ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<calculate_uv>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movapd %xmm0,%xmm6
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %rcx,-0x8(%rsp)
  mov    %r9,%rdi
  cmp    $0x1,%ebx
  jle    L24
  mov    %rdx,%r10
  movslq %esi,%r14
  shl    $0x3,%r14
  lea    0x8(%r8),%r9
  mov    0x38(%rsp),%rax
  lea    0x8(%rax),%rdx
  add    $0x8,%r10
  lea    0x8(%rdi),%r12
  lea    0x8(%rcx),%r11
  lea    -0x2(%rbx),%eax
  lea    0x10(%r8,%rax,8),%r15
  lea    -0x2(%rsi),%eax
  lea    0x10(,%rax,8),%r13
  mov    %ebx,%esi
  jmp    L28
L22:
  movapd %xmm6,%xmm5
  divsd  %xmm1,%xmm5
  movapd %xmm6,%xmm4
  divsd  %xmm2,%xmm4
  mov    $0x8,%eax
  mov    %r14,%r8
L66:
  mov    (%r10),%rcx
  mov    0x8(%rdx),%r14
  mov    (%rdx),%rbx
  movsd  (%r14,%rax,1),%xmm0
  subsd  (%rbx,%rax,1),%xmm0
  mulsd  %xmm5,%xmm0
  mov    (%r9),%rbx
  movsd  (%rbx,%rax,1),%xmm3
  subsd  %xmm0,%xmm3
  movsd  %xmm3,(%rcx,%rax,1)
  mov    (%r12),%rcx
  movsd  (%rcx,%rax,1),%xmm3
  mov    (%rdx),%rbx
  mov    %rax,%rcx
  add    $0x8,%rax
  mov    (%r11),%r14
  movsd  (%rbx,%rax,1),%xmm0
  subsd  (%rbx,%rcx,1),%xmm0
  mulsd  %xmm4,%xmm0
  subsd  %xmm0,%xmm3
  movsd  %xmm3,(%r14,%rcx,1)
  cmp    %r13,%rax
  jne    L66
  mov    %r8,%r14
L55:
  mov    (%r10),%rax
  mov    (%r9),%rcx
  mov    0x8(%rdx),%r8
  mov    (%rdx),%rbx
  movsd  (%r8,%r14,1),%xmm0
  subsd  (%rbx,%r14,1),%xmm0
  movapd %xmm6,%xmm3
  divsd  %xmm1,%xmm3
  mulsd  %xmm3,%xmm0
  movsd  (%rcx,%r14,1),%xmm3
  subsd  %xmm0,%xmm3
  movsd  %xmm3,(%rax,%r14,1)
  add    $0x8,%r9
  add    $0x8,%rdx
  add    $0x8,%r10
  add    $0x8,%r12
  add    $0x8,%r11
  cmp    %r15,%r9
  je     L70
L28:
  cmp    $0x1,%ebp
  jg     L22
  jmp    L55
L70:
  mov    %esi,%ebx
L24:
  cmp    $0x1,%ebp
  jle    L45
  movslq %ebx,%rax
  shl    $0x3,%rax
  lea    (%rdi,%rax,1),%r8
  divsd  %xmm2,%xmm6
  mov    %rax,%r10
  add    0x38(%rsp),%r10
  mov    -0x8(%rsp),%r9
  add    %rax,%r9
  mov    %ebp,%esi
  shl    $0x3,%rsi
  mov    $0x8,%eax
L33:
  mov    (%r8),%rdx
  movsd  (%rdx,%rax,1),%xmm1
  mov    (%r10),%rcx
  mov    %rax,%rdx
  add    $0x8,%rax
  mov    (%r9),%rdi
  movsd  (%rcx,%rax,1),%xmm0
  subsd  (%rcx,%rdx,1),%xmm0
  mulsd  %xmm6,%xmm0
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdi,%rdx,1)
  cmp    %rax,%rsi
  jne    L33
L45:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void calculate_uv(double dt, double dx, double dy, int imax, int jmax, double **U, double **V, double **F, double **G, double **P) {
  int i;
  int j;
  for (i = 1; i < imax; i++) {
    for (j = 1; j < jmax; j++) {
      U[i][j] = F[i][j] - dt / dx * (P[i + 1][j] - P[i][j]);
      V[i][j] = G[i][j] - dt / dy * (P[i][j + 1] - P[i][j]);
    }
    j = jmax;
    U[i][j] = F[i][j] - dt / dx * (P[i + 1][j] - P[i][j]);
  }
  i = imax;
  for (j = 1; j < jmax; j++) {
    V[i][j] = G[i][j] - dt / dy * (P[i][j + 1] - P[i][j]);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<calculate_uv>:
  endbr64
  push   %r15
  movapd %xmm0,%xmm5
