ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<clk_ratechange_after>:
  endbr64
  sub    $0x8,%rsp
  mov    %edx,%ecx
  lea    D6(%rip),%rdi
  mov    %rsi,%rdx
  xor    %eax,%eax
  lea    D13(%rip),%rsi
  call   <printk@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
data_type: string
name: parse_data
arguments: tag: D13
data_type: string

tool: D6: [%s] :%s   failed:%d


tool: D13: clk_ratechange_after

assistant: ```c
int clk_ratechange_after(unsigned long newrate, void *privdata, int failed) {
  printk("[%s] :%s   failed:%d\n", __func__, privdata, failed);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<clk_ratechange_after>:
  endbr64
  sub    $0x8,%rsp
  mov    %edx,%ecx
  lea    D13(%rip),%rdi
  mov    %rsi,%rdx
  xor    %eax,%eax
  lea    D15(%rip),%rsi
  call   <printk@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D13
type_guessed: string
name: retrieve_data
arguments: label: D15
type_guessed: string

tool: D13: [%s] :%s   failed:%d


tool: D15: clk_ratechange_after

assistant: ```c
int clk_ratechange_after(unsigned long newrate, void *privdata, int failed) {
  printk("[%s] :%s   failed:%d\n", __func__, privdata, failed);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<try_down>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L1
  mov    $0x0,%eax
  jmp    L39
L1:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  jle    L8
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  jmp    L39
L8:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
L39:
  pop    %rbp
  ret
```

assistant: ```c
int try_down(int *sem) {
  if (*sem)
    return 0;
  (*sem)++;
  if (*sem > 1) {
    (*sem)--;
    return 0;
  }
  return *sem;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<try_down>:
  endbr64
  mov    $0x0,%eax
  cmpl   $0x0,(%rdi)
  jne    L0
  movl   $0x1,(%rdi)
  mov    $0x1,%eax
L0:
  ret
```

assistant: ```c
int try_down(int *sem) {
  if (*sem)
    return 0;
  (*sem)++;
  if (*sem > 1) {
    (*sem)--;
    return 0;
  }
  return *sem;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<try_down>:
  endbr64
  mov    (%rdi),%edx
  xor    %eax,%eax
  test   %edx,%edx
  jne    L4
  movl   $0x1,(%rdi)
  mov    $0x1,%eax
L4:
  ret
```

assistant: ```c
int try_down(int *sem) {
  if (*sem)
    return 0;
  (*sem)++;
  if (*sem > 1) {
    (*sem)--;
    return 0;
  }
  return *sem;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<try_down>:
  endbr64
  mov    (%rdi),%edx
  xor    %eax,%eax
  test   %edx,%edx
  jne    L2
  movl   $0x1,(%rdi)
  mov    $0x1,%eax
L2:
  ret
```

assistant: ```c
int try_down(int *sem) {
  if (*sem)
    return 0;
  (*sem)++;
  if (*sem > 1) {
    (*sem)--;
    return 0;
  }
  return *sem;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movq   $0x0,-0x8(%rbp)
  movq   $0x2,-0x18(%rbp)
  movq   $0x1,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    $0x1,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    $0x2,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x3,-0x1c(%rbp)
  jmp    L13
L10:
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  addl   $0x1,-0x1c(%rbp)
L13:
  cmpl   $0x64,-0x1c(%rbp)
  jle    L10
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_guessed: string

tool: D3: %d - %lu


assistant: ```c
int main() {
  unsigned long int a, b, c;
  int i;
  c = 0;
  a = 2;
  b = 1;
  printf("%d - %lu\n", 1, a);
  printf("%d - %lu\n", 2, b);
  for (i = 3; i <= 100; i++) {
    c = b + a;
    printf("%d - %lu\n", i, c);
    a = b;
    b = c;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x2,%edx
  mov    $0x1,%esi
  lea    D3(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x1,%edx
  mov    $0x2,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3,%ebp
  mov    $0x1,%ebx
  mov    $0x2,%eax
  lea    D3(%rip),%r13
L5:
  mov    %rbx,%r12
  add    %rax,%rbx
  mov    %rbx,%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%ebp
  mov    %r12,%rax
  cmp    $0x65,%ebp
  jne    L5
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

