Big endian test
Possible duplicate:
Small vs. Large Endians: How to interpret the test
Is there an easy way to test the code using gcc or any online compiler like ideone for big endian? I don't want to use qemu or virtual machines
EDIT
Can someone explain the behavior of this piece of code on the system using big end?
#include <stdio.h>
#include <string.h>
#include <stdint.h>
int main (void)
{
int32_t i;
unsigned char u[4] = {'a', 'b', 'c', 'd'};
memcpy(&i, u, sizeof(u));
printf("%d\n", i);
memcpy(u, &i, sizeof(i));
for (i = 0; i < 4; i++) {
printf("%c", u[i]);
}
printf("\n");
return 0;
}
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How is the program?
#include <stdio.h>
#include <stdint.h>
int main(int argc, char** argv) {
union {
uint32_t word;
uint8_t bytes[4];
} test_struct;
test_struct.word = 0x1;
if (test_struct.bytes[0] != 0)
printf("little-endian\n");
else
printf("big-endian\n");
return 0;
}
In little-endian architecture, the least significant byte is stored first. In a big end architecture, the most significant byte is stored first. That way, by superimposing a uint32_t
on a uint8_t[4]
, I can check that the first byte comes first. See: http://en.wikipedia.org/wiki/Big_endian
GCC specifically defines a macro __BYTE_ORDER__
as an extension. You can test against __ORDER_BIG_ENDIAN__
, __ORDER_LITTLE_ENDIAN__
and __ORDER_PDP_ENDIAN__
(which I didn't know!) - see http://gcc.gnu.org/onlinedocs/cpp/Common-Predefined-Macros.html
See also http://en.wikipedia.org/wiki/Big_endian
As for running code in an endianness that doesn't fit your intrinsic nature of your computer, then you have to compile and run it on an architecture with such varying degrees of precision. Thus, you will need to cross-compile and run on an emulator or virtual machine.
edit : ah, I haven't seen the first one printf()
.
The first printf
will print "1633837924" since the big end machine will interpret the character 'a'
as the most significant byte in an int.
The second one printf
will just print "abcd" since the value u
was copied byte by byte back and forth from i
.
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