  mov    %r9,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  mov    0x38(%rsp),%r13
  mov    %edi,-0x4(%rsp)
  mov    %rcx,-0x10(%rsp)
  cmp    $0x1,%edi
  jle    L44
  movapd %xmm0,%xmm3
  movslq %edi,%rax
  movslq %ebp,%r9
  mov    %r8,%r14
  divsd  %xmm1,%xmm3
  mov    %rax,-0x18(%rsp)
  mov    %rdx,%r12
  lea    0x0(,%r9,8),%rbx
  mov    $0x1,%r8d
  cs nopw 0x0(%rax,%rax,1)
L13:
  mov    (%r14,%r8,8),%rcx
  mov    0x8(%r13,%r8,8),%rsi
  mov    0x0(%r13,%r8,8),%rdx
  mov    (%r12,%r8,8),%rdi
  cmp    $0x1,%ebp
  jle    L54
  movapd %xmm5,%xmm4
  mov    -0x10(%rsp),%rax
  mov    (%r15,%r8,8),%r11
  mov    (%rax,%r8,8),%r10
  mov    $0x1,%eax
  divsd  %xmm2,%xmm4
  nopl   0x0(%rax)
L51:
  movsd  (%rsi,%rax,8),%xmm0
  subsd  (%rdx,%rax,8),%xmm0
  movsd  (%rcx,%rax,8),%xmm1
  mulsd  %xmm3,%xmm0
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
  movsd  0x8(%rdx,%rax,8),%xmm0
  subsd  (%rdx,%rax,8),%xmm0
  movsd  (%r11,%rax,8),%xmm1
  mulsd  %xmm4,%xmm0
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%r10,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%r9
  jne    L51
L54:
  movsd  (%rsi,%rbx,1),%xmm0
  subsd  (%rdx,%rbx,1),%xmm0
  add    $0x1,%r8
  movsd  (%rcx,%rbx,1),%xmm1
  mulsd  %xmm3,%xmm0
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdi,%rbx,1)
  cmp    %r8,-0x18(%rsp)
  jne    L13
L44:
  cmp    $0x1,%ebp
  jle    L36
  divsd  %xmm2,%xmm5
  movslq -0x4(%rsp),%rax
  mov    -0x10(%rsp),%rbx
  mov    (%r15,%rax,8),%rsi
  mov    0x0(%r13,%rax,8),%rdx
  mov    (%rbx,%rax,8),%rcx
  mov    $0x1,%eax
  nopl   0x0(%rax)
L17:
  movsd  0x8(%rdx,%rax,8),%xmm0
  subsd  (%rdx,%rax,8),%xmm0
  movsd  (%rsi,%rax,8),%xmm1
  mulsd  %xmm5,%xmm0
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%rcx,%rax,8)
  add    $0x1,%rax
  cmp    %rbp,%rax
  jne    L17
L36:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void calculate_uv(double dt, double dx, double dy, int imax, int jmax, double **U, double **V, double **F, double **G, double **P) {
  int i;
  int j;
  for (i = 1; i < imax; i++) {
    for (j = 1; j < jmax; j++) {
      U[i][j] = F[i][j] - dt / dx * (P[i + 1][j] - P[i][j]);
      V[i][j] = G[i][j] - dt / dy * (P[i][j + 1] - P[i][j]);
    }
    j = jmax;
    U[i][j] = F[i][j] - dt / dx * (P[i + 1][j] - P[i][j]);
  }
  i = imax;
  for (j = 1; j < jmax; j++) {
    V[i][j] = G[i][j] - dt / dy * (P[i][j + 1] - P[i][j]);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Platform_Absent>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  sete   %al
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
char Platform_Absent(int e) {
  char _o_result;
  _o_result = e == ENOENT;
  return _o_result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Platform_Absent>:
  endbr64
  mov    D8(%rip),%rax
  cmp    %edi,(%rax)
  sete   %al
  ret
```

assistant: ```c
char Platform_Absent(int e) {
  char _o_result;
  _o_result = e == ENOENT;
  return _o_result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Platform_Absent>:
  endbr64
  mov    D9(%rip),%rax
  cmp    %edi,(%rax)
  sete   %al
  ret
```

