ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<inet_ntoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x20,%edx
  mov    $0x0,%esi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%rax
  movzbl %al,%esi
  mov    -0x8(%rbp),%rax
  shr    $0x8,%rax
  movzbl %al,%ecx
  mov    -0x8(%rbp),%rax
  shr    $0x10,%rax
  movzbl %al,%edx
  mov    -0x8(%rbp),%rax
  shr    $0x18,%rax
  movzbl %al,%eax
  mov    %esi,%r9d
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    %eax,%edx
  lea    D29(%rip),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    D32(%rip),%rax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D29
datatype: string

tool: D29: %d.%d.%d.%d

assistant: ```c
char *inet_ntoa(unsigned long addr) {
  static char addr_str[32];
  memset(addr_str, 0, 32);
  sprintf(addr_str, "%d.%d.%d.%d", (int)(addr >> 24 & 0xFF), (int)(addr >> 16 & 0xFF), (int)(addr >> 8 & 0xFF), (int)(addr & 0xFF));
  return addr_str;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<inet_ntoa>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    $0x20,%edx
  mov    $0x0,%esi
  lea    D10(%rip),%rbp
  mov    %rbp,%rdi
  call   <memset@plt>
  movzbl %bh,%eax
  mov    %eax,%r8d
  mov    %rbx,%rcx
  shr    $0x10,%rcx
  movzbl %cl,%ecx
  mov    %ebx,%edx
  shr    $0x18,%edx
  movzbl %bl,%r9d
  lea    D7(%rip),%rsi
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
datatype: string

tool: D7: %d.%d.%d.%d

assistant: ```c
char *inet_ntoa(unsigned long addr) {
  static char addr_str[32];
  memset(addr_str, 0, 32);
  sprintf(addr_str, "%d.%d.%d.%d", (int)(addr >> 24 & 0xFF), (int)(addr >> 16 & 0xFF), (int)(addr >> 8 & 0xFF), (int)(addr & 0xFF));
  return addr_str;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<inet_ntoa>:
  endbr64
  push   %r12
  lea    D1(%rip),%r12
  mov    $0x20,%edx
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  mov    %r12,%rdi
  sub    $0x8,%rsp
  call   <memset@plt>
  mov    %rbx,%rcx
  movzbl %bh,%eax
  mov    %ebx,%edx
  shr    $0x10,%rcx
  movzbl %bl,%r9d
  mov    %r12,%rdi
  mov    %eax,%r8d
  movzbl %cl,%ecx
  shr    $0x18,%edx
  lea    D4(%rip),%rsi
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D4
data_type: string

tool: D4: %d.%d.%d.%d

assistant: ```c
char *inet_ntoa(unsigned long addr) {
  static char addr_str[32];
  memset(addr_str, 0, 32);
  sprintf(addr_str, "%d.%d.%d.%d", (int)(addr >> 24 & 0xFF), (int)(addr >> 16 & 0xFF), (int)(addr >> 8 & 0xFF), (int)(addr & 0xFF));
  return addr_str;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<inet_ntoa>:
  endbr64
  push   %r12
  lea    D19(%rip),%r12
  mov    $0x20,%edx
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  mov    %r12,%rdi
  sub    $0x8,%rsp
  call   <memset@plt>
  mov    %rbx,%rcx
  movzbl %bh,%eax
  mov    %ebx,%edx
  shr    $0x10,%rcx
  movzbl %bl,%r9d
  mov    %r12,%rdi
  mov    %eax,%r8d
  movzbl %cl,%ecx
  shr    $0x18,%edx
  lea    D8(%rip),%rsi
  call   <sprintf@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
datatype: string

