My goal is to send a integer value over TCP as little endian bytes to a client on windows, the client is written in x86 assembly.
If I were to send htonl() encoded bytes to the client would it be necessary being the client is compiled as x86? For instance wouldn't it be redundant to call ntohl() within my client's assembly code?
My overarching question is do I need to call htonl() server-side and ntohl() client side (x86 windows client)? Or should I just let the server do the work by checking if the server's architecture is big endian and if so swap the integer bytes via __builtin_bswap32() and send the little endian bytes to the client? I'm asking because I've read x86 is always little endian so it seems redundant if I know the client is always going to be written in x86 assembly.
My overarching question is do I need to call htonl() server-side and ntohl() client side (x86 windows client)?
No, that would convert to big-endian ("network" byte order), but you said you wanted to send data over the network in little-endian format. On x86, that already is the h ("host") order.
In x86 asm, your data in memory will already be little-endian integers / floats unless you did something unusual (like using bswap, movbe, or pshufb, or byte-at-a-time shift / store.)
To be compatible with that in C, use le32toh (on receive) and htole32 (before send) from GCC's / BSD <endian.h> instead of ntohl / htonl. i.e. use LE as your network format instead of the traditional BE.
call ntohl() within my client's assembly code?
That would be insane. If you did want that, just use the bswap or movbe instructions instead of actually setting up args for a function call. Normally those functions inline when you use them in C, although there is a stand-alone definition of ntohl in libc.
Also, no, you wouldn't want to do that. Your client doesn't want to have anything to do with big-endian, which is what those traditional functions call "network" byte order.
x86 asm with AVX2 vpshufb can byte-swap at memcpy speed (including on small buffers that fit in L1d cache), but it's even more efficient not to have to swap at all as part of the first step that reads the data.
The htonl function converts a 32 bit value from [h]ost byte order to [n]etwork byte order. Host byte order may be either big endian or little endian, and network byte order is big endian. The ntohl function does the reverse.
The purpose of these functions is to abstract away any possible conversion of the host's byte order and to have a known format for the network. So on a little endian system these functions reverse the byte order, while on a big endian system they return the original value unchanged.
I would recommend using the standard network byte order, i.e. big endian, for sending values over the network. They your code would use htonl to prepare values for sending and ntohl to read received values regardless of the host architecture.
Two binaries were compiled. One is on a 32-bit Windows 7 machine and other is on a 64-bit Windows 10. All the source files and dependencies are the same; however, after compilation the checksum was compared and they are different. Could someone provide an explanation as to why?
If you're comparing a 32-bit build and a 64-bit build you'll need to keep in mind that the compiled code will be almost completely different. 64-bit x86_64 and 32-bit x86 machine code will not only vary considerably, but even if that was not the case, remember that pointers are twice the size in 64-bit code, so a lot of code will be structured differently and addresses in the code will be expressed as bigger pointers.
The technical reason is the machine code is not the same between different architectures. Intel x86, and x86_64 have considerable variation in how the opcodes are expressed, and in how programs are structured internally. If you want to read up on the differences there's a lot of published material that can explain, like Intel's own references.
As others have pointed out building twice on the same machine may not even produce a byte-for-byte matching binary. There will be slight differences in it, especially if there's a code-signing step.
In short, you can't do this and expect them to match.
All the source files and dependencies are the same; however, after compilation the checksum was compared and they are different.
The checksum is calculated from the binary code that the compiler produces, not the source code. If anything is different, you should get a different checksum. Changes in the source code will change the binary, so a different checksum. But using a different compiler will also produce a different binary. Using the same compiler with different options will produce a different binary. Compiling the same program with the same compiler and the same compiler options on a different machine might produce a different binary. Compiling the same program for different processor architectures will definitely produce a different binary.
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This question: How many bits does a WORD contain in 32/64 bit OS respectively?, mentions that word size refers to the bit size of a processor register - which i take to mean the number of bits that a computer processor operates on / i.e. the smallest 'indivisible' amount of bits that a processor operates on.
