I've come across a problem with interop between C# and C++ where I'm sharing memory between the two 'sides' of my application via a struct defined in both native and managed code. The struct on the native side is defined as such:
#pragma pack(push, 1)
struct RayTestCollisionDesc {
btVector3 hitPosition;
btRigidBody* hitBody;
RayTestCollisionDesc(btRigidBody* body, btVector3& position)
: hitBody(body), hitPosition(position) { }
};
#pragma pack(pop)
And a similar struct is defined on the managed (C#) side. On C# the struct size is 20 bytes (as I would expect on a 32-bit system). However, despite the pragma pack directive, the struct size on the C++ size is still 32. For clarity here's the sizeof() from C++ of each of those types:
sizeof(btVector3) : 16
sizeof(btRigidBody*) : 4
sizeof(RayTestCollisionDesc) : 32
Clearly pragma pack is only referring to packing between members of the struct, and not padding at the end of the struct (i.e. alignment). I also tried adding __declspec(align(1)) but that had no effect, and MSDN itself does say "__declspec(align(#)) can only increase alignment restrictions."
And FWIW I'm using the VS2013 compiler (Platform Toolset v120).
Is there a way to 'force' the struct size to 20 bytes?
You are transferring data between two different compilers. In general, it is impossible to make this match. Especially if you transfer data from one computer to another.
First write a spec for the data that you are transferring. The spec CANNOT be a C++ or C# struct. The spec must be something like "four bytes this, four bytes that, two bytes a third thing..." and so on "with a total of 20 bytes", for example.
Then whether you use C++ or C#, or whether a totally different developer uses Objective-C code to read the data, you read an array of 20 bytes, take the 20 bytes, look at them, and fill whatever struct you want with these 20 bytes. And for writing, you do the opposite. Now it does not matter what C++ compiler you use, what weird pragmas you are using, it just works.
Or use something portable like JSON.
You can't do interop this way. A C# object and a C++ struct are different. I suggest you tu use a serialization library like Capt'n Proto or Protobuf
Related
If I have a struct in C++, is there no way to safely read/write it to a file that is cross-platform/compiler compatible?
Because if I understand correctly, every compiler 'pads' differently based on the target platform.
No. That is not possible. It's because of lack of standardization of C++ at the binary level.
Don Box writes (quoting from his book Essential COM, chapter COM As A Better C++)
C++ and Portability
Once the decision is made to
distribute a C++ class as a DLL, one
is faced with one of the fundamental
weaknesses of C++, that is, lack of
standardization at the binary level.
Although the ISO/ANSI C++ Draft
Working Paper attempts to codify which
programs will compile and what the
semantic effects of running them will
be, it makes no attempt to standardize
the binary runtime model of C++. The
first time this problem will become
evident is when a client tries to link
against the FastString DLL's import library from
a C++ developement environment other
than the one used to build the
FastString DLL.
Struct padding is done differently by different compilers. Even if you use the same compiler, the packing alignment for structs can be different based on what pragma pack you're using.
Not only that if you write two structs whose members are exactly same, the only difference is that the order in which they're declared is different, then the size of each struct can be (and often is) different.
For example, see this,
struct A
{
char c;
char d;
int i;
};
struct B
{
char c;
int i;
char d;
};
int main() {
cout << sizeof(A) << endl;
cout << sizeof(B) << endl;
}
Compile it with gcc-4.3.4, and you get this output:
8
12
That is, sizes are different even though both structs have the same members!
The bottom line is that the standard doesn't talk about how padding should be done, and so the compilers are free to make any decision and you cannot assume all compilers make the same decision.
If you have the opportunity to design the struct yourself, it should be possible. The basic idea is that you should design it so that there would be no need to insert pad bytes into it. the second trick is that you must handle differences in endianess.
I'll describe how to construct the struct using scalars, but the you should be able to use nested structs, as long as you would apply the same design for each included struct.
First, a basic fact in C and C++ is that the alignment of a type can not exceed the size of the type. If it would, then it would not be possible to allocate memory using malloc(N*sizeof(the_type)).
Layout the struct, starting with the largest types.
struct
{
uint64_t alpha;
uint32_t beta;
uint32_t gamma;
uint8_t delta;
Next, pad out the struct manually, so that in the end you will match up the largest type:
uint8_t pad8[3]; // Match uint32_t
uint32_t pad32; // Even number of uint32_t
}
Next step is to decide if the struct should be stored in little or big endian format. The best way is to "swap" all the element in situ before writing or after reading the struct, if the storage format does not match the endianess of the host system.
No, there's no safe way. In addition to padding, you have to deal with different byte ordering, and different sizes of builtin types.