assistant: ```c
char Platform_Absent(int e) {
  char _o_result;
  _o_result = e == ENOENT;
  return _o_result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Platform_Absent>:
  endbr64
  mov    D8(%rip),%rax
  cmp    %edi,(%rax)
  sete   %al
  ret
```

assistant: ```c
char Platform_Absent(int e) {
  char _o_result;
  _o_result = e == ENOENT;
  return _o_result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<write_quaddobl_target_Laurent_system>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x302,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  leave
  ret
```

assistant: ```c
int write_quaddobl_target_Laurent_system(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(770, a, b, c, 0);
  return fail;
}
```
user: Translate the below assembly code into C programming language:

```asm
<write_quaddobl_target_Laurent_system>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x302,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int write_quaddobl_target_Laurent_system(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(770, a, b, c, 0);
  return fail;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<write_quaddobl_target_Laurent_system>:
  endbr64
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x302,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int write_quaddobl_target_Laurent_system(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(770, a, b, c, 0);
  return fail;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<write_quaddobl_target_Laurent_system>:
  endbr64
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x302,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int write_quaddobl_target_Laurent_system(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(770, a, b, c, 0);
  return fail;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<cmd_clrtoeol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <check_arg_count@plt>
  cmp    $0x1,%eax
  je     L0
  mov    $0x1,%edi
  call   <report_count@plt>
  mov    $0x0,%eax
  call   <clrtoeol@plt>
  mov    %eax,%edi
  call   <report_return@plt>
  jmp    L19
L0:
  nop
L19:
  leave
  ret
```

assistant: ```c
void cmd_clrtoeol(int nargs, char **args) {
  if (check_arg_count(nargs, 0) == 1)
    return;
  report_count(1);
  report_return(clrtoeol());
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<cmd_clrtoeol>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  call   <check_arg_count@plt>
  cmp    $0x1,%eax
  je     L9
  mov    $0x1,%edi
  call   <report_count@plt>
  mov    $0x0,%eax
  call   <clrtoeol@plt>
  mov    %eax,%edi
  call   <report_return@plt>
L9:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void cmd_clrtoeol(int nargs, char **args) {
  if (check_arg_count(nargs, 0) == 1)
    return;
  report_count(1);
  report_return(clrtoeol());
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cmd_clrtoeol>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  call   <check_arg_count@plt>
  cmp    $0x1,%eax
  je     L7
  mov    $0x1,%edi
  call   <report_count@plt>
  xor    %eax,%eax
  call   <clrtoeol@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    1090 <report_return@plt>
L7:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void cmd_clrtoeol(int nargs, char **args) {
  if (check_arg_count(nargs, 0) == 1)
    return;
  report_count(1);
  report_return(clrtoeol());
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<cmd_clrtoeol>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  call   <check_arg_count@plt>
  cmp    $0x1,%eax
  je     L4
  mov    $0x1,%edi
  call   <report_count@plt>
  xor    %eax,%eax
  call   <clrtoeol@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    1090 <report_return@plt>
L4:
  add    $0x8,%rsp
  ret
```

assistant: ```c
void cmd_clrtoeol(int nargs, char **args) {
  if (check_arg_count(nargs, 0) == 1)
    return;
  report_count(1);
  report_return(clrtoeol());
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<display_matrix0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  je     L9
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L9:
  movq   $0x0,-0x10(%rbp)
  jmp    L34
L6:
  movq   $0x0,-0x8(%rbp)
  jmp    L39
L29:
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x30(%rbp),%rax
  movd   %edx,%xmm0
  mov    %rax,%rdi
  call   <printf@plt>
  addq   $0x1,-0x8(%rbp)
L39:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jl     L29
  mov    $0xa,%edi
  call   <putchar@plt>
  addq   $0x1,-0x10(%rbp)
L34:
  mov    -0x10(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jl     L6
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

