ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<mainmemc>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0xa,%edi
  mov    $0x0,%eax
  call   <semget@plt>
  mov    %eax,%ebp
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0xa,%edi
  mov    $0x0,%eax
  call   <shmget@plt>
  mov    %eax,%edi
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%eax
  call   <shmat@plt>
  movslq %eax,%r12
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x0,%edi
  mov    $0x0,%eax
  call   <shmat@plt>
  movslq %eax,%rbx
  mov    $0x1,%esi
  mov    %ebp,%edi
  mov    $0x0,%eax
  call   <P@plt>
  mov    $0x0,%esi
  mov    %ebp,%edi
  mov    $0x0,%eax
  call   <V@plt>
  cmpl   $0x0,(%rbx)
  jne    L2
L7:
  mov    $0x1,%esi
  mov    %ebp,%edi
  mov    $0x0,%eax
  call   <P@plt>
  movsbl (%r12),%edi
  call   <putchar@plt>
  subl   $0x1,(%rbx)
  mov    $0x0,%esi
  mov    %ebp,%edi
  mov    $0x0,%eax
  call   <V@plt>
  cmpl   $0x0,(%rbx)
  je     L7
L2:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <shmdt@plt>
  mov    $0x5,%edi
  mov    $0x0,%eax
  call   <shmdt@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void mainmemc() {
  int shmid2;
  int shmid;
  int semid = semget(0XA, 0, 0);
  shmid = shmget(0XA, 0, 0);
  char *c;
  int *contador;
  char *aux;
  int *aux2;
  c = (char *)shmat(shmid, 0, 0);
  contador = (int *)shmat(shmid2, 0, 0);
  aux = c;
  aux2 = 5;
  P(semid, 1);
  V(semid, 0);
  int sale = 0;
  while (*contador == 0) {
    P(semid, 1);
    printf("%c", c[sale]);
    *contador = *contador - 1;
    V(semid, 0);
  }
  shmdt(aux);
  shmdt(aux2);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mainmemc>:
  endbr64
  push   %r12
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0xa,%edi
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  call   <semget@plt>
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0xa,%edi
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <shmget@plt>
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %eax,%edi
  xor    %eax,%eax
  call   <shmat@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  xor    %edx,%edx
  movslq %eax,%r12
  xor    %eax,%eax
  call   <shmat@plt>
  mov    $0x1,%esi
  mov    %ebp,%edi
  movslq %eax,%rbx
  xor    %eax,%eax
  call   <P@plt>
  jmp    L16
  nopw   0x0(%rax,%rax,1)
L0:
  mov    $0x1,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <P@plt>
  movsbl (%r12),%edi
  call   <putchar@plt>
  subl   $0x1,(%rbx)
L16:
  xor    %esi,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <V@plt>
  mov    (%rbx),%eax
  test   %eax,%eax
  je     L0
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <shmdt@plt>
  pop    %rbx
  mov    $0x5,%edi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    10f0 <shmdt@plt>
```

assistant: ```c
void mainmemc() {
  int shmid2;
  int shmid;
  int semid = semget(0XA, 0, 0);
  shmid = shmget(0XA, 0, 0);
  char *c;
  int *contador;
  char *aux;
  int *aux2;
  c = (char *)shmat(shmid, 0, 0);
  contador = (int *)shmat(shmid2, 0, 0);
  aux = c;
  aux2 = 5;
  P(semid, 1);
  V(semid, 0);
  int sale = 0;
  while (*contador == 0) {
    P(semid, 1);
    printf("%c", c[sale]);
    *contador = *contador - 1;
    V(semid, 0);
  }
  shmdt(aux);
  shmdt(aux2);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mainmemc>:
  endbr64
  push   %r12
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0xa,%edi
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  call   <semget@plt>
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0xa,%edi
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <shmget@plt>
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %eax,%edi
  xor    %eax,%eax
  call   <shmat@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  xor    %edx,%edx
  movslq %eax,%r12
  xor    %eax,%eax
  call   <shmat@plt>
  mov    $0x1,%esi
  mov    %ebp,%edi
  movslq %eax,%rbx
  xor    %eax,%eax
  call   <P@plt>
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L19:
  mov    $0x1,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <P@plt>
  movsbl (%r12),%edi
  call   <putchar@plt>
  subl   $0x1,(%rbx)
L10:
  xor    %esi,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <V@plt>
  mov    (%rbx),%eax
  test   %eax,%eax
  je     L19
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <shmdt@plt>
  pop    %rbx
  mov    $0x5,%edi
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  jmp    10f0 <shmdt@plt>
```

