Why endianness required

In a system with several discrete hardware components — such as a host processor and external devices connected to it via a PCI bus, for example — the hardware components may support different endianness modes. Device driver developers need to make the data transfers among these hardware components endianness-proof. Endianness effect in software From a software standpoint, the 'endianness effect' comes into play when writing and reading data from memory.

To fully understand the implications we need to explain the use of data types in embedded software. All high-level programming languages support several data types. For example, C supports data types such as char, int, long, float, and so on, each having different memory storage length requirements for data elements of its type.

The lengths of data types of C are part standard such as defined by ANSI C and part compiler implementation dependent. For example, one compiler may implement char data type to be one byte in length and an int to be two bytes in length. Another compiler may implement char data type to be two bytes in length and an int to be four bytes in length.

To avoid confusion on lengths and to ensure portability across C compilers, embedded software programmers often define their own data types that explicitly give the number of bytes for the data type.

The software then uses these user defined types rather than C standard types. Guess what the value of var8 would be at Bookmark D? If you guessed 0x44 or any other value, read on. Remember that in a little endian system, when an element of multi-byte length data type is written to memory, the least significant byte is stored in the lowest address offset of memory. Whereas, in a big endian system the most significant byte is stored in the lowest address offset of memory.

Let us say in the above code, ptr32 takes the address value 0x from the malloc. The content of byte addressable locations starting at address 0x would look as follows at Bookmark B :.

From the above layout in memory, it is clear that var8 gets the value 0x11 in a little endian system and value 0x44 in a big endian system. At Bookmark H, var16 will have value 0x in a little endian system and 0x in a big endian system. When var16 get its value from address 0x using a bit two byte access, the two bytes starting at location 0x are read out. These turn out to be 0x11 and 0x22 in little endian system and 0x44 and 0x33 in big endian system.

Further, in a little endian system, of the two bytes that are read out, the byte stored at the lower address is interpreted as the least significant byte. That is, 0x11 is interpreted as the least significant byte and 0x22 is interpreted as the most significant byte resulting in var16 getting a value of 0x Likewise, in a big endian system, the byte stored in the lower address is interpreted as the most significant byte.

How do programmers ensure that they do not get surprises when accessing selective bytes of multi-byte length elements? A better way around the problem is to keep in mind that software sees the endianness effect only when mixing data types — specifically, when storing a certain byte-length element into memory and reading the same memory as a different byte-length element.

So, the solution is that, if a bit element was stored at a memory address, the content at that memory address needs to be read out as bit element only. Once it is read out of memory and is in a CPU register, the required bytes can be extracted from it. Hardware implications of endianness Today's complex SoCs have many hardware IP blocks integrated with the CPU core, which all communicate via interconnected buses. Each bus may serve a specific purpose for the hardware IP block, such as obtaining its configuration parameters, obtaining input data for processing, or giving out the output data after processing.

A bus may be designed in different widths such as , or bit lines, depending on the transfer bandwidth requirements of its hardware end points. Data transfer is done over a bus between the end points in units called transactions. A transaction can be a read type or a write type. So the LSB Byte will store at the lowest memory address. Note: Some processor has the ability to switch one endianness to other endianness using the software that means it can perform like both big endian or little endian at a time.

This processor is known as the Bi-endian. As we know that in little endian machine least significant byte of any multibyte data field is stored at the lowest memory address. So in the below program, we are checking the value of the lowest address. If the value is 1 then it will little endian either it will big endian. If your machine is little endian, the data in the memory will be something like the below expression:. We can also check the endianness of the machine using the union.

We need to create a union that has an integer variable and an array of 4 characters. If the first element au8DataBuff [0] of the character array is equal to the LSB Bytes of integer, then the system will be little endian otherwise big-endian. We can convert little endian to big endian or vice versa using the C programs. So let us see few ways to convert one endian to another. After the bit-wise operation if the machine is little endian the LSB store at lower address either the LSB stored at a higher address.

