ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%edi
  call   <omp_set_num_threads@plt>
  xor    %eax,%eax
  call   <omp_get_wtime@plt>
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  call   <omp_get_thread_num@plt>
  pxor   %xmm2,%xmm2
  lea    D22(%rip),%rdi
  cvtsi2sd %eax,%xmm2
  mov    $0x1,%eax
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x8(%rsp)
  call   <printf@plt>
  xor    %eax,%eax
  call   <tarea_uno@plt>
  movsd  0x8(%rsp),%xmm2
  mov    $0x1,%eax
  lea    D29(%rip),%rdi
  movapd %xmm2,%xmm0
  call   <printf@plt>
  xor    %eax,%eax
  call   <tarea_dos@plt>
  xor    %eax,%eax
  call   <omp_get_wtime@plt>
  movsd  0x8(%rsp),%xmm2
  mov    $0x2,%eax
  lea    D20(%rip),%rdi
  movapd %xmm0,%xmm1
  subsd  (%rsp),%xmm1
  movapd %xmm2,%xmm0
  call   <printf@plt>
  xor    %eax,%eax
  call   <omp_get_wtime@plt>
  subsd  (%rsp),%xmm0
  mov    $0x1,%eax
  lea    D27(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D22
type_guessed: string
name: unpack_data
arguments: data_label: D29
type_guessed: string
name: unpack_data
arguments: data_label: D20
type_guessed: string
name: unpack_data
arguments: data_label: D27
type_guessed: string

tool: D22: Hilo %i ejecutando tarea 1 


tool: D29: Hilo %i ejecutando tarea 2



tool: D20: Tiempo de salida hilo %i = %.4lf segundos.



tool: D27: Tiempo tardado = %.4lf segundos



assistant: ```c
int main() {
  double timeIni, timeFin, time_s1, time_s2;
  int nthread;
  omp_set_num_threads(4);
  timeIni = omp_get_wtime();
#pragma omp parallel private(nthread)
  {
    nthread = omp_get_thread_num();
#pragma omp sections nowait
    {
#pragma omp section
      {
        printf("Hilo %i ejecutando tarea 1 \n", nthread);
        tarea_uno();
      }
#pragma omp section
      {
        printf("Hilo %i ejecutando tarea 2\n\n", nthread);
        tarea_dos();
      }
    }
    printf("Tiempo de salida hilo %i = %.4lf segundos.\n\n", nthread, (omp_get_wtime() - timeIni));
  }
  timeFin = omp_get_wtime();
  printf("Tiempo tardado = %.4lf segundos\n\n", (timeFin - timeIni));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x4,%edi
  call   <omp_set_num_threads@plt>
  xor    %eax,%eax
  call   <omp_get_wtime@plt>
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  call   <omp_get_thread_num@plt>
  pxor   %xmm2,%xmm2
  lea    D9(%rip),%rdi
  cvtsi2sd %eax,%xmm2
  mov    $0x1,%eax
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x8(%rsp)
  call   <printf@plt>
  xor    %eax,%eax
  call   <tarea_uno@plt>
  movsd  0x8(%rsp),%xmm2
  mov    $0x1,%eax
  lea    D23(%rip),%rdi
  movapd %xmm2,%xmm0
  call   <printf@plt>
  xor    %eax,%eax
  call   <tarea_dos@plt>
  xor    %eax,%eax
  call   <omp_get_wtime@plt>
  movsd  0x8(%rsp),%xmm2
  mov    $0x2,%eax
  lea    D30(%rip),%rdi
  movapd %xmm0,%xmm1
  subsd  (%rsp),%xmm1
  movapd %xmm2,%xmm0
  call   <printf@plt>
  xor    %eax,%eax
  call   <omp_get_wtime@plt>
  subsd  (%rsp),%xmm0
  mov    $0x1,%eax
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D9
type: string
name: read_data
arguments: data_label: D23
type: string
name: read_data
arguments: data_label: D30
type: string
name: read_data
arguments: data_label: D8
type: string

tool: D9: Hilo %i ejecutando tarea 1 


tool: D23: Hilo %i ejecutando tarea 2



tool: D30: Tiempo de salida hilo %i = %.4lf segundos.



