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

```asm
<strupper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movl   $0x0,-0x20(%rbp)
  jmp    L5
L14:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x20(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
  addl   $0x1,-0x20(%rbp)
L5:
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L14
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *strupper(const char *str) {
  int length = strlen(str);
  char *new_str = malloc(length * sizeof(char));
  strcpy(new_str, str);
  for (int i = 0; i < length; i++) {
    new_str[i] = toupper(new_str[i]);
  }
  return new_str;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strupper>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%ebp
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  test   %ebp,%ebp
  jle    L6
  mov    %r12,%rbx
  lea    -0x1(%rbp),%eax
  lea    0x1(%r12,%rax,1),%rbp
L9:
  movsbl (%rbx),%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L9
L6:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strupper(const char *str) {
  int length = strlen(str);
  char *new_str = malloc(length * sizeof(char));
  strcpy(new_str, str);
  for (int i = 0; i < length; i++) {
    new_str[i] = toupper(new_str[i]);
  }
  return new_str;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strupper>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  test   %ebp,%ebp
  jle    L7
  lea    -0x1(%rbp),%eax
  mov    %r12,%rbx
  lea    0x1(%r12,%rax,1),%rbp
  nopl   0x0(%rax)
L12:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <toupper@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L12
L7:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strupper(const char *str) {
  int length = strlen(str);
  char *new_str = malloc(length * sizeof(char));
  strcpy(new_str, str);
  for (int i = 0; i < length; i++) {
    new_str[i] = toupper(new_str[i]);
  }
  return new_str;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strupper>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <malloc@plt>
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  test   %ebp,%ebp
  jle    L12
  lea    -0x1(%rbp),%eax
  mov    %r12,%rbx
  lea    0x1(%r12,%rax,1),%rbp
  nopl   0x0(%rax)
L18:
  movsbl (%rbx),%edi
  add    $0x1,%rbx
  call   <toupper@plt>
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L18
L12:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strupper(const char *str) {
  int length = strlen(str);
  char *new_str = malloc(length * sizeof(char));
  strcpy(new_str, str);
  for (int i = 0; i < length; i++) {
    new_str[i] = toupper(new_str[i]);
  }
  return new_str;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<compute_multiple_spans_min_max>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8d,-0x2c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L14:
  mov    -0x28(%rbp),%ecx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <compute_span_min_max@plt>
  addq   $0x2,-0x18(%rbp)
  mov    -0x28(%rbp),%eax
  imul   -0x24(%rbp),%eax
  cltq
  add    %rax,-0x20(%rbp)
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L14
  nop
  nop
  leave
  ret
```

assistant: ```c
void compute_multiple_spans_min_max(unsigned char *min_max_dest, const unsigned char *source, int source_stride, int vector_size, int span_count) {
  int i;
  for (i = 0; i < span_count; ++i) {
    compute_span_min_max(min_max_dest, source, source_stride, vector_size);
    min_max_dest += 2;
    source += vector_size * source_stride;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<compute_multiple_spans_min_max>:
  endbr64
  test   %r8d,%r8d
