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

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x4,%esi
  lea    D18(%rip),%rdi
  push   %r14
  lea    D31(%rip),%r14
  push   %r13
  lea    D19(%rip),%r13
  push   %r12
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  lea    D28(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x2000000011,%rax
  mov    %rsp,%r12
  lea    0x14(%rsp),%r15
  mov    %rax,0x4(%rsp)
  movabs $0x790000002b,%rax
  mov    %rax,0xc(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L5
  nopl   (%rax)
L0:
  mov    (%r12),%ebp
L5:
  mov    %ebp,%edi
  call   <is_power_of_two@plt>
  mov    %r13,%rdi
  test   %rax,%rax
  cmovne %r14,%rdi
  xor    %eax,%eax
  add    $0x4,%r12
  call   <printf@plt>
  mov    %ebp,%edi
  call   <next_power_of_two1@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <next_power_of_two2@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <next_power_of_two3@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <next_power_of_two4@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %r15,%r12
  jne    L0
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D18
type: string
name: disassemble_data
arguments: tag: D31
type: string
name: disassemble_data
arguments: tag: D19
type: string
name: disassemble_data
arguments: tag: D28
type: string

tool: D18: sizeof int: %ld


tool: D31: true


tool: D19: false


tool: D28: %d -> %d


assistant: ```c
int main(int argc, char **argv) {
  unsigned int x[] = {1, 17, 32, 43, 121};
  int i;
  printf("sizeof int: %ld\n", sizeof(unsigned int));
  for (i = 0; i < sizeof(x) / sizeof(x[0]); i++) {
    printf(is_power_of_two(x[i]) ? "true\n" : "false\n");
    printf("%d -> %d\n", x[i], next_power_of_two1(x[i]));
    printf("%d -> %d\n", x[i], next_power_of_two2(x[i]));
    printf("%d -> %d\n", x[i], next_power_of_two3(x[i]));
    printf("%d -> %d\n", x[i], next_power_of_two4(x[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   %r15
  mov    $0x4,%esi
  lea    D1(%rip),%rdi
  lea    D23(%rip),%r15
  push   %r14
  lea    D24(%rip),%r14
  push   %r13
  push   %r12
  lea    D5(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  movdqa D10(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x14(%rsp),%r13
  movl   $0x79,0x10(%rsp)
  movaps %xmm0,(%rsp)
  call   <printf@plt>
  nopl   0x0(%rax)
L22:
  mov    (%rbx),%ebp
  mov    %ebp,%edi
  call   <is_power_of_two@plt>
  mov    %r14,%rdi
  test   %rax,%rax
  jne    L28
  mov    %r15,%rdi
L28:
  call   <puts@plt>
  mov    %ebp,%edi
  add    $0x4,%rbx
  call   <next_power_of_two1@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <next_power_of_two2@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <next_power_of_two3@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <next_power_of_two4@plt>
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  cmp    %rbx,%r13
  jne    L22
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D1
type: string
name: parse_data
arguments: tag: D23
type: string
name: parse_data
arguments: tag: D24
type: string
name: parse_data
arguments: tag: D5
type: string
name: parse_data
arguments: tag: D10
type: byte[16]

tool: D1: sizeof int: %ld


tool: D23: false

tool: D24: true

tool: D5: %d -> %d


tool: {"D10": [0x01,0x00,0x00,0x00,0x11,0x00,0x00,0x00,0x20,0x00,0x00,0x00,0x2b,0x00,0x00,0x00]}