tool: D8: Tiempo tardado = %.4lf segundos



assistant: ```c
int main() {
  double timeIni, timeFin, time_s1, time_s2;
  int nthread;
  omp_set_num_threads(4);
  timeIni = omp_get_wtime();
#pragma omp parallel private(nthread)
  {
    nthread = omp_get_thread_num();
#pragma omp sections nowait
    {
#pragma omp section
      {
        printf("Hilo %i ejecutando tarea 1 \n", nthread);
        tarea_uno();
      }
#pragma omp section
      {
        printf("Hilo %i ejecutando tarea 2\n\n", nthread);
        tarea_dos();
      }
    }
    printf("Tiempo de salida hilo %i = %.4lf segundos.\n\n", nthread, (omp_get_wtime() - timeIni));
  }
  timeFin = omp_get_wtime();
  printf("Tiempo tardado = %.4lf segundos\n\n", (timeFin - timeIni));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<getItemN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x58(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1e,%eax
  mov    %eax,-0x4c(%rbp)
  mov    -0x4c(%rbp),%eax
  add    $0x1,%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x48(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L34:
  cmp    %rdx,%rsp
  je     L37
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L34
L37:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L29
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L29:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x40(%rbp)
  mov    -0x58(%rbp),%rax
  lea    0x1e(%rax),%rcx
  mov    -0x4c(%rbp),%edx
  mov    -0x40(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x40(%rbp),%rdx
  mov    -0x4c(%rbp),%eax
  cltq
  movb   $0x0,(%rdx,%rax,1)
  lea    -0x50(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  lea    D4(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  mov    -0x50(%rbp),%eax
  mov    %rbx,%rsp
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
data_type: string

tool: D4: %d 

assistant: ```c
int getItemN(char *strLine) {
  int result, len;
  len = strlen(strLine) - 30;
  char tmp[len + 1];
  strncpy(tmp, strLine + 30, len);
  tmp[len] = '\0';
  sscanf(tmp, "%d ", &result);
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<getItemN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  lea    -0x1e(%rax),%r12d
  sub    $0x1d,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L22:
  cmp    %rdx,%rsp
  je     L18
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L22
L18:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L9
  orq    $0x0,-0x8(%rsp,%rax,1)
L9:
  lea    0x1e(%rbx),%rsi
  mov    %r12d,%edx
  mov    %rsp,%rdi
  call   <strncpy@plt>
  movslq %r12d,%r12
  movb   $0x0,(%rsp,%r12,1)
  lea    -0x1c(%rbp),%rdx
  lea    D3(%rip),%rsi
  mov    %rsp,%rdi
  call   <sscanf@plt>
  mov    -0x1c(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D3
type_infered: string

tool: D3: %d 

assistant: ```c
int getItemN(char *strLine) {
  int result, len;
  len = strlen(strLine) - 30;
  char tmp[len + 1];
  strncpy(tmp, strLine + 30, len);
  tmp[len] = '\0';
  sscanf(tmp, "%d ", &result);
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<getItemN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  lea    -0x1e(%rax),%r12d
  sub    $0x1d,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L8
L44:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L44
L8:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L48
L32:
  mov    %r12d,%edx
  lea    0x1e(%rbx),%rsi
  mov    %rsp,%rdi
  movslq %r12d,%r12
  call   <strncpy@plt>
  lea    -0x1c(%rbp),%rdx
  mov    %rsp,%rdi
  lea    D4(%rip),%rsi
  movb   $0x0,(%rsp,%r12,1)
  call   <sscanf@plt>
  mov    -0x1c(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L48:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L32
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D4
type_guessed: string