assistant: ```c
void display_matrix0(float **m, long nrow, long ncol, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = 0; i < nrow; i++) {
    for (j = 0; j < ncol; j++)
      printf(format, m[i][j]);
    putchar('\n');
  }
  putchar('\n');
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<display_matrix0>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r15
  mov    %rdx,%r12
  mov    %rcx,%r13
  test   %r8,%r8
  je     L44
  mov    %r8,%rdi
  call   <puts@plt>
L44:
  test   %r15,%r15
  jle    L33
  mov    $0x0,%r14d
  jmp    L1
L22:
  mov    0x0(%rbp),%rax
  movss  (%rax,%rbx,4),%xmm0
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jne    L22
L18:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r14
  add    $0x8,%rbp
  cmp    %r14,%r15
  je     L33
L1:
  mov    $0x0,%ebx
  test   %r12,%r12
  jg     L22
  jmp    L18
L33:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void display_matrix0(float **m, long nrow, long ncol, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = 0; i < nrow; i++) {
    for (j = 0; j < ncol; j++)
      printf(format, m[i][j]);
    putchar('\n');
  }
  putchar('\n');
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<display_matrix0>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %r8,%r8
  je     L38
  mov    %r8,%rdi
  call   <puts@plt>
L38:
  test   %r15,%r15
  jle    L20
  xor    %r14d,%r14d
  nopl   0x0(%rax,%rax,1)
L13:
  xor    %ebx,%ebx
  test   %r12,%r12
  jle    L9
  nop
L12:
  mov    0x0(%rbp),%rax
  mov    %r13,%rdi
  movss  (%rax,%rbx,4),%xmm0
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L12
L9:
  mov    $0xa,%edi
  add    $0x1,%r14
  add    $0x8,%rbp
  call   <putchar@plt>
  cmp    %r14,%r15
  jne    L13
L20:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <putchar@plt>
```

assistant: ```c
void display_matrix0(float **m, long nrow, long ncol, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = 0; i < nrow; i++) {
    for (j = 0; j < ncol; j++)
      printf(format, m[i][j]);
    putchar('\n');
  }
  putchar('\n');
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<display_matrix0>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %r8,%r8
  je     L43
  mov    %r8,%rdi
  call   <puts@plt>
L43:
  test   %r15,%r15
  jle    L11
  xor    %r14d,%r14d
  nopl   0x0(%rax,%rax,1)
L37:
  xor    %ebx,%ebx
  test   %r12,%r12
  jle    L15
  nop
L20:
  mov    0x0(%rbp),%rax
  mov    %r13,%rdi
  movss  (%rax,%rbx,4),%xmm0
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L20
L15:
  mov    $0xa,%edi
  add    $0x1,%r14
  add    $0x8,%rbp
  call   <putchar@plt>
  cmp    %r14,%r15
  jne    L37
L11:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <putchar@plt>
```

assistant: ```c
void display_matrix0(float **m, long nrow, long ncol, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = 0; i < nrow; i++) {
    for (j = 0; j < ncol; j++)
      printf(format, m[i][j]);
    putchar('\n');
  }
  putchar('\n');
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x40(%rbp)
  movl   $0x6,-0x3c(%rbp)
  movl   $0x3,-0x38(%rbp)
  movl   $0x8,-0x34(%rbp)
  movl   $0x2d,-0x30(%rbp)
  movl   $0x17,-0x2c(%rbp)
  movl   $0x3,-0x28(%rbp)
  movl   $0x7,-0x24(%rbp)
  movl   $0x2d,-0x20(%rbp)
  movl   $0x57,-0x1c(%rbp)
  movl   $0x2d,-0x18(%rbp)
  movl   $0x38,-0x14(%rbp)
  movl   $0xb,-0x10(%rbp)
  movl   $0x154,-0xc(%rbp)
  movl   $0xe,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  lea    -0x1(%rax),%edx
  lea    -0x40(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <divide@plt>
  movl   $0x0,-0x48(%rbp)
  jmp    L27
L29:
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x48(%rbp)
L27:
  mov    -0x48(%rbp),%eax
  cmp    -0x44(%rbp),%eax
  jl     L29
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
type_infered: string

tool: D0: %d 

assistant: ```c
int main(void) {
  int i, size, input[] = {1, 6, 3, 8, 45, 23, 3, 7, 45, 87, 45, 56, 11, 340};
  size = (sizeof(input) / sizeof(int));
  divide(input, 0, size - 1);
  for (i = 0; i < size; i++)
    printf("%d ", input[i]);
  printf("\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x6,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x8,0xc(%rsp)
  movl   $0x2d,0x10(%rsp)
  movl   $0x17,0x14(%rsp)
  movl   $0x3,0x18(%rsp)
  movl   $0x7,0x1c(%rsp)
  movl   $0x2d,0x20(%rsp)
  movl   $0x57,0x24(%rsp)
  movl   $0x2d,0x28(%rsp)
  movl   $0x38,0x2c(%rsp)
  movl   $0xb,0x30(%rsp)
  movl   $0x154,0x34(%rsp)
  mov    %rsp,%rbx
  mov    $0xd,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <divide@plt>
  lea    0x38(%rsp),%r12
  lea    D4(%rip),%rbp
L6:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L6
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D4
type_guessed: string