  jle    L17
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r13d
  mov    %ecx,%r12d
  mov    %edx,%r14d
  imul   %ecx,%r14d
  movslq %r14d,%r14
  mov    %r8d,%r8d
  lea    (%rdi,%r8,2),%r15
L14:
  mov    %r12d,%ecx
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <compute_span_min_max@plt>
  add    $0x2,%rbx
  add    %r14,%rbp
  cmp    %r15,%rbx
  jne    L14
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L17:
  ret
```

assistant: ```c
void compute_multiple_spans_min_max(unsigned char *min_max_dest, const unsigned char *source, int source_stride, int vector_size, int span_count) {
  int i;
  for (i = 0; i < span_count; ++i) {
    compute_span_min_max(min_max_dest, source, source_stride, vector_size);
    min_max_dest += 2;
    source += vector_size * source_stride;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<compute_multiple_spans_min_max>:
  endbr64
  test   %r8d,%r8d
  jle    L8
  push   %r15
  movslq %r8d,%r8
  push   %r14
  mov    %edx,%r14d
  lea    (%rdi,%r8,2),%r15
  push   %r13
  imul   %ecx,%r14d
  mov    %edx,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movslq %r14d,%r14
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L0:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %r12d,%ecx
  mov    %r13d,%edx
  call   <compute_span_min_max@plt>
  add    $0x2,%rbx
  add    %r14,%rbp
  cmp    %r15,%rbx
  jne    L0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L8:
  ret
```

assistant: ```c
void compute_multiple_spans_min_max(unsigned char *min_max_dest, const unsigned char *source, int source_stride, int vector_size, int span_count) {
  int i;
  for (i = 0; i < span_count; ++i) {
    compute_span_min_max(min_max_dest, source, source_stride, vector_size);
    min_max_dest += 2;
    source += vector_size * source_stride;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<compute_multiple_spans_min_max>:
  endbr64
  test   %r8d,%r8d
  jle    L8
  push   %r15
  movslq %r8d,%r8
  push   %r14
  mov    %edx,%r14d
  lea    (%rdi,%r8,2),%r15
  push   %r13
  imul   %ecx,%r14d
  mov    %edx,%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  movslq %r14d,%r14
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L7:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    %r12d,%ecx
  mov    %r13d,%edx
  call   <compute_span_min_max@plt>
  add    $0x2,%rbx
  add    %r14,%rbp
  cmp    %r15,%rbx
  jne    L7
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L8:
  ret
```

assistant: ```c
void compute_multiple_spans_min_max(unsigned char *min_max_dest, const unsigned char *source, int source_stride, int vector_size, int span_count) {
  int i;
  for (i = 0; i < span_count; ++i) {
    compute_span_min_max(min_max_dest, source, source_stride, vector_size);
    min_max_dest += 2;
    source += vector_size * source_stride;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<thread_kmalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0x0,%eax
  call   <putcurrent@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <kzmem_alloc@plt>
  leave
  ret
```

assistant: ```c
void *thread_kmalloc(int size) {
  putcurrent();
  return kzmem_alloc(size);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<thread_kmalloc>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  mov    $0x0,%eax
  call   <putcurrent@plt>
  mov    %ebx,%edi
  call   <kzmem_alloc@plt>
  pop    %rbx
  ret
```

assistant: ```c
void *thread_kmalloc(int size) {
  putcurrent();
  return kzmem_alloc(size);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<thread_kmalloc>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %edi,%ebp
