ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L17
L9:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
L17:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L9
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c;
  while ((c = getchar()) != EOF) {
    putchar(c);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  jmp    L17
L11:
  call   <putchar@plt>
L17:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  mov    D2(%rip),%rax
  cmp    %edi,(%rax)
  jne    L11
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c;
  while ((c = getchar()) != EOF) {
    putchar(c);
  }
}
```
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   %rbx
  mov    D7(%rip),%rbx
  jmp    L0
  xchg   %ax,%ax
L19:
  call   <putchar@plt>
L0:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L19
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c;
  while ((c = getchar()) != EOF) {
    putchar(c);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D0(%rip),%rbx
  jmp    L11
  xchg   %ax,%ax
L4:
  call   <putchar@plt>
L11:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L4
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c;
  while ((c = getchar()) != EOF) {
    putchar(c);
  }
}
```
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    $0x10,%rsp
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L6
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L6:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_infered: string

tool: D5: Ptr e 0

assistant: ```c
int main(void) {
  int *ptr = NULL;
  if (ptr == 0)
    printf("Ptr e 0");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D3: Ptr e 0

assistant: ```c
int main(void) {
  int *ptr = NULL;
  if (ptr == 0)
    printf("Ptr e 0");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D2: Ptr e 0

assistant: ```c
int main(void) {
  int *ptr = NULL;
  if (ptr == 0)
    printf("Ptr e 0");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D3
data_type: string

tool: D3: Ptr e 0

assistant: ```c
int main(void) {
  int *ptr = NULL;
  if (ptr == 0)
    printf("Ptr e 0");
  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
<test_float_times64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <u2f@plt>
  movd   %xmm0,%eax
  mov    %eax,-0x8(%rbp)
  movss  -0x8(%rbp),%xmm1
  movss  D8(%rip),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  movd   %eax,%xmm0
  call   <isnan@plt>
  test   %rax,%rax
  je     L6
  mov    -0x14(%rbp),%eax
  jmp    L8
L6:
  mov    -0x4(%rbp),%eax
  movd   %eax,%xmm0
  call   <f2u@plt>
L8:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
type_infered: f32

tool: D8: 64

assistant: ```c
unsigned test_float_times64(unsigned uf) {
  float f = u2f(uf);
  float tenf = 64 * f;
  if (isnan(f))
    return uf;
  else
    return f2u(tenf);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<test_float_times64>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebx
  call   <u2f@plt>
  movss  %xmm0,0xc(%rsp)
  call   <isnan@plt>
  test   %rax,%rax
  je     L16
L17:
  mov    %ebx,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L16:
  movss  0xc(%rsp),%xmm0
  mulss  D4(%rip),%xmm0
  call   <f2u@plt>
  mov    %eax,%ebx
  jmp    L17
```

tool_calls: name: parse_data
arguments: data_tag: D4
type_infered: f32

tool: D4: 64

assistant: ```c
unsigned test_float_times64(unsigned uf) {
  float f = u2f(uf);
  float tenf = 64 * f;
  if (isnan(f))
    return uf;
  else
    return f2u(tenf);
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_float_times64>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  sub    $0x10,%rsp
  call   <u2f@plt>
  movss  %xmm0,0xc(%rsp)
  call   <isnan@plt>
  movss  0xc(%rsp),%xmm0
  test   %rax,%rax
  je     L4
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  mulss  D4(%rip),%xmm0
  add    $0x10,%rsp
  pop    %r12
  jmp    1090 <f2u@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
datatype: f32

tool: D4: 64

assistant: ```c
unsigned test_float_times64(unsigned uf) {
  float f = u2f(uf);
  float tenf = 64 * f;
  if (isnan(f))
    return uf;
  else
    return f2u(tenf);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_float_times64>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  sub    $0x10,%rsp
  call   <u2f@plt>
  movss  %xmm0,0xc(%rsp)
  call   <isnan@plt>
  movss  0xc(%rsp),%xmm0
  test   %rax,%rax
  je     L1
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mulss  D6(%rip),%xmm0
  add    $0x10,%rsp
  pop    %r12
  jmp    1090 <f2u@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_infered: f32

tool: D6: 64

assistant: ```c
unsigned test_float_times64(unsigned uf) {
  float f = u2f(uf);
  float tenf = 64 * f;
  if (isnan(f))
    return uf;
  else
    return f2u(tenf);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<left>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L0
L14:
  mov    -0x8(%rbp),%rax
  mov    0x4(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  addq   $0x4,-0x8(%rbp)
L0:
  mov    -0x18(%rbp),%rax
  add    $0x24,%rax
  cmp    %rax,-0x8(%rbp)
  jb     L14
  mov    -0x8(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void left(int a[]) {
  int temp = *a, *p;
  for (p = a; p < a + 9; p++)
    *p = *(p + 1);
  *p = temp;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<left>:
  endbr64
  mov    (%rdi),%esi
  lea    0x24(%rdi),%rcx
  mov    %rdi,%rax
L4:
  mov    0x4(%rax),%edx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L4
  mov    %esi,0x24(%rdi)
  ret
```

