ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<geodist>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  mov    $0x3,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <norm@plt>
  movsd  D1(%rip),%xmm2
  movapd %xmm0,%xmm1
  movsd  D35(%rip),%xmm0
  comisd %xmm1,%xmm2
  ja     L2
  xor    %eax,%eax
L6:
  movsd  0x0(%rbp,%rax,1),%xmm0
  subsd  (%r12,%rax,1),%xmm0
  movsd  %xmm0,(%rbx,%rax,1)
  add    $0x8,%rax
  cmp    $0x18,%rax
  jne    L6
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <norm@plt>
  movapd %xmm0,%xmm1
  movsd  (%rbx),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rbx)
  movsd  0x8(%rbx),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%rbx)
  movsd  0x10(%rbx),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%rbx)
  movsd  0x0(%rbp),%xmm0
  movsd  0x8(%rbp),%xmm2
  mulsd  0x8(%r12),%xmm0
  mulsd  (%r12),%xmm2
  subsd  %xmm2,%xmm0
  mulsd  D20(%rip),%xmm0
  divsd  D16(%rip),%xmm0
  addsd  %xmm1,%xmm0
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D1
type_infered: f64
name: disassemble_data
arguments: tag: D35
type_infered: byte[8]
name: disassemble_data
arguments: tag: D20
type_infered: byte[8]
name: disassemble_data
arguments: tag: D16
type_infered: f64

tool: D1: 6378137

tool: {"D35": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

tool: {"D20": [0x5b,0x8d,0xcb,0xd7,0xa7,0x1d,0x13,0x3f]}

tool: D16: 299792458

assistant: ```c
extern double geodist(const double *rs, const double *rr, double *e) {
  double r;
  int i;
  if (norm(rs, 3) < 6378137.0) {
    return -1.0;
  }
  for (i = 0; i < 3; ++i) {
    e[i] = rs[i] - rr[i];
  }
  r = norm(e, 3);
  for (i = 0; i < 3; ++i) {
    e[i] /= r;
  }
  return r + 7.2921151467E-5 * (rs[0] * rr[1] - rs[1] * rr[0]) / 299792458.0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<geodist>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  mov    $0x3,%esi
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <norm@plt>
  movsd  D18(%rip),%xmm2
  movapd %xmm0,%xmm1
  movsd  D32(%rip),%xmm0
  comisd %xmm1,%xmm2
  ja     L13
  lea    0x8(%r12),%rax
  cmp    %rax,%rbx
  je     L0
  lea    0x8(%rbp),%rax
  cmp    %rax,%rbx
  je     L0
  movupd 0x0(%rbp),%xmm0
  movupd (%r12),%xmm3
  subpd  %xmm3,%xmm0
  movups %xmm0,(%rbx)
  movsd  0x10(%rbp),%xmm0
  subsd  0x10(%r12),%xmm0
  movsd  %xmm0,0x10(%rbx)
L10:
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <norm@plt>
  movapd %xmm0,%xmm1
  movapd %xmm0,%xmm2
  movupd (%rbx),%xmm0
  unpcklpd %xmm2,%xmm2
  divpd  %xmm2,%xmm0
  movups %xmm0,(%rbx)
  movsd  0x10(%rbx),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,0x10(%rbx)
  movsd  0x0(%rbp),%xmm0
  movsd  0x8(%rbp),%xmm2
  mulsd  0x8(%r12),%xmm0
  mulsd  (%r12),%xmm2
  subsd  %xmm2,%xmm0
  mulsd  D13(%rip),%xmm0
  divsd  D17(%rip),%xmm0
  addsd  %xmm1,%xmm0
L13:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L0:
  movsd  0x0(%rbp),%xmm0
  subsd  (%r12),%xmm0
  movsd  %xmm0,(%rbx)
  movsd  0x8(%rbp),%xmm0
  subsd  0x8(%r12),%xmm0
  movsd  %xmm0,0x8(%rbx)
  movsd  0x10(%rbp),%xmm0
  subsd  0x10(%r12),%xmm0
  movsd  %xmm0,0x10(%rbx)
  jmp    L10
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
data_type: f64
name: extract_symbolic_data
arguments: tag: D32
data_type: byte[8]
name: extract_symbolic_data
arguments: tag: D13
data_type: byte[8]
name: extract_symbolic_data
arguments: tag: D17
data_type: f64