You need to define a file format, and convert your struct to and from that format. Serialization libraries (e.g. boost::serialization, or google's protocolbuffers) can help with this.
Long story short, no. There is no platform-independent, Standard-conformant way to deal with padding.
Padding is called "alignment" in the Standard, and it begins discussing it in 3.9/5:
Object types have alignment
requirements (3.9.1, 3.9.2). The
alignment of a complete object type is
an implementation-defined integer
value representing a number of bytes;
an object is allocated at an address
that meets the alignment requirements
of its object type.
But it goes on from there and winds off to many dark corners of the Standard. Alignment is "implementation-defined" meaning it can be different across different compilers, or even across address models (ie 32-bit/64-bit) under the same compiler.
Unless you have truly harsh performance requirements, you might consider storing your data to disc in a different format, like char strings. Many high-performance protocols send everything using strings when the natural format might be something else. For example, a low-latency exchange feed I recently worked on sends dates as strings formatted like this: "20110321" and times are sent similarly: "141055.200". Even though this exchange feed sends 5 million messages per second all day long, they still use strings for everything because that way they can avoid endian-ness and other issues.
If I have a struct in C++, is there no way to safely read/write it to a file that is cross-platform/compiler compatible?
Because if I understand correctly, every compiler 'pads' differently based on the target platform.
No. That is not possible. It's because of lack of standardization of C++ at the binary level.
Don Box writes (quoting from his book Essential COM, chapter COM As A Better C++)
C++ and Portability
Once the decision is made to
distribute a C++ class as a DLL, one
is faced with one of the fundamental
weaknesses of C++, that is, lack of
standardization at the binary level.
Although the ISO/ANSI C++ Draft
Working Paper attempts to codify which
programs will compile and what the
semantic effects of running them will
be, it makes no attempt to standardize
the binary runtime model of C++. The
first time this problem will become
evident is when a client tries to link
against the FastString DLL's import library from
a C++ developement environment other
than the one used to build the
FastString DLL.
Struct padding is done differently by different compilers. Even if you use the same compiler, the packing alignment for structs can be different based on what pragma pack you're using.
Not only that if you write two structs whose members are exactly same, the only difference is that the order in which they're declared is different, then the size of each struct can be (and often is) different.
For example, see this,
struct A
{
char c;
char d;
int i;
};
struct B
{
char c;
int i;
char d;
};
int main() {
cout << sizeof(A) << endl;
cout << sizeof(B) << endl;
}
Compile it with gcc-4.3.4, and you get this output:
8
12
That is, sizes are different even though both structs have the same members!
The bottom line is that the standard doesn't talk about how padding should be done, and so the compilers are free to make any decision and you cannot assume all compilers make the same decision.
If you have the opportunity to design the struct yourself, it should be possible. The basic idea is that you should design it so that there would be no need to insert pad bytes into it. the second trick is that you must handle differences in endianess.
I'll describe how to construct the struct using scalars, but the you should be able to use nested structs, as long as you would apply the same design for each included struct.
First, a basic fact in C and C++ is that the alignment of a type can not exceed the size of the type. If it would, then it would not be possible to allocate memory using malloc(N*sizeof(the_type)).
Layout the struct, starting with the largest types.
struct
{
uint64_t alpha;
uint32_t beta;
uint32_t gamma;
uint8_t delta;
Next, pad out the struct manually, so that in the end you will match up the largest type:
uint8_t pad8[3]; // Match uint32_t
uint32_t pad32; // Even number of uint32_t
}
Next step is to decide if the struct should be stored in little or big endian format. The best way is to "swap" all the element in situ before writing or after reading the struct, if the storage format does not match the endianess of the host system.
No, there's no safe way. In addition to padding, you have to deal with different byte ordering, and different sizes of builtin types.
You need to define a file format, and convert your struct to and from that format. Serialization libraries (e.g. boost::serialization, or google's protocolbuffers) can help with this.
Long story short, no. There is no platform-independent, Standard-conformant way to deal with padding.
Padding is called "alignment" in the Standard, and it begins discussing it in 3.9/5:
Object types have alignment
requirements (3.9.1, 3.9.2). The
alignment of a complete object type is
an implementation-defined integer
value representing a number of bytes;
an object is allocated at an address
that meets the alignment requirements
of its object type.
But it goes on from there and winds off to many dark corners of the Standard. Alignment is "implementation-defined" meaning it can be different across different compilers, or even across address models (ie 32-bit/64-bit) under the same compiler.
Unless you have truly harsh performance requirements, you might consider storing your data to disc in a different format, like char strings. Many high-performance protocols send everything using strings when the natural format might be something else. For example, a low-latency exchange feed I recently worked on sends dates as strings formatted like this: "20110321" and times are sent similarly: "141055.200". Even though this exchange feed sends 5 million messages per second all day long, they still use strings for everything because that way they can avoid endian-ness and other issues.
