ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <add@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D7
type: string

tool: D7: add(1, 2) = %d


assistant: ```c
int main(int argc, char **argv) {
  printf("add(1, 2) = %d\n", add(1, 2));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%esi
  mov    $0x1,%edi
  call   <add@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
type_infered: string

tool: D5: add(1, 2) = %d


assistant: ```c
int main(int argc, char **argv) {
  printf("add(1, 2) = %d\n", add(1, 2));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ilog>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L19:
  addl   $0x1,-0x4(%rbp)
  sarl   -0x14(%rbp)
L13:
  cmpl   $0x0,-0x14(%rbp)
  jg     L19
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int ilog(int x) {
  unsigned int r = 0;
  while (x > 0) {
    r += 1;
    x >>= 1;
  }
  return r;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ilog>:
  endbr64
  test   %edi,%edi
  jle    L13
  mov    $0x0,%eax
L0:
  add    $0x1,%eax
  sar    %edi
  test   %edi,%edi
  jg     L0
  ret
L13:
  mov    $0x0,%eax
  ret
```

assistant: ```c
unsigned int ilog(int x) {
  unsigned int r = 0;
  while (x > 0) {
    r += 1;
    x >>= 1;
  }
  return r;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ilog>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  jle    L15
  nopw   0x0(%rax,%rax,1)
L11:
  add    $0x1,%eax
  sar    %edi
  jne    L11
  ret
  nopl   0x0(%rax,%rax,1)
L15:
  ret
```

assistant: ```c
unsigned int ilog(int x) {
  unsigned int r = 0;
  while (x > 0) {
    r += 1;
    x >>= 1;
  }
  return r;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ilog>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  jle    L3
  nopw   0x0(%rax,%rax,1)
L9:
  add    $0x1,%eax
  sar    %edi
  jne    L9
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  ret
```

assistant: ```c
unsigned int ilog(int x) {
  unsigned int r = 0;
  while (x > 0) {
    r += 1;
    x >>= 1;
  }
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ok_at_loc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %esi,-0xcc(%rbp)
  mov    %edx,-0xd0(%rbp)
  mov    %rcx,-0xd8(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L0
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L0:
  mov    -0xd8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edi
  mov    -0xd8(%rbp),%rcx
  mov    -0xd0(%rbp),%edx
  mov    -0xcc(%rbp),%esi
  mov    -0xc8(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <vok_at_loc@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xd0(%rbp),%eax
  leave
  ret
```

assistant: ```c
int ok_at_loc(const char *file, int line, int test, const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  vok_at_loc(file, line, test, fmt, args);
  va_end(args);
  return test;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ok_at_loc>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %edx,%ebx
  mov    %rcx,%rbp
  mov    %rcx,%rsi
  mov    $0x0,%r14d
  mov    %r14d,%edi
  call   <va_start@plt>
  mov    %r14d,%r8d
  mov    %rbp,%rcx
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <vok_at_loc@plt>
  mov    %r14d,%edi
  call   <va_end@plt>
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int ok_at_loc(const char *file, int line, int test, const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  vok_at_loc(file, line, test, fmt, args);
  va_end(args);
  return test;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ok_at_loc>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  mov    %rcx,%rsi
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%r8d
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %r15d,%esi
  mov    %r14,%rdi
  call   <vok_at_loc@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int ok_at_loc(const char *file, int line, int test, const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  vok_at_loc(file, line, test, fmt, args);
  va_end(args);
  return test;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ok_at_loc>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  mov    %rcx,%rsi
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%r8d
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %r15d,%esi
  mov    %r14,%rdi
  call   <vok_at_loc@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int ok_at_loc(const char *file, int line, int test, const char *fmt, ...) {
  va_list args;
  va_start(args, fmt);
  vok_at_loc(file, line, test, fmt, args);
  va_end(args);
  return test;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x60(%rbp),%rax
  lea    D1(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  movl   $0x18,-0x10(%rbp)
  lea    -0x60(%rbp),%rax
  mov    %rax,%rsi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D1
type: string
name: parse_data
arguments: label: D42
type: string
name: parse_data
arguments: label: D7
type: string
name: parse_data
arguments: label: D46
type: string
name: parse_data
arguments: label: D11
type: string