assistant: ```c
int main(int argc, char **argv) {
  unsigned int x[] = {1, 17, 32, 43, 121};
  int i;
  printf("sizeof int: %ld\n", sizeof(unsigned int));
  for (i = 0; i < sizeof(x) / sizeof(x[0]); i++) {
    printf(is_power_of_two(x[i]) ? "true\n" : "false\n");
    printf("%d -> %d\n", x[i], next_power_of_two1(x[i]));
    printf("%d -> %d\n", x[i], next_power_of_two2(x[i]));
    printf("%d -> %d\n", x[i], next_power_of_two3(x[i]));
    printf("%d -> %d\n", x[i], next_power_of_two4(x[i]));
  }
  printf("\n");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<librpip_device_node_exists>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0xa,%edi
  ja     L56
  mov    %edi,%edi
  lea    D83(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    0x4(%rsp),%rsi
  lea    D43(%rip),%rdi
  call   <stat@plt>
L82:
  not    %eax
  shr    $0x1f,%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L104
  add    $0x18,%rsp
  ret
  lea    0x4(%rsp),%rsi
  lea    D60(%rip),%rdi
  call   <stat@plt>
  jmp    L82
  lea    0x4(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <stat@plt>
  jmp    L82
  lea    0x4(%rsp),%rsi
  lea    D27(%rip),%rdi
  call   <stat@plt>
  jmp    L82
  lea    0x4(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <stat@plt>
  jmp    L82
  lea    0x4(%rsp),%rsi
  lea    D21(%rip),%rdi
  call   <stat@plt>
  jmp    L82
  lea    0x4(%rsp),%rsi
  lea    D32(%rip),%rdi
  call   <stat@plt>
  jmp    L82
  lea    0x4(%rsp),%rsi
  lea    D55(%rip),%rdi
  call   <stat@plt>
  jmp    L82
  lea    0x4(%rsp),%rsi
  lea    D64(%rip),%rdi
  call   <stat@plt>
  jmp    L82
  lea    0x4(%rsp),%rsi
  lea    D64(%rip),%rdi
  call   <stat@plt>
  jmp    L82
L56:
  mov    $0xffffffff,%eax
  jmp    L82
L104:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D43
type_guessed: string
name: unpack_data
arguments: label: D60
type_guessed: string
name: unpack_data
arguments: label: D6
type_guessed: string
name: unpack_data
arguments: label: D27
type_guessed: string
name: unpack_data
arguments: label: D9
type_guessed: string
name: unpack_data
arguments: label: D21
type_guessed: string
name: unpack_data
arguments: label: D32
type_guessed: string
name: unpack_data
arguments: label: D55
type_guessed: string
name: unpack_data
arguments: label: D64
type_guessed: string

tool: D43: /dev/gpiomem

tool: D60: /dev/ttyAMA0

tool: D6: /dev/ttyS0

tool: D27: /dev/i2c-0

tool: D9: /dev/i2c-1

tool: D21: /dev/spidev0.0

tool: D32: /dev/spidev0.1

tool: D55: /dev/spidev1.0

tool: D64: /dev/spidev1.1

assistant: ```c
uint32_t librpip_device_node_exists(uint32_t devid) {
  struct stat fileStat;
  int result;
  switch (devid) {
  case 1:
    result = stat("/dev/gpiomem", &fileStat);
    break;
  case 2:
    result = stat("/dev/ttyAMA0", &fileStat);
    break;
  case 3:
    result = stat("/dev/ttyS0", &fileStat);
    break;
  case 4:
    result = stat("/dev/i2c-0", &fileStat);
    break;
  case 5:
    result = stat("/dev/i2c-1", &fileStat);
    break;
  case 6:
    result = stat("/dev/spidev0.0", &fileStat);
    break;
  case 7:
    result = stat("/dev/spidev0.1", &fileStat);
    break;
  case 8:
    result = stat("/dev/spidev1.0", &fileStat);
    break;
  case 9:
    result = stat("/dev/spidev1.1", &fileStat);
    break;
  case 10:
    result = stat("/dev/spidev1.1", &fileStat);
    break;
  default:
    result = -1;
  }
  return result < 0 ? 0 : 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<pentatonic_frequency>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%esi
  sar    $0x1f,%esi
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %ecx,%edx
  shl    $0x2,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  lea    D3(%rip),%rax
  movsd  (%rdx,%rax,1),%xmm1
  mov    -0x4(%rbp),%eax
  mov    $0x2,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double pentatonic_frequency(int y) {
  static double frequencies[] = {69.2957, 77.7817, 92.4986, 103.826, 116.541};
  int pitch = y % 5;
  int octave = y / 5;
  return frequencies[pitch] * (2 << octave);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pentatonic_frequency>:
  endbr64
  movslq %edi,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x21,%rcx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%ecx
  mov    $0x2,%eax
  shl    %cl,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  lea    (%rcx,%rcx,4),%eax
  sub    %eax,%edi
  movslq %edi,%rdi
  lea    D3(%rip),%rax
  mulsd  (%rax,%rdi,8),%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: label: D3
type_infered: f64