Mainly endianness affects the result when you perform the typecasting in your program, suppose you are creating a character array of four elements and you need to convert character array in a single integer element then what will be the result?

It depends on the endianness of the processor. If you compile this code on a little-endian processor then the output will be 0x01 but if you compiled it on the big-endian processor then the output will be 0x Perhaps at one time the designers of a given architecture decided that big-endianness was preferable to little-endianness, and as the architecture evolved over the years the endianness stayed the same.

JPEG is big-endian format, despite the fact that virtually all the machines that consume it are little-endian. While one can ask what are the benefits to JPEG being big-endian, I would venture out and say that for all intents and purposes the performance arguments mentioned above don't make a shred of difference. The fact is that JPEG was designed that way, and so long as it remains in use, that way it shall stay.

Once established, and for compatibility reasons, the endianness was more or less carried on to later generations of hardware; which would support the 'legacy' argument for why still both kinds exist today. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Collectives on Stack Overflow. Learn more. The reason behind endianness? Ask Question. Asked 8 years, 11 months ago.

Active 7 days ago. Viewed 12k times. Also I know that: The little-endian system has the property that the same value can be read from memory at different lengths without using different addresses. Matthias Braun Valentin Radu Valentin Radu 8, 8 8 gold badges 58 58 silver badges 90 90 bronze badges. It appears you're quoting the Wikipedia article on Endianness.

It can all be done in a simple loop with a simple end condition. Now try the same in BE order. Usually you need another divisor that holds the largest power of 10 for the specific number here You first need to find this, of course.

Much more stuff to do. The string to int conversion is easier to do in BE, when it is done as the reverse write operation. Write stores the most significant digit last, so it should be read first. Now do the same in LE order. Again, you'd need an additional factor starting with 1 and being multiplied by 10 for each digit. Thus I usually prefer to use BE for storage, because a value is written exactly once, but read at least once and maybe many times. For its simpler structure, I usually also go the route to convert to LE and then reverse the result, even if it writes the value a second time.

Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. What is the advantage of little endian format? Ask Question. Asked 10 years, 3 months ago. Active 1 year, 5 months ago. Viewed 87k times. Intel processors and maybe some others use the little endian format for storage.

Improve this question. Radu Murzea 1, 2 2 gold badges 17 17 silver badges 24 24 bronze badges. Cracker Cracker 2, 5 5 gold badges 17 17 silver badges 20 20 bronze badges. The was an early the first? I seem to remember some claim about it being little-endian for some performance-related issue due to the pipeline - but I have no idea now what that issue might have been. Any suggestions? Little-endian, big-endian - you must choose one or the other.

Like driving on the left or the right side of the road. I suggest you to write some code in ASM, preferably for an "old-school" architecture such as or Z You will immediately see why these use little endian. Architectures that use big endian have certain characteristics to their instruction set that make that format preferable instead. It's not an arbitrary decision to make! Each byte-order system has its advantages.

Little-endian machines let you read the lowest-byte first, without reading the others. You can check whether a number is odd or even last bit is 0 very easily, which is cool if you're into that kind of thing.

Big-endian systems store data in memory the same way we humans think about data left-to-right , which makes low-level debugging easier. Actually it's the Big Endian which has the bytes swapped. Why would you ever want to read as "a thousand"?. Show 9 more comments. Active Oldest Votes. In other words, if you have in memory a two byte value: 0x00f0 16 0x00f1 0 taking that '16' as a bit value c 'short' on most bit systems or as an 8-bit value generally c 'char' changes only the fetch instruction you use — not the address you fetch from.

On a big-endian system, with the above layed out as: 0x00f0 0 0x00f1 16 you would need to increment the pointer and then perform the narrower fetch operation on the new value. Improve this answer. Neil G 2 2 silver badges 14 14 bronze badges. Assuming, of course, that the high-order bytes that you didn't read can reasonably be ignored e.