tool: D18: 6378137

tool: {"D32": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0xbf]}

tool: {"D13": [0x5b,0x8d,0xcb,0xd7,0xa7,0x1d,0x13,0x3f]}

tool: D17: 299792458

assistant: ```c
extern double geodist(const double *rs, const double *rr, double *e) {
  double r;
  int i;
  if (norm(rs, 3) < 6378137.0) {
    return -1.0;
  }
  for (i = 0; i < 3; ++i) {
    e[i] = rs[i] - rr[i];
  }
  r = norm(e, 3);
  for (i = 0; i < 3; ++i) {
    e[i] /= r;
  }
  return r + 7.2921151467E-5 * (rs[0] * rr[1] - rs[1] * rr[0]) / 299792458.0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<string_from_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L24
L12:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x27,%al
  jne    L5
  addl   $0x1,-0xc(%rbp)
L5:
  addl   $0x1,-0xc(%rbp)
L24:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x22,%al
  jne    L12
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *string_from_string(char **string) {
  int length = 0;
  for (; (*string)[length] != '"'; length++) {
    if ((*string)[length] == '\'') {
      length++;
    }
  }
  char *result = malloc(length + 1);
  memcpy(result, *string, length);
  result[length] = 0;
  (*string) += (length + 1);
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<string_from_string>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    (%rdi),%rdx
  movzbl (%rdx),%eax
  cmp    $0x22,%al
  je     L12
  mov    $0x0,%ebx
L3:
  cmp    $0x27,%al
  sete   %al
  movzbl %al,%eax
  lea    0x1(%rbx,%rax,1),%ebx
  movslq %ebx,%rbp
  movzbl (%rdx,%rbp,1),%eax
  cmp    $0x22,%al
  jne    L3
L18:
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    %ebx,%edx
  mov    (%r12),%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,0x0(%r13,%rbp,1)
  add    $0x1,%rbp
  add    %rbp,(%r12)
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  jmp    L18
```

assistant: ```c
char *string_from_string(char **string) {
  int length = 0;
  for (; (*string)[length] != '"'; length++) {
    if ((*string)[length] == '\'') {
      length++;
    }
  }
  char *result = malloc(length + 1);
  memcpy(result, *string, length);
  result[length] = 0;
  (*string) += (length + 1);
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<string_from_string>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    (%rdi),%rdx
  movzbl (%rdx),%eax
  cmp    $0x22,%al
  je     L23
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L29:
  xor    %edi,%edi
  cmp    $0x27,%al
  sete   %dil
  add    %r12d,%edi
  lea    0x1(%rdi),%r12d
  movslq %r12d,%rbx
  movzbl (%rdx,%rbx,1),%eax
  cmp    $0x22,%al
  jne    L29
  add    $0x2,%edi
  lea    0x1(%rbx),%r14
L22:
  call   <malloc@plt>
  mov    0x0(%rbp),%rsi
  mov    %r12d,%edx
  mov    %rax,%r13
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,0x0(%r13,%rbx,1)
  mov    %r13,%rax
  pop    %rbx
  add    %r14,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L23:
  mov    $0x1,%r14d
  mov    $0x1,%edi
  xor    %r12d,%r12d
  xor    %ebx,%ebx
  jmp    L22
```