Is that correct? Using software like Word/Excel/etc, the installers have the option for a 32bit or a 64bit installation. What is the difference?
Since the computer architecture is fixed, it would seem to me that software that is '32 bit' would be designed to align with a computer architecture that has a 32 bit architecture. And a 64 bit program would make efforts to align instruction sets with 64 bit word sizes.
Is that correct?
A very similar question is asked here: From a programming point of view, what does it mean when a program is 32 or 64 bit? - and the accepted answer mentions that the difference is the amount of memory that can be allocated to an application. But this is too vague - unless 32 bit / 64 bit software as a concept is completely unrelated to 32 bit / 64 bit word processor size?
Word size is a major difference, but it's not the only one. It tends to define the number of bits a CPU is "rated" for, but word size and overall capability are only loosely related. And overall capability is what matters.
On an Intel or AMD CPU, 32-bit vs. 64-bit software really refers to the mode in which the CPU operates when running it. 32-bit mode has fewer/smaller registers and instructions available, but the most important limitation is the amount of memory available. 32-bit software is generally limited to using between 2GB and just under 4GB of memory.
Each byte of memory has a unique address, which is not very different from each house having a unique postal address. A memory address is just a number that a program can use to find a piece of data again once it has saved it in memory, and each byte of memory has to have an address. If an address is 32 bits, then there are 2^32 possible addresses, and that means 2^32 addressable bytes of memory. On today's Intel/AMD CPUs, the size of a memory address is the same as the size of the registers (although this wasn't always true).
With 32 bit addresses, 4GB (2^32 bytes) can be addressed by the program, however up to half of that space is reserved by the OS. Into the available memory space must fit program code, data, and often also files being accessed. In today's PCs, with many gigabytes of RAM, this fails to take advantage of available memory. That is the main reason why 64-bit has become popular. 64-bit CPUs were available and widely used (typically in 32-bit mode) for several years, until memory sizes larger than 2GB became common, at which point 64-bit mode started to offer real-world advantages and it became popular. 64 bits of memory address space provides 16 exabytes of addressable memory (~18 quintillion bytes), which is more than any current software can use, and certainly no PC has anywhere near that much RAM.
The majority of data used in typical applications, even in 64-bit mode, does not need to be 64-bit and so most of it is still stored in 32-bit (or even smaller) formats. The common ASCII and UTF-8 representations of text use 8-bit data formats. If the program needs to move a large block of text from one place to another in memory, it may try to do it 64 bits at a time, but if it needs to interpret the text, it will probably do it 8 bits at a time. Similarly, 32 bits is a common size for integers (maximum range of +/- 2^31, or approximately +/- 2.1 billion). 2.1 billion is enough range for many uses. Graphics data is usually naturally represented pixel by pixel, and each pixel, usually, contains at most 32 bits of data.
There are disadvantages to using 64-bit data needlessly. 64-bit data takes up more space in memory, and more space in the CPU cache (very fast memory used by the CPU for short-term storage). Memory can only transfer data at a maximum rate, and 64-bit data is twice as big. This can reduce performance if used wastefully. And if it's necessary to support both 32-bit and 64-bit versions of software, using 32-bit values where possible can reduce the differences between the two versions and make development easier (doesn't always work out that way, though).
Prior to 32-bit, the address and word size were usually different (e.g. 16-bit 8086/88 with 20-bit memory addresses but 16-bit registers, or 8-bit 6502 with 16-bit memory addresses, or even early 32-bit ARM with 26-bit addresses). While no programmer ever turned up their nose at better registers, memory space was usually the real driving force for each advancing generation of technology. This is because most programmers rarely work directly with registers, but do work directly with memory, and memory limitations directly cause unpleasantness for the programmer, and in the 32-bit to 64-bit case, for the user as well.
To sum up, while there are real and important technological differences between the various bit sizes, what 32-bit or 64-bit (or 16-bit or 8-bit) really means is simply a collection of capabilities that tend to be associated with CPUs of a particular technological generation, and/or software that takes advantage of those capabilities. Word length is a part of that, but not the only, or necessarily the most important part.
Source: Have been programmer through all these technological eras.