tool: D8: %d.%d.%d.%d

assistant: ```c
char *inet_ntoa(unsigned long addr) {
  static char addr_str[32];
  memset(addr_str, 0, 32);
  sprintf(addr_str, "%d.%d.%d.%d", (int)(addr >> 24 & 0xFF), (int)(addr >> 16 & 0xFF), (int)(addr >> 8 & 0xFF), (int)(addr & 0xFF));
  return addr_str;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<alldigits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L14
L6:
  addq   $0x1,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L6
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  sete   %al
  movzbl %al,%eax
  leave
  ret
```

assistant: ```c
int alldigits(const char *s) {
  while (isdigit(*s))
    s++;
  return *s == '\0';
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<alldigits>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L11
L2:
  add    $0x1,%rbx
L11:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L2
  cmpb   $0x0,(%rbx)
  sete   %al
  movzbl %al,%eax
  pop    %rbx
  ret
```

assistant: ```c
int alldigits(const char *s) {
  while (isdigit(*s))
    s++;
  return *s == '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<alldigits>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L11:
  add    $0x1,%rbx
L1:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L11
  xor    %eax,%eax
  cmpb   $0x0,(%rbx)
  pop    %rbx
  sete   %al
  ret
```

assistant: ```c
int alldigits(const char *s) {
  while (isdigit(*s))
    s++;
  return *s == '\0';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<alldigits>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L1:
  add    $0x1,%rbx
L12:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L1
  xor    %eax,%eax
  cmpb   $0x0,(%rbx)
  pop    %rbx
  sete   %al
  ret
```

assistant: ```c
int alldigits(const char *s) {
  while (isdigit(*s))
    s++;
  return *s == '\0';
}
```
user: What is the c source code of the assembly code below:

```asm
<FB_Ability_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <ChangPingJun_WuMing_bind@plt>
  mov    $0x0,%eax
  call   <ChangWenJun_LongJuanQiangFa_bind@plt>
  mov    $0x0,%eax
  call   <LaoAi_XieWuShi_bind@plt>
  mov    $0x0,%eax
  call   <HuanQi_KuangNuQiangFa_bind@plt>
  mov    $0x0,%eax
  call   <ZhongXieJianFa_bind@plt>
  mov    $0x0,%eax
  call   <HuBenBiFa_bind@plt>
  mov    $0x0,%eax
  call   <QinJunQiangFa_bind@plt>
  mov    $0x0,%eax
  call   <ZhuiYingQuan_bind@plt>
  mov    $0x0,%eax
  call   <WuLaiGunFa_bind@plt>
  mov    $0x0,%eax
  call   <XiuWuDuanHun_bind@plt>
  mov    $0x0,%eax
  call   <ZhenWuWoJianQi_bind@plt>
  mov    $0x0,%eax
  call   <JiBenBianFa_bind@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void FB_Ability_init() {
  ChangPingJun_WuMing_bind();
  ChangWenJun_LongJuanQiangFa_bind();
  LaoAi_XieWuShi_bind();
  HuanQi_KuangNuQiangFa_bind();
  ZhongXieJianFa_bind();
  HuBenBiFa_bind();
  QinJunQiangFa_bind();
  ZhuiYingQuan_bind();
  WuLaiGunFa_bind();
  XiuWuDuanHun_bind();
  ZhenWuWoJianQi_bind();
  JiBenBianFa_bind();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<FB_Ability_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <ChangPingJun_WuMing_bind@plt>
  mov    $0x0,%eax
  call   <ChangWenJun_LongJuanQiangFa_bind@plt>
  mov    $0x0,%eax
  call   <LaoAi_XieWuShi_bind@plt>
  mov    $0x0,%eax
  call   <HuanQi_KuangNuQiangFa_bind@plt>
  mov    $0x0,%eax
  call   <ZhongXieJianFa_bind@plt>
  mov    $0x0,%eax
  call   <HuBenBiFa_bind@plt>
  mov    $0x0,%eax
  call   <QinJunQiangFa_bind@plt>
  mov    $0x0,%eax
  call   <ZhuiYingQuan_bind@plt>
  mov    $0x0,%eax
  call   <WuLaiGunFa_bind@plt>
  mov    $0x0,%eax
  call   <XiuWuDuanHun_bind@plt>
  mov    $0x0,%eax
  call   <ZhenWuWoJianQi_bind@plt>
  mov    $0x0,%eax
  call   <JiBenBianFa_bind@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void FB_Ability_init() {
  ChangPingJun_WuMing_bind();
  ChangWenJun_LongJuanQiangFa_bind();
  LaoAi_XieWuShi_bind();
  HuanQi_KuangNuQiangFa_bind();
  ZhongXieJianFa_bind();
  HuBenBiFa_bind();
  QinJunQiangFa_bind();
  ZhuiYingQuan_bind();
  WuLaiGunFa_bind();
  XiuWuDuanHun_bind();
  ZhenWuWoJianQi_bind();
  JiBenBianFa_bind();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<FB_Ability_init>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ChangPingJun_WuMing_bind@plt>
  xor    %eax,%eax
  call   <ChangWenJun_LongJuanQiangFa_bind@plt>
  xor    %eax,%eax
  call   <LaoAi_XieWuShi_bind@plt>
  xor    %eax,%eax
  call   <HuanQi_KuangNuQiangFa_bind@plt>
  xor    %eax,%eax
  call   <ZhongXieJianFa_bind@plt>
  xor    %eax,%eax
  call   <HuBenBiFa_bind@plt>
  xor    %eax,%eax
  call   <QinJunQiangFa_bind@plt>
  xor    %eax,%eax
  call   <ZhuiYingQuan_bind@plt>
  xor    %eax,%eax
  call   <WuLaiGunFa_bind@plt>
  xor    %eax,%eax
  call   <XiuWuDuanHun_bind@plt>
  xor    %eax,%eax
  call   <ZhenWuWoJianQi_bind@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    11a0 <JiBenBianFa_bind@plt>
```

assistant: ```c
void FB_Ability_init() {
  ChangPingJun_WuMing_bind();
  ChangWenJun_LongJuanQiangFa_bind();
  LaoAi_XieWuShi_bind();
  HuanQi_KuangNuQiangFa_bind();
  ZhongXieJianFa_bind();
  HuBenBiFa_bind();
  QinJunQiangFa_bind();
  ZhuiYingQuan_bind();
  WuLaiGunFa_bind();
  XiuWuDuanHun_bind();
  ZhenWuWoJianQi_bind();
  JiBenBianFa_bind();
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<FB_Ability_init>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ChangPingJun_WuMing_bind@plt>
  xor    %eax,%eax
  call   <ChangWenJun_LongJuanQiangFa_bind@plt>
  xor    %eax,%eax
  call   <LaoAi_XieWuShi_bind@plt>
  xor    %eax,%eax
  call   <HuanQi_KuangNuQiangFa_bind@plt>
  xor    %eax,%eax
  call   <ZhongXieJianFa_bind@plt>
  xor    %eax,%eax
  call   <HuBenBiFa_bind@plt>
  xor    %eax,%eax
  call   <QinJunQiangFa_bind@plt>
  xor    %eax,%eax
  call   <ZhuiYingQuan_bind@plt>
  xor    %eax,%eax
  call   <WuLaiGunFa_bind@plt>
  xor    %eax,%eax
  call   <XiuWuDuanHun_bind@plt>
  xor    %eax,%eax
  call   <ZhenWuWoJianQi_bind@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    11a0 <JiBenBianFa_bind@plt>
```

assistant: ```c
void FB_Ability_init() {
  ChangPingJun_WuMing_bind();
  ChangWenJun_LongJuanQiangFa_bind();
  LaoAi_XieWuShi_bind();
  HuanQi_KuangNuQiangFa_bind();
  ZhongXieJianFa_bind();
  HuBenBiFa_bind();
  QinJunQiangFa_bind();
  ZhuiYingQuan_bind();
  WuLaiGunFa_bind();
  XiuWuDuanHun_bind();
  ZhenWuWoJianQi_bind();
  JiBenBianFa_bind();
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<del_delaunay_reduce_2D>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  movq   %rsi,%xmm0
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <delaunay_reduce_2D@plt>
  leave
  ret
```

assistant: ```c
int del_delaunay_reduce_2D(double min_lattice[3][3], double lattice[3][3], const int unique_axis, const double symprec) {
  ;
  return delaunay_reduce_2D(min_lattice, lattice, unique_axis, symprec);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<del_delaunay_reduce_2D>:
  endbr64
  sub    $0x8,%rsp
  call   <delaunay_reduce_2D@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int del_delaunay_reduce_2D(double min_lattice[3][3], double lattice[3][3], const int unique_axis, const double symprec) {
  ;
  return delaunay_reduce_2D(min_lattice, lattice, unique_axis, symprec);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<del_delaunay_reduce_2D>:
  endbr64
  jmp    1050 <delaunay_reduce_2D@plt>
```