assistant: ```c
char *string_from_string(char **string) {
  int length = 0;
  for (; (*string)[length] != '"'; length++) {
    if ((*string)[length] == '\'') {
      length++;
    }
  }
  char *result = malloc(length + 1);
  memcpy(result, *string, length);
  result[length] = 0;
  (*string) += (length + 1);
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<string_from_string>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    (%rdi),%rdx
  movzbl (%rdx),%eax
  cmp    $0x22,%al
  je     L6
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L29:
  xor    %edi,%edi
  cmp    $0x27,%al
  sete   %dil
  add    %r12d,%edi
  lea    0x1(%rdi),%r12d
  movslq %r12d,%rbx
  movzbl (%rdx,%rbx,1),%eax
  cmp    $0x22,%al
  jne    L29
  add    $0x2,%edi
  lea    0x1(%rbx),%r14
L13:
  call   <malloc@plt>
  mov    0x0(%rbp),%rsi
  mov    %r12d,%edx
  mov    %rax,%r13
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,0x0(%r13,%rbx,1)
  mov    %r13,%rax
  pop    %rbx
  add    %r14,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L6:
  mov    $0x1,%r14d
  mov    $0x1,%edi
  xor    %r12d,%r12d
  xor    %ebx,%ebx
  jmp    L13
```

assistant: ```c
char *string_from_string(char **string) {
  int length = 0;
  for (; (*string)[length] != '"'; length++) {
    if ((*string)[length] == '\'') {
      length++;
    }
  }
  char *result = malloc(length + 1);
  memcpy(result, *string, length);
  result[length] = 0;
  (*string) += (length + 1);
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<u_swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  je     L9
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L9:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void u_swap(int *items, int i, int j) {
  if (i != j) {
    int temp = items[i];
    items[i] = items[j];
    items[j] = temp;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<u_swap>:
  endbr64
  cmp    %edx,%esi
  je     L6
  movslq %esi,%rsi
  lea    (%rdi,%rsi,4),%rcx
  mov    (%rcx),%esi
  movslq %edx,%rdx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
L6:
  ret
```

assistant: ```c
void u_swap(int *items, int i, int j) {
  if (i != j) {
    int temp = items[i];
    items[i] = items[j];
    items[j] = temp;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<u_swap>:
  endbr64
  cmp    %edx,%esi
  je     L6
  movslq %esi,%rsi
  movslq %edx,%rdx
  lea    (%rdi,%rsi,4),%rcx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
L6:
  ret
```

assistant: ```c
void u_swap(int *items, int i, int j) {
  if (i != j) {
    int temp = items[i];
    items[i] = items[j];
    items[j] = temp;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<u_swap>:
  endbr64
  cmp    %edx,%esi
  je     L3
  movslq %esi,%rsi
  movslq %edx,%rdx
  lea    (%rdi,%rsi,4),%rcx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rcx),%esi
  mov    (%rax),%edx
  mov    %edx,(%rcx)
  mov    %esi,(%rax)
L3:
  ret
```

assistant: ```c
void u_swap(int *items, int i, int j) {
  if (i != j) {
    int temp = items[i];
    items[i] = items[j];
    items[j] = temp;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gpio_print_header>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  sub    $0x8,%rsp
  lea    D30(%rip),%rdx
  push   %rdx
  lea    D0(%rip),%r9
  lea    D41(%rip),%r8
  lea    D1(%rip),%rdx
  mov    %rdx,%rcx
  lea    D25(%rip),%rdx
  lea    D58(%rip),%rsi
  mov    %rax,%rdi
  call   <asprintf@plt>
  add    $0x10,%rsp
  test   %rax,%rax
  jns    L8
  mov    $0xffffffff,%eax
  jmp    L5
L8:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <display_column_name@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x14(%rbp),%eax
L5:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D30
type: string
name: extract_symbolic_data
arguments: data_tag: D0
type: string
name: extract_symbolic_data
arguments: data_tag: D41
type: string
name: extract_symbolic_data
arguments: data_tag: D1
type: string
name: extract_symbolic_data
arguments: data_tag: D25
type: string
name: extract_symbolic_data
arguments: data_tag: D58
type: string