The answer you reference describes benefits of 64-bit over 32-bit. As far as what's actually different about the program itself, it depends on your perspective.
Generally speaking, the program source code does not have to be different at all. Most programs can be written so that they compile perfectly well as either 32-bit or 64-bit programs, as controlled by appropriate choice of compiler and / or compiler options. There is often some impact on the source, however, in that a (C) compiler targeting 64-bit may choose to define its types differently. In particular, long int is ubiquitously 32 bits wide on 32-bit platforms, but it is 64 bits wide on many (but not all) 64-bit platforms. This can be a source of bugs in code that makes unwarranted assumptions about such details.
The main differences are all in the binary. 64-bit programs make use of the full instruction sets of their 64-bit target CPUs, which invariably contain instructions that 32-bit counterpart CPUs do not contain. They will use registers that 32-bit counterpart CPUs do not have. They will use function-call conventions appropriate for their target CPU, which often means passing more arguments in registers than 32-bit programs do. Use of these and other facilities of 64-bit CPUs affords functional advantages such as the ability to use more memory and (sometimes) improved performance.
A program runs on top of a given architecture (arch, or ISA), which is implemented by processors. Typically, an architecture defines a "main" word size, which is the size most of the registers and operations on those registers run (although you can design architectures that work differently). This is usually called the "native" word size, although an architecture may allow operations using different sized registers.
Further, processors use memory, and need to address that memory somehow -- this means operating with those addresses. Therefore, the addresses are typically able to be stored and manipulated like any other number, which means you have registers capable of holding them. Although it is not required that those registers to match the word size nor it is required that an address is computed out of a single register, in some architectures this is the case.
Throughout history, there have been many architectures of different word sizes, even weird ones. Nowadays, you can easily find processors around you that are not just 32-bit and 64-bit, but also e.g. 8-bit and 16-bit (typically in embedded devices). In the typical desktop computer, you are using x86 or x64, which are 32-bit and 64-bit respectively.
Therefore, when you say that a program is 32-bit or 64-bit, you are referring to a particular architecture. In the popular desktop scenario, you are referring to x86 vs. x64. There are many questions, articles and books discussing the differences between the two.
Now, a final note: for compatibility reasons, x64 processors can operate in different modes, one of which is capable of running the 32-bit code from x86. This means that if your computer is x64 (likely) and if your operating system has support for it (also likely, e.g. Windows 64-bit), it can still run programs compiled for x86.
Using software like Word/Excel/etc, the installers have the option for a 32bit or a 64bit installation. What is the difference?
This depends on the CPU used:
On SPARC CPUs, the difference between "32-bit" and "64-bit" programs is exactly what you think:
64-bit programs use additional operations that are not supported by 32-bit SPARC CPUs. On the other hand the Solaris or Linux operating system places the data accessed by 64-bit programs in memory areas which can only be accessed using 64-bit instructions. This means that a 64-bit program even MUST use instructions not supported by 32-bit CPUs.
For x86 CPUs this is different:
Modern x86 CPUs have different operating modes and they can execute different types of code. In the different modes, they can execute 16-, 32- or 64-bit code.
In 16-, 32- and 64-bit code, the CPU interprets the bytes differently:
The bytes (hexadecimal) b8 4e 61 bc 00 c3 would be interpreted as:
mov eax,0xbc614e
ret
... in 32-bit code and as:
mov ax,0x614e
mov sp,0xc300
... in 16 bit code.
The bytes in the EXE file of the "64-bit installation" and of the "32-bit installation" must be interpreted differently by the CPU.
And a 64 bit program would make efforts to align instruction sets with 64 bit word sizes.
16-bit code (see above) can access 32-bit registers when the CPU is not a 16-bit CPU.
So a "16-bit program" can access 32-bit registers on a 32- or 64-bit x86 CPU.
mentions that word size refers to the bit size of a processor register
Generally yes (though there are some exceptions/complications)
which I take to mean the number of bits that a computer processor operates on / i.e. the smallest 'indivisible' amount of bits that a processor operates on.
