Sunday, July 8, 2007

What Little-Endian and Big-Endian? How can I determine whether a machine's byte order is big-endian or little endian? How can we convert from one to a

First of all, Do you know what Little-Endian and Big-Endian mean?

Little Endian means that the lower order byte of the number is stored in memory at the lowest address, and the higher order byte is stored at the highest address. That is, the little end comes first.

For example, a 4 byte, 32-bit integer

Byte3 Byte2 Byte1 Byte0

will be arranged in memory as follows:

Base_Address+0 Byte0
Base_Address+1 Byte1
Base_Address+2 Byte2
Base_Address+3 Byte3

Intel processors use "Little Endian" byte order.

"Big Endian" means that the higher order byte of the number is stored in memory at the lowest address, and the lower order byte at the highest address. The big end comes first.

Base_Address+0 Byte3
Base_Address+1 Byte2
Base_Address+2 Byte1
Base_Address+3 Byte0

Motorola, Solaris processors use "Big Endian" byte order.

In "Little Endian" form, code which picks up a 1, 2, 4, or longer byte number proceed in the same way for all formats. They first pick up the lowest order byte at offset 0 and proceed from there. Also, because of the 1:1 relationship between address offset and byte number (offset 0 is byte 0), multiple precision mathematic routines are easy to code. In "Big Endian" form, since the high-order byte comes first, the code can test whether the number is positive or negative by looking at the byte at offset zero. Its not required to know how long the number is, nor does the code have to skip over any bytes to find the byte containing the sign information. The numbers are also stored in the order in which they are printed out, so binary to decimal routines are particularly efficient.

Here is some code to determine what is the type of your machine

int num = 1;
if(*(char *)&num == 1)

And here is some code to convert from one Endian to another.

int myreversefunc(int num)
int byte0, byte1, byte2, byte3;

byte0 = (num & x000000FF) >> 0 ;
byte1 = (num & x0000FF00) >> 8 ;
byte2 = (num & x00FF0000) >> 16 ;
byte3 = (num & xFF000000) >> 24 ;

return((byte0 << 24) | (byte1 << 16) | (byte2 << 8) | (byte3 << 0));


Anonymous said...

Thanks Vijay..
Good Explaination..
Its easy for me to understand this now...

kavan said...

explanation is to the point
I couldn't find better explanation than this.

Anonymous said...

Union Solution

An alternative solution also exists where
a union is created of char[4] and int.

Pardon me for not writing exact C code.
unionvar.int_member = 0xABCDEF12 ;

if ( unionvar.char_member[0] == 0xAB )
print Biggie ; // MSB First
print little

Ivan Novick said...

For your reverse question it would also be nice if the function was more generic to accept multiple sizes of integer variables like 2, 4 or 8 byte integers:

void reverse(char* variable, unsigned short length){
for (unsigned short i = 0, j= length-1; i < length/2; ++i, --j){
char tmp = variable[i];
variable[i] = variable[j];
variable[j] = tmp;

sneha said...

nice explaination :)

Vijay Agrawal said...

Thanks guys for appreciating my work!

Scorpion said...

Copied directly from CrackTheInterview pdf file. Shame on you.
Show some creativity once in a while.

Dheeraj said...


dude what creativity you want ???
you give me better explanation than this.
i am sure you wont be able to ...

Ronak said...

unsigned char endian[2] = {1, 0};
short x;

x = *(short *) endian;

machine is little endian system
machine is big endian system

I think this may also work

Anonymous said...

@Vijay Your explanation was good and check methods were good.

However i feel this code is simple and gets the desired result:-


int main(){

char *p="12";

if((p - ++p) == -1)
printf("Little endian\n");
printf("Big endian\n");

return 0;

Anonymous said...

Certainly. I join told all above.

subburajr said...

unsigned long LITTLETOBIG(unsigned long x)
return ((x << 24) | ((x << 8) & 0x00ff0000) | ((x >> 8) & 0x0000ff00) | (x >> 24));
unsigned long BIGTOLITTLE(unsigned long x)

Anonymous said...

Nice explanation dude....

Anonymous said...

short & sweet explanation..
n easy to understand..

Adeel Chaudary said...

Excellent way to explain.After reading your post i have been able to completely understand the concept of big and little endian.

Anonymous said...

Thanks !

I had an assignment and all the other examples I found were written / documented like crap. Kudos!

Anonymous said...

@ Dheeraj and Vijay
guys its totally wrong to declare someone else work as your..
Direct copy from the book u stupid

Anonymous said...

i would report it to
guys .....may be it would make you upside down :D

JABoye47x said...

Thanks so much!!!

saravanan said...


Dhaval said...

Good explanation..

Anonymous said...


Anonymous said...


Anonymous said...

Really nice article...

Anonymous said...

Nice One

Anonymous said...

I tried a lot of places , yours is pretty a neat solution and explanation .... Good work Vijay !

vikas said...

good job keep going.

Anonymous said...

@ this program

int main(){

char *p="12";

if((p - ++p) == -1)
printf("Little endian\n");
printf("Big endian\n");

return 0;
in the if statement p refers to address of character 1 and ++p refers to character 2. As character is 1 byte it will always give -1. The correct statement is
if ((*p - *++p)=-1)

Aakhilesh said...


glatrix said...

gr8 work,, thanku so much.. :)

Anonymous said...

Best!! :)

Sourish Jana said...

Well said Vijay. This is proper and helps soooo much. Thanks a lot...

Anonymous said...

I have been taking operating systems class for the whole semester, and finally!! nice to meet you little endian and big endian. Thanks

Anonymous said...

It must have been a lot of effort to copy paste the whole crack the coding interview..

deep said...

According to what you explained, then the output of following code must depend on endianess of the machine:

int main()
int x=300;
/* greater than any value that a
signed or an unsigned char can
hold */
char *cptr=(char *)&x;
return 0;

In a big endian machine, value will be displayed according to the bit pattern in MSB while in a little endian machine, value will be displayed according to the bit pattern in LSB(300%256).

Is it true?

Anonymous said...

Never knew about the Solaris processors. ;-)

PRAVIN said...

nicely explained...............thanks.

Anonymous said...

Abhinandan Satpute said...

good one

Anonymous said...

@ Anonymous said...

i agree for your solution but it is totally compiler dependent means this method may drop SIDE_EFFECT.

Anonymous said...

Good Explanation .. thanks

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Anonymous said...

Hi Vijay,
What's your reasoning behind "*(char *)&num"? I "sort of" get it but I don't think it would help me very much by "kind of" knowing things.

If you want to do "*(char *)&num" why not declare "num" as a char instead of an int?

Any detail explanation of your logic is greatly appreciated.

Abinaya said...

Hi Vijay,

Please walk me through your code which determines the system being big endian or little endian.

Couldn't get it.

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