tool: D30: Direction

tool: D0: Edge

tool: D41: Active_low

tool: D1: Value

tool: D25: Name

tool: D58: %-20s %-10s %-10s %-10s %-10s

assistant: ```c
int gpio_print_header(void) {
  char *buf;
  int ret;
  if (asprintf(&buf, "%-20s %-10s %-10s %-10s %-10s", "Name", "Value", "Active_low", "Edge", "Direction") < 0)
    return -1;
  ret = display_column_name(buf);
  free(buf);
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<gpio_print_header>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  sub    $0x8,%rsp
  lea    D26(%rip),%rax
  push   %rax
  lea    D53(%rip),%r9
  lea    D1(%rip),%r8
  lea    D22(%rip),%rcx
  lea    D23(%rip),%rdx
  lea    D9(%rip),%rsi
  call   <asprintf@plt>
  add    $0x10,%rsp
  test   %rax,%rax
  js     L9
  mov    (%rsp),%rdi
  call   <display_column_name@plt>
  mov    %eax,%ebx
  mov    (%rsp),%rdi
  call   <free@plt>
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L9:
  mov    $0xffffffff,%ebx
  jmp    L7
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D26
type_guessed: string
name: retrieve_data
arguments: label: D53
type_guessed: string
name: retrieve_data
arguments: label: D1
type_guessed: string
name: retrieve_data
arguments: label: D22
type_guessed: string
name: retrieve_data
arguments: label: D23
type_guessed: string
name: retrieve_data
arguments: label: D9
type_guessed: string

tool: D26: Direction

tool: D53: Edge

tool: D1: Active_low

tool: D22: Value

tool: D23: Name

tool: D9: %-20s %-10s %-10s %-10s %-10s

assistant: ```c
int gpio_print_header(void) {
  char *buf;
  int ret;
  if (asprintf(&buf, "%-20s %-10s %-10s %-10s %-10s", "Name", "Value", "Active_low", "Edge", "Direction") < 0)
    return -1;
  ret = display_column_name(buf);
  free(buf);
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<gpio_print_header>:
  endbr64
  push   %r12
  lea    D2(%rip),%rcx
  lea    D58(%rip),%rdx
  lea    D11(%rip),%r9
  lea    D9(%rip),%r8
  lea    D45(%rip),%rsi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    D21(%rip),%rax
  sub    $0x8,%rsp
  push   %rax
  call   <asprintf@plt>
  pop    %rdx
  pop    %rcx
  test   %rax,%rax
  js     L20
  mov    (%rsp),%rdi
  call   <display_column_name@plt>
  mov    (%rsp),%rdi
  mov    %eax,%r12d
  call   <free@plt>
L22:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  mov    $0xffffffff,%r12d
  jmp    L22
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
data_type: string
name: extract_symbolic_data
arguments: data_label: D58
data_type: string
name: extract_symbolic_data
arguments: data_label: D11
data_type: string
name: extract_symbolic_data
arguments: data_label: D9
data_type: string
name: extract_symbolic_data
arguments: data_label: D45
data_type: string
name: extract_symbolic_data
arguments: data_label: D21
data_type: string

tool: D2: Value

tool: D58: Name

tool: D11: Edge

tool: D9: Active_low

tool: D45: %-20s %-10s %-10s %-10s %-10s

tool: D21: Direction

assistant: ```c
int gpio_print_header(void) {
  char *buf;
  int ret;
  if (asprintf(&buf, "%-20s %-10s %-10s %-10s %-10s", "Name", "Value", "Active_low", "Edge", "Direction") < 0)
    return -1;
  ret = display_column_name(buf);
  free(buf);
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<gpio_print_header>:
  endbr64
  push   %r12
  lea    D40(%rip),%rcx
  lea    D19(%rip),%rdx
  lea    D25(%rip),%r9
  lea    D43(%rip),%r8
  lea    D27(%rip),%rsi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    D5(%rip),%rax
  sub    $0x8,%rsp
  push   %rax
  call   <asprintf@plt>
  pop    %rdx
  pop    %rcx
  test   %rax,%rax
  js     L2
  mov    (%rsp),%rdi
  call   <display_column_name@plt>
  mov    (%rsp),%rdi
  mov    %eax,%r12d
  call   <free@plt>
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mov    $0xffffffff,%r12d
  jmp    L10
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D40
type_infered: string
name: extract_symbolic_data
arguments: label: D19
type_infered: string
name: extract_symbolic_data
arguments: label: D25
type_infered: string
name: extract_symbolic_data
arguments: label: D43
type_infered: string
name: extract_symbolic_data
arguments: label: D27
type_infered: string
name: extract_symbolic_data
arguments: label: D5
type_infered: string