No, most processor architectures can work on values smaller than their native word size. A better (but not perfect) definition would be the largest piece of data that the processor can process (through the main integer datapath) as a single unit.
In general on modern 32-bit and 64-bit systems pointers are the same size as the word size, though on many 64-bit systems not all bits of said pointer are actually usable. It is possible to have a memory model where addressable memory is greater than the system's native word size and it was common to do so in the 8-bit and 16-bit eras, but it has fallen out of favour since the introduction of 32-bit CPUs.
Since the computer architecture is fixed
While the physical architecture is of course fixed, many processors have multiple operating modes with different instructions and registers available to the programmer. In 64-bit mode, the full features of the CPU are available, in 32-bit mode, the processor presents a backwards compatible interface which limits the features and the address space. The modes are sufficiently different that code must be compiled for a particular mode.
As a general rule, an OS running in 64-bit mode can support applications running in 32-bit mode but not vice-versa.
So a 32-bit application runs in 32-bit mode on either a 32-bit processor running a 32-bit OS, a 64-bit processor running a 32-bit OS or a 64-bit processor running a 64-bit OS.
A 64-bit application on the other hand normally runs only on a 64-bit processor running a 64-bit OS.
The information you have is a good part of the picture, but not all of it. I'm not a processor expert, so there are likely some details that my answer will be missing.
The 32 bit vs 64 bit is related to the processor architecture. An increase in word size does a few things:
Larger word size enables more instructions to be defined. For instance, and 8-bit processor that does a single load instruction can only have 256 total instructions, where a larger word size allows more instructions to be defined in the processors micro-code. Obviously, there is a limit to how many truly useful instructions are defined.
More data can be processed with a single instruction cycle as there are more bits available. This speeds up execution.
Like you stated, it also allows access to a larger memory space without having to do things like multiple address cycles, or multiplexing high/low data words.
When the processor architecture moves from 32-bit to 64-bit, the chip manufacturer will likely maintain compatibility with the previous instruction set, so that all the software that was developed previously will still run on the new architecture. When you target the 64-bit architecture, the compiler will have new instructions available and memory addressing schemes with which to process data more efficiently.
Short answer: This is a convention based solely on the width of the underlying data bus
An n-bit program is a program that is optimized for an n-bit CPU. Said otherwise a 64-bit program is a binary program compiled for a 64 bit CPU. A 64 bit CPU, in turn, is one taking advantage of a 64-bit data bus for the exchange of data between CPU and memory.
That's as simple, but you can read more below.
The definition actually redirects to understanding what is a 32/64 bit CPU, indirectly to what is a 32/64 bit operating system, and how compilers optimize binaries for a given architecture.
Optimization here encompasses the format of the binary itself. 32 bit and 64-bit binaries for a given OS, e.g. a Windows binary, have different formats. However, a given 64 bit OS, e.g. Windows 64, will be able to read and launch a 32-bit binary file written for the 32-bit version and a 32-bit wide data bus.
32/64 bit CPU, first definition
The CPU can store/recall a certain quantity of data in memory in a single instruction. A 32-bit CPU can transfer 4 bytes (32 bits) at once and a 64-bit CPU can transfer 8 bytes (64 bits) at once. So "32/64 bit" prefix comes from the quantity of RAM transferred in a single read/write cycle.
This quantity impacts the execution time: The fewer transfer cycles are required, the less the CPU waits for the memory, the program executes faster. It's like carrying a large quantity of water with a small or a large bucket.
The size of the bucket (the number of bits used for data transfer) is used to indicate how efficient the architecture is, hence for the same CPU, a 32-bit application is less efficient than a 64-bit application.
32/64 bit CPU, technical definition
Obviously, the RAM and the CPU must be both able to manage a 32/64-bit data transfer, which in turn determines the number of wires used to connect the CPU to the RAM (system bus). 32/64 bit is actually the number of wires/tracks composing the data bus (usually named the bus "width").
(Wikipedia: System bus - The data bus width determines the prefix 32/64 bit for a CPU, a program, an OS, ...)