tool: D3: 69.2957

assistant: ```c
double pentatonic_frequency(int y) {
  static double frequencies[] = {69.2957, 77.7817, 92.4986, 103.826, 116.541};
  int pitch = y % 5;
  int octave = y / 5;
  return frequencies[pitch] * (2 << octave);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pentatonic_frequency>:
  endbr64
  movslq %edi,%rcx
  mov    %edi,%eax
  pxor   %xmm0,%xmm0
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%eax
  sar    $0x21,%rcx
  sub    %eax,%ecx
  mov    $0x2,%eax
  shl    %cl,%eax
  cvtsi2sd %eax,%xmm0
  lea    (%rcx,%rcx,4),%eax
  sub    %eax,%edi
  lea    D7(%rip),%rax
  movslq %edi,%rdi
  mulsd  (%rax,%rdi,8),%xmm0
  ret
```

tool_calls: name: read_data
arguments: label: D7
data_type: f64

tool: D7: 69.2957

assistant: ```c
double pentatonic_frequency(int y) {
  static double frequencies[] = {69.2957, 77.7817, 92.4986, 103.826, 116.541};
  int pitch = y % 5;
  int octave = y / 5;
  return frequencies[pitch] * (2 << octave);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<pentatonic_frequency>:
  endbr64
  movslq %edi,%rcx
  mov    %edi,%eax
  pxor   %xmm0,%xmm0
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%eax
  sar    $0x21,%rcx
  sub    %eax,%ecx
  mov    $0x2,%eax
  shl    %cl,%eax
  cvtsi2sd %eax,%xmm0
  lea    (%rcx,%rcx,4),%eax
  sub    %eax,%edi
  lea    D0(%rip),%rax
  movslq %edi,%rdi
  mulsd  (%rax,%rdi,8),%xmm0
  ret
```

tool_calls: name: parse_data
arguments: data_label: D0
data_type: f64

tool: D0: 69.2957

assistant: ```c
double pentatonic_frequency(int y) {
  static double frequencies[] = {69.2957, 77.7817, 92.4986, 103.826, 116.541};
  int pitch = y % 5;
  int octave = y / 5;
  return frequencies[pitch] * (2 << octave);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L3:
  addl   $0x1,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L3
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void reverse(char rtol[], char ltor[], int u) {
  int i = 0;
  while ((rtol[(u - 1) - i] = ltor[i]) != '\n')
    ++i;
  rtol[u] = '\0';
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<reverse>:
  endbr64
  lea    -0x1(%rdx),%eax
  cltq
  add    %rdi,%rax
  movzbl (%rsi),%ecx
  mov    %cl,(%rax)
  cmp    $0xa,%cl
  je     L10
  add    $0x1,%rsi
L9:
  movzbl (%rsi),%ecx
  mov    %cl,-0x1(%rax)
  add    $0x1,%rsi
  sub    $0x1,%rax
  cmp    $0xa,%cl
  jne    L9
L10:
  movslq %edx,%rdx
  movb   $0x0,(%rdi,%rdx,1)
  ret
```

assistant: ```c
void reverse(char rtol[], char ltor[], int u) {
  int i = 0;
  while ((rtol[(u - 1) - i] = ltor[i]) != '\n')
    ++i;
  rtol[u] = '\0';
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<reverse>:
  endbr64
  lea    -0x1(%rdx),%eax
  movzbl (%rsi),%ecx
  cltq
  add    %rdi,%rax
  mov    %cl,(%rax)
  cmp    $0xa,%cl
  je     L18
  add    $0x1,%rsi
  nopw   0x0(%rax,%rax,1)
L12:
  movzbl (%rsi),%ecx
  sub    $0x1,%rax
  add    $0x1,%rsi
  mov    %cl,(%rax)
  cmp    $0xa,%cl
  jne    L12
L18:
  movslq %edx,%rdx
  movb   $0x0,(%rdi,%rdx,1)
  ret
```

assistant: ```c
void reverse(char rtol[], char ltor[], int u) {
  int i = 0;
  while ((rtol[(u - 1) - i] = ltor[i]) != '\n')
    ++i;
  rtol[u] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<reverse>:
  endbr64
  lea    -0x1(%rdx),%eax
  movzbl (%rsi),%ecx
  cltq
  add    %rdi,%rax
  mov    %cl,(%rax)
  cmp    $0xa,%cl
  je     L14
  add    $0x1,%rsi
  nopw   0x0(%rax,%rax,1)
L0:
  movzbl (%rsi),%ecx
  sub    $0x1,%rax
  add    $0x1,%rsi
  mov    %cl,(%rax)
  cmp    $0xa,%cl
  jne    L0
L14:
  movslq %edx,%rdx
  movb   $0x0,(%rdi,%rdx,1)
  ret
```

assistant: ```c
void reverse(char rtol[], char ltor[], int u) {
  int i = 0;
  while ((rtol[(u - 1) - i] = ltor[i]) != '\n')
    ++i;
  rtol[u] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<print_tagname>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <print_indent@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_guessed: string

tool: D3: %s packet


assistant: ```c
void print_tagname(int indent, const char *str) {
  print_indent(indent);
  printf("%s packet\n", str);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<print_tagname>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  call   <print_indent@plt>
  mov    %rbx,%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
datatype: string

tool: D5: %s packet


assistant: ```c
void print_tagname(int indent, const char *str) {
  print_indent(indent);
  printf("%s packet\n", str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_tagname>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  call   <print_indent@plt>
  mov    %rbp,%rsi
  lea    D5(%rip),%rdi
  pop    %rbp
  jmp    1070 <printf@plt>
```

