ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<v4l2_init>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <open_camera@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <init_camera@plt>
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <start_capturing@plt>
  lea    D41(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <init_encoder@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <init_file@plt>
  lea    D35(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: tag: D9
type_guessed: string
name: read_data
arguments: tag: D38
type_guessed: string
name: read_data
arguments: tag: D41
type_guessed: string
name: read_data
arguments: tag: D29
type_guessed: string
name: read_data
arguments: tag: D35
type_guessed: string

tool: D9: open_camera 


tool: D38: init_camera 


tool: D41: start_capturing 


tool: D29: init_encoder 


tool: D35: init_file 


assistant: ```c
void v4l2_init(struct camera *cam) {
  open_camera(cam);
  printf("open_camera \n");
  init_camera(cam);
  printf("init_camera \n");
  start_capturing(cam);
  printf("start_capturing \n");
  init_encoder(cam);
  printf("init_encoder \n");
  init_file();
  printf("init_file \n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<v4l2_init>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <open_camera@plt>
  lea    D20(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <init_camera@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <start_capturing@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <init_encoder@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <init_file@plt>
  lea    D32(%rip),%rdi
  pop    %rbp
  jmp    10c0 <printf@plt>
```

tool_calls: name: parse_data
arguments: label: D20
datatype: string
name: parse_data
arguments: label: D0
datatype: string
name: parse_data
arguments: label: D15
datatype: string
name: parse_data
arguments: label: D1
datatype: string
name: parse_data
arguments: label: D32
datatype: string

tool: D20: open_camera 


tool: D0: init_camera 


tool: D15: start_capturing 


tool: D1: init_encoder 


tool: D32: init_file 


assistant: ```c
void v4l2_init(struct camera *cam) {
  open_camera(cam);
  printf("open_camera \n");
  init_camera(cam);
  printf("init_camera \n");
  start_capturing(cam);
  printf("start_capturing \n");
  init_encoder(cam);
  printf("init_encoder \n");
  init_file();
  printf("init_file \n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<v4l2_init>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <open_camera@plt>
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <init_camera@plt>
  lea    D44(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <start_capturing@plt>
  lea    D17(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <init_encoder@plt>
  lea    D30(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  call   <init_file@plt>
  lea    D33(%rip),%rdi
  pop    %rbp
  jmp    10c0 <printf@plt>
```

tool_calls: name: read_data
arguments: tag: D19
type_infered: string
name: read_data
arguments: tag: D44
type_infered: string
name: read_data
arguments: tag: D17
type_infered: string
name: read_data
arguments: tag: D30
type_infered: string
name: read_data
arguments: tag: D33
type_infered: string

tool: D19: open_camera 


tool: D44: init_camera 


tool: D17: start_capturing 


tool: D30: init_encoder 


tool: D33: init_file 


assistant: ```c
void v4l2_init(struct camera *cam) {
  open_camera(cam);
  printf("open_camera \n");
  init_camera(cam);
  printf("init_camera \n");
  start_capturing(cam);
  printf("start_capturing \n");
  init_encoder(cam);
  printf("init_encoder \n");
  init_file();
  printf("init_file \n");
}
```
user: What is the c source code of the assembly code below?