assistant: ```c
void left(int a[]) {
  int temp = *a, *p;
  for (p = a; p < a + 9; p++)
    *p = *(p + 1);
  *p = temp;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<left>:
  endbr64
  push   %rbx
  lea    0x4(%rdi),%rsi
  mov    (%rdi),%ebx
  mov    $0x24,%edx
  call   <memmove@plt>
  mov    %ebx,0x24(%rax)
  pop    %rbx
  ret
```

assistant: ```c
void left(int a[]) {
  int temp = *a, *p;
  for (p = a; p < a + 9; p++)
    *p = *(p + 1);
  *p = temp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<left>:
  endbr64
  push   %rbx
  lea    0x4(%rdi),%rsi
  mov    (%rdi),%ebx
  mov    $0x24,%edx
  call   <memmove@plt>
  mov    %ebx,0x24(%rax)
  pop    %rbx
  ret
```

assistant: ```c
void left(int a[]) {
  int temp = *a, *p;
  for (p = a; p < a + 9; p++)
    *p = *(p + 1);
  *p = temp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<address_metaphone>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L11
  jmp    L48
L33:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rcx),%edx
  mov    %dl,(%rax)
L48:
  mov    -0xc(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jge    L9
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L9
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L33
L9:
  mov    -0x20(%rbp),%rax
  movb   $0x0,(%rax)
  mov    $0x1,%eax
  jmp    L23
L11:
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <metaphone@plt>
L23:
  leave
  ret
```

assistant: ```c
int address_metaphone(const char *input, char *output, int max_phones) {
  const char *n = input;
  int i = 0;
  if (isdigit(*n)) {
    while (i < max_phones && isdigit(n[i]) && n[i] != '\0')
      *output++ = n[i++];
    *output = '\0';
    return 1;
  } else {
    return metaphone(input, output, max_phones);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<address_metaphone>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r12d
  movsbl (%rdi),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L44
  test   %r12d,%r12d
  jle    L53
  mov    %r12d,%r12d
  add    %rbp,%r12
L37:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L53
  movzbl (%rbx),%eax
  test   %al,%al
  je     L53
  add    $0x1,%rbp
  mov    %al,-0x1(%rbp)
  add    $0x1,%rbx
  cmp    %r12,%rbp
  jne    L37
  mov    %r12,%rbp
L53:
  movb   $0x0,0x0(%rbp)
  mov    $0x1,%eax
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L44:
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <metaphone@plt>
  jmp    L39
```

assistant: ```c
int address_metaphone(const char *input, char *output, int max_phones) {
  const char *n = input;
  int i = 0;
  if (isdigit(*n)) {
    while (i < max_phones && isdigit(n[i]) && n[i] != '\0')
      *output++ = n[i++];
    *output = '\0';
    return 1;
  } else {
    return metaphone(input, output, max_phones);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<address_metaphone>:
  endbr64
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L26
  test   %r12d,%r12d
  jle    L5
  add    %rbp,%r12
  jmp    L8
  nopl   0x0(%rax,%rax,1)
L57:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L5
  add    $0x1,%rbp
  add    $0x1,%rbx
  mov    %al,-0x1(%rbp)
  cmp    %r12,%rbp
  je     L5
L8:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L57
L5:
  pop    %rbx
  movb   $0x0,0x0(%rbp)
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <metaphone@plt>
```

assistant: ```c
int address_metaphone(const char *input, char *output, int max_phones) {
  const char *n = input;
  int i = 0;
  if (isdigit(*n)) {
    while (i < max_phones && isdigit(n[i]) && n[i] != '\0')
      *output++ = n[i++];
    *output = '\0';
    return 1;
  } else {
    return metaphone(input, output, max_phones);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<address_metaphone>:
  endbr64
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L53
  test   %r12d,%r12d
  jle    L10
  add    %rbp,%r12
  jmp    L22
  nopl   0x0(%rax,%rax,1)
L4:
  movzbl (%rbx),%eax
  test   %al,%al
  je     L10
  add    $0x1,%rbp
  add    $0x1,%rbx
  mov    %al,-0x1(%rbp)
  cmp    %r12,%rbp
  je     L10
L22:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L4
L10:
  pop    %rbx
  movb   $0x0,0x0(%rbp)
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L53:
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <metaphone@plt>
```