tool: D4: %d 

assistant: ```c
int getItemN(char *strLine) {
  int result, len;
  len = strlen(strLine) - 30;
  char tmp[len + 1];
  strncpy(tmp, strLine + 30, len);
  tmp[len] = '\0';
  sscanf(tmp, "%d ", &result);
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getItemN>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rsp,%rcx
  lea    -0x1e(%rax),%r12d
  sub    $0x1d,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L4
L44:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L44
L4:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L10
L32:
  mov    %r12d,%edx
  lea    0x1e(%rbx),%rsi
  mov    %rsp,%rdi
  movslq %r12d,%r12
  call   <strncpy@plt>
  lea    -0x1c(%rbp),%rdx
  mov    %rsp,%rdi
  lea    D8(%rip),%rsi
  movb   $0x0,(%rsp,%r12,1)
  call   <sscanf@plt>
  mov    -0x1c(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L32
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D8
type_infered: string

tool: D8: %d 

assistant: ```c
int getItemN(char *strLine) {
  int result, len;
  len = strlen(strLine) - 30;
  char tmp[len + 1];
  strncpy(tmp, strLine + 30, len);
  tmp[len] = '\0';
  sscanf(tmp, "%d ", &result);
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<solve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    -0x3c(%rbp),%edx
  mov    -0x3c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <new_matrix2@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x3c(%rbp),%edx
  mov    -0x30(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <inv@plt>
  movl   $0x0,-0x1c(%rbp)
  jmp    L15
L2:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    -0x3c(%rbp),%edx
  mov    -0x38(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <dot@plt>
  movq   %xmm0,%rax
  mov    %rax,(%rbx)
  addl   $0x1,-0x1c(%rbp)
L15:
  mov    -0x1c(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L2
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free_matrix2@plt>
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void solve(double *x, double **A, double *b, int n) {
  double **A_inv = new_matrix2(n, n);
  inv(A_inv, A, n);
  int i;
  for (i = 0; i < n; i++)
    x[i] = dot(A_inv[i], b, n);
  free_matrix2(A_inv);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<solve>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbx
  mov    %rdx,%r14
  mov    %ecx,%ebp
  mov    %ecx,%esi
  mov    %ecx,%edi
  call   <new_matrix2@plt>
  mov    %rax,%r12
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <inv@plt>
  test   %ebp,%ebp
  jle    L1
  mov    %ebp,%r15d
  mov    $0x0,%ebx
L16:
  mov    (%r12,%rbx,8),%rdi
  mov    %ebp,%edx
  mov    %r14,%rsi
  call   <dot@plt>
  movsd  %xmm0,0x0(%r13,%rbx,8)
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L16
L1:
  mov    %r12,%rdi
  call   <free_matrix2@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void solve(double *x, double **A, double *b, int n) {
  double **A_inv = new_matrix2(n, n);
  inv(A_inv, A, n);
  int i;
  for (i = 0; i < n; i++)
    x[i] = dot(A_inv[i], b, n);
  free_matrix2(A_inv);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<solve>:
  endbr64
  push   %r15
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %rsi,%r13
  mov    %ecx,%esi
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %ecx,%edi
  push   %rbx
  sub    $0x8,%rsp
  call   <new_matrix2@plt>
  mov    %r14d,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r15
  call   <inv@plt>
  test   %r14d,%r14d
  jle    L6
  movslq %r14d,%r13
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L3:
  mov    (%r15,%rbx,8),%rdi
  mov    %r14d,%edx
  mov    %r12,%rsi
  call   <dot@plt>
  movsd  %xmm0,0x0(%rbp,%rbx,8)
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L3
L6:
  add    $0x8,%rsp
  mov    %r15,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1090 <free_matrix2@plt>
```

assistant: ```c
void solve(double *x, double **A, double *b, int n) {
  double **A_inv = new_matrix2(n, n);
  inv(A_inv, A, n);
  int i;
  for (i = 0; i < n; i++)
    x[i] = dot(A_inv[i], b, n);
  free_matrix2(A_inv);
}
```
user: Translate the below assembly code into C programming language:

```asm
<solve>:
  endbr64
  push   %r15
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %rsi,%r13
  mov    %ecx,%esi
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %ecx,%edi
  push   %rbx
  sub    $0x8,%rsp
  call   <new_matrix2@plt>
  mov    %r14d,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r15
  call   <inv@plt>
  test   %r14d,%r14d
  jle    L16
  movslq %r14d,%r13
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L11:
  mov    (%r15,%rbx,8),%rdi
  mov    %r14d,%edx
  mov    %r12,%rsi
  call   <dot@plt>
  movsd  %xmm0,0x0(%rbp,%rbx,8)
  add    $0x1,%rbx
  cmp    %rbx,%r13
  jne    L11
L16:
  add    $0x8,%rsp
  mov    %r15,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1090 <free_matrix2@plt>
```