assistant: ```c
void mainmemc() {
  int shmid2;
  int shmid;
  int semid = semget(0XA, 0, 0);
  shmid = shmget(0XA, 0, 0);
  char *c;
  int *contador;
  char *aux;
  int *aux2;
  c = (char *)shmat(shmid, 0, 0);
  contador = (int *)shmat(shmid2, 0, 0);
  aux = c;
  aux2 = 5;
  P(semid, 1);
  V(semid, 0);
  int sale = 0;
  while (*contador == 0) {
    P(semid, 1);
    printf("%c", c[sale]);
    *contador = *contador - 1;
    V(semid, 0);
  }
  shmdt(aux);
  shmdt(aux2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  mov    $0x64,%esi
  lea    D83(%rip),%rdi
  call   <printf@plt>
  movsd  D55(%rip),%xmm0
  lea    D22(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  D19(%rip),%xmm0
  lea    D53(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D13(%rip),%rbx
  movsbl %bl,%ebx
  mov    %ebx,%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xf(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D91(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x13(%rsp),%r12
  mov    %r12,%rsi
  lea    D80(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rcx
  mov    %ebx,%edx
  mov    %rbp,%rsi
  lea    D51(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D83
type_infered: string
name: parse_data
arguments: tag: D55
type_infered: byte[8]
name: parse_data
arguments: tag: D22
type_infered: string
name: parse_data
arguments: tag: D19
type_infered: f64
name: parse_data
arguments: tag: D53
type_infered: string
name: parse_data
arguments: tag: D13
type_infered: string
name: parse_data
arguments: tag: D16
type_infered: string
name: parse_data
arguments: tag: D91
type_infered: string
name: parse_data
arguments: tag: D80
type_infered: string
name: parse_data
arguments: tag: D51
type_infered: string

tool: D83: You are %d miles away. 


tool: {"D55": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D22: You have %f levels of power. 


tool: D19: 56789.121344

tool: D53: You have %f awesome super powers. 


tool: D13: A

tool: D16: I have an initial %c. 


tool: D91: I have a first name %s.


tool: D80: I have a last name %s.


tool: D51: My whole name is %s %c.  %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.121344;
  char initial = "A";
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d miles away. \n", distance);
  printf("You have %f levels of power. \n", power);
  printf("You have %f awesome super powers. \n", super_power);
  printf("I have an initial %c. \n", initial);
  printf("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c.  %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x64,%esi
  lea    D71(%rip),%rdi
  push   %r12
  lea    D31(%rip),%r12
  push   %rbp
  movsbl %r12b,%r12d
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  lea    0xf(%rsp),%rbp
  lea    0x13(%rsp),%r13
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D79(%rip),%xmm0
  mov    $0x1,%eax
  lea    D54(%rip),%rdi
  call   <printf@plt>
  movsd  D94(%rip),%xmm0
  mov    $0x1,%eax
  lea    D51(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D55(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D97(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  lea    D64(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  lea    D99(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D71
type_infered: string
name: parse_data
arguments: data_label: D31
type_infered: string
name: parse_data
arguments: data_label: D79
type_infered: byte[8]
name: parse_data
arguments: data_label: D54
type_infered: string
name: parse_data
arguments: data_label: D94
type_infered: f64
name: parse_data
arguments: data_label: D51
type_infered: string
name: parse_data
arguments: data_label: D55
type_infered: string
name: parse_data
arguments: data_label: D97
type_infered: string
name: parse_data
arguments: data_label: D64
type_infered: string
name: parse_data
arguments: data_label: D99
type_infered: string

tool: D71: You are %d miles away. 


tool: D31: A

tool: {"D79": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D54: You have %f levels of power. 


tool: D94: 56789.121344

tool: D51: You have %f awesome super powers. 


tool: D55: I have an initial %c. 


tool: D97: I have a first name %s.


tool: D64: I have a last name %s.