```asm
<aml_array_dbl_1d_new>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L11:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L3:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L11
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
double *aml_array_dbl_1d_new(int s1, double val) {
  int i;
  double *arr = (double *)malloc(s1 * sizeof(double));
  for (i = 0; i < s1; i++)
    arr[i] = val;
  return arr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<aml_array_dbl_1d_new>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  movq   %xmm0,%rbx
  lea    0x0(,%rdi,8),%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  jle    L10
  mov    %rax,%rdx
  mov    %ebp,%ebp
  lea    (%rax,%rbp,8),%rcx
L0:
  mov    %rbx,(%rdx)
  add    $0x8,%rdx
  cmp    %rcx,%rdx
  jne    L0
L10:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
double *aml_array_dbl_1d_new(int s1, double val) {
  int i;
  double *arr = (double *)malloc(s1 * sizeof(double));
  for (i = 0; i < s1; i++)
    arr[i] = val;
  return arr;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<aml_array_dbl_1d_new>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%rbx,8),%edi
  sub    $0x10,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L1
  movsd  0x8(%rsp),%xmm0
  mov    %rax,%rdx
  lea    (%rax,%rbx,8),%rcx
  xchg   %ax,%ax
L11:
  movsd  %xmm0,(%rdx)
  add    $0x8,%rdx
  cmp    %rcx,%rdx
  jne    L11
L1:
  add    $0x10,%rsp
  pop    %rbx
  ret
```

assistant: ```c
double *aml_array_dbl_1d_new(int s1, double val) {
  int i;
  double *arr = (double *)malloc(s1 * sizeof(double));
  for (i = 0; i < s1; i++)
    arr[i] = val;
  return arr;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<aml_array_dbl_1d_new>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  shl    $0x3,%edi
  sub    $0x10,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L12
  cmp    $0x1,%ebx
  movsd  0x8(%rsp),%xmm0
  je     L44
  mov    %ebx,%ecx
  movapd %xmm0,%xmm1
  mov    %rax,%rdx
  shr    %ecx
  unpcklpd %xmm1,%xmm1
  shl    $0x4,%rcx
  add    %rax,%rcx
  xchg   %ax,%ax
L32:
  movups %xmm1,(%rdx)
  add    $0x10,%rdx
  cmp    %rcx,%rdx
  jne    L32
  mov    %ebx,%edx
  and    $0xfffffffe,%edx
  and    $0x1,%ebx
  je     L12
L2:
  movslq %edx,%rdx
  movsd  %xmm0,(%rax,%rdx,8)
L12:
  add    $0x10,%rsp
  pop    %rbx
  ret
L44:
  xor    %edx,%edx
  jmp    L2
```

assistant: ```c
double *aml_array_dbl_1d_new(int s1, double val) {
  int i;
  double *arr = (double *)malloc(s1 * sizeof(double));
  for (i = 0; i < s1; i++)
    arr[i] = val;
  return arr;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<digit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  cmpl   $0x39,-0x14(%rbp)
  jg     L47
  mov    -0x14(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L56
L47:
  cmpl   $0x5a,-0x14(%rbp)
  jg     L18
  mov    -0x14(%rbp),%eax
  sub    $0x37,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L56
L18:
  cmpl   $0x7a,-0x14(%rbp)
  jg     L56
  mov    -0x14(%rbp),%eax
  sub    $0x57,%eax
  mov    %eax,-0x4(%rbp)
L56:
  mov    -0x4(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  jbe    L1
  mov    -0x4(%rbp),%eax
  jmp    L58
L1:
  mov    $0xffffffff,%eax
L58:
  pop    %rbp
  ret
```

assistant: ```c
int digit(int c, unsigned int base) {
  int n = -1;
  if (c <= '9') {
    n = c - '0';
  } else if (c <= 'Z') {
    n = c - 'A' + 10;
  } else if (c <= 'z') {
    n = c - 'a' + 10;
  }
  if ((unsigned int)n < base) {
    return n;
  }
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<digit>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x39,%edi
  jle    L3
  lea    -0x37(%rdi),%eax
  cmp    $0x5a,%edi
  jle    L3
  mov    $0xffffffff,%eax
  cmp    $0x7a,%edi
  jg     L29
  lea    -0x57(%rdi),%eax
L3:
  cmp    %esi,%eax
  jae    L42
L29:
  ret
L42:
  mov    $0xffffffff,%eax
  jmp    L29
```

assistant: ```c
int digit(int c, unsigned int base) {
  int n = -1;
  if (c <= '9') {
    n = c - '0';
  } else if (c <= 'Z') {
    n = c - 'A' + 10;
  } else if (c <= 'z') {
    n = c - 'a' + 10;
  }
  if ((unsigned int)n < base) {
    return n;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below:

```asm
<digit>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x39,%edi
  jle    L19
  lea    -0x37(%rdi),%eax
  cmp    $0x5a,%edi
  jle    L19
  mov    $0xffffffff,%eax
  cmp    $0x7a,%edi
  jg     L8
  lea    -0x57(%rdi),%eax
  nopl   0x0(%rax)
L19:
  cmp    %esi,%eax
  jae    L36
L8:
  ret
L36:
  or     $0xffffffff,%eax
  ret
```