assistant: ```c
void solve(double *x, double **A, double *b, int n) {
  double **A_inv = new_matrix2(n, n);
  inv(A_inv, A, n);
  int i;
  for (i = 0; i < n; i++)
    x[i] = dot(A_inv[i], b, n);
  free_matrix2(A_inv);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<parse_hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L15
  mov    $0x1,%eax
  jmp    L53
L15:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x8(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L47
  mov    $0x1,%eax
  jmp    L53
L47:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x10(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <hexcpy@plt>
  test   %rax,%rax
  je     L59
  mov    $0x1,%eax
  jmp    L53
L59:
  mov    $0x0,%eax
L53:
  leave
  ret
```

assistant: ```c
int parse_hex(uint8_t **dst, const char *src, uintmax_t *len) {
  *len = strlen(src);
  if (*len % 2)
    return 1;
  *len /= 2;
  if ((*dst = malloc(*len)) == NULL)
    return 1;
  if (hexcpy(*dst, src, *len * 2))
    return 1;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<parse_hex>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%rbp
  mov    %rsi,%rdi
  call   <strlen@plt>
  test   $0x1,%al
  je     L2
  mov    %eax,0x0(%rbp)
  mov    $0x1,%eax
L27:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  mov    %eax,%edi
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  mov    %edi,0x0(%rbp)
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,(%r12)
  mov    $0x1,%eax
  test   %rdi,%rdi
  je     L27
  mov    0x0(%rbp),%edx
  add    %edx,%edx
  mov    %rbx,%rsi
  call   <hexcpy@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L27
```