tool: D40: Value

tool: D19: Name

tool: D25: Edge

tool: D43: Active_low

tool: D27: %-20s %-10s %-10s %-10s %-10s

tool: D5: Direction

assistant: ```c
int gpio_print_header(void) {
  char *buf;
  int ret;
  if (asprintf(&buf, "%-20s %-10s %-10s %-10s %-10s", "Name", "Value", "Active_low", "Edge", "Direction") < 0)
    return -1;
  ret = display_column_name(buf);
  free(buf);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<base_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %esi,-0x18(%rbp)
  mov    %al,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  movzbl -0x14(%rbp),%eax
  sub    $0x41,%eax
  cmp    $0x18,%eax
  ja     L44
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D16(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D16(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  movl   $0x1,-0x4(%rbp)
  movb   $0x0,-0x14(%rbp)
  jmp    L56
  movl   $0x1,-0x4(%rbp)
  movb   $0x1,-0x14(%rbp)
  jmp    L56
  movb   $0x2,-0x14(%rbp)
  jmp    L56
  movl   $0x1,-0x4(%rbp)
  movb   $0x3,-0x14(%rbp)
  jmp    L56
  movl   $0x1,-0x4(%rbp)
  movb   $0x0,-0x14(%rbp)
  jmp    L56
L44:
  movl   $0x2,-0x4(%rbp)
  movb   $0x0,-0x14(%rbp)
L56:
  shll   $0x2,-0x18(%rbp)
  movzbl -0x14(%rbp),%eax
  add    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int base_hash(unsigned char a, int b) {
  int base_status = 0;
  switch (a) {
  case 'A':
  case 'M':
  case 'R':
    base_status = 1;
  case 'W':
    a = 0;
    break;
  case 'C':
  case 'Y':
    base_status = 1;
  case 'S':
    a = 1;
    break;
  case 'G':
    a = 2;
    break;
  case 'T':
    base_status = 1;
  case 'K':
    a = 3;
    break;
  case 'N':
    base_status = 1;
    a = 0;
    break;
  default:
    base_status = 2;
    a = 0;
  }
  b = b << 2;
  b = b + a;
  return b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<base_hash>:
  endbr64
  lea    -0x43(%rdi),%ecx
  cmp    $0x16,%cl
  ja     L14
  mov    $0x1,%eax
  shl    %cl,%rax
  test   $0x410001,%eax
  jne    L5
  test   $0x20100,%eax
  jne    L33
  cmp    $0x4,%cl
  sete   %al
  add    %eax,%eax
L3:
  movzbl %al,%eax
  lea    (%rax,%rsi,4),%eax
  ret
L14:
  mov    $0x0,%eax
  jmp    L3
L5:
  mov    $0x1,%eax
  jmp    L3
L33:
  mov    $0x3,%eax
  jmp    L3
```

assistant: ```c
int base_hash(unsigned char a, int b) {
  int base_status = 0;
  switch (a) {
  case 'A':
  case 'M':
  case 'R':
    base_status = 1;
  case 'W':
    a = 0;
    break;
  case 'C':
  case 'Y':
    base_status = 1;
  case 'S':
    a = 1;
    break;
  case 'G':
    a = 2;
    break;
  case 'T':
    base_status = 1;
  case 'K':
    a = 3;
    break;
  case 'N':
    base_status = 1;
    a = 0;
    break;
  default:
    base_status = 2;
    a = 0;
  }
  b = b << 2;
  b = b + a;
  return b;
}
```
user: Translate the below assembly code into C programming language:

```asm
<base_hash>:
  endbr64
  lea    -0x43(%rdi),%ecx
  cmp    $0x16,%cl
  ja     L3
  mov    $0x1,%eax
  shl    %cl,%rax
  test   $0x410001,%eax
  jne    L39
  test   $0x20100,%eax
  jne    L2
  xor    %eax,%eax
  cmp    $0x4,%cl
  sete   %al
  add    %eax,%eax
L33:
  lea    (%rax,%rsi,4),%eax
  ret
L3:
  xor    %eax,%eax
  lea    (%rax,%rsi,4),%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L39:
  mov    $0x1,%eax
  lea    (%rax,%rsi,4),%eax
  ret
  nopl   0x0(%rax)
L2:
  mov    $0x3,%eax
  jmp    L33
```