assistant: ```c
int del_delaunay_reduce_2D(double min_lattice[3][3], double lattice[3][3], const int unique_axis, const double symprec) {
  ;
  return delaunay_reduce_2D(min_lattice, lattice, unique_axis, symprec);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<del_delaunay_reduce_2D>:
  endbr64
  jmp    1050 <delaunay_reduce_2D@plt>
```

assistant: ```c
int del_delaunay_reduce_2D(double min_lattice[3][3], double lattice[3][3], const int unique_axis, const double symprec) {
  ;
  return delaunay_reduce_2D(min_lattice, lattice, unique_axis, symprec);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_ts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    -0x18(%rbp),%rax
  imul   $0xf4240,%rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int64_t get_ts(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (int64_t)tv.tv_sec * 1000000LL + tv.tv_usec;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_ts>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x28,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t get_ts(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (int64_t)tv.tv_sec * 1000000LL + tv.tv_usec;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_ts>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t get_ts(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (int64_t)tv.tv_sec * 1000000LL + tv.tv_usec;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_ts>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x28,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int64_t get_ts(void) {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return (int64_t)tv.tv_sec * 1000000LL + tv.tv_usec;
}
```
user: What is the c source code of the assembly code below:

```asm
<rop3_218_1_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x1c(%rbp),%al
  mov    %al,-0x3(%rbp)
  notb   -0x3(%rbp)
  movzbl -0x20(%rbp),%eax
  and    -0x3(%rbp),%al
  mov    %al,-0x2(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  xor    -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %eax,%edx
  movzbl -0x1(%rbp),%eax
  and    %eax,%edx
  movzbl -0x20(%rbp),%eax
  not    %eax
  and    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x1c(%rbp),%al
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x20(%rbp),%al
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_218_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S & *D;
  stk3 = ~stk3;
  stk2 = T & stk3;
  stk1 = *D ^ stk2;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rop3_218_1_1>:
  endbr64
  mov    %esi,%eax
  and    (%rdi),%sil
  or     %edx,%eax
  not    %esi
  and    %edx,%esi
  xor    (%rdi),%sil
  mov    %eax,%edx
  not    %edx
  and    %edx,%esi
  and    (%rdi),%al
  or     %eax,%esi
  mov    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_218_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S & *D;
  stk3 = ~stk3;
  stk2 = T & stk3;
  stk1 = *D ^ stk2;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<rop3_218_1_1>:
  endbr64
  mov    %esi,%eax
  movzbl (%rdi),%esi
  and    %eax,%esi
  or     %edx,%eax
  not    %esi
  and    %edx,%esi
  mov    %eax,%edx
  xor    (%rdi),%sil
  and    (%rdi),%al
  not    %edx
  and    %edx,%esi
  or     %eax,%esi
  mov    %sil,(%rdi)
  ret
```

assistant: ```c
void rop3_218_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = S & *D;
  stk3 = ~stk3;
  stk2 = T & stk3;
  stk1 = *D ^ stk2;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<problem36>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L48
L6:
  cmpl   $0x3,-0x8(%rbp)
  jne    L36
  mov    $0x3,%edi
  call   <isPalindromic@plt>
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L36:
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <isPalindromic@plt>
  test   %eax,%eax
  je     L0
  cmpl   $0x3,-0x8(%rbp)
  jne    L11
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L11:
  mov    -0x8(%rbp),%eax
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <intToString@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
  cmpl   $0x3,-0x8(%rbp)
  jne    L0
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L0:
  addl   $0x1,-0x8(%rbp)
L48:
  cmpl   $0xf423f,-0x8(%rbp)
  jle    L6
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3,%edi
  call   <isPalindromic@plt>
  mov    %eax,%esi
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D29
type_infered: string
name: unpack_data
arguments: label: D11
type_infered: string
name: unpack_data
arguments: label: D23
type_infered: string
name: unpack_data
arguments: label: D51
type_infered: string
name: unpack_data
arguments: label: D25
type_infered: string
name: unpack_data
arguments: label: D53
type_infered: string