  call   <putcurrent@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1060 <kzmem_alloc@plt>
```

assistant: ```c
void *thread_kmalloc(int size) {
  putcurrent();
  return kzmem_alloc(size);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<thread_kmalloc>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %edi,%ebp
  call   <putcurrent@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1060 <kzmem_alloc@plt>
```

assistant: ```c
void *thread_kmalloc(int size) {
  putcurrent();
  return kzmem_alloc(size);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  movss  D56(%rip),%xmm0
  movss  %xmm0,-0x8(%rbp)
  movss  D73(%rip),%xmm0
  movss  %xmm0,-0x4(%rbp)
  lea    D19(%rip),%rax
  mov    %rax,%rdx
  lea    D41(%rip),%rax
  mov    %rax,%rsi
  lea    D75(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movss  -0x8(%rbp),%xmm0
  movss  %xmm0,-0x14(%rbp)
  jmp    L16
L12:
  movss  -0x14(%rbp),%xmm0
  movss  D28(%rip),%xmm2
  movaps %xmm0,%xmm1
  subss  %xmm2,%xmm1
  movss  D35(%rip),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0x10(%rbp),%xmm0
  pxor   %xmm3,%xmm3
  cvtss2sd -0x14(%rbp),%xmm3
  movq   %xmm3,%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  movss  -0x14(%rbp),%xmm0
  subss  -0x4(%rbp),%xmm0
  movss  %xmm0,-0x14(%rbp)
L16:
  movss  -0x14(%rbp),%xmm0
  comiss -0xc(%rbp),%xmm0
  jae    L12
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D56
data_type: byte[4]
name: parse_data
arguments: label: D73
data_type: byte[4]
name: parse_data
arguments: label: D19
data_type: string
name: parse_data
arguments: label: D41
data_type: string
name: parse_data
arguments: label: D75
data_type: string
name: parse_data
arguments: label: D28
data_type: byte[4]
name: parse_data
arguments: label: D35
data_type: byte[4]
name: parse_data
arguments: label: D50
data_type: string

tool: {"D56": [0x00,0x00,0x96,0x43]}

tool: {"D73": [0x00,0x00,0xa0,0x41]}

tool: D19: C^o

tool: D41: F^o

tool: D75: %3s %6s


tool: {"D28": [0x00,0x00,0x00,0x42]}

tool: {"D35": [0xe4,0x38,0x0e,0x3f]}

tool: D50: %3.0f %6.1f


assistant: ```c
int main(int argc, char *argv[]) {
  float f = 0.0f;
  float c = 0.0f;
  float L = 0.0f;
  float U = 300.0f;
  float dT = 20.0f;
  printf("%3s %6s\n", "F^o", "C^o");
  for (f = U; f >= L; f -= dT) {
    c = (5.0f / 9.0f) * (f - 32.0f);
    printf("%3.0f %6.1f\n", f, c);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  lea    D65(%rip),%rdx
  lea    D20(%rip),%rsi
  lea    D29(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x10,%ebx
  movss  D4(%rip),%xmm3
  movss  %xmm3,0xc(%rsp)
  lea    D12(%rip),%rbp
L8:
  movss  0xc(%rsp),%xmm2
  movaps %xmm2,%xmm1
  subss  D27(%rip),%xmm1
  mulss  D25(%rip),%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd %xmm2,%xmm0
  cvtss2sd %xmm1,%xmm1
  mov    %rbp,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  movss  0xc(%rsp),%xmm2
  subss  D48(%rip),%xmm2
  movss  %xmm2,0xc(%rsp)
  sub    $0x1,%ebx
  jne    L8
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D65
data_type: string
name: parse_data
arguments: label: D20
data_type: string
name: parse_data
arguments: label: D29
data_type: string
name: parse_data
arguments: label: D4
data_type: byte[4]
name: parse_data
arguments: label: D12
data_type: string
name: parse_data
arguments: label: D27
data_type: byte[4]
name: parse_data
arguments: label: D25
data_type: byte[4]
name: parse_data
arguments: label: D48
data_type: byte[4]

tool: D65: C^o

tool: D20: F^o

tool: D29: %3s %6s


tool: {"D4": [0x00,0x00,0x96,0x43]}

tool: D12: %3.0f %6.1f


tool: {"D27": [0x00,0x00,0x00,0x42]}

tool: {"D25": [0xe4,0x38,0x0e,0x3f]}

