#define NULL ((void *)0)
typedef unsigned long size_t; // Customize by platform.
typedef long intptr_t;
typedef unsigned long uintptr_t;
typedef long scalar_t__; // Either arithmetic or pointer type.
/* By default, we understand bool (as a convenience). */
typedef int bool;
#define false 0
#define true 1

/* Forward declarations */

/* Type definitions */

/* Variables and functions */
int /*<<< orphan*/ printf(char *);
void howto() {
  printf("\n\n\n\t\t\tHow to Play COW-BULL\n");
  printf("\t\t================================ \n\n");
  printf("\n\n\t -Cow-Bull is a code-breaking game.");
  printf("\n\n\t -The computer will hold a secret number.");
  printf("\n\n\t -All the digits in the number will be different.");
  printf("\n\n\t -You have to guess the secret number, who gives the number of "
         "matches,");
  printf("\n\n\t\t within 10 chances.");
  printf("\n\n\t -If the matching digits are on their exact positions, they "
         "are BULLs,");
  printf("\n\n\t\t  and if on different positions, they are COWs.");
  printf("\n\n\t -Example for 4-digits: \n\n\t\t Secret number:  4263 \n\n\t\t "
         "Opponent's try: 1234");
  printf("\n\n\t\t -Answer: 1 bull and 2 cows.\n\n\t\t -The bull is '2' and "
         "the cows are '4' & '3'.");
  printf(
      "\n\n\t Note : You can  quit the game any time by entering 'Q' or 'q'.");
}


x86 assembly (clang, -O0):
<howto>:
push   %rbp
mov    %rsp,%rbp
lea    0xee5(%rip),%rdi
call   1030 <printf@plt>
lea    0xef5(%rip),%rdi
call   1030 <printf@plt>
lea    0xf0f(%rip),%rdi
call   1030 <printf@plt>
lea    0xf2a(%rip),%rdi
call   1030 <printf@plt>
lea    0xf4b(%rip),%rdi
call   1030 <printf@plt>
lea    0xf74(%rip),%rdi
call   1030 <printf@plt>
lea    0xfb3(%rip),%rdi
call   1030 <printf@plt>
lea    0xfbf(%rip),%rdi
call   1030 <printf@plt>
lea    0xffd(%rip),%rdi
call   1030 <printf@plt>
lea    0x1025(%rip),%rdi
call   1030 <printf@plt>
lea    0x1068(%rip),%rdi
call   1030 <printf@plt>
lea    0x10ae(%rip),%rdi
call   1030 <printf@plt>
pop    %rbp
ret

x86 assembly (clang, -O2):
<howto>:
push   %rax
lea    0x1126(%rip),%rdi
call   1030 <puts@plt>
lea    0x1135(%rip),%rdi
call   1030 <puts@plt>
lea    0xec0(%rip),%rdi
call   1040 <printf@plt>
lea    0xedb(%rip),%rdi
call   1040 <printf@plt>
lea    0xefc(%rip),%rdi
call   1040 <printf@plt>
lea    0xf25(%rip),%rdi
call   1040 <printf@plt>
lea    0xf64(%rip),%rdi
call   1040 <printf@plt>
lea    0xf70(%rip),%rdi
call   1040 <printf@plt>
lea    0xfae(%rip),%rdi
call   1040 <printf@plt>
lea    0xfd6(%rip),%rdi
call   1040 <printf@plt>
lea    0x1019(%rip),%rdi
call   1040 <printf@plt>
lea    0x105f(%rip),%rdi
pop    %rax
jmp    1040 <printf@plt>