tool_calls: name: read_data
arguments: label: D5
type: string

tool: D5: %s packet


assistant: ```c
void print_tagname(int indent, const char *str) {
  print_indent(indent);
  printf("%s packet\n", str);
}
```
user: What is the c source code of the assembly code below:

```asm
<print_tagname>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  call   <print_indent@plt>
  mov    %rbp,%rsi
  lea    D2(%rip),%rdi
  pop    %rbp
  jmp    1070 <printf@plt>
```

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

tool: D2: %s packet


assistant: ```c
void print_tagname(int indent, const char *str) {
  print_indent(indent);
  printf("%s packet\n", str);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_rows>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L19
L6:
  mov    $0x5b,%edi
  call   <putchar@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L37
L22:
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L37:
  cmpl   $0x2,-0x4(%rbp)
  jle    L22
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addl   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L6
  nop
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D4
data_type: string
name: unpack_data
arguments: data_label: D13
data_type: string

tool: D4: %f 

tool: D13: ]

assistant: ```c
void print_rows(double *m, int n) {
  int i, j;
  for (i = 0; i < n; i++) {
    printf("[");
    for (j = 0; j < 3; j++)
      printf("%f ", m[i * 3 + j]);
    printf("]\n");
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<print_rows>:
  endbr64
  test   %esi,%esi
  jle    L39
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  lea    (%rsi,%rsi,2),%r15d
  mov    $0x0,%r13d
  lea    D3(%rip),%r12
  lea    D1(%rip),%r14
  jmp    L31
L3:
  mov    %r14,%rdi
  call   <puts@plt>
  add    $0x3,%r13d
  add    $0x18,%rbp
  cmp    %r15d,%r13d
  je     L15
L31:
  mov    $0x5b,%edi
  call   <putchar@plt>
  mov    $0x0,%ebx
L25:
  movsd  0x0(%rbp,%rbx,8),%xmm0
  mov    %r12,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x3,%rbx
  jne    L25
  jmp    L3
L15:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L39:
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
type_guessed: string
name: unpack_data
arguments: tag: D1
type_guessed: string

tool: D3: %f 

tool: D1: ]

assistant: ```c
void print_rows(double *m, int n) {
  int i, j;
  for (i = 0; i < n; i++) {
    printf("[");
    for (j = 0; j < 3; j++)
      printf("%f ", m[i * 3 + j]);
    printf("]\n");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<print_rows>:
  endbr64
  test   %esi,%esi
  jle    L7
  push   %r15
  lea    (%rsi,%rsi,2),%r15d
  push   %r14
  lea    D9(%rip),%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L5:
  mov    $0x5b,%edi
  xor    %ebx,%ebx
  call   <putchar@plt>
L12:
  movsd  0x0(%rbp,%rbx,8),%xmm0
  mov    %r12,%rdi
  mov    $0x1,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x3,%rbx
  jne    L12
  mov    %r14,%rdi
  add    $0x3,%r13d
  add    $0x18,%rbp
  call   <puts@plt>
  cmp    %r15d,%r13d
  jne    L5
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
data_type: string
name: retrieve_data
arguments: tag: D4
data_type: string

tool: D9: ]

tool: D4: %f 

assistant: ```c
void print_rows(double *m, int n) {
  int i, j;
  for (i = 0; i < n; i++) {
    printf("[");
    for (j = 0; j < 3; j++)
      printf("%f ", m[i * 3 + j]);
    printf("]\n");
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<print_rows>:
  endbr64
  test   %esi,%esi
  jle    L26
  push   %r15
  lea    (%rsi,%rsi,2),%r15d
  push   %r14
  lea    D1(%rip),%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  lea    D18(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  xchg   %ax,%ax
L17:
  mov    $0x5b,%edi
  xor    %ebx,%ebx
  call   <putchar@plt>
L13:
  movsd  0x0(%rbp,%rbx,8),%xmm0
  mov    %r12,%rdi
  mov    $0x1,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x3,%rbx
  jne    L13
  mov    %r14,%rdi
  add    $0x3,%r13d
  add    $0x18,%rbp
  call   <puts@plt>
  cmp    %r13d,%r15d
  jne    L17
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  ret
```