tool: D1: Maria

tool: D42: %s

tool: D7:  tem 

tool: D46: %d

tool: D11:  anos

assistant: ```c
int main() {
  struct {
    char nome[80];
    int idade;
  } reg;
  strcpy(reg.nome, "Maria");
  reg.idade = 24;
  printf("%s", reg.nome);
  printf(" tem ");
  printf("%d", reg.idade);
  printf(" anos");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    D28(%rip),%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  movl   $0x18,0x50(%rsp)
  mov    %rbx,%rsi
  lea    D49(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D34(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x50(%rsp),%esi
  lea    D39(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D35(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x60,%rsp
  pop    %rbx
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D28
datatype: string
name: unpack_data
arguments: data_tag: D49
datatype: string
name: unpack_data
arguments: data_tag: D34
datatype: string
name: unpack_data
arguments: data_tag: D39
datatype: string
name: unpack_data
arguments: data_tag: D35
datatype: string

tool: D28: Maria

tool: D49: %s

tool: D34:  tem 

tool: D39: %d

tool: D35:  anos

assistant: ```c
int main() {
  struct {
    char nome[80];
    int idade;
  } reg;
  strcpy(reg.nome, "Maria");
  reg.idade = 24;
  printf("%s", reg.nome);
  printf(" tem ");
  printf("%d", reg.idade);
  printf(" anos");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D34(%rip),%rsi
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  xor    %eax,%eax
  lea    D37(%rip),%rdi
  movl   $0x18,0x50(%rsp)
  call   <printf@plt>
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x50(%rsp),%esi
  lea    D47(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D34
data_type: string
name: retrieve_data
arguments: label: D37
data_type: string
name: retrieve_data
arguments: label: D42
data_type: string
name: retrieve_data
arguments: label: D47
data_type: string
name: retrieve_data
arguments: label: D16
data_type: string

tool: D34: Maria

tool: D37: %s

tool: D42:  tem 

tool: D47: %d

tool: D16:  anos

assistant: ```c
int main() {
  struct {
    char nome[80];
    int idade;
  } reg;
  strcpy(reg.nome, "Maria");
  reg.idade = 24;
  printf("%s", reg.nome);
  printf(" tem ");
  printf("%d", reg.idade);
  printf(" anos");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D13(%rip),%rsi
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbp,%rsi
  xor    %eax,%eax
  lea    D31(%rip),%rdi
  movl   $0x18,0x50(%rsp)
  call   <printf@plt>
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x50(%rsp),%esi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x60,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D13
datatype: string
name: extract_symbolic_data
arguments: data_label: D31
datatype: string
name: extract_symbolic_data
arguments: data_label: D42
datatype: string
name: extract_symbolic_data
arguments: data_label: D6
datatype: string
name: extract_symbolic_data
arguments: data_label: D12
datatype: string