assistant: ```c
int digit(int c, unsigned int base) {
  int n = -1;
  if (c <= '9') {
    n = c - '0';
  } else if (c <= 'Z') {
    n = c - 'A' + 10;
  } else if (c <= 'z') {
    n = c - 'a' + 10;
  }
  if ((unsigned int)n < base) {
    return n;
  }
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<digit>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x39,%edi
  jle    L39
  lea    -0x37(%rdi),%eax
  cmp    $0x5a,%edi
  jle    L39
  mov    $0xffffffff,%eax
  cmp    $0x7a,%edi
  jg     L36
  lea    -0x57(%rdi),%eax
  nopl   0x0(%rax)
L39:
  cmp    %esi,%eax
  jae    L5
L36:
  ret
L5:
  or     $0xffffffff,%eax
  ret
```

assistant: ```c
int digit(int c, unsigned int base) {
  int n = -1;
  if (c <= '9') {
    n = c - '0';
  } else if (c <= 'Z') {
    n = c - 'A' + 10;
  } else if (c <= 'z') {
    n = c - 'a' + 10;
  }
  if ((unsigned int)n < base) {
    return n;
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<deserialize_ip_address>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x10(%rbp),%rcx
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  leave
  ret
```

assistant: ```c
unsigned char *deserialize_ip_address(struct in6_addr *ip_address, unsigned char *data) {
  memcpy(ip_address->s6_addr, data, sizeof(struct in6_addr));
  return data + sizeof(struct in6_addr);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<deserialize_ip_address>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x4,%edx
  mov    (%rdi),%edi
  call   <memcpy@plt>
  lea    0x4(%rbx),%rax
  pop    %rbx
  ret
```

assistant: ```c
unsigned char *deserialize_ip_address(struct in6_addr *ip_address, unsigned char *data) {
  memcpy(ip_address->s6_addr, data, sizeof(struct in6_addr));
  return data + sizeof(struct in6_addr);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<deserialize_ip_address>:
  endbr64
  push   %rbx
  mov    (%rdi),%edi
  mov    %rsi,%rbx
  mov    $0x4,%edx
  call   <memcpy@plt>
  lea    0x4(%rbx),%rax
  pop    %rbx
  ret
```