If I have a struct in C++, is there no way to safely read/write it to a file that is cross-platform/compiler compatible?
Because if I understand correctly, every compiler 'pads' differently based on the target platform.
No. That is not possible. It's because of lack of standardization of C++ at the binary level.
Don Box writes (quoting from his book Essential COM, chapter COM As A Better C++)
C++ and Portability
Once the decision is made to
distribute a C++ class as a DLL, one
is faced with one of the fundamental
weaknesses of C++, that is, lack of
standardization at the binary level.
Although the ISO/ANSI C++ Draft
Working Paper attempts to codify which
programs will compile and what the
semantic effects of running them will
be, it makes no attempt to standardize
the binary runtime model of C++. The
first time this problem will become
evident is when a client tries to link
against the FastString DLL's import library from
a C++ developement environment other
than the one used to build the
FastString DLL.
Struct padding is done differently by different compilers. Even if you use the same compiler, the packing alignment for structs can be different based on what pragma pack you're using.
Not only that if you write two structs whose members are exactly same, the only difference is that the order in which they're declared is different, then the size of each struct can be (and often is) different.
For example, see this,
struct A
{
char c;
char d;
int i;
};
struct B
{
char c;
int i;
char d;
};
int main() {
cout << sizeof(A) << endl;
cout << sizeof(B) << endl;
}
Compile it with gcc-4.3.4, and you get this output:
8
12
That is, sizes are different even though both structs have the same members!
The bottom line is that the standard doesn't talk about how padding should be done, and so the compilers are free to make any decision and you cannot assume all compilers make the same decision.
If you have the opportunity to design the struct yourself, it should be possible. The basic idea is that you should design it so that there would be no need to insert pad bytes into it. the second trick is that you must handle differences in endianess.
I'll describe how to construct the struct using scalars, but the you should be able to use nested structs, as long as you would apply the same design for each included struct.
First, a basic fact in C and C++ is that the alignment of a type can not exceed the size of the type. If it would, then it would not be possible to allocate memory using malloc(N*sizeof(the_type)).
Layout the struct, starting with the largest types.
struct
{
uint64_t alpha;
uint32_t beta;
uint32_t gamma;
uint8_t delta;
Next, pad out the struct manually, so that in the end you will match up the largest type:
uint8_t pad8[3]; // Match uint32_t
uint32_t pad32; // Even number of uint32_t
}
Next step is to decide if the struct should be stored in little or big endian format. The best way is to "swap" all the element in situ before writing or after reading the struct, if the storage format does not match the endianess of the host system.
No, there's no safe way. In addition to padding, you have to deal with different byte ordering, and different sizes of builtin types.
You need to define a file format, and convert your struct to and from that format. Serialization libraries (e.g. boost::serialization, or google's protocolbuffers) can help with this.
Long story short, no. There is no platform-independent, Standard-conformant way to deal with padding.
Padding is called "alignment" in the Standard, and it begins discussing it in 3.9/5:
Object types have alignment
requirements (3.9.1, 3.9.2). The
alignment of a complete object type is
an implementation-defined integer
value representing a number of bytes;
an object is allocated at an address
that meets the alignment requirements
of its object type.
But it goes on from there and winds off to many dark corners of the Standard. Alignment is "implementation-defined" meaning it can be different across different compilers, or even across address models (ie 32-bit/64-bit) under the same compiler.
Unless you have truly harsh performance requirements, you might consider storing your data to disc in a different format, like char strings. Many high-performance protocols send everything using strings when the natural format might be something else. For example, a low-latency exchange feed I recently worked on sends dates as strings formatted like this: "20110321" and times are sent similarly: "141055.200". Even though this exchange feed sends 5 million messages per second all day long, they still use strings for everything because that way they can avoid endian-ness and other issues.
I am attempting to read in a binary file. The problem is that the creator of the file took no time to properly align data structures to their natural boundaries and everything is packed tight. This makes it difficult to read the data using C++ structs.
Is there a way to force a struct to be packed tight?
Example:
struct {
short a;
int b;
}
The above structure is 8 bytes: 2 for short a, 2 for padding, 4 for int b. However, on disk, the data is only 6 bytes (not having the 2 bytes of padding for alignment)
Please be aware the actual data structures are thousands of bytes and many fields, including a couple arrays, so I would prefer not to read each field individually.
If you're using GCC, you can do struct __attribute__ ((packed)) { short a; int b; }
On VC++ you can do #pragma pack(1). This option is also supported by GCC.
#pragma pack(push, 1)
struct { short a; int b; }
#pragma pack(pop)
Other compilers may have options to do a tight packing of the structure with no padding.
