CODEWITHSUNDEEP
cendianness
MasterMind
MasterMind

Reputation: 61

Address operator on big endian system

Many people know the following example to find out whether a system is little endian or big endian:

#include <stdio.h> 
int main()  
{ 
   unsigned int i = 1; 
   char *c = (char*)&i; 
   if (*c)     
       printf("Little-endian"); 
   else
       printf("Big-endian"); 
   getchar(); 
   return 0; 
}

However, I would like to know whether the following piece of code is also still valid and functional:

#include <stdio.h>
int main()
{
   unsigned int i = 1;
   char *c = &i;
   if (*c)
       printf("Little-endian");
   else
       printf("Big-endian");
   getchar();
   return 0;
}

I want to ask whether the address of operator, &, works the same on both big-endian and little-endian systems. Will it in both cases point to the lowest addressed byte of the object when a pointer to an object is converted to a pointer to a character type?

My worry is the following. Suppose that you do not do the type conversion as above and also suppose a big-endian system returns you the highest bit address (basically where the int starts), then this piece of code will still print "Little-endian" although you are really dealing with a big-endian system.

Any comment on this would be really helpful. Thank you in advance :D.

Upvotes: 1

Views: 192

Answers (2)

Eric Postpischil
Eric Postpischil

Reputation: 223484

The assignment char *c = &i; violates the constraints in C 2018 6.5.16.1 1. That paragraph says “One of the following shall hold,” and the closest match in the list of options is:

  • the left operand has atomic, qualified, or unqualified pointer type, and (considering the type the left operand would have after lvalue conversion) both operands are pointers to qualified or unqualified versions of compatible types, and the type pointed to by the left has all the qualifiers of the type pointed to by the right;

char * and int * are not compatible, so the assignment fails the constraints.

That said, if the compiler did accept this statement (which it may, since the C standard permits extensions) and converts the pointer on the right to the type on the left (as assignments should), the result should be a pointer to the lowest addressed byte of the object, regardless of endianness. This is normal for pointer conversions per C 2018 6.3.2.3 7:

… When a pointer to an object is converted to a pointer to a character type, the result points to the lowest addressed byte of the object.

Upvotes: 4

Aganju
Aganju

Reputation: 6405

The second example will give you a compiler warning that the assignment is invalid without a casting. If you ignore the warning, you might get the same result or not, as always when trying undefined behavior.

Upvotes: 0

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