(Another bus is the address bus, which is usually wider, but the address bus width is irrelevant in naming a CPU as 32 or 64 bit CPU. This address bus width determines the total quantity of RAM which can be reached / "addressed" by the CPU, e.g. 2 GB or 32 GB. As for the control bus, it is a small bus used to synchronize everything connected to the data bus, in particular, it indicates when the data bus is stable and ready to be sampled in a data transfer operation).
When bits are transferred between the CPU and the RAM, the voltage on the different copper tracks of the data bus must be stable prior to reading data on the bus, else one or more bit values would be wrong. It takes less time to stabilize 8 bits than 64 bits, so increasing the data bus width is not without problems to solve.
32/64 bit program: A compiler matter
Programs don't always need to transfer 4 bytes (32-bit data bus) or 8 bytes (64-bit data bus), so they use different instructions to read 1 byte, 2 bytes, 4 bytes, and 8 bits, for performance reasons.
Binaries (native assembly language programs) are written either with the 32-bit architecture in mind, or the 64-bit architecture, and the associated instruction set. So the name 32/64 bit program.
The choice of the target architecture is a matter of compiler/compiler options used when converting the source program into a binary. Most compilers are able to produce a 32 bit or a 64 binary from the same source program. That's why you'll find both versions of an application when downloading your preferred program or tool.
However, most programs rely on ready-made libraries written by other programmers (e.g. a video editing program may use FFmpeg library). To produce a fully 64-bit application, the compiler (actually the link editor, but let's keep it simple) needs to access a 64-bit version of any library used, which may not be possible.
This also applies to operating systems themselves, as an OS is just a suite of individual programs and libraries. However, an OS is itself a kind of big library for the user programs, acting as a gateway between the computer hardware and the user programs, for efficiency and security reasons. The way OS is written car prevent the user programs to access the full potential of the underlying CPU architecture.
32-bit program compatibility with 64-bit CPU
A 64-bit operating system is able to run a 32-bit binary on a 64-bit architecture, as the 64 bit CPU instruction set is retro-compatible. However, some adjustments are required.
In addition of the data bus width and read/write instructions subset, there are many other differences between 32 bit and 64 bit CPU (register operations, memory caches, data alignment/boundaries, timing, ...).
Running a 32-bit program on a 64-bit architecture:
is more efficient than running it on an older 32-bit architecture (almost solely due to CPU clock speed improvement compared to older 32/64 bit CPU generations)
is less efficient than running the same application compiled into a 64-bit binary to take advantage of the 64-bit architecture, in particular, the ability to transfer 64 bits at once from/to memory.
When compiling a source into a 32-bit binary, the compiler will still use small buckets, instead of the larger available with the 64-bit data bus. This has the largest impact on execution speed, compared to the same application compiled to use large buckets.
For information, the applications compiled into 16 bit Windows binaries (earlier versions of Windows running on 80-286 CPU with a 16-bit data bus) are not fully supported anymore, though there is still a possibility on Windows 10 to activate NTVDM.
The case of .NET, Java and other interpreted "byte-code"
While until recent years, compilers were used to translate a source program (e.g. a C++ source) into a machine language program, this method is now in regression.
The main problem is that machine language for some CPU is not the same than for another (think about differences between a smartphone using an ARM chip and a server using an Intel chip). You definitely can't use the same binary on both hardware, they are not talking the same language, and even if this were possible it would be inefficient on both machines due to the huge differences in how they work.
The current idea is to use an intermediate representation (IR) of the instructions, derived from the source. Java (Sun, sadly now Oracle) and IL (Microsoft) are such intermediate representations. The same IR file can be used on any OS supporting the IR.
Once the OS opens the file, it performs the final compilation into the "local" machine language understood by the actual CPU and taking into account the final architecture on which to run the program. For example, for Microsoft .NET, the universal version is executed by a CoreCLR virtual machine located on the final computer. There is usually no notion of data bus width in such intermediate languages, hence less and less application will have this n-bit prefix.
However we cannot forget the actual architecture, so there will be still 32 and 64 bit versions produced for the CoreCLR to optimize the final code, even if the application itself, at the IR level, is not optimized for a given architecture (only one IR version to download and install).