assistant: ```c
int address_metaphone(const char *input, char *output, int max_phones) {
  const char *n = input;
  int i = 0;
  if (isdigit(*n)) {
    while (i < max_phones && isdigit(n[i]) && n[i] != '\0')
      *output++ = n[i++];
    *output = '\0';
    return 1;
  } else {
    return metaphone(input, output, max_phones);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<enumerate_objects_raw>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L17
  mov    -0x20(%rbp),%rax
  and    $0xfffffffffffffffe,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <enumerate_string@plt>
  jmp    L14
L17:
  mov    -0x20(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <enumerate_object@plt>
L14:
  nop
  leave
  ret
```

assistant: ```c
void enumerate_objects_raw(int indent, void *data) {
  if (((uintptr_t)data) & 1) {
    uint8_t *ptr = (uint8_t *)(((uintptr_t)data) & ~1);
    enumerate_string(indent, ptr);
  } else {
    enumerate_object(indent, (uint8_t *)data);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<enumerate_objects_raw>:
  endbr64
  sub    $0x8,%rsp
  test   $0x1,%sil
  je     L3
  and    $0xfffffffffffffffe,%rsi
  call   <enumerate_string@plt>
L11:
  add    $0x8,%rsp
  ret
L3:
  call   <enumerate_object@plt>
  jmp    L11
```

assistant: ```c
void enumerate_objects_raw(int indent, void *data) {
  if (((uintptr_t)data) & 1) {
    uint8_t *ptr = (uint8_t *)(((uintptr_t)data) & ~1);
    enumerate_string(indent, ptr);
  } else {
    enumerate_object(indent, (uint8_t *)data);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<enumerate_objects_raw>:
  endbr64
  test   $0x1,%sil
  je     L5
  and    $0xfffffffffffffffe,%rsi
  jmp    1060 <enumerate_string@plt>
  nopl   0x0(%rax,%rax,1)
L5:
  jmp    1070 <enumerate_object@plt>
```

assistant: ```c
void enumerate_objects_raw(int indent, void *data) {
  if (((uintptr_t)data) & 1) {
    uint8_t *ptr = (uint8_t *)(((uintptr_t)data) & ~1);
    enumerate_string(indent, ptr);
  } else {
    enumerate_object(indent, (uint8_t *)data);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<enumerate_objects_raw>:
  endbr64
  test   $0x1,%sil
  je     L3
  and    $0xfffffffffffffffe,%rsi
  jmp    1060 <enumerate_string@plt>
  nopl   0x0(%rax,%rax,1)
L3:
  jmp    1070 <enumerate_object@plt>
```

assistant: ```c
void enumerate_objects_raw(int indent, void *data) {
  if (((uintptr_t)data) & 1) {
    uint8_t *ptr = (uint8_t *)(((uintptr_t)data) & ~1);
    enumerate_string(indent, ptr);
  } else {
    enumerate_object(indent, (uint8_t *)data);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_dobldobl_Newton_step>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <syscon_read_dobldobl_system@plt>
  mov    %eax,-0xc(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <syscon_number_of_dobldobl_polynomials@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <solcon_read_dobldobl_solutions@plt>
  mov    %eax,-0xc(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <solcon_number_of_dobldobl_solutions@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <solcon_dimension_of_dobldobl_solutions@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <dobldobl_Newton_step@plt>
  mov    %eax,-0xc(%rbp)
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <solcon_write_dobldobl_solutions@plt>
  mov    %eax,-0xc(%rbp)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D19
type: string
name: disassemble_data
arguments: tag: D27
type: string
name: disassemble_data
arguments: tag: D9
type: string
name: disassemble_data
arguments: tag: D36
type: string
name: disassemble_data
arguments: tag: D29
type: string

tool: D19: 
Running Newton step with double double arithmetic ...

tool: D27: The system container has %d polynomials.


tool: D9: The solution container has size %d.


tool: D36: The solutions in the container have dimension %d.