assistant: ```c
unsigned char *deserialize_ip_address(struct in6_addr *ip_address, unsigned char *data) {
  memcpy(ip_address->s6_addr, data, sizeof(struct in6_addr));
  return data + sizeof(struct in6_addr);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<deserialize_ip_address>:
  endbr64
  push   %rbx
  mov    (%rdi),%edi
  mov    %rsi,%rbx
  mov    $0x4,%edx
  call   <memcpy@plt>
  lea    0x4(%rbx),%rax
  pop    %rbx
  ret
```

assistant: ```c
unsigned char *deserialize_ip_address(struct in6_addr *ip_address, unsigned char *data) {
  memcpy(ip_address->s6_addr, data, sizeof(struct in6_addr));
  return data + sizeof(struct in6_addr);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<plot6>:
  endbr64
  sub    $0xd8,%rsp
  mov    %rdx,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0xc8(%rsp)
  xor    %eax,%eax
  movslq %esi,%rsi
  movl   $0x0,(%rsp)
  movl   $0x0,0x4(%rsp)
  movl   $0x0,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0x14(%rsp)
  movl   $0x1,0x18(%rsp)
  movl   $0x2,0x1c(%rsp)
  movl   $0x3,0x20(%rsp)
  movl   $0x3,0x24(%rsp)
  movl   $0x3,0x28(%rsp)
  movl   $0x3,0x2c(%rsp)
  movl   $0x3,0x30(%rsp)
  movl   $0x3,0x34(%rsp)
  movl   $0x4,0x38(%rsp)
  movl   $0x4,0x3c(%rsp)
  movl   $0x5,0x40(%rsp)
  movl   $0x5,0x44(%rsp)
  movl   $0x6,0x48(%rsp)
  movl   $0x6,0x4c(%rsp)
  movl   $0x6,0x50(%rsp)
  movl   $0x6,0x54(%rsp)
  movl   $0x6,0x58(%rsp)
  lea    0x1(%rdi),%r10d
  mov    %r10d,0x60(%rsp)
  lea    0x2(%rdi),%r9d
  mov    %r9d,0x64(%rsp)
  lea    0x3(%rdi),%ecx
  mov    %ecx,0x68(%rsp)
  lea    0x4(%rdi),%edx
  mov    %edx,0x6c(%rsp)
  lea    0x5(%rdi),%eax
  mov    %eax,0x70(%rsp)
  lea    0x6(%rdi),%r11d
  mov    %r11d,0x74(%rsp)
  mov    %edi,0x78(%rsp)
  mov    %edi,0x7c(%rsp)
  mov    %edi,0x80(%rsp)
  mov    %r10d,0x84(%rsp)
  mov    %r9d,0x88(%rsp)
  mov    %ecx,0x8c(%rsp)
  mov    %edx,0x90(%rsp)
  mov    %eax,0x94(%rsp)
  mov    %edi,0x98(%rsp)
  mov    %r11d,0x9c(%rsp)
  mov    %edi,0xa0(%rsp)
  mov    %r11d,0xa4(%rsp)
  mov    %r10d,0xa8(%rsp)
  mov    %r9d,0xac(%rsp)
  mov    %ecx,0xb0(%rsp)
  mov    %edx,0xb4(%rsp)
  mov    %eax,0xb8(%rsp)
  mov    $0x0,%edx
L17:
  movslq (%rsp,%rdx,1),%rax
  imul   %rsi,%rax
  movslq 0x60(%rsp,%rdx,1),%rcx
  add    %r8,%rax
  movb   $0x36,(%rax,%rcx,1)
  add    $0x4,%rdx
  cmp    $0x5c,%rdx
  jne    L17
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0xd8,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void plot6(int posicao, int colunas, char matriz[7][colunas]) {
  int cx = 0;
  int cy = 0;
  int p = posicao;
  int x[] = {0, 0, 0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6};
  int y[] = {p + 1, p + 2, p + 3, p + 4, p + 5, p + 6, p + 0, p + 0, p + 0, p + 1, p + 2, p + 3, p + 4, p + 5, p + 0, p + 6, p + 0, p + 6, p + 1, p + 2, p + 3, p + 4, p + 5};
  int points = sizeof(x) / sizeof(int);
  for (cx = 0; cx < points; ++cx) {
    matriz[x[cx]][y[cx]] = '6';
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<plot6>:
  endbr64
  sub    $0xd8,%rsp
  mov    %edi,%eax
  mov    %rdx,%rdi
  movslq %esi,%rsi
  mov    %fs:0x28,%rdx
  mov    %rdx,0xc8(%rsp)
  xor    %edx,%edx
  lea    0x2(%rax),%r10d
  lea    0x3(%rax),%r9d
  mov    %eax,0x78(%rsp)
  lea    0x4(%rax),%r8d
  lea    0x6(%rax),%r11d
  mov    %eax,0x7c(%rsp)
  movabs $0x100000000,%rcx
  movabs $0x300000003,%rdx
  mov    %rcx,0x14(%rsp)
  movabs $0x300000002,%rcx
  mov    %rcx,0x1c(%rsp)
  movabs $0x400000003,%rcx
  mov    %rcx,0x34(%rsp)
  movabs $0x500000004,%rcx
  mov    %rdx,0x24(%rsp)
  mov    %rdx,0x2c(%rsp)
  movabs $0x600000006,%rdx
  mov    %rcx,0x3c(%rsp)
  movabs $0x600000005,%rcx
  mov    %rcx,0x44(%rsp)
  lea    0x1(%rax),%ecx
  mov    %rdx,0x4c(%rsp)
  mov    %rdx,0x54(%rsp)
  lea    0x5(%rax),%edx
  mov    %edx,0x70(%rsp)
  mov    %eax,0x80(%rsp)
  mov    %edx,0x94(%rsp)
  mov    %eax,0x98(%rsp)
  mov    %eax,0xa0(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x4(%rsp)
  movq   $0x0,0xc(%rsp)
  mov    %r10d,0x64(%rsp)
  mov    %r9d,0x68(%rsp)
  mov    %r8d,0x6c(%rsp)
  mov    %r11d,0x74(%rsp)
  mov    %ecx,0x84(%rsp)
  mov    %r10d,0x88(%rsp)
  mov    %r9d,0x8c(%rsp)
  mov    %r8d,0x90(%rsp)
  mov    %r11d,0x9c(%rsp)
  mov    %r11d,0xa4(%rsp)
  mov    %ecx,0xa8(%rsp)
  mov    %r10d,0xac(%rsp)
  mov    %edx,0xb8(%rsp)
  xor    %edx,%edx
  mov    %r9d,0xb0(%rsp)
  mov    %r8d,0xb4(%rsp)
  jmp    L2
  nopl   0x0(%rax,%rax,1)
L16:
  movslq (%rsp,%rdx,1),%rax
  mov    0x60(%rsp,%rdx,1),%ecx
L2:
  imul   %rsi,%rax
  movslq %ecx,%rcx
  add    $0x4,%rdx
  add    %rdi,%rax
  movb   $0x36,(%rax,%rcx,1)
  cmp    $0x5c,%rdx
  jne    L16
  mov    0xc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0xd8,%rsp
  ret
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void plot6(int posicao, int colunas, char matriz[7][colunas]) {
  int cx = 0;
  int cy = 0;
  int p = posicao;
  int x[] = {0, 0, 0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6};
  int y[] = {p + 1, p + 2, p + 3, p + 4, p + 5, p + 6, p + 0, p + 0, p + 0, p + 1, p + 2, p + 3, p + 4, p + 5, p + 0, p + 6, p + 0, p + 6, p + 1, p + 2, p + 3, p + 4, p + 5};
  int points = sizeof(x) / sizeof(int);
  for (cx = 0; cx < points; ++cx) {
    matriz[x[cx]][y[cx]] = '6';
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<SET_b_mem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    $0x4,%edi
  call   <update_all_cycles@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <get_mem@plt>
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    %edx,%edi
  call   <SET_b_r@plt>
  mov    %eax,-0x4(%rbp)
  mov    $0x4,%edi
  call   <update_all_cycles@plt>
  mov    -0x4(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <set_mem@plt>
  nop
  leave
  ret
```

