How to find out if a processor is "little endian" or "big endian"

[Ashli]

Hi all,

How to find out if a processor is "little endian" or "big endian" by writing C code???

[Silent1Mezzo]

Hi all,

How to find out if a processor is "little endian" or "big endian" by writing C code???

Just ask your processor which it is...wouldn't this be easiest?

Basically what you should do is research the subject, think of how you're going to code this, start coding it....THEN come back here and ask about any problems you have.

Thanks

[weaknessforcats]

Put a 1 in an unsigned int.

Add three bytes to the address of the int.

Assign the contents of that address to an unsigned char.

If your 1 is there, you are big endian.

[RedSon]

Put a 1 in an unsigned int.

Add three bytes to the address of the int.

Assign the contents of that address to an unsigned char.

If your 1 is there, you are big endian.

Explanation needed !

[AdrianH]

Put a 1 in an unsigned int.

Add three bytes to the address of the int.

Assign the contents of that address to an unsigned char.

If your 1 is there, you are big endian.

Not necessarily, you could just have a random 1 on the stack.

Try using a union with a 4 byte char and a long. Put 1 in the long and then check where the one showed up on the 4 byte char. Zeroth byte is little endian, third byte is big endian.

BTW, this trick only sorta works. There are processors that you can map different regions of memory to be big or little endian. The hardware does everything for you.

Also there are some processors out there that do it on a bit rather then on a byte level. I don't know which, but there is literature stating that this design exists.

To further complicate things, apparently floating point and integer endianness don't have to coincide.


Adrian

[weaknessforcats]

BTW, this trick only sorta works. There are processors that you can map different regions of memory to be big or little endian. The hardware does everything for you.

True. But at the byte level there is no endian-ess.

Your union solution is a good one. It is equivalent to my suggestion.

I didn't know floating point had its own endian.

[AdrianH]

True. But at the byte level there is no endian-ess.

Yes, usually HW does magic on what is actually stored.

Your union solution is a good one. It is equivalent to my suggestion.

Now that I think about it, your right. I think that the union is somewhat cleaner, but they are equivialant.

I didn't know floating point had its own endian.

It apperently can, I read that somewhere. Maybe on wikipidia.


Adrian

[narasimha435]

i also want the reply interms of c or c++ code format

[weaknessforcats]

Just follow the steps I outlined. The code is simple and is the same for C and C++.

[aalam]

int a = 10;
char *ch = (char*)a;
printf("%d",*ch);

Execute this code, if output is "10" then your processor is "Little Endian" and if output is "0" then your processor is "Big Endian".

[JosAH]

There are a few nice macros/functions available: hton*() and ntoh*(). The following
does the job for you:

#define BIGENDIAN (htons(1) == 1)
 

kind regards,

Jos

[archonmagnus]

Here's a code that'll do it for you (for educational purposes):

#include <iostream>
 
using namespace std;
 
int main (int argc, char *argv[])
{
    unsigned int i;
    unsigned char *p = (unsigned char *)&i;
 
    cout<<"Enter an unsigned integer value: ";
    cin>>i;
 
    cout<<"i = "<<i<<endl
        <<"Comparison: &i = "<<&i<<endl
        <<"            *p = "<<(int *)p<<endl<<endl
        <<"Byte 1: "<<(int)p[0]<<endl
        <<"Byte 2: "<<(int)p[1]<<endl
        <<"Byte 3: "<<(int)p[2]<<endl
        <<"Byte 4: "<<(int)p[3]<<endl;
 
    unsigned int foo = ((i % (1 << 24)) % (1 << 16)) % (1 << 8);
 
    if ((unsigned int)p[0] == foo)
        cout<<endl<<"    === Little Endian ==="<<endl;
    else
        cout<<endl<<"    === Big Endian ==="<<endl;
 
    cout<<endl;
 
    return 0;
}