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

tool: D3: %d - %lu


assistant: ```c
int main() {
  unsigned long int a, b, c;
  int i;
  c = 0;
  a = 2;
  b = 1;
  printf("%d - %lu\n", 1, a);
  printf("%d - %lu\n", 2, b);
  for (i = 3; i <= 100; i++) {
    c = b + a;
    printf("%d - %lu\n", i, c);
    a = b;
    b = c;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  lea    D7(%rip),%r13
  mov    $0x2,%edx
  mov    $0x1,%esi
  push   %r12
  mov    %r13,%rdi
  push   %rbp
  mov    $0x3,%ebp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    $0x1,%edx
  mov    $0x2,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  mov    $0x2,%eax
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %rbx,%r12
  add    %rax,%rbx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    %rbx,%rdx
  add    $0x1,%ebp
  call   <printf@plt>
  mov    %r12,%rax
  cmp    $0x65,%ebp
  jne    L8
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: tag: D7
type_guessed: string

tool: D7: %d - %lu


assistant: ```c
int main() {
  unsigned long int a, b, c;
  int i;
  c = 0;
  a = 2;
  b = 1;
  printf("%d - %lu\n", 1, a);
  printf("%d - %lu\n", 2, b);
  for (i = 3; i <= 100; i++) {
    c = b + a;
    printf("%d - %lu\n", i, c);
    a = b;
    b = c;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r13
  lea    D0(%rip),%r13
  mov    $0x2,%edx
  mov    $0x1,%esi
  push   %r12
  mov    %r13,%rdi
  push   %rbp
  mov    $0x3,%ebp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  call   <printf@plt>
  mov    $0x1,%edx
  mov    $0x2,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  mov    $0x2,%eax
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %rbx,%r12
  add    %rax,%rbx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    %rbx,%rdx
  add    $0x1,%ebp
  call   <printf@plt>
  mov    %r12,%rax
  cmp    $0x65,%ebp
  jne    L3
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: tag: D0
type_guessed: string

tool: D0: %d - %lu


assistant: ```c
int main() {
  unsigned long int a, b, c;
  int i;
  c = 0;
  a = 2;
  b = 1;
  printf("%d - %lu\n", 1, a);
  printf("%d - %lu\n", 2, b);
  for (i = 3; i <= 100; i++) {
    c = b + a;
    printf("%d - %lu\n", i, c);
    a = b;
    b = c;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<access>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x3,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    $0x8,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L25
L28:
  movl   $0x0,-0x8(%rbp)
  jmp    L26
L31:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L26:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L31
  addl   $0x1,-0xc(%rbp)
L25:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L28
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D19
type_guessed: string
name: extract_symbolic_data
arguments: tag: D3
type_guessed: string

tool: D19: sizeof(a) = %d


tool: D3: %d


assistant: ```c
void access(int a[][3], int row) {
  int col = sizeof(*a) / sizeof(int);
  int i = 0;
  int j = 0;
  printf("sizeof(a) = %d\n", sizeof(a));
  for (i = 0; i < row; i++)
    for (j = 0; j < col; j++)
      printf("%d\n", a[i][j]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<access>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0x8,%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L8
  mov    %r12,%rbx
  lea    -0x1(%rbp),%eax
  lea    (%rax,%rax,2),%rax
  lea    0xc(%r12,%rax,4),%r12
  lea    D17(%rip),%rbp
L11:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x4(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x8(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0xc,%rbx
  cmp    %rbx,%r12
  jne    L11
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D15
type: string
name: disassemble_data
arguments: tag: D17
type: string

