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
<strcpy>:
  endbr64
  movzbl (%rsi),%ecx
  mov    %rdi,%rax
  test   %cl,%cl
  je     L0
  xor    %edx,%edx
L4:
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  movzbl (%rsi,%rdx,1),%ecx
  test   %cl,%cl
  jne    L4
  add    %rax,%rdx
  movb   $0x0,(%rdx)
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  mov    %rdi,%rdx
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
char *strcpy(char *dst, const char *src) {
  int i;
  i = 0;
  while (src[i]) {
    dst[i] = src[i];
    i++;
  }
  dst[i] = '\0';
  return (dst);
}
```
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
  movl   $0x3,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <bar@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <rec1@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <foo@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x9,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <__CRAB_assert@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x = 3;
  int y = bar(x);
  int z = rec1(y);
  int w = foo(y);
  __CRAB_assert(w == 9);
  return z + w;
}
```
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   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x3,%edi
  call   <bar@plt>
  mov    %eax,%ebp
  mov    %eax,%edi
  call   <rec1@plt>
  mov    %eax,%ebx
  mov    %ebp,%edi
  call   <foo@plt>
  mov    %eax,%ebp
  cmp    $0x9,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <__CRAB_assert@plt>
  lea    (%rbx,%rbp,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int x = 3;
  int y = bar(x);
  int z = rec1(y);
  int w = foo(y);
  __CRAB_assert(w == 9);
  return z + w;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x3,%edi
  push   %rbx
  sub    $0x8,%rsp
  call   <bar@plt>
  mov    %eax,%ebp
  mov    %eax,%edi
  call   <rec1@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <foo@plt>
  xor    %edi,%edi
  cmp    $0x9,%eax
  mov    %eax,%ebp
  sete   %dil
  call   <__CRAB_assert@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int x = 3;
  int y = bar(x);
  int z = rec1(y);
  int w = foo(y);
  __CRAB_assert(w == 9);
  return z + w;
}
```
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   %rbp
  mov    $0x3,%edi
  push   %rbx
  sub    $0x8,%rsp
  call   <bar@plt>
  mov    %eax,%ebp
  mov    %eax,%edi
  call   <rec1@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <foo@plt>
  xor    %edi,%edi
  cmp    $0x9,%eax
  mov    %eax,%ebp
  sete   %dil
  call   <__CRAB_assert@plt>
  add    $0x8,%rsp
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int x = 3;
  int y = bar(x);
  int z = rec1(y);
  int w = foo(y);
  __CRAB_assert(w == 9);
  return z + w;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0xff,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <reverse@plt>
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%ecx
  mov    -0x10(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <reverse@plt>
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <reverse@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%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: data_label: D37
type: string
name: parse_data
arguments: data_label: D14
type: string
name: parse_data
arguments: data_label: D32
type: string
name: parse_data
arguments: data_label: D44
type: string
name: parse_data
arguments: data_label: D22
type: string

tool: D37: Enter a string: 

tool: D14: %s

tool: D32: Enter how many characters to rotate: 

tool: D44: %d

tool: D22: Rotated string: %s


assistant: ```c
int main() {
  char *ch = malloc(255);
  printf("Enter a string: ");
  scanf("%s", ch);
  printf("Enter how many characters to rotate: ");
  int i, n = strlen(ch);
  scanf("%d", &i);
  reverse(ch, 0, i - 1);
  reverse(ch, i, n - 1);
  reverse(ch, 0, n - 1);
  printf("Rotated string: %s\n", ch);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0xff,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D24(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <reverse@plt>
  sub    $0x1,%ebp
  mov    %ebp,%edx
  mov    0x4(%rsp),%esi
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    %ebp,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <reverse@plt>
  mov    %rbx,%rsi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
type: string
name: parse_data
arguments: data_tag: D5
type: string
name: parse_data
arguments: data_tag: D24
type: string
name: parse_data
arguments: data_tag: D14
type: string
name: parse_data
arguments: data_tag: D2
type: string

tool: D0: Enter a string: 

tool: D5: %s

tool: D24: Enter how many characters to rotate: 

tool: D14: %d

tool: D2: Rotated string: %s


assistant: ```c
int main() {
  char *ch = malloc(255);
  printf("Enter a string: ");
  scanf("%s", ch);
  printf("Enter how many characters to rotate: ");
  int i, n = strlen(ch);
  scanf("%d", &i);
  reverse(ch, 0, i - 1);
  reverse(ch, i, n - 1);
  reverse(ch, 0, n - 1);
  printf("Rotated string: %s\n", ch);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xff,%edi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  lea    D24(%rip),%rdi
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D48(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x4(%rsp),%rsi
  lea    D5(%rip),%rdi
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  xor    %esi,%esi
  mov    %rbp,%rdi
  sub    $0x1,%r12d
  lea    -0x1(%rax),%edx
  call   <reverse@plt>
  mov    0x4(%rsp),%esi
  mov    %r12d,%edx
  mov    %rbp,%rdi