LLVM IR (-O0):
@.str = [28 x i8] c"\0A\0A\0A\09\09\09How to Play COW-BULL\0A\00"
@.str.1 = [38 x i8] c"\09\09================================ \0A\0A\00"
@.str.2 = [39 x i8] c"\0A\0A\09 -Cow-Bull is a code-breaking game.\00"
@.str.3 = [45 x i8] c"\0A\0A\09 -The computer will hold a secret number.\00"
@.str.4 = [53 x i8] c"\0A\0A\09 -All the digits in the number will be different.\00"
@.str.5 = [75 x i8] c"\0A\0A\09 -You have to guess the secret number, who gives the number of matches,\00"
@.str.6 = [24 x i8] c"\0A\0A\09\09 within 10 chances.\00"
@.str.7 = [74 x i8] c"\0A\0A\09 -If the matching digits are on their exact positions, they are BULLs,\00"
@.str.8 = [52 x i8] c"\0A\0A\09\09 and if on different positions, they are COWs.\00"
@.str.9 = [79 x i8] c"\0A\0A\09 -Example for 4-digits: \0A\0A\09\09 Secret number: 4263 \0A\0A\09\09 Opponent's try: 1234\00"
@.str.10 = [82 x i8] c"\0A\0A\09\09 -Answer: 1 bull and 2 cows.\0A\0A\09\09 -The bull is '2' and the cows are '4' & '3'.\00"
@.str.11 = [67 x i8] c"\0A\0A\09 Note : You can quit the game any time by entering 'Q' or 'q'.\00"
define void @howto() {
%1 = call i32 @printf(i8* getelementptr inbounds ([28 x i8], [28 x i8]* @.str, i64 0, i64 0))
%2 = call i32 @printf(i8* getelementptr inbounds ([38 x i8], [38 x i8]* @.str.1, i64 0, i64 0))
%3 = call i32 @printf(i8* getelementptr inbounds ([39 x i8], [39 x i8]* @.str.2, i64 0, i64 0))
%4 = call i32 @printf(i8* getelementptr inbounds ([45 x i8], [45 x i8]* @.str.3, i64 0, i64 0))
%5 = call i32 @printf(i8* getelementptr inbounds ([53 x i8], [53 x i8]* @.str.4, i64 0, i64 0))
%6 = call i32 @printf(i8* getelementptr inbounds ([75 x i8], [75 x i8]* @.str.5, i64 0, i64 0))
%7 = call i32 @printf(i8* getelementptr inbounds ([24 x i8], [24 x i8]* @.str.6, i64 0, i64 0))
%8 = call i32 @printf(i8* getelementptr inbounds ([74 x i8], [74 x i8]* @.str.7, i64 0, i64 0))
%9 = call i32 @printf(i8* getelementptr inbounds ([52 x i8], [52 x i8]* @.str.8, i64 0, i64 0))
%10 = call i32 @printf(i8* getelementptr inbounds ([79 x i8], [79 x i8]* @.str.9, i64 0, i64 0))
%11 = call i32 @printf(i8* getelementptr inbounds ([82 x i8], [82 x i8]* @.str.10, i64 0, i64 0))
%12 = call i32 @printf(i8* getelementptr inbounds ([67 x i8], [67 x i8]* @.str.11, i64 0, i64 0))
ret void
}

LLVM IR (-O2):
@.str.2 = [39 x i8] c"\0A\0A\09 -Cow-Bull is a code-breaking game.\00"
@.str.3 = [45 x i8] c"\0A\0A\09 -The computer will hold a secret number.\00"
@.str.4 = [53 x i8] c"\0A\0A\09 -All the digits in the number will be different.\00"
@.str.5 = [75 x i8] c"\0A\0A\09 -You have to guess the secret number, who gives the number of matches,\00"
@.str.6 = [24 x i8] c"\0A\0A\09\09 within 10 chances.\00"
@.str.7 = [74 x i8] c"\0A\0A\09 -If the matching digits are on their exact positions, they are BULLs,\00"
@.str.8 = [52 x i8] c"\0A\0A\09\09 and if on different positions, they are COWs.\00"
@.str.9 = [79 x i8] c"\0A\0A\09 -Example for 4-digits: \0A\0A\09\09 Secret number: 4263 \0A\0A\09\09 Opponent's try: 1234\00"
@.str.10 = [82 x i8] c"\0A\0A\09\09 -Answer: 1 bull and 2 cows.\0A\0A\09\09 -The bull is '2' and the cows are '4' & '3'.\00"
@.str.11 = [67 x i8] c"\0A\0A\09 Note : You can quit the game any time by entering 'Q' or 'q'.\00"
@str = [27 x i8] c"\0A\0A\0A\09\09\09How to Play COW-BULL\00"
@str.12 = [37 x i8] c"\09\09================================ \0A\00"
define void @howto() {
%1 = tail call i32 @puts(i8* dereferenceable(1) getelementptr inbounds ([27 x i8], [27 x i8]* @str, i64 0, i64 0))
%2 = tail call i32 @puts(i8* dereferenceable(1) getelementptr inbounds ([37 x i8], [37 x i8]* @str.12, i64 0, i64 0))
%3 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([39 x i8], [39 x i8]* @.str.2, i64 0, i64 0))
%4 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([45 x i8], [45 x i8]* @.str.3, i64 0, i64 0))
%5 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([53 x i8], [53 x i8]* @.str.4, i64 0, i64 0))
%6 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([75 x i8], [75 x i8]* @.str.5, i64 0, i64 0))
%7 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([24 x i8], [24 x i8]* @.str.6, i64 0, i64 0))
%8 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([74 x i8], [74 x i8]* @.str.7, i64 0, i64 0))
%9 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([52 x i8], [52 x i8]* @.str.8, i64 0, i64 0))
%10 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([79 x i8], [79 x i8]* @.str.9, i64 0, i64 0))
%11 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([82 x i8], [82 x i8]* @.str.10, i64 0, i64 0))
%12 = tail call i32 @printf(i8* dereferenceable(1) getelementptr inbounds ([67 x i8], [67 x i8]* @.str.11, i64 0, i64 0))
ret void
}