assistant: ```c
int base_hash(unsigned char a, int b) {
  int base_status = 0;
  switch (a) {
  case 'A':
  case 'M':
  case 'R':
    base_status = 1;
  case 'W':
    a = 0;
    break;
  case 'C':
  case 'Y':
    base_status = 1;
  case 'S':
    a = 1;
    break;
  case 'G':
    a = 2;
    break;
  case 'T':
    base_status = 1;
  case 'K':
    a = 3;
    break;
  case 'N':
    base_status = 1;
    a = 0;
    break;
  default:
    base_status = 2;
    a = 0;
  }
  b = b << 2;
  b = b + a;
  return b;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<base_hash>:
  endbr64
  lea    -0x43(%rdi),%ecx
  cmp    $0x16,%cl
  ja     L34
  mov    $0x1,%eax
  shl    %cl,%rax
  test   $0x410001,%eax
  jne    L29
  test   $0x20100,%eax
  jne    L2
  xor    %eax,%eax
  cmp    $0x4,%cl
  sete   %al
  add    %eax,%eax
L31:
  lea    (%rax,%rsi,4),%eax
  ret
L34:
  xor    %eax,%eax
  lea    (%rax,%rsi,4),%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L29:
  mov    $0x1,%eax
  lea    (%rax,%rsi,4),%eax
  ret
  nopl   0x0(%rax)
L2:
  mov    $0x3,%eax
  jmp    L31
```

assistant: ```c
int base_hash(unsigned char a, int b) {
  int base_status = 0;
  switch (a) {
  case 'A':
  case 'M':
  case 'R':
    base_status = 1;
  case 'W':
    a = 0;
    break;
  case 'C':
  case 'Y':
    base_status = 1;
  case 'S':
    a = 1;
    break;
  case 'G':
    a = 2;
    break;
  case 'T':
    base_status = 1;
  case 'K':
    a = 3;
    break;
  case 'N':
    base_status = 1;
    a = 0;
    break;
  default:
    base_status = 2;
    a = 0;
  }
  b = b << 2;
  b = b + a;
  return b;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Setsockopt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %rcx,-0x18(%rbp)
  mov    %r8d,-0x10(%rbp)
  mov    -0x10(%rbp),%edi
  mov    -0x18(%rbp),%rcx
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%esi
  mov    -0x4(%rbp),%eax
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <setsockopt@plt>
  test   %rax,%rax
  jns    L1
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <err_sys@plt>
L1:
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
type: string

tool: D3: setsockopt error

assistant: ```c
void Setsockopt(int fd, int level, int optname, const void *optval, socklen_t optlen) {
  if (setsockopt(fd, level, optname, optval, optlen) < 0)
    err_sys("setsockopt error");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Setsockopt>:
  endbr64
  sub    $0x8,%rsp
  call   <setsockopt@plt>
  test   %rax,%rax
  js     L2
L17:
  add    $0x8,%rsp
  ret
L2:
  lea    D4(%rip),%rdi
  call   <err_sys@plt>
  jmp    L17
```

tool_calls: name: unpack_data
arguments: tag: D4
datatype: string

tool: D4: setsockopt error

assistant: ```c
void Setsockopt(int fd, int level, int optname, const void *optval, socklen_t optlen) {
  if (setsockopt(fd, level, optname, optval, optlen) < 0)
    err_sys("setsockopt error");
}
```