tool: D99: My whole name is %s %c.  %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.121344;
  char initial = "A";
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d miles away. \n", distance);
  printf("You have %f levels of power. \n", power);
  printf("You have %f awesome super powers. \n", super_power);
  printf("I have an initial %c. \n", initial);
  printf("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c.  %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x64,%esi
  lea    D57(%rip),%rdi
  push   %r12
  lea    D59(%rip),%r12
  push   %rbp
  movsbl %r12b,%r12d
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  lea    0xf(%rsp),%rbp
  lea    0x13(%rsp),%r13
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D85(%rip),%xmm0
  mov    $0x1,%eax
  lea    D33(%rip),%rdi
  call   <printf@plt>
  movsd  D20(%rip),%xmm0
  mov    $0x1,%eax
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D66(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D56(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  lea    D67(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D57
type_infered: string
name: unpack_data
arguments: data_label: D59
type_infered: string
name: unpack_data
arguments: data_label: D85
type_infered: byte[8]
name: unpack_data
arguments: data_label: D33
type_infered: string
name: unpack_data
arguments: data_label: D20
type_infered: f64
name: unpack_data
arguments: data_label: D19
type_infered: string
name: unpack_data
arguments: data_label: D66
type_infered: string
name: unpack_data
arguments: data_label: D56
type_infered: string
name: unpack_data
arguments: data_label: D67
type_infered: string
name: unpack_data
arguments: data_label: D28
type_infered: string

tool: D57: You are %d miles away. 


tool: D59: A

tool: {"D85": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D33: You have %f levels of power. 


tool: D20: 56789.121344

tool: D19: You have %f awesome super powers. 


tool: D66: I have an initial %c. 


tool: D56: I have a first name %s.


tool: D67: I have a last name %s.


tool: D28: My whole name is %s %c.  %s.


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.121344;
  char initial = "A";
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("You are %d miles away. \n", distance);
  printf("You have %f levels of power. \n", power);
  printf("You have %f awesome super powers. \n", super_power);
  printf("I have an initial %c. \n", initial);
  printf("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c.  %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bound>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L32
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0xc(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <bound>
  jmp    L34
L32:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L17
  mov    -0x4(%rbp),%eax
  jmp    L34
L17:
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L41
  mov    -0xc(%rbp),%eax
  jmp    L34
L41:
  mov    -0x8(%rbp),%eax
L34:
  leave
  ret
```

assistant: ```c
int bound(int a, int x, int b) {
  if (b < a)
    return bound(b, x, a);
  if (x < a)
    return a;
  if (x > b)
    return b;
  return x;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bound>:
  endbr64
  mov    %edi,%ecx
  mov    %edx,%edi
  cmp    %ecx,%edx
  jl     L0
  cmp    %esi,%edx
  mov    %edx,%eax
  cmovg  %esi,%eax
  cmp    %esi,%ecx
  cmovg  %ecx,%eax
  ret
L0:
  sub    $0x8,%rsp
  mov    %ecx,%edx
  call   <bound>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int bound(int a, int x, int b) {
  if (b < a)
    return bound(b, x, a);
  if (x < a)
    return a;
  if (x > b)
    return b;
  return x;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<bound>:
  endbr64
  jmp    L6
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    %edi,%eax
  mov    %edx,%edi
  mov    %eax,%edx
L6:
  cmp    %edi,%edx
  jl     L1
  cmp    %esi,%edx
  mov    %edi,%eax
  cmovg  %esi,%edx
  cmp    %esi,%edi
  cmovle %edx,%eax
  ret
```