assistant: ```c
void SET_b_mem(uint16_t const addr, uint8_t const bit) {
  update_all_cycles(4);
  uint8_t result = SET_b_r(get_mem(addr), bit);
  update_all_cycles(4);
  set_mem(addr, result);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<SET_b_mem>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    $0x4,%edi
  call   <update_all_cycles@plt>
  mov    %ebx,%edi
  call   <get_mem@plt>
  mov    %eax,%edi
  mov    %ebp,%esi
  call   <SET_b_r@plt>
  mov    %eax,%ebp
  mov    $0x4,%edi
  call   <update_all_cycles@plt>
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <set_mem@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void SET_b_mem(uint16_t const addr, uint8_t const bit) {
  update_all_cycles(4);
  uint8_t result = SET_b_r(get_mem(addr), bit);
  update_all_cycles(4);
  set_mem(addr, result);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<SET_b_mem>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  mov    $0x4,%edi
  sub    $0x8,%rsp
  call   <update_all_cycles@plt>
  mov    %ebp,%edi
  call   <get_mem@plt>
  mov    %r12d,%esi
  mov    %eax,%edi
  call   <SET_b_r@plt>
  mov    $0x4,%edi
  mov    %eax,%r12d
  call   <update_all_cycles@plt>
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1090 <set_mem@plt>
```

assistant: ```c
void SET_b_mem(uint16_t const addr, uint8_t const bit) {
  update_all_cycles(4);
  uint8_t result = SET_b_r(get_mem(addr), bit);
  update_all_cycles(4);
  set_mem(addr, result);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<SET_b_mem>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  mov    $0x4,%edi
  sub    $0x8,%rsp
  call   <update_all_cycles@plt>
  mov    %ebp,%edi
  call   <get_mem@plt>
  mov    %r12d,%esi
  mov    %eax,%edi
  call   <SET_b_r@plt>
  mov    $0x4,%edi
  mov    %eax,%r12d
  call   <update_all_cycles@plt>
  add    $0x8,%rsp
  mov    %r12d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    1090 <set_mem@plt>
```

assistant: ```c
void SET_b_mem(uint16_t const addr, uint8_t const bit) {
  update_all_cycles(4);
  uint8_t result = SET_b_r(get_mem(addr), bit);
  update_all_cycles(4);
  set_mem(addr, result);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<TC0_IrqHandler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
L4:
  jmp    L4
```

assistant: ```c
__attribute__((weak)) void TC0_IrqHandler(void) {
  while (1)
    ;
}
```
user: What is the c source code of the assembly code below:

```asm
<TC0_IrqHandler>:
  endbr64
L5:
  jmp    L5
```