tool: D13: Maria

tool: D31: %s

tool: D42:  tem 

tool: D6: %d

tool: D12:  anos

assistant: ```c
int main() {
  struct {
    char nome[80];
    int idade;
  } reg;
  strcpy(reg.nome, "Maria");
  reg.idade = 24;
  printf("%s", reg.nome);
  printf(" tem ");
  printf("%d", reg.idade);
  printf(" anos");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<htonll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x1c(%rbp)
  lea    -0x1c(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x1,%al
  jne    L27
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <htonl@plt>
  mov    %eax,%ebx
  mov    $0x20,%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,%edi
  call   <htonl@plt>
  or     %ebx,%eax
  jmp    L12
L27:
  mov    -0x24(%rbp),%eax
L12:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
uint64_t htonll(uint64_t value) {
  int num = 1;
  if (*(char *)&num == 1)
    return htonl((uint32_t)(value & 0xFFFFFFFF)) | htonl((uint32_t)(value >> 32));
  return value;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<htonll>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  call   <htonl@plt>
  mov    %eax,%ebp
  mov    $0x20,%ecx
  sar    %cl,%ebx
  mov    %ebx,%edi
  call   <htonl@plt>
  or     %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
uint64_t htonll(uint64_t value) {
  int num = 1;
  if (*(char *)&num == 1)
    return htonl((uint32_t)(value & 0xFFFFFFFF)) | htonl((uint32_t)(value >> 32));
  return value;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<htonll>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <htonl@plt>
  mov    $0x20,%ecx
  sar    %cl,%ebx
  mov    %eax,%ebp
  mov    %ebx,%edi
  call   <htonl@plt>
  add    $0x8,%rsp
  or     %ebp,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
uint64_t htonll(uint64_t value) {
  int num = 1;
  if (*(char *)&num == 1)
    return htonl((uint32_t)(value & 0xFFFFFFFF)) | htonl((uint32_t)(value >> 32));
  return value;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<htonll>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <htonl@plt>
  mov    $0x20,%ecx
  sar    %cl,%ebx
  mov    %eax,%ebp
  mov    %ebx,%edi
  call   <htonl@plt>
  add    $0x8,%rsp
  or     %ebp,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
uint64_t htonll(uint64_t value) {
  int num = 1;
  if (*(char *)&num == 1)
    return htonl((uint32_t)(value & 0xFFFFFFFF)) | htonl((uint32_t)(value >> 32));
  return value;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getRoot_g>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  movsd  %xmm0,-0x38(%rbp)
  mov    %edi,-0x3c(%rbp)
  movl   $0x1,-0x24(%rbp)
  mov    -0x38(%rbp),%rax
  movq   %rax,%xmm0
  call   <g@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  movq   %rax,%xmm0
  call   <gprime@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  movq   %rax,%xmm0
  call   <gprime2@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L35
L31:
  movsd  -0x20(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x18(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movsd  -0x20(%rbp),%xmm2
  mulsd  -0x10(%rbp),%xmm2
  subsd  %xmm2,%xmm0
  divsd  %xmm0,%xmm1
  movsd  -0x38(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movsd  -0x8(%rbp),%xmm0
  subsd  -0x38(%rbp),%xmm0
  movq   D29(%rip),%xmm1
  andpd  %xmm0,%xmm1
  movsd  D33(%rip),%xmm0
  comisd %xmm1,%xmm0
  jbe    L4
  movsd  -0x8(%rbp),%xmm0
  jmp    L15
L4:
  addl   $0x1,-0x24(%rbp)
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <g@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <gprime@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <gprime2@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
L35:
  mov    -0x24(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jle    L31
  movsd  D27(%rip),%xmm0
L15:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type: string
name: retrieve_data
arguments: data_tag: D29
type: qword
name: retrieve_data
arguments: data_tag: D33
type: f64
name: retrieve_data
arguments: data_tag: D27
type: f32[2]

tool: D4: Iter %d: %.9lf


tool: D29: 9223372036854775807

tool: D33: 1e-05

tool: {"D27": [0.0,-1.875]}

assistant: ```c
double getRoot_g(double p0, int maxiter) {
  int i = 1;
  double fp0 = g(p0);
  double f1p0 = gprime(p0);
  double f2p0 = gprime2(p0);
  double p1 = 0;
  while (i <= maxiter) {
    p1 = p0 - (fp0 * f1p0) / (f1p0 * f1p0 - fp0 * f2p0);
    printf("Iter %d: %.9lf\n", i, p1);
    if (fabs(p1 - p0) < 0.00001)
      return p1;
    i++;
    p0 = p1;
    fp0 = g(p1);
    f1p0 = gprime(p1);
    f2p0 = gprime2(p1);
  }
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<getRoot_g>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movq   %xmm0,%rbx
  mov    %edi,%r12d
  call   <g@plt>
  movsd  %xmm0,(%rsp)
  movq   %rbx,%xmm0
  call   <gprime@plt>
  movsd  %xmm0,0x8(%rsp)
  movq   %rbx,%xmm0
  call   <gprime2@plt>
  test   %r12d,%r12d
  jle    L9
  mov    $0x1,%ebp
  lea    D43(%rip),%r13
L19:
  mov    %rbx,%r14
  movsd  (%rsp),%xmm3
  movapd %xmm3,%xmm1
  movsd  0x8(%rsp),%xmm2
  mulsd  %xmm2,%xmm1
  mulsd  %xmm2,%xmm2
  mulsd  %xmm3,%xmm0
  subsd  %xmm0,%xmm2
  divsd  %xmm2,%xmm1
  movq   %rbx,%xmm4
  subsd  %xmm1,%xmm4
  movq   %xmm4,%rbx
  movapd %xmm4,%xmm0
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  movq   %rbx,%xmm0
  movq   %r14,%xmm5
  subsd  %xmm5,%xmm0
  andpd  D9(%rip),%xmm0
  movsd  D45(%rip),%xmm6
  comisd %xmm0,%xmm6
  ja     L14
  add    $0x1,%ebp
  movq   %rbx,%xmm0
  call   <g@plt>
  movsd  %xmm0,(%rsp)
  movq   %rbx,%xmm0
  call   <gprime@plt>
  movsd  %xmm0,0x8(%rsp)
  movq   %rbx,%xmm0
  call   <gprime2@plt>
  cmp    %ebp,%r12d
  jge    L19
  movsd  D42(%rip),%xmm0
  jmp    L1
L9:
  movsd  D42(%rip),%xmm0
  jmp    L1
L14:
  movq   %rbx,%xmm0
L1:
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: tag: D43
type_guessed: string
name: read_data
arguments: tag: D9
type_guessed: qword[2]
name: read_data
arguments: tag: D45
type_guessed: f64
name: read_data
arguments: tag: D42
type_guessed: f32[2]