assistant: ```c
int bound(int a, int x, int b) {
  if (b < a)
    return bound(b, x, a);
  if (x < a)
    return a;
  if (x > b)
    return b;
  return x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bound>:
  endbr64
  jmp    L14
  cs nopw 0x0(%rax,%rax,1)
L18:
  mov    %edi,%eax
  mov    %edx,%edi
  mov    %eax,%edx
L14:
  cmp    %edi,%edx
  jl     L18
  cmp    %esi,%edx
  mov    %edi,%eax
  cmovg  %esi,%edx
  cmp    %esi,%edi
  cmovle %edx,%eax
  ret
```

assistant: ```c
int bound(int a, int x, int b) {
  if (b < a)
    return bound(b, x, a);
  if (x < a)
    return a;
  if (x > b)
    return b;
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<__imlib_ReCopyToRGB>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  lea    -0x14(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  sub    $0x7f,%eax
  add    %eax,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  or     %eax,%ecx
  mov    -0x4(%rbp),%eax
  shr    $0x9,%eax
  not    %eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  and    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  lea    -0x14(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  sub    $0x7f,%eax
  add    %eax,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  or     %eax,%ecx
  mov    -0x4(%rbp),%eax
  shr    $0x9,%eax
  not    %eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  and    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  lea    -0x14(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  sub    $0x7f,%eax
  add    %eax,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  neg    %eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  or     %eax,%edx
  mov    -0x4(%rbp),%eax
  shr    $0x9,%eax
  not    %eax
  and    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void __imlib_ReCopyToRGB(unsigned int color, unsigned int *dst) {
  unsigned int tmp;
  do {
    tmp = (((unsigned char *)(dst))[2]) + (((((unsigned char *)(&color))[2]) - 127) << 1);
    ((unsigned char *)(dst))[2] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
    tmp = (((unsigned char *)(dst))[1]) + (((((unsigned char *)(&color))[1]) - 127) << 1);
    ((unsigned char *)(dst))[1] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
    tmp = (((unsigned char *)(dst))[0]) + (((((unsigned char *)(&color))[0]) - 127) << 1);
    ((unsigned char *)(dst))[0] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
  } while (0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<__imlib_ReCopyToRGB>:
  endbr64
  mov    %edi,%edx
  mov    %rsi,%rax
  mov    %edi,%ecx
  shr    $0x10,%ecx
  movzbl %cl,%ecx
  movzbl 0x2(%rsi),%esi
  lea    -0xfe(%rsi,%rcx,2),%esi
  mov    %esi,%ecx
  shr    $0x8,%ecx
  neg    %ecx
  or     %esi,%ecx
  shr    $0x9,%esi
  not    %esi
  and    %esi,%ecx
  mov    %cl,0x2(%rax)
  movzbl %dh,%esi
  movzbl 0x1(%rax),%ecx
  lea    -0xfe(%rcx,%rsi,2),%esi
  mov    %esi,%ecx
  shr    $0x8,%ecx
  neg    %ecx
  or     %esi,%ecx
  shr    $0x9,%esi
  not    %esi
  and    %esi,%ecx
  mov    %cl,0x1(%rax)
  movzbl %dil,%edx
  movzbl (%rax),%ecx
  lea    -0xfe(%rcx,%rdx,2),%ecx
  mov    %ecx,%edx
  shr    $0x8,%edx
  neg    %edx
  or     %ecx,%edx
  shr    $0x9,%ecx
  not    %ecx
  and    %ecx,%edx
  mov    %dl,(%rax)
  ret
```