assistant: ```c
__attribute__((weak)) void TC0_IrqHandler(void) {
  while (1)
    ;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<TC0_IrqHandler>:
  endbr64
  nopl   0x0(%rax)
L5:
  jmp    L5
```

assistant: ```c
__attribute__((weak)) void TC0_IrqHandler(void) {
  while (1)
    ;
}
```
user: Translate the below assembly code into C programming language:

```asm
<TC0_IrqHandler>:
  endbr64
  nopl   0x0(%rax)
L1:
  jmp    L1
```

assistant: ```c
__attribute__((weak)) void TC0_IrqHandler(void) {
  while (1)
    ;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<witness>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <power@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L4
L42:
  mov    -0x10(%rbp),%rax
  imul   %rax,%rax
  mov    $0x0,%edx
  divq   -0x18(%rbp)
  mov    %rdx,-0x8(%rbp)
  cmpq   $0x1,-0x8(%rbp)
  jne    L41
  cmpq   $0x1,-0x10(%rbp)
  je     L41
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  cmp    %rax,-0x10(%rbp)
  je     L41
  mov    $0x0,%eax
  jmp    L1
L41:
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  subq   $0x1,-0x20(%rbp)
L4:
  cmpq   $0x0,-0x20(%rbp)
  jne    L42
  cmpq   $0x1,-0x8(%rbp)
  je     L40
  mov    $0x0,%eax
  jmp    L1
L40:
  mov    $0x1,%eax
L1:
  leave
  ret
```

assistant: ```c
bool witness(size_t n, size_t s, size_t d, size_t a) {
  size_t x = power(a, d, n);
  size_t y;
  while (s) {
    y = (x * x) % n;
    if (y == 1 && x != 1 && x != n - 1)
      return false;
    x = y;
    --s;
  }
  if (y != 1)
    return false;
  return true;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<witness>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%rsi
  mov    %rcx,%rdi
  mov    %rbp,%rdx
  call   <power@plt>
  test   %rbx,%rbx
  je     L37
  lea    -0x1(%rbp),%rsi
  jmp    L29
L59:
  sub    $0x1,%rbx
  je     L18
L29:
  mov    %rax,%rcx
  imul   %rax,%rax
  mov    $0x0,%edx
  div    %rbp
  mov    %rdx,%rax
  cmp    $0x1,%rdx
  jne    L59
  cmp    $0x1,%rcx
  je     L59
  cmp    %rcx,%rsi
  je     L59
  mov    $0x0,%eax
  jmp    L11
L37:
  mov    $0x0,%eax
L18:
  cmp    $0x1,%rax
  sete   %al
  movzbl %al,%eax
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
bool witness(size_t n, size_t s, size_t d, size_t a) {
  size_t x = power(a, d, n);
  size_t y;
  while (s) {
    y = (x * x) % n;
    if (y == 1 && x != 1 && x != n - 1)
      return false;
    x = y;
    --s;
  }
  if (y != 1)
    return false;
  return true;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<witness>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  mov    %rcx,%rdi
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdx,%rsi
  mov    %rbp,%rdx
  sub    $0x8,%rsp
  call   <power@plt>
  test   %rbx,%rbx
  je     L45
  mov    %rax,%rdx
  lea    -0x1(%rbp),%rdi
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %rdx,%rax
  mov    %rdx,%rcx
  imul   %rdx,%rax
  xor    %edx,%edx
  div    %rbp
  cmp    $0x1,%rdx
  sete   %sil
  movzbl %sil,%eax
  cmp    $0x1,%rcx
  je     L26
  test   %sil,%sil
  je     L26
  cmp    %rcx,%rdi
  jne    L45
L26:
  sub    $0x1,%rbx
  jne    L5
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L45:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
bool witness(size_t n, size_t s, size_t d, size_t a) {
  size_t x = power(a, d, n);
  size_t y;
  while (s) {
    y = (x * x) % n;
    if (y == 1 && x != 1 && x != n - 1)
      return false;
    x = y;
    --s;
  }
  if (y != 1)
    return false;
  return true;
}
```