tool: {"D48": [0x00,0x00,0xa0,0x41]}

assistant: ```c
int main(int argc, char *argv[]) {
  float f = 0.0f;
  float c = 0.0f;
  float L = 0.0f;
  float U = 300.0f;
  float dT = 20.0f;
  printf("%3s %6s\n", "F^o", "C^o");
  for (f = U; f >= L; f -= dT) {
    c = (5.0f / 9.0f) * (f - 32.0f);
    printf("%3.0f %6.1f\n", f, c);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D77(%rip),%rdx
  lea    D78(%rip),%rsi
  xor    %eax,%eax
  push   %rbx
  lea    D38(%rip),%rdi
  mov    $0x10,%ebx
  lea    D60(%rip),%rbp
  sub    $0x18,%rsp
  call   <printf@plt>
  movss  D8(%rip),%xmm2
  nopw   0x0(%rax,%rax,1)
L6:
  movaps %xmm2,%xmm1
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  mov    $0x2,%eax
  subss  D75(%rip),%xmm1
  mulss  D5(%rip),%xmm1
  movss  %xmm2,0xc(%rsp)
  cvtss2sd %xmm2,%xmm0
  cvtss2sd %xmm1,%xmm1
  call   <printf@plt>
  movss  0xc(%rsp),%xmm2
  subss  D62(%rip),%xmm2
  sub    $0x1,%ebx
  jne    L6
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D77
type_infered: string
name: parse_data
arguments: tag: D78
type_infered: string
name: parse_data
arguments: tag: D38
type_infered: string
name: parse_data
arguments: tag: D60
type_infered: string
name: parse_data
arguments: tag: D8
type_infered: byte[4]
name: parse_data
arguments: tag: D75
type_infered: byte[4]
name: parse_data
arguments: tag: D5
type_infered: byte[4]
name: parse_data
arguments: tag: D62
type_infered: byte[4]

tool: D77: C^o

tool: D78: F^o

tool: D38: %3s %6s


tool: D60: %3.0f %6.1f


tool: {"D8": [0x00,0x00,0x96,0x43]}

tool: {"D75": [0x00,0x00,0x00,0x42]}

tool: {"D5": [0xe4,0x38,0x0e,0x3f]}

tool: {"D62": [0x00,0x00,0xa0,0x41]}

assistant: ```c
int main(int argc, char *argv[]) {
  float f = 0.0f;
  float c = 0.0f;
  float L = 0.0f;
  float U = 300.0f;
  float dT = 20.0f;
  printf("%3s %6s\n", "F^o", "C^o");
  for (f = U; f >= L; f -= dT) {
    c = (5.0f / 9.0f) * (f - 32.0f);
    printf("%3.0f %6.1f\n", f, c);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  lea    D63(%rip),%rdx
  lea    D30(%rip),%rsi
  xor    %eax,%eax
  push   %rbx
  lea    D20(%rip),%rdi
  mov    $0x10,%ebx
  lea    D7(%rip),%rbp
  sub    $0x18,%rsp
  call   <printf@plt>
  movss  D68(%rip),%xmm2
  nopw   0x0(%rax,%rax,1)
L2:
  movaps %xmm2,%xmm1
  pxor   %xmm0,%xmm0
  mov    %rbp,%rdi
  mov    $0x2,%eax
  subss  D53(%rip),%xmm1
  mulss  D47(%rip),%xmm1
  movss  %xmm2,0xc(%rsp)
  cvtss2sd %xmm2,%xmm0
  cvtss2sd %xmm1,%xmm1
  call   <printf@plt>
  movss  0xc(%rsp),%xmm2
  subss  D46(%rip),%xmm2
  sub    $0x1,%ebx
  jne    L2
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D63
datatype: string
name: unpack_data
arguments: tag: D30
datatype: string
name: unpack_data
arguments: tag: D20
datatype: string
name: unpack_data
arguments: tag: D7
datatype: string
name: unpack_data
arguments: tag: D68
datatype: byte[4]
name: unpack_data
arguments: tag: D53
datatype: byte[4]
name: unpack_data
arguments: tag: D47
datatype: byte[4]
name: unpack_data
arguments: tag: D46
datatype: byte[4]

tool: D63: C^o

tool: D30: F^o

tool: D20: %3s %6s


tool: D7: %3.0f %6.1f


tool: {"D68": [0x00,0x00,0x96,0x43]}

tool: {"D53": [0x00,0x00,0x00,0x42]}

tool: {"D47": [0xe4,0x38,0x0e,0x3f]}

tool: {"D46": [0x00,0x00,0xa0,0x41]}

assistant: ```c
int main(int argc, char *argv[]) {
  float f = 0.0f;
  float c = 0.0f;
  float L = 0.0f;
  float U = 300.0f;
  float dT = 20.0f;
  printf("%3s %6s\n", "F^o", "C^o");
  for (f = U; f >= L; f -= dT) {
    c = (5.0f / 9.0f) * (f - 32.0f);
    printf("%3.0f %6.1f\n", f, c);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ptree>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L0