tool: D29: %d	 


tool: D11: adding

tool: D23: Decimal: %d	Binary: %s


tool: D51: added

tool: D25: 
Sum of all numbers, less than one million, which are palindromic in base 10 and base 2 == %d


tool: D53: 
isPalindromic: %d


assistant: ```c
void problem36() {
  int i, sum = 0;
  for (i = 1; i < 1000000; i++) {
    if (i == 3)
      printf("%d\t \n", isPalindromic(3));
    if (isPalindromic(i)) {
      if (i == 3)
        printf("adding\n");
      printf("Decimal: %d\tBinary: %s\n", i, intToString(i, 2));
      sum += i;
      if (i == 3)
        printf("added\n");
    }
  }
  printf("\nSum of all numbers, less than one million, which are palindromic in base 10 and base 2 == %d\n", sum);
  printf("\nisPalindromic: %d\n", isPalindromic(3));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<problem36>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x1,%ebx
  mov    $0x0,%r12d
  lea    D33(%rip),%r13
  lea    D26(%rip),%r14
  lea    D43(%rip),%r15
  jmp    L32
L9:
  mov    $0x3,%edi
  call   <isPalindromic@plt>
  mov    %eax,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3,%edi
  call   <isPalindromic@plt>
  test   %eax,%eax
  je     L58
  mov    %r15,%rdi
  call   <puts@plt>
  mov    $0x2,%esi
  mov    %ebx,%edi
  call   <intToString@plt>
  mov    %rax,%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    %ebx,%r12d
  lea    D12(%rip),%rdi
  call   <puts@plt>
  jmp    L58
L27:
  lea    0x1(%rbx),%eax
  cmp    $0xf423f,%eax
  jg     L12
L58:
  add    $0x1,%ebx
L32:
  cmp    $0x3,%ebx
  je     L9
  mov    %ebx,%edi
  call   <isPalindromic@plt>
  test   %eax,%eax
  je     L27
  mov    $0x2,%esi
  mov    %ebx,%edi
  call   <intToString@plt>
  mov    %rax,%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    %ebx,%r12d
  jmp    L27
L12:
  mov    %r12d,%esi
  lea    D40(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3,%edi
  call   <isPalindromic@plt>
  mov    %eax,%esi
  lea    D32(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D33
type_guessed: string
name: parse_data
arguments: label: D26
type_guessed: string
name: parse_data
arguments: label: D43
type_guessed: string
name: parse_data
arguments: label: D12
type_guessed: string
name: parse_data
arguments: label: D40
type_guessed: string
name: parse_data
arguments: label: D32
type_guessed: string

tool: D33: Decimal: %d	Binary: %s


tool: D26: %d	 


tool: D43: adding

tool: D12: added

tool: D40: 
Sum of all numbers, less than one million, which are palindromic in base 10 and base 2 == %d


tool: D32: 
isPalindromic: %d


assistant: ```c
void problem36() {
  int i, sum = 0;
  for (i = 1; i < 1000000; i++) {
    if (i == 3)
      printf("%d\t \n", isPalindromic(3));
    if (isPalindromic(i)) {
      if (i == 3)
        printf("adding\n");
      printf("Decimal: %d\tBinary: %s\n", i, intToString(i, 2));
      sum += i;
      if (i == 3)
        printf("added\n");
    }
  }
  printf("\nSum of all numbers, less than one million, which are palindromic in base 10 and base 2 == %d\n", sum);
  printf("\nisPalindromic: %d\n", isPalindromic(3));
}
```