You need to use a compiler-specific, non-Standard directive to specify 1-byte packing. Such as under Windows:
#pragma pack (push, 1)
The problem is that the creator of the file took no time to properly
byte align the data structures and everything is packed tight.
Actually, the designer did the right thing. Padding is something that the Standard says can be applied, but it doesn't say how much padding should be applied in what cases. The Standard doesn't even say how many bits are in a byte. Even though you might assume that even though these things aren't specified they should still be the same reasonable value on modern machines, that's simply not true. On a 32-bit Windows machine for example the padding might be one thing whereas on the 64-bit version of Windows is might be something else. Maybe it will be the same -- that's not the point. The point is you don't know what the padding will be on different systems.
So by "packing it tight" the developer did the only thing they could -- use some packing that he can be reasonably sure that every system will be able to understand. In that case that commonly-understood packing is to use no padding in structures saved to disk or sent down a wire.
If I have a struct in C++, is there no way to safely read/write it to a file that is cross-platform/compiler compatible?
Because if I understand correctly, every compiler 'pads' differently based on the target platform.
No. That is not possible. It's because of lack of standardization of C++ at the binary level.
Don Box writes (quoting from his book Essential COM, chapter COM As A Better C++)
C++ and Portability
Once the decision is made to
distribute a C++ class as a DLL, one
is faced with one of the fundamental
weaknesses of C++, that is, lack of
standardization at the binary level.
Although the ISO/ANSI C++ Draft
Working Paper attempts to codify which
programs will compile and what the
semantic effects of running them will
be, it makes no attempt to standardize
the binary runtime model of C++. The
first time this problem will become
evident is when a client tries to link
against the FastString DLL's import library from
a C++ developement environment other
than the one used to build the
FastString DLL.
Struct padding is done differently by different compilers. Even if you use the same compiler, the packing alignment for structs can be different based on what pragma pack you're using.
Not only that if you write two structs whose members are exactly same, the only difference is that the order in which they're declared is different, then the size of each struct can be (and often is) different.
For example, see this,
struct A
{
char c;
char d;
int i;
};
struct B
{
char c;
int i;
char d;
};
int main() {
cout << sizeof(A) << endl;
cout << sizeof(B) << endl;
}
Compile it with gcc-4.3.4, and you get this output:
8
12
That is, sizes are different even though both structs have the same members!
The bottom line is that the standard doesn't talk about how padding should be done, and so the compilers are free to make any decision and you cannot assume all compilers make the same decision.
If you have the opportunity to design the struct yourself, it should be possible. The basic idea is that you should design it so that there would be no need to insert pad bytes into it. the second trick is that you must handle differences in endianess.
I'll describe how to construct the struct using scalars, but the you should be able to use nested structs, as long as you would apply the same design for each included struct.
First, a basic fact in C and C++ is that the alignment of a type can not exceed the size of the type. If it would, then it would not be possible to allocate memory using malloc(N*sizeof(the_type)).
Layout the struct, starting with the largest types.
struct
{
uint64_t alpha;
uint32_t beta;
uint32_t gamma;
uint8_t delta;
Next, pad out the struct manually, so that in the end you will match up the largest type:
uint8_t pad8[3]; // Match uint32_t
uint32_t pad32; // Even number of uint32_t
}
Next step is to decide if the struct should be stored in little or big endian format. The best way is to "swap" all the element in situ before writing or after reading the struct, if the storage format does not match the endianess of the host system.
No, there's no safe way. In addition to padding, you have to deal with different byte ordering, and different sizes of builtin types.
You need to define a file format, and convert your struct to and from that format. Serialization libraries (e.g. boost::serialization, or google's protocolbuffers) can help with this.
Long story short, no. There is no platform-independent, Standard-conformant way to deal with padding.
Padding is called "alignment" in the Standard, and it begins discussing it in 3.9/5:
Object types have alignment
requirements (3.9.1, 3.9.2). The
alignment of a complete object type is
an implementation-defined integer
value representing a number of bytes;
an object is allocated at an address
that meets the alignment requirements
of its object type.
But it goes on from there and winds off to many dark corners of the Standard. Alignment is "implementation-defined" meaning it can be different across different compilers, or even across address models (ie 32-bit/64-bit) under the same compiler.
Unless you have truly harsh performance requirements, you might consider storing your data to disc in a different format, like char strings. Many high-performance protocols send everything using strings when the natural format might be something else. For example, a low-latency exchange feed I recently worked on sends dates as strings formatted like this: "20110321" and times are sent similarly: "141055.200". Even though this exchange feed sends 5 million messages per second all day long, they still use strings for everything because that way they can avoid endian-ness and other issues.