assistant: ```c
int parse_hex(uint8_t **dst, const char *src, uintmax_t *len) {
  *len = strlen(src);
  if (*len % 2)
    return 1;
  *len /= 2;
  if ((*dst = malloc(*len)) == NULL)
    return 1;
  if (hexcpy(*dst, src, *len * 2))
    return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_hex>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  test   $0x1,%al
  je     L7
  mov    %eax,(%rbx)
L5:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L7:
  mov    %eax,%edi
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  mov    %edi,(%rbx)
  call   <malloc@plt>
  mov    %rax,(%r12)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L5
  mov    (%rbx),%edx
  mov    %rbp,%rsi
  add    %edx,%edx
  call   <hexcpy@plt>
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int parse_hex(uint8_t **dst, const char *src, uintmax_t *len) {
  *len = strlen(src);
  if (*len % 2)
    return 1;
  *len /= 2;
  if ((*dst = malloc(*len)) == NULL)
    return 1;
  if (hexcpy(*dst, src, *len * 2))
    return 1;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<parse_hex>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %rsi,%rdi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  call   <strlen@plt>
  test   $0x1,%al
  je     L0
  mov    %eax,(%rbx)
L3:
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L0:
  mov    %eax,%edi
  shr    $0x1f,%edi
  add    %eax,%edi
  sar    %edi
  mov    %edi,(%rbx)
  call   <malloc@plt>
  mov    %rax,(%r12)
  mov    %rax,%rdi
  test   %rax,%rax
  je     L3
  mov    (%rbx),%edx
  mov    %rbp,%rsi
  add    %edx,%edx
  call   <hexcpy@plt>
  pop    %rbx
  pop    %rbp
  test   %rax,%rax
  pop    %r12
  setne  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int parse_hex(uint8_t **dst, const char *src, uintmax_t *len) {
  *len = strlen(src);
  if (*len % 2)
    return 1;
  *len /= 2;
  if ((*dst = malloc(*len)) == NULL)
    return 1;
  if (hexcpy(*dst, src, *len * 2))
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<UARTModemControlSet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x4000d000,-0x14(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x18(%rbp),%eax
  and    $0xf3,%ah
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x14(%rbp),%eax
  add    $0x30,%eax
  cltq
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%eax
  and    $0xc00,%eax
  or     %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x30,%eax
  cltq
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void UARTModemControlSet(uint32_t ui32Base, uint32_t ui32Control) {
  uint32_t ui32Temp;
  ASSERT(ui32Base == 0x4000D000);
  ASSERT((ui32Control & ~(0x00000800 | 0x00000400)) == 0);
  ui32Temp = (*((volatile uint32_t *)(ui32Base + 0x00000030)));
  ui32Temp |= (ui32Control & (0x00000800 | 0x00000400));
  (*((volatile uint32_t *)(ui32Base + 0x00000030))) = ui32Temp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<UARTModemControlSet>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  cmp    $0x4000d000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  test   $0xfffff3ff,%ebp
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  add    $0x30,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  and    $0xc00,%ebp
  or     %eax,%ebp
  mov    %ebp,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void UARTModemControlSet(uint32_t ui32Base, uint32_t ui32Control) {
  uint32_t ui32Temp;
  ASSERT(ui32Base == 0x4000D000);
  ASSERT((ui32Control & ~(0x00000800 | 0x00000400)) == 0);
  ui32Temp = (*((volatile uint32_t *)(ui32Base + 0x00000030)));
  ui32Temp |= (ui32Control & (0x00000800 | 0x00000400));
  (*((volatile uint32_t *)(ui32Base + 0x00000030))) = ui32Temp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<UARTModemControlSet>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x8,%rsp
  cmp    $0x4000d000,%ebx
  sete   %dil
  call   <ASSERT@plt>
  xor    %edi,%edi
  test   $0xfffff3ff,%ebp
  sete   %dil
  add    $0x30,%ebx
  and    $0xc00,%ebp
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  or     %eax,%ebp
  mov    %ebp,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void UARTModemControlSet(uint32_t ui32Base, uint32_t ui32Control) {
  uint32_t ui32Temp;
  ASSERT(ui32Base == 0x4000D000);
  ASSERT((ui32Control & ~(0x00000800 | 0x00000400)) == 0);
  ui32Temp = (*((volatile uint32_t *)(ui32Base + 0x00000030)));
  ui32Temp |= (ui32Control & (0x00000800 | 0x00000400));
  (*((volatile uint32_t *)(ui32Base + 0x00000030))) = ui32Temp;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<UARTModemControlSet>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x8,%rsp
  cmp    $0x4000d000,%ebx
  sete   %dil
  call   <ASSERT@plt>
  xor    %edi,%edi
  test   $0xfffff3ff,%ebp
  sete   %dil
  add    $0x30,%ebx
  and    $0xc00,%ebp
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  or     %eax,%ebp
  mov    %ebp,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void UARTModemControlSet(uint32_t ui32Base, uint32_t ui32Control) {
  uint32_t ui32Temp;
  ASSERT(ui32Base == 0x4000D000);
  ASSERT((ui32Control & ~(0x00000800 | 0x00000400)) == 0);
  ui32Temp = (*((volatile uint32_t *)(ui32Base + 0x00000030)));
  ui32Temp |= (ui32Control & (0x00000800 | 0x00000400));
  (*((volatile uint32_t *)(ui32Base + 0x00000030))) = ui32Temp;
}
```