tool: D4: %d 

assistant: ```c
int main(void) {
  int i, size, input[] = {1, 6, 3, 8, 45, 23, 3, 7, 45, 87, 45, 56, 11, 340};
  size = (sizeof(input) / sizeof(int));
  divide(input, 0, size - 1);
  for (i = 0; i < size; i++)
    printf("%d ", input[i]);
  printf("\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xd,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x600000001,%rax
  mov    %rsp,%rbx
  lea    0x38(%rsp),%r12
  mov    %rbx,%rdi
  mov    %rax,(%rsp)
  movabs $0x800000003,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x170000002d,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x700000003,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x570000002d,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x380000002d,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x1540000000b,%rax
  mov    %rax,0x30(%rsp)
  call   <divide@plt>
L13:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L13
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D2
data_type: string

tool: D2: %d 

assistant: ```c
int main(void) {
  int i, size, input[] = {1, 6, 3, 8, 45, 23, 3, 7, 45, 87, 45, 56, 11, 340};
  size = (sizeof(input) / sizeof(int));
  divide(input, 0, size - 1);
  for (i = 0; i < size; i++)
    printf("%d ", input[i]);
  printf("\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xd,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D30(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  movdqa D25(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    D48(%rip),%rax
  lea    0x38(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D39(%rip),%xmm0
  mov    %rbx,%rdi
  mov    %rax,0x30(%rsp)
  movaps %xmm0,0x10(%rsp)
  movdqa D21(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  call   <divide@plt>
  nopl   0x0(%rax)
L11:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L11
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D30
data_type: string
name: read_data
arguments: label: D25
data_type: byte[16]
name: read_data
arguments: label: D48
data_type: byte[8]
name: read_data
arguments: label: D39
data_type: byte[16]
name: read_data
arguments: label: D21
data_type: byte[16]

tool: D30: %d 

tool: {"D25": [0x01,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x08,0x00,0x00,0x00]}

tool: {"D48": [0x0b,0x00,0x00,0x00,0x54,0x01,0x00,0x00]}

tool: {"D39": [0x2d,0x00,0x00,0x00,0x17,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: {"D21": [0x2d,0x00,0x00,0x00,0x57,0x00,0x00,0x00,0x2d,0x00,0x00,0x00,0x38,0x00,0x00,0x00]}

assistant: ```c
int main(void) {
  int i, size, input[] = {1, 6, 3, 8, 45, 23, 3, 7, 45, 87, 45, 56, 11, 340};
  size = (sizeof(input) / sizeof(int));
  divide(input, 0, size - 1);
  for (i = 0; i < size; i++)
    printf("%d ", input[i]);
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fill>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L14
L12:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  lea    D0(%rip),%rcx
  movzbl (%rax,%rcx,1),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L14:
  mov    -0xc(%rbp),%eax
  cltq
  cmp    %rax,-0x20(%rbp)
  jg     L12
  nop
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
data_type: string

tool: D0: The quick brown dingo jumped over the layzy dog.

assistant: ```c
void fill(void *buf, long sizeof_buf) {
  char *p = buf;
  int i;
  for (i = 0; i < sizeof_buf; i++)
    p[i] = "The quick brown dingo jumped over the layzy dog."[i];
}
```
user: What is the c source code of the assembly code below?

```asm
<fill>:
  endbr64
  test   %rsi,%rsi
  jle    L18
  mov    $0x0,%eax
  lea    D3(%rip),%rcx
L3:
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L3
L18:
  ret
```

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

tool: D3: The quick brown dingo jumped over the layzy dog.

assistant: ```c
void fill(void *buf, long sizeof_buf) {
  char *p = buf;
  int i;
  for (i = 0; i < sizeof_buf; i++)
    p[i] = "The quick brown dingo jumped over the layzy dog."[i];
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<fill>:
  endbr64
  test   %rsi,%rsi
  jle    L4
  xor    %eax,%eax
  lea    D2(%rip),%rcx
  nopw   0x0(%rax,%rax,1)
L14:
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L14
L4:
  ret
```

tool_calls: name: disassemble_data
arguments: label: D2
type_guessed: string

tool: D2: The quick brown dingo jumped over the layzy dog.

assistant: ```c
void fill(void *buf, long sizeof_buf) {
  char *p = buf;
  int i;
  for (i = 0; i < sizeof_buf; i++)
    p[i] = "The quick brown dingo jumped over the layzy dog."[i];
}
```