tool: D29: The solutions after the Newton step :

assistant: ```c
void test_dobldobl_Newton_step(void) {
  int fail, dim, len;
  printf("\nRunning Newton step with double double arithmetic ...\n");
  fail = syscon_read_dobldobl_system();
  fail = syscon_number_of_dobldobl_polynomials(&dim);
  printf("The system container has %d polynomials.\n", dim);
  fail = solcon_read_dobldobl_solutions();
  fail = solcon_number_of_dobldobl_solutions(&len);
  printf("The solution container has size %d.\n", len);
  fail = solcon_dimension_of_dobldobl_solutions(&dim);
  printf("The solutions in the container have dimension %d.\n", dim);
  fail = dobldobl_Newton_step();
  printf("The solutions after the Newton step :\n");
  fail = solcon_write_dobldobl_solutions();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_dobldobl_Newton_step>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <syscon_read_dobldobl_system@plt>
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <syscon_number_of_dobldobl_polynomials@plt>
  mov    (%rsp),%esi
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <solcon_read_dobldobl_solutions@plt>
  lea    0x4(%rsp),%rdi
  call   <solcon_number_of_dobldobl_solutions@plt>
  mov    0x4(%rsp),%esi
  lea    D21(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <solcon_dimension_of_dobldobl_solutions@plt>
  mov    (%rsp),%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <dobldobl_Newton_step@plt>
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <solcon_write_dobldobl_solutions@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D10
data_type: string
name: retrieve_data
arguments: label: D20
data_type: string
name: retrieve_data
arguments: label: D21
data_type: string
name: retrieve_data
arguments: label: D6
data_type: string
name: retrieve_data
arguments: label: D34
data_type: string

tool: D10: 
Running Newton step with double double arithmetic ...

tool: D20: The system container has %d polynomials.


tool: D21: The solution container has size %d.


tool: D6: The solutions in the container have dimension %d.


tool: D34: The solutions after the Newton step :

assistant: ```c
void test_dobldobl_Newton_step(void) {
  int fail, dim, len;
  printf("\nRunning Newton step with double double arithmetic ...\n");
  fail = syscon_read_dobldobl_system();
  fail = syscon_number_of_dobldobl_polynomials(&dim);
  printf("The system container has %d polynomials.\n", dim);
  fail = solcon_read_dobldobl_solutions();
  fail = solcon_number_of_dobldobl_solutions(&len);
  printf("The solution container has size %d.\n", len);
  fail = solcon_dimension_of_dobldobl_solutions(&dim);
  printf("The solutions in the container have dimension %d.\n", dim);
  fail = dobldobl_Newton_step();
  printf("The solutions after the Newton step :\n");
  fail = solcon_write_dobldobl_solutions();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test_dobldobl_Newton_step>:
  endbr64
  push   %rbp
  lea    D14(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  call   <puts@plt>
  xor    %eax,%eax
  call   <syscon_read_dobldobl_system@plt>
  mov    %rbp,%rdi
  call   <syscon_number_of_dobldobl_polynomials@plt>
  mov    (%rsp),%esi
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <solcon_read_dobldobl_solutions@plt>
  lea    0x4(%rsp),%rdi
  call   <solcon_number_of_dobldobl_solutions@plt>
  mov    0x4(%rsp),%esi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <solcon_dimension_of_dobldobl_solutions@plt>
  mov    (%rsp),%esi
  lea    D29(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  call   <dobldobl_Newton_step@plt>
  lea    D45(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <solcon_write_dobldobl_solutions@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x10,%rsp
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D14
type_infered: string
name: unpack_data
arguments: tag: D32
type_infered: string
name: unpack_data
arguments: tag: D6
type_infered: string
name: unpack_data
arguments: tag: D29
type_infered: string
name: unpack_data
arguments: tag: D45
type_infered: string

tool: D14: 
Running Newton step with double double arithmetic ...

tool: D32: The system container has %d polynomials.


tool: D6: The solution container has size %d.


tool: D29: The solutions in the container have dimension %d.


tool: D45: The solutions after the Newton step :

assistant: ```c
void test_dobldobl_Newton_step(void) {
  int fail, dim, len;
  printf("\nRunning Newton step with double double arithmetic ...\n");
  fail = syscon_read_dobldobl_system();
  fail = syscon_number_of_dobldobl_polynomials(&dim);
  printf("The system container has %d polynomials.\n", dim);
  fail = solcon_read_dobldobl_solutions();
  fail = solcon_number_of_dobldobl_solutions(&len);
  printf("The solution container has size %d.\n", len);
  fail = solcon_dimension_of_dobldobl_solutions(&dim);
  printf("The solutions in the container have dimension %d.\n", dim);
  fail = dobldobl_Newton_step();
  printf("The solutions after the Newton step :\n");
  fail = solcon_write_dobldobl_solutions();
}
```