tool_calls: name: retrieve_data
arguments: label: D1
type_infered: string
name: retrieve_data
arguments: label: D18
type_infered: string

tool: D1: ]

tool: D18: %f 

assistant: ```c
void print_rows(double *m, int n) {
  int i, j;
  for (i = 0; i < n; i++) {
    printf("[");
    for (j = 0; j < 3; j++)
      printf("%f ", m[i * 3 + j]);
    printf("]\n");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<inw_p>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %ax,-0x14(%rbp)
  movzwl -0x14(%rbp),%eax
  mov    %eax,%edx
  in     (%dx),%ax
  out    %al,$0x80
  mov    %ax,-0x2(%rbp)
  movzwl -0x2(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned short int inw_p(unsigned short int port) {
  unsigned short int value;
  __asm__ __volatile__("inw %w1,%0\noutb %%al,$0x80" : "=a"(value) : "Nd"(port));
  return value;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<inw_p>:
  endbr64
  mov    %edi,%edx
  in     (%dx),%ax
  out    %al,$0x80
  ret
```

assistant: ```c
unsigned short int inw_p(unsigned short int port) {
  unsigned short int value;
  __asm__ __volatile__("inw %w1,%0\noutb %%al,$0x80" : "=a"(value) : "Nd"(port));
  return value;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<inw_p>:
  endbr64
  mov    %edi,%edx
  in     (%dx),%ax
  out    %al,$0x80
  ret
```

assistant: ```c
unsigned short int inw_p(unsigned short int port) {
  unsigned short int value;
  __asm__ __volatile__("inw %w1,%0\noutb %%al,$0x80" : "=a"(value) : "Nd"(port));
  return value;
}
```
user: What is the c source code of the assembly code below?

```asm
<inw_p>:
  endbr64
  mov    %edi,%edx
  in     (%dx),%ax
  out    %al,$0x80
  ret
```

assistant: ```c
unsigned short int inw_p(unsigned short int port) {
  unsigned short int value;
  __asm__ __volatile__("inw %w1,%0\noutb %%al,$0x80" : "=a"(value) : "Nd"(port));
  return value;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<R_DestroyVolatileAssets>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <R_ShutdownCustomColors@plt>
  mov    $0x0,%eax
  call   <R_ShutdownShadows@plt>
  mov    $0x0,%eax
  call   <R_ShutdownOutlines@plt>
  mov    $0x0,%eax
  call   <R_ShutdownCoronas@plt>
  mov    $0x0,%eax
  call   <R_ShutdownOcclusionQueries@plt>
  mov    $0x0,%eax
  call   <R_ShutdownSkeletalCache@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void R_DestroyVolatileAssets(void) {
  R_ShutdownCustomColors();
  R_ShutdownShadows();
  R_ShutdownOutlines();
  R_ShutdownCoronas();
  R_ShutdownOcclusionQueries();
  R_ShutdownSkeletalCache();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<R_DestroyVolatileAssets>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <R_ShutdownCustomColors@plt>
  mov    $0x0,%eax
  call   <R_ShutdownShadows@plt>
  mov    $0x0,%eax
  call   <R_ShutdownOutlines@plt>
  mov    $0x0,%eax
  call   <R_ShutdownCoronas@plt>
  mov    $0x0,%eax
  call   <R_ShutdownOcclusionQueries@plt>
  mov    $0x0,%eax
  call   <R_ShutdownSkeletalCache@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void R_DestroyVolatileAssets(void) {
  R_ShutdownCustomColors();
  R_ShutdownShadows();
  R_ShutdownOutlines();
  R_ShutdownCoronas();
  R_ShutdownOcclusionQueries();
  R_ShutdownSkeletalCache();
}
```
user: What is the c source code of the assembly code below?

```asm
<R_DestroyVolatileAssets>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <R_ShutdownCustomColors@plt>
  xor    %eax,%eax
  call   <R_ShutdownShadows@plt>
  xor    %eax,%eax
  call   <R_ShutdownOutlines@plt>
  xor    %eax,%eax
  call   <R_ShutdownCoronas@plt>
  xor    %eax,%eax
  call   <R_ShutdownOcclusionQueries@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    10e0 <R_ShutdownSkeletalCache@plt>
```

assistant: ```c
void R_DestroyVolatileAssets(void) {
  R_ShutdownCustomColors();
  R_ShutdownShadows();
  R_ShutdownOutlines();
  R_ShutdownCoronas();
  R_ShutdownOcclusionQueries();
  R_ShutdownSkeletalCache();
}
```