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/* Explanation of byte-reversing in ARM assembler
for PC only (uses cursor control codes)
*/
#include <stdio.h>
static void cls()
{
printf("\033[2J");
}
static void up(int n)
{
while (n > 0)
{ printf("\033[1A"); --n;
}
}
static void pause()
{
fprintf(stderr, "Please press <Return> to proceeed ");
while( fgetc(stdin) != '\n' );
}
static void prologue()
{
printf("\
AMAZING FACTS ABOUT THE ARM - Reversing the bytes in a word\n\
\n\
This function reverses the bytes in a word. The method was discovered in\n\
1986 following a competition between ARM programmers; it requires just 4\n\
instructions and 1 work register. A method using only 3 instructions per\n\
word reversed was also found, but it has some set-up overhead and uses a\n\
2nd register. Can you re-discover this method?\n\
\n\
Later, the C compiler was 'taught' to generate exactly the instructions\n\
needed, from C source. Check this claim using armcc -S -DREV byterev.c\n\
and examining the assembly code file produced.\n\
\n\
unsigned long reverse(unsigned long v)\n\
{ unsigned long t;\n\
t = v ^ ((v << 16) | (v >> 16)); /* EOR r1,r0,r0,ROR #16 */\n\
t &= ~0xff0000; /* BIC r1,r1,#&ff0000 */\n\
v = (v << 24) | (v >> 8); /* MOV r0,r0,ROR #8 */\n\
return v ^ (t >> 8); /* EOR r0,r0,r1,LSR #8 */\n\
}\n\
\n\
To see the method in action, press <Return>. Each time you press <Return>\n\
one step of the reversal process will be executed. The values displayed\n\
are symbolic, starting with the input word D C B A.\n\
\n");
}
static void prelude()
{
printf("\
AMAZING FACTS ABOUT THE ARM - Reversing the bytes in a word\n\
\n\
unsigned long reverse(unsigned long v)\n\
{ unsigned long t;\n\
t = v ^ ((v << 16) | (v >> 16)); /* EOR r1,r0,r0,ROR #16 */\n\
t &= ~0xff0000; /* BIC r1,r1,#&ff0000 */\n\
v = (v << 24) | (v >> 8); /* MOV r0,r0,ROR #8 */\n\
return v ^ (t >> 8); /* EOR r0,r0,r1,LSR #8 */\n\
}\n\
\n");
}
static void show_state_1()
{
printf("\
v / r0 t / r1 original input in v/r0\n\
\n\
+---+---+---+---+ +---+---+---+---+\n\
| D | C | B | A | | x | x | x | x |\n\
+---+---+---+---+ +---+---+---+---+\n\
\n\n\n\n\n\n\n");
}
static void show_state_2()
{
printf("\
v / r0 t / r1 state after executing \n\
\n\
+---+---+---+---+ +---+---+---+---+ t = v ^ ((v<<16) | (v>>16));\n\
| D | C | B | A | |D^B|C^A|B^D|A^C|\n\
+---+---+---+---+ +---+---+---+---+ EOR r1,r0,r0,ROR #16\n\
\n\n\n\n\n\n\n");
}
static void show_state_3()
{
printf("\
+---+---+---+---+ +---+---+---+---+ t &= ~0xff0000; \n\
| D | C | B | A | |D^B| 0 |B^D|A^C|\n\
+---+---+---+---+ +---+---+---+---+ BIC r1,r1,#&ff0000 \n\
\n\n\n\n\n\n\n");
}
static void show_state_4()
{
printf("\
+---+---+---+---+ +---+---+---+---+ v = (v << 24) | (v >> 8);\n\
| A | D | C | B | |D^B| 0 |B^D|A^C|\n\
+---+---+---+---+ +---+---+---+---+ MOV r0,r0,ROR #8 \n\
\n\n\n\n\n\n\n");
}
static void show_state_5()
{
printf("\
+---+---+---+---+ +---+---+---+---+ v = v ^ (t >> 8); \n\
\n\
+---+-^-+-^-+-^-+---+ +---+---+---+---+ EOR r0,r0,r1,LSR #8\n\
|D^B| 0 |B^D|A^C|\n\
+---+---+---+---+\n\
\n\
+---+---+---+---+\n\
| A | B | C | D | (original input was: D C B A)\n\
+---+---+---+---+\n\
\n");
}
int main()
{
cls();
prologue();
pause();
cls();
prelude();
show_state_1();
pause();
up(13);
show_state_2();
pause();
up(11);
show_state_3();
pause();
up(11);
show_state_4();
pause();
up(11);
show_state_5();
pause();
fputc('\n', stdout);
return( 0 );
}
|
