why are these outputs different?
var data = File.ReadAllBytes(testFile);
var pos = 0;
var read8 = new Func<uint>(() => data[pos++]);
var read32 = new Func<uint>(() => (read8() << 24)
using (var reader = new BinaryReader(File.Open(testFile, FileMode.Open)))
Console.WriteLine(reader.ReadUInt32());
c# binaryreader
c# binaryreader
edited Nov 15 '18 at 18:58
GSerg
60.1k15106231
asked Nov 15 '18 at 18:56
Brandon PrudentBrandon Prudent
1057
edited Nov 15 '18 at 18:58
GSerg
60.1k15106231
asked Nov 15 '18 at 18:56
Brandon PrudentBrandon Prudent
1057
edited Nov 15 '18 at 18:58
GSerg
60.1k15106231
edited Nov 15 '18 at 18:58
GSerg
60.1k15106231
edited Nov 15 '18 at 18:58
GSerg
60.1k15106231
60.1k15106231
asked Nov 15 '18 at 18:56
Brandon PrudentBrandon Prudent
1057
asked Nov 15 '18 at 18:56
Brandon PrudentBrandon Prudent
1057
asked Nov 15 '18 at 18:56
Brandon PrudentBrandon Prudent
1057
1057
add a comment |
add a comment |
1 Answer
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active
oldest
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Endiannes.
Use:
var read32 = new Func<uint>(() => (read8() | (read8() << 8) | (read8() << 16) | read8() << 24));
On a side note, please don't write code with such side effects.
You are getting away with them here because the order of evaluation is guaranteed, but still please don't.
answered Nov 15 '18 at 19:03
GSergGSerg
60.1k15106231
Right, of course. Thank you. On the side note I'm not sure I'm trackin' - is the order of evaluation not always (parens), functions(), and then operators? What is the side effect?
– Brandon Prudent
Nov 15 '18 at 20:09
The side effect is the++inside theread8. That same code would look completely safe if it was a series of statements terminated by a semicolon, but a single expression where each member has a side effect that affects calculation of other members of the same expression is something you want to avoid - even though it does not lead to undefined behaviour like in C++ (see the link), it may still be very confusing.
– GSerg
Nov 15 '18 at 20:58
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Endiannes.
Use:
var read32 = new Func<uint>(() => (read8() | (read8() << 8) | (read8() << 16) | read8() << 24));
On a side note, please don't write code with such side effects.
You are getting away with them here because the order of evaluation is guaranteed, but still please don't.
answered Nov 15 '18 at 19:03
GSergGSerg
60.1k15106231
Right, of course. Thank you. On the side note I'm not sure I'm trackin' - is the order of evaluation not always (parens), functions(), and then operators? What is the side effect?
– Brandon Prudent
Nov 15 '18 at 20:09
The side effect is the++inside theread8. That same code would look completely safe if it was a series of statements terminated by a semicolon, but a single expression where each member has a side effect that affects calculation of other members of the same expression is something you want to avoid - even though it does not lead to undefined behaviour like in C++ (see the link), it may still be very confusing.
– GSerg
Nov 15 '18 at 20:58
add a comment |
2
Endiannes.
Use:
var read32 = new Func<uint>(() => (read8() | (read8() << 8) | (read8() << 16) | read8() << 24));
On a side note, please don't write code with such side effects.
You are getting away with them here because the order of evaluation is guaranteed, but still please don't.
answered Nov 15 '18 at 19:03
GSergGSerg
60.1k15106231
Right, of course. Thank you. On the side note I'm not sure I'm trackin' - is the order of evaluation not always (parens), functions(), and then operators? What is the side effect?
– Brandon Prudent
Nov 15 '18 at 20:09
The side effect is the++inside theread8. That same code would look completely safe if it was a series of statements terminated by a semicolon, but a single expression where each member has a side effect that affects calculation of other members of the same expression is something you want to avoid - even though it does not lead to undefined behaviour like in C++ (see the link), it may still be very confusing.
– GSerg
Nov 15 '18 at 20:58
add a comment |
2
2
2
Endiannes.
Use:
var read32 = new Func<uint>(() => (read8() | (read8() << 8) | (read8() << 16) | read8() << 24));
On a side note, please don't write code with such side effects.
You are getting away with them here because the order of evaluation is guaranteed, but still please don't.
answered Nov 15 '18 at 19:03
GSergGSerg
60.1k15106231
Endiannes.
Use:
var read32 = new Func<uint>(() => (read8() | (read8() << 8) | (read8() << 16) | read8() << 24));
On a side note, please don't write code with such side effects.
You are getting away with them here because the order of evaluation is guaranteed, but still please don't.
answered Nov 15 '18 at 19:03
GSergGSerg
60.1k15106231
answered Nov 15 '18 at 19:03
GSergGSerg
60.1k15106231
answered Nov 15 '18 at 19:03
GSergGSerg
60.1k15106231
answered Nov 15 '18 at 19:03
GSergGSerg
60.1k15106231
60.1k15106231
Right, of course. Thank you. On the side note I'm not sure I'm trackin' - is the order of evaluation not always (parens), functions(), and then operators? What is the side effect?
– Brandon Prudent
Nov 15 '18 at 20:09
The side effect is the++inside theread8. That same code would look completely safe if it was a series of statements terminated by a semicolon, but a single expression where each member has a side effect that affects calculation of other members of the same expression is something you want to avoid - even though it does not lead to undefined behaviour like in C++ (see the link), it may still be very confusing.
– GSerg
Nov 15 '18 at 20:58
add a comment |
Right, of course. Thank you. On the side note I'm not sure I'm trackin' - is the order of evaluation not always (parens), functions(), and then operators? What is the side effect?
– Brandon Prudent
Nov 15 '18 at 20:09
The side effect is the++inside theread8. That same code would look completely safe if it was a series of statements terminated by a semicolon, but a single expression where each member has a side effect that affects calculation of other members of the same expression is something you want to avoid - even though it does not lead to undefined behaviour like in C++ (see the link), it may still be very confusing.
– GSerg
Nov 15 '18 at 20:58
Right, of course. Thank you. On the side note I'm not sure I'm trackin' - is the order of evaluation not always (parens), functions(), and then operators? What is the side effect?
– Brandon Prudent
Nov 15 '18 at 20:09
Right, of course. Thank you. On the side note I'm not sure I'm trackin' - is the order of evaluation not always (parens), functions(), and then operators? What is the side effect?
– Brandon Prudent
Nov 15 '18 at 20:09
The side effect is the
++ inside the read8. That same code would look completely safe if it was a series of statements terminated by a semicolon, but a single expression where each member has a side effect that affects calculation of other members of the same expression is something you want to avoid - even though it does not lead to undefined behaviour like in C++ (see the link), it may still be very confusing.– GSerg
Nov 15 '18 at 20:58
The side effect is the
++ inside the read8. That same code would look completely safe if it was a series of statements terminated by a semicolon, but a single expression where each member has a side effect that affects calculation of other members of the same expression is something you want to avoid - even though it does not lead to undefined behaviour like in C++ (see the link), it may still be very confusing.– GSerg
Nov 15 '18 at 20:58
add a comment |
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