assistant: ```c
void __imlib_ReCopyToRGB(unsigned int color, unsigned int *dst) {
  unsigned int tmp;
  do {
    tmp = (((unsigned char *)(dst))[2]) + (((((unsigned char *)(&color))[2]) - 127) << 1);
    ((unsigned char *)(dst))[2] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
    tmp = (((unsigned char *)(dst))[1]) + (((((unsigned char *)(&color))[1]) - 127) << 1);
    ((unsigned char *)(dst))[1] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
    tmp = (((unsigned char *)(dst))[0]) + (((((unsigned char *)(&color))[0]) - 127) << 1);
    ((unsigned char *)(dst))[0] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
  } while (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__imlib_ReCopyToRGB>:
  endbr64
  mov    %edi,%ecx
  mov    %rsi,%rax
  movzbl 0x2(%rsi),%esi
  mov    %edi,%edx
  shr    $0x10,%ecx
  movzbl %cl,%ecx
  lea    -0xfe(%rsi,%rcx,2),%esi
  mov    %esi,%ecx
  shr    $0x8,%ecx
  neg    %ecx
  or     %esi,%ecx
  shr    $0x9,%esi
  not    %esi
  and    %esi,%ecx
  movzbl %dh,%esi
  movzbl %dil,%edx
  mov    %cl,0x2(%rax)
  movzbl 0x1(%rax),%ecx
  lea    -0xfe(%rcx,%rsi,2),%esi
  mov    %esi,%ecx
  shr    $0x8,%ecx
  neg    %ecx
  or     %esi,%ecx
  shr    $0x9,%esi
  not    %esi
  and    %esi,%ecx
  mov    %cl,0x1(%rax)
  movzbl (%rax),%ecx
  lea    -0xfe(%rcx,%rdx,2),%ecx
  mov    %ecx,%edx
  shr    $0x8,%edx
  neg    %edx
  or     %ecx,%edx
  shr    $0x9,%ecx
  not    %ecx
  and    %ecx,%edx
  mov    %dl,(%rax)
  ret
```

assistant: ```c
void __imlib_ReCopyToRGB(unsigned int color, unsigned int *dst) {
  unsigned int tmp;
  do {
    tmp = (((unsigned char *)(dst))[2]) + (((((unsigned char *)(&color))[2]) - 127) << 1);
    ((unsigned char *)(dst))[2] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
    tmp = (((unsigned char *)(dst))[1]) + (((((unsigned char *)(&color))[1]) - 127) << 1);
    ((unsigned char *)(dst))[1] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
    tmp = (((unsigned char *)(dst))[0]) + (((((unsigned char *)(&color))[0]) - 127) << 1);
    ((unsigned char *)(dst))[0] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
  } while (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<__imlib_ReCopyToRGB>:
  endbr64
  mov    %edi,%ecx
  mov    %rsi,%rax
  movzbl 0x2(%rsi),%esi
  mov    %edi,%edx
  shr    $0x10,%ecx
  movzbl %cl,%ecx
  lea    -0xfe(%rsi,%rcx,2),%esi
  mov    %esi,%ecx
  shr    $0x8,%ecx
  neg    %ecx
  or     %esi,%ecx
  shr    $0x9,%esi
  not    %esi
  and    %esi,%ecx
  movzbl %dh,%esi
  movzbl %dil,%edx
  mov    %cl,0x2(%rax)
  movzbl 0x1(%rax),%ecx
  lea    -0xfe(%rcx,%rsi,2),%esi
  mov    %esi,%ecx
  shr    $0x8,%ecx
  neg    %ecx
  or     %esi,%ecx
  shr    $0x9,%esi
  not    %esi
  and    %esi,%ecx
  mov    %cl,0x1(%rax)
  movzbl (%rax),%ecx
  lea    -0xfe(%rcx,%rdx,2),%ecx
  mov    %ecx,%edx
  shr    $0x8,%edx
  neg    %edx
  or     %ecx,%edx
  shr    $0x9,%ecx
  not    %ecx
  and    %ecx,%edx
  mov    %dl,(%rax)
  ret
```

assistant: ```c
void __imlib_ReCopyToRGB(unsigned int color, unsigned int *dst) {
  unsigned int tmp;
  do {
    tmp = (((unsigned char *)(dst))[2]) + (((((unsigned char *)(&color))[2]) - 127) << 1);
    ((unsigned char *)(dst))[2] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
    tmp = (((unsigned char *)(dst))[1]) + (((((unsigned char *)(&color))[1]) - 127) << 1);
    ((unsigned char *)(dst))[1] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
    tmp = (((unsigned char *)(dst))[0]) + (((((unsigned char *)(&color))[0]) - 127) << 1);
    ((unsigned char *)(dst))[0] = (tmp | (-(tmp >> 8))) & (~(tmp >> 9));
    ;
  } while (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
<cut_and_print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  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
  mov    %rax,%rdi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L36
L19:
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L10
L21:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
L10:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jb     L21
  addq   $0x1,-0x10(%rbp)
  cmpb   $0x0,-0x20(%rbp)
  je     L22
  mov    -0x10(%rbp),%rax
  jmp    L4
L22:
  mov    -0x18(%rbp),%rax
L4:
  mov    %rax,-0x8(%rbp)
L36:
  movsbl -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L19
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_infered: string
name: retrieve_data
arguments: data_tag: D19
type_infered: string

tool: D7: %s

tool: D19:  %s

assistant: ```c
void cut_and_print(char *string, char delimiter, char opt_short) {
  char *p, *s;
  assert(string);
  p = s = string;
  printf("%s", string);
  while ((p = strchr(p, delimiter))) {
    putchar(' ');
    while (s < p)
      putchar(*(s++));
    p++;
    s = (opt_short) ? p : string;
  }
  printf(" %s", s);
}
```