L11:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jne    L22
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
L22:
  addl   $0x1,-0xc(%rbp)
L0:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L11
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_infered: string

tool: D1: %d	

assistant: ```c
void ptree(int a[], int len) {
  int i, level, m;
  for (i = 0, level = 1, m = 1; i < len; i++) {
    printf("%d\t", a[i]);
    if (i == m) {
      printf("\n");
      level++;
      m = 1 << level;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ptree>:
  endbr64
  test   %esi,%esi
  jle    L3
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%ebp
  mov    (%rdi),%esi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%ebp
  mov    $0x1,%ebx
  mov    $0x1,%r12d
  mov    $0x1,%r15d
  lea    D18(%rip),%r14
  jmp    L6
L3:
  ret
L44:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r15d
  mov    $0x1,%r12d
  mov    %r15d,%ecx
  shl    %cl,%r12d
L15:
  add    $0x1,%rbx
L6:
  cmp    %rbx,%rbp
  je     L21
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %ebx,%r12d
  jne    L15
  jmp    L44
L21:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D18
type: string

tool: D18: %d	

assistant: ```c
void ptree(int a[], int len) {
  int i, level, m;
  for (i = 0, level = 1, m = 1; i < len; i++) {
    printf("%d\t", a[i]);
    if (i == m) {
      printf("\n");
      level++;
      m = 1 << level;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ptree>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %esi,%rbx
  sub    $0x8,%rsp
  test   %ebx,%ebx
  jle    L7
  mov    (%rdi),%esi
  lea    D8(%rip),%r13
  mov    %rdi,%rbp
  xor    %eax,%eax
  mov    %r13,%rdi
  mov    $0x1,%r15d
  mov    $0x1,%r12d
  mov    $0x1,%r14d
  call   <printf@plt>
  jmp    L20
  nopl   0x0(%rax,%rax,1)
L0:
  add    $0x1,%r15
L20:
  cmp    %r15,%rbx
  je     L7
  mov    0x0(%rbp,%r15,4),%esi
  xor    %eax,%eax
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    %r15d,%r12d
  jne    L0
  mov    $0xa,%edi
  add    $0x1,%r14d
  mov    $0x1,%r12d
  call   <putchar@plt>
  mov    %r14d,%ecx
  shl    %cl,%r12d
  jmp    L0
  nopl   0x0(%rax)
L7:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D8
data_type: string

tool: D8: %d	

assistant: ```c
void ptree(int a[], int len) {
  int i, level, m;
  for (i = 0, level = 1, m = 1; i < len; i++) {
    printf("%d\t", a[i]);
    if (i == m) {
      printf("\n");
      level++;
      m = 1 << level;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ptree>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %esi,%rbx
  sub    $0x8,%rsp
  test   %ebx,%ebx
  jle    L36
  mov    (%rdi),%esi
  lea    D6(%rip),%r13
  mov    %rdi,%rbp
  xor    %eax,%eax
  mov    %r13,%rdi
  mov    $0x1,%r15d
  mov    $0x1,%r12d
  mov    $0x1,%r14d
  call   <printf@plt>
  cmp    $0x1,%ebx
  je     L36
  cs nopw 0x0(%rax,%rax,1)
L17:
  mov    0x0(%rbp,%r15,4),%esi
  xor    %eax,%eax
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    %r15d,%r12d
  je     L3
  add    $0x1,%r15
  cmp    %r15,%rbx
  jne    L17
L36:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L3:
  mov    $0xa,%edi
  add    $0x1,%r14d
  mov    $0x1,%r12d
  add    $0x1,%r15
  call   <putchar@plt>
  mov    %r14d,%ecx
  shl    %cl,%r12d
  cmp    %rbx,%r15
  jne    L17
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_infered: string

tool: D6: %d	

assistant: ```c
void ptree(int a[], int len) {
  int i, level, m;
  for (i = 0, level = 1, m = 1; i < len; i++) {
    printf("%d\t", a[i]);
    if (i == m) {
      printf("\n");
      level++;
      m = 1 << level;
    }
  }
}
```