tool: D15: sizeof(a) = %d


tool: D17: %d


assistant: ```c
void access(int a[][3], int row) {
  int col = sizeof(*a) / sizeof(int);
  int i = 0;
  int j = 0;
  printf("sizeof(a) = %d\n", sizeof(a));
  for (i = 0; i < row; i++)
    for (j = 0; j < col; j++)
      printf("%d\n", a[i][j]);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<access>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  mov    $0x8,%esi
  push   %rbx
  mov    %rdi,%rbx
  lea    D0(%rip),%rdi
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L1
  lea    -0x1(%rbp),%eax
  lea    D11(%rip),%rbp
  lea    (%rax,%rax,2),%rax
  lea    0xc(%rbx,%rax,4),%r12
  nopl   (%rax)
L17:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0xc,%rbx
  call   <printf@plt>
  mov    -0x8(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    -0x4(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L17
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D0
data_type: string
name: extract_symbolic_data
arguments: data_tag: D11
data_type: string

tool: D0: sizeof(a) = %d


tool: D11: %d


assistant: ```c
void access(int a[][3], int row) {
  int col = sizeof(*a) / sizeof(int);
  int i = 0;
  int j = 0;
  printf("sizeof(a) = %d\n", sizeof(a));
  for (i = 0; i < row; i++)
    for (j = 0; j < col; j++)
      printf("%d\n", a[i][j]);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<access>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  mov    $0x8,%esi
  push   %rbx
  mov    %rdi,%rbx
  lea    D12(%rip),%rdi
  call   <printf@plt>
  test   %ebp,%ebp
  jle    L10
  lea    -0x1(%rbp),%eax
  lea    D8(%rip),%rbp
  lea    (%rax,%rax,2),%rax
  lea    0xc(%rbx,%rax,4),%r12
  nopl   (%rax)
L13:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0xc,%rbx
  call   <printf@plt>
  mov    -0x8(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    -0x4(%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L13
L10:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D12
type_guessed: string
name: read_data
arguments: data_tag: D8
type_guessed: string

tool: D12: sizeof(a) = %d


tool: D8: %d


assistant: ```c
void access(int a[][3], int row) {
  int col = sizeof(*a) / sizeof(int);
  int i = 0;
  int j = 0;
  printf("sizeof(a) = %d\n", sizeof(a));
  for (i = 0; i < row; i++)
    for (j = 0; j < col; j++)
      printf("%d\n", a[i][j]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_test_3>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    $0x0,%eax
  call   <rewind_tape@plt>
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block_position@plt>
  mov    %eax,%edx
  lea    D16(%rip),%rbp
  mov    %rbp,%rsi
  lea    D7(%rip),%r12
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x2000,%esi
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block@plt>
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block_position@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <space_forward_filemark@plt>
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block_position@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x10000,%esi
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block@plt>
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block_position@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <space_back_block@plt>
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block_position@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x10000,%esi
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block@plt>
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block_position@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D22(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <space_forward_block@plt>
  mov    %ebx,%edi
  mov    $0x0,%eax
  call   <read_block_position@plt>
  mov    %eax,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
type: string
name: extract_symbolic_data
arguments: tag: D7
type: string
name: extract_symbolic_data
arguments: tag: D22
type: string

tool: D16: read_test_3

tool: D7: %s: Block %d


tool: D22: %s: Note: following space_forward_block() should error


assistant: ```c
int read_test_3(int fd) {
  rewind_tape(fd);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  read_block(fd, 8 * 1024);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  space_forward_filemark(fd, 3);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  read_block(fd, 64 * 1024);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  space_back_block(fd, 1);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  read_block(fd, 64 * 1024);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  printf("%s: Note: following space_forward_block() should error\n", __func__);
  space_forward_block(fd, 1);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<read_test_3>:
  endbr64
  push   %r13
  xor    %eax,%eax
  lea    D21(%rip),%r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  call   <rewind_tape@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x2000,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <space_forward_filemark@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x10000,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <space_back_block@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x10000,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block@plt>
  xor    %eax,%eax
  mov    %ebp,%edi
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <space_forward_block@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D21
type: string
name: extract_symbolic_data
arguments: label: D0
type: string
name: extract_symbolic_data
arguments: label: D17
type: string