I want my program to use big end by default (now it is little).
That means,everytime I declare an uint_32/int,the value which be transformed to it will be on big end order.
Is that possible? (Without calling each time to ntohl()).
I have researched on google and s.o for 3 days and haven't got any anawer
I will strongly appreciate any help!
edit:
I have server and client.
The client works with big endian,and the server is little endian.
Now,i am sending to the server md5 value for a byte array,ntohl()ing it on the server and get the valid md5 value.
In the server,when I call the md5.c function (which is dll,by the way),I am getting a different value.
This value is not even similair on any way to the recived value.
I assume it happens because of the endiannes.
The byte array I send to the function is not changing because those are single bytes,which are not seensetive the endianess,but the vars I declare on the function,and with them I manipulate my byte arr,can make a problem.
That is the reason big endian is so important to me.
I want my program to use big end by default
You cannot. The endianness is a fixed property of the target processor (of your C compiler) and related to its instruction set architecture.
You might in principle (but with C, that is really not worth the trouble) use a C compiler for some virtual machine (like WebAssembly, or MMIX) of the appropriate endianness, or even define a bytecode VM for a machine of desired endianness.
A variant of that might be to cross-compile for some other instruction set architecture (e.g. MIPS, ARM, etc...) and use some emulator to run the executable (and the necessary OS).
Regarding your edit, you could consider sending on the network the md5sum in alphanumerical form (e.g. like the output of the md5sum command).
You can compile the program for different processors, some of which use big endian instead of little endian, or you can control how your code will be compiled in that aspect. For example, for MIPS: https://gcc.gnu.org/onlinedocs/gcc/MIPS-Options.html
Endianness is purely about how a processor performs multi-byte arithmetic. The only time a programmer need to be aware is when serializing or addressing parts of an integer.
So unless you can change how a processor performs arithmetic on multi-byte words (ARM allows you to change endianness) you are stuck with how the processor works.
I have a C program last compiled in 1990, that reads and writes some binary files. The executable still works, reading and writing them perfectly. I need to recompile the source, add some features, and then use the code, reading in some of the old data, and outputting it with additional information.
When I recompile the code, with no changes, and execute it, it fails reading in the old files, giving segmentation faults when I try to process the data read into an area of memory. I believe that the problem may be that the binary files written earlier used 4 8-bit byte integers, 8 byte longs, and 4 byte floats. The architecture on my machine now uses 64-bit words instead of 32. Thus when I extract an integer from the data read in, it is aligned incorrectly and sets an array index that is out of range for the program space.
On the Mac OS X 10.12.6, using its C compiler which might be:
Apple LLVM version 8.0.0 (clang-800.0.33.1)
Target: x86_64-apple-darwin16.7.0
Is there a compiler switch that would set the compiled lengths of integers and floats to the above values? If not, how do I approach getting the code to correctly read the data?
Welcome to the world of portability headaches!
If your program was compiled in 1990, there is a good chance it uses 4 byte longs, and it is even possible that it use 2 byte int, depending on the architecture it was compiled for.
The size of basic C types is heavily system dependent, among a number of more subtle portability issues. long is now 64-bit on both 64-bit linux and 64-bit OS/X, but still 32-bit on Windows (for both 32-bit and 64-bit versions!).
Reading binary files, you must also deal with endianness, that changed from big-endian in 1990 MacOS to little-endian on today's OS/X, but still big-endian on other systems.
To make matters worse, the C language evolved over this long period and some non trivial semantic changes occurred between pre-ANSI C and Standard C. Some old syntaxes are no longer supported either...
There is no magic flag to address these issues, you will need to dive into the C code and understand what is does and try and modernize the code and make it more portable, independent on the target architecture. You can use the fixed width types from <stdint.h> to ease this process (int32_t, ...).
People answering C questions on Stackoverflow are usually careful to post portable code that works correctly for all target architectures, even some purposely vicious ones such as the DS9K (a ficticious computer that does everything in correct but unexpected ways).