tool: D43: Iter %d: %.9lf


tool: {"D9": [0x7fffffffffffffff,0x0000000000000000]}

tool: D45: 1e-05

tool: {"D42": [0.0,-1.875]}

assistant: ```c
double getRoot_g(double p0, int maxiter) {
  int i = 1;
  double fp0 = g(p0);
  double f1p0 = gprime(p0);
  double f2p0 = gprime2(p0);
  double p1 = 0;
  while (i <= maxiter) {
    p1 = p0 - (fp0 * f1p0) / (f1p0 * f1p0 - fp0 * f2p0);
    printf("Iter %d: %.9lf\n", i, p1);
    if (fabs(p1 - p0) < 0.00001)
      return p1;
    i++;
    p0 = p1;
    fp0 = g(p1);
    f1p0 = gprime(p1);
    f2p0 = gprime2(p1);
  }
  return -1;
}
```
user: What is the c source code of the assembly code below:

```asm
<getRoot_g>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x20,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <g@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  %xmm0,0x18(%rsp)
  movapd %xmm2,%xmm0
  call   <gprime@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  %xmm0,0x10(%rsp)
  movapd %xmm2,%xmm0
  call   <gprime2@plt>
  test   %ebx,%ebx
  jle    L35
  movsd  0x18(%rsp),%xmm3
  movsd  0x10(%rsp),%xmm1
  mov    $0x1,%r12d
  lea    D36(%rip),%rbp
  movsd  0x8(%rsp),%xmm2
  jmp    L5
  nopl   0x0(%rax)
L17:
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x8(%rsp)
  add    $0x1,%r12d
  call   <g@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  %xmm0,0x18(%rsp)
  movapd %xmm2,%xmm0
  call   <gprime@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  %xmm0,0x10(%rsp)
  movapd %xmm2,%xmm0
  call   <gprime2@plt>
  cmp    %r12d,%ebx
  movsd  0x8(%rsp),%xmm2
  movsd  0x10(%rsp),%xmm1
  movsd  0x18(%rsp),%xmm3
  jl     L35
L5:
  movapd %xmm3,%xmm4
  movsd  %xmm2,0x8(%rsp)
  mov    %r12d,%esi
  mov    %rbp,%rdi
  mulsd  %xmm0,%xmm3
  mulsd  %xmm1,%xmm4
  mulsd  %xmm1,%xmm1
  subsd  %xmm3,%xmm1
  divsd  %xmm1,%xmm4
  subsd  %xmm4,%xmm2
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x10(%rsp)
  call   <printf@plt>
  movsd  0x10(%rsp),%xmm2
  movsd  D32(%rip),%xmm6
  movapd %xmm2,%xmm0
  subsd  0x8(%rsp),%xmm0
  andpd  D25(%rip),%xmm0
  comisd %xmm0,%xmm6
  jbe    L17
  add    $0x20,%rsp
  movapd %xmm2,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L35:
  movsd  D35(%rip),%xmm0
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: tag: D36
type_infered: string
name: unpack_data
arguments: tag: D32
type_infered: f64
name: unpack_data
arguments: tag: D25
type_infered: qword[2]
name: unpack_data
arguments: tag: D35
type_infered: f32[2]