tool: D21: %s: Block %d


tool: D0: read_test_3

tool: D17: %s: Note: following space_forward_block() should error


assistant: ```c
int read_test_3(int fd) {
  rewind_tape(fd);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  read_block(fd, 8 * 1024);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  space_forward_filemark(fd, 3);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  read_block(fd, 64 * 1024);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  space_back_block(fd, 1);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  read_block(fd, 64 * 1024);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  printf("%s: Note: following space_forward_block() should error\n", __func__);
  space_forward_block(fd, 1);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<read_test_3>:
  endbr64
  push   %r13
  xor    %eax,%eax
  lea    D12(%rip),%r13
  push   %r12
  lea    D26(%rip),%r12
  push   %rbp
  mov    %edi,%ebp
  call   <rewind_tape@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x2000,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <space_forward_filemark@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x10000,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <space_back_block@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x10000,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block@plt>
  xor    %eax,%eax
  mov    %ebp,%edi
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%esi
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <space_forward_block@plt>
  mov    %ebp,%edi
  xor    %eax,%eax
  call   <read_block_position@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %eax,%edx
  xor    %eax,%eax
  call   <printf@plt>
  pop    %rbp
  xor    %eax,%eax
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: label: D12
type_infered: string
name: parse_data
arguments: label: D26
type_infered: string
name: parse_data
arguments: label: D29
type_infered: string

tool: D12: %s: Block %d


tool: D26: read_test_3

tool: D29: %s: Note: following space_forward_block() should error


assistant: ```c
int read_test_3(int fd) {
  rewind_tape(fd);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  read_block(fd, 8 * 1024);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  space_forward_filemark(fd, 3);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  read_block(fd, 64 * 1024);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  space_back_block(fd, 1);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  read_block(fd, 64 * 1024);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  printf("%s: Note: following space_forward_block() should error\n", __func__);
  space_forward_block(fd, 1);
  printf("%s: Block %d\n", __func__, read_block_position(fd));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Multp_Rec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x1,-0x8(%rbp)
  jne    L14
  mov    -0x4(%rbp),%eax
  jmp    L7
L14:
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <Multp_Rec>
  mov    -0x4(%rbp),%edx
  add    %edx,%eax
L7:
  leave
  ret
```

assistant: ```c
int Multp_Rec(int n1, int n2) {
  if (n2 == 1)
    return n1;
  else
    return n1 + Multp_Rec(n1, n2 - 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Multp_Rec>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x1,%esi
  jne    L11
L6:
  mov    %ebx,%eax
  pop    %rbx
  ret
L11:
  sub    $0x1,%esi
  call   <Multp_Rec>
  add    %eax,%ebx
  jmp    L6
```

assistant: ```c
int Multp_Rec(int n1, int n2) {
  if (n2 == 1)
    return n1;
  else
    return n1 + Multp_Rec(n1, n2 - 1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Multp_Rec>:
  endbr64
  cmp    $0x1,%esi
  je     L0
  sub    $0x2,%esi
  imul   %edi,%esi
  lea    (%rsi,%rdi,2),%eax
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %edi,%eax
  ret
```

assistant: ```c
int Multp_Rec(int n1, int n2) {
  if (n2 == 1)
    return n1;
  else
    return n1 + Multp_Rec(n1, n2 - 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Multp_Rec>:
  endbr64
  cmp    $0x1,%esi
  je     L6
  sub    $0x2,%esi
  imul   %edi,%esi
  lea    (%rsi,%rdi,2),%eax
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %edi,%eax
  ret
```

assistant: ```c
int Multp_Rec(int n1, int n2) {
  if (n2 == 1)
    return n1;
  else
    return n1 + Multp_Rec(n1, n2 - 1);
}
```