tool: D36: Iter %d: %.9lf


tool: D32: 1e-05

tool: {"D25": [0x7fffffffffffffff,0x0000000000000000]}

tool: {"D35": [0.0,-1.875]}

assistant: ```c
double getRoot_g(double p0, int maxiter) {
  int i = 1;
  double fp0 = g(p0);
  double f1p0 = gprime(p0);
  double f2p0 = gprime2(p0);
  double p1 = 0;
  while (i <= maxiter) {
    p1 = p0 - (fp0 * f1p0) / (f1p0 * f1p0 - fp0 * f2p0);
    printf("Iter %d: %.9lf\n", i, p1);
    if (fabs(p1 - p0) < 0.00001)
      return p1;
    i++;
    p0 = p1;
    fp0 = g(p1);
    f1p0 = gprime(p1);
    f2p0 = gprime2(p1);
  }
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getRoot_g>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x20,%rsp
  movsd  %xmm0,0x8(%rsp)
  call   <g@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  %xmm0,0x18(%rsp)
  movapd %xmm2,%xmm0
  call   <gprime@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  %xmm0,0x10(%rsp)
  movapd %xmm2,%xmm0
  call   <gprime2@plt>
  test   %ebx,%ebx
  jle    L5
  movsd  0x18(%rsp),%xmm3
  movsd  0x10(%rsp),%xmm1
  mov    $0x1,%r12d
  lea    D32(%rip),%rbp
  movsd  0x8(%rsp),%xmm2
  jmp    L3
  nopl   0x0(%rax)
L2:
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x8(%rsp)
  add    $0x1,%r12d
  call   <g@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  %xmm0,0x18(%rsp)
  movapd %xmm2,%xmm0
  call   <gprime@plt>
  movsd  0x8(%rsp),%xmm2
  movsd  %xmm0,0x10(%rsp)
  movapd %xmm2,%xmm0
  call   <gprime2@plt>
  cmp    %r12d,%ebx
  movsd  0x8(%rsp),%xmm2
  movsd  0x10(%rsp),%xmm1
  movsd  0x18(%rsp),%xmm3
  jl     L5
L3:
  movapd %xmm3,%xmm4
  movsd  %xmm2,0x8(%rsp)
  mov    %r12d,%esi
  mov    %rbp,%rdi
  mulsd  %xmm0,%xmm3
  mulsd  %xmm1,%xmm4
  mulsd  %xmm1,%xmm1
  subsd  %xmm3,%xmm1
  divsd  %xmm1,%xmm4
  subsd  %xmm4,%xmm2
  movapd %xmm2,%xmm0
  movsd  %xmm2,0x10(%rsp)
  call   <printf@plt>
  movsd  0x10(%rsp),%xmm2
  movsd  D10(%rip),%xmm6
  movapd %xmm2,%xmm0
  subsd  0x8(%rsp),%xmm0
  andpd  D1(%rip),%xmm0
  comisd %xmm0,%xmm6
  jbe    L2
  add    $0x20,%rsp
  movapd %xmm2,%xmm0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  movsd  D31(%rip),%xmm0
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_tag: D32
type: string
name: read_data
arguments: data_tag: D10
type: f64
name: read_data
arguments: data_tag: D1
type: qword[2]
name: read_data
arguments: data_tag: D31
type: f32[2]

tool: D32: Iter %d: %.9lf


tool: D10: 1e-05

tool: {"D1": [0x7fffffffffffffff,0x0000000000000000]}

tool: {"D31": [0.0,-1.875]}

assistant: ```c
double getRoot_g(double p0, int maxiter) {
  int i = 1;
  double fp0 = g(p0);
  double f1p0 = gprime(p0);
  double f2p0 = gprime2(p0);
  double p1 = 0;
  while (i <= maxiter) {
    p1 = p0 - (fp0 * f1p0) / (f1p0 * f1p0 - fp0 * f2p0);
    printf("Iter %d: %.9lf\n", i, p1);
    if (fabs(p1 - p0) < 0.00001)
      return p1;
    i++;
    p0 = p1;
    fp0 = g(p1);
    f1p0 = gprime(p1);
    f2p0 = gprime2(p1);
  }
  return -1;
}
```