  call   <reverse@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    %r12d,%edx
  call   <reverse@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D24
type_guessed: string
name: retrieve_data
arguments: data_label: D48
type_guessed: string
name: retrieve_data
arguments: data_label: D14
type_guessed: string
name: retrieve_data
arguments: data_label: D5
type_guessed: string
name: retrieve_data
arguments: data_label: D38
type_guessed: string

tool: D24: Enter a string: 

tool: D48: %s

tool: D14: Enter how many characters to rotate: 

tool: D5: %d

tool: D38: Rotated string: %s


assistant: ```c
int main() {
  char *ch = malloc(255);
  printf("Enter a string: ");
  scanf("%s", ch);
  printf("Enter how many characters to rotate: ");
  int i, n = strlen(ch);
  scanf("%d", &i);
  reverse(ch, 0, i - 1);
  reverse(ch, i, n - 1);
  reverse(ch, 0, n - 1);
  printf("Rotated string: %s\n", ch);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xff,%edi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  lea    D35(%rip),%rdi
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D45(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D31(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x4(%rsp),%rsi
  lea    D26(%rip),%rdi
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  xor    %esi,%esi
  mov    %rbp,%rdi
  sub    $0x1,%r12d
  lea    -0x1(%rax),%edx
  call   <reverse@plt>
  mov    0x4(%rsp),%esi
  mov    %r12d,%edx
  mov    %rbp,%rdi
  call   <reverse@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  mov    %r12d,%edx
  call   <reverse@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D29(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D35
type_infered: string
name: parse_data
arguments: tag: D45
type_infered: string
name: parse_data
arguments: tag: D31
type_infered: string
name: parse_data
arguments: tag: D26
type_infered: string
name: parse_data
arguments: tag: D29
type_infered: string

tool: D35: Enter a string: 

tool: D45: %s

tool: D31: Enter how many characters to rotate: 

tool: D26: %d

tool: D29: Rotated string: %s


assistant: ```c
int main() {
  char *ch = malloc(255);
  printf("Enter a string: ");
  scanf("%s", ch);
  printf("Enter how many characters to rotate: ");
  int i, n = strlen(ch);
  scanf("%d", &i);
  reverse(ch, 0, i - 1);
  reverse(ch, i, n - 1);
  reverse(ch, 0, n - 1);
  printf("Rotated string: %s\n", ch);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rpi_gpio_toggle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <rpi_gpio_get_val@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L19
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <rpi_gpio_set_val@plt>
  jmp    L9
L19:
  mov    -0x18(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <rpi_gpio_set_val@plt>
L9:
  nop
  leave
  ret
```

assistant: ```c
void rpi_gpio_toggle(uint32_t pin) {
  uint32_t result = rpi_gpio_get_val(pin);
  if (result) {
    rpi_gpio_set_val(pin, 0);
  } else {
    rpi_gpio_set_val(pin, 1);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rpi_gpio_toggle>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <rpi_gpio_get_val@plt>
  test   %rax,%rax
  je     L13
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <rpi_gpio_set_val@plt>
L8:
  pop    %rbx
  ret
L13:
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <rpi_gpio_set_val@plt>
  jmp    L8
```

assistant: ```c
void rpi_gpio_toggle(uint32_t pin) {
  uint32_t result = rpi_gpio_get_val(pin);
  if (result) {
    rpi_gpio_set_val(pin, 0);
  } else {
    rpi_gpio_set_val(pin, 1);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rpi_gpio_toggle>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <rpi_gpio_get_val@plt>
  test   %rax,%rax
  je     L4
  mov    %rbp,%rdi
  xor    %esi,%esi
  pop    %rbp
  jmp    1060 <rpi_gpio_set_val@plt>
  nopl   (%rax)
L4:
  mov    %rbp,%rdi
  mov    $0x1,%esi
  pop    %rbp
  jmp    1060 <rpi_gpio_set_val@plt>
```

assistant: ```c
void rpi_gpio_toggle(uint32_t pin) {
  uint32_t result = rpi_gpio_get_val(pin);
  if (result) {
    rpi_gpio_set_val(pin, 0);
  } else {
    rpi_gpio_set_val(pin, 1);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<rpi_gpio_toggle>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <rpi_gpio_get_val@plt>
  test   %rax,%rax
  je     L2
  mov    %rbp,%rdi
  xor    %esi,%esi
  pop    %rbp
  jmp    1060 <rpi_gpio_set_val@plt>
  nopl   (%rax)
L2:
  mov    %rbp,%rdi
  mov    $0x1,%esi
  pop    %rbp
  jmp    1060 <rpi_gpio_set_val@plt>
```

assistant: ```c
void rpi_gpio_toggle(uint32_t pin) {
  uint32_t result = rpi_gpio_get_val(pin);
  if (result) {
    rpi_gpio_set_val(pin, 0);
  } else {
    rpi_gpio_set_val(pin, 1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hash_gd5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L24
L25:
  cmpl   $0x3f,-0xc(%rbp)
  je     L31
  cmpl   $0xa,-0xc(%rbp)
  je     L31
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  shl    $0x13,%rax
  sub    %rax,%rdx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rcx
  mov    %rdx,%rax
  sub    %rcx,%rax
  mov    %rax,-0x8(%rbp)
L24:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L25
L31:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long hash_gd5(char *uri) {
  unsigned long hash = 0;
  int c;
  while ((c = *uri++)) {
    if (c == '?' || c == '\n')
      break;
    hash = hash + (hash << 2) - (hash << 19) - c;
  }
  return hash;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<hash_gd5>:
  endbr64
  lea    0x1(%rdi),%rsi
  movzbl (%rdi),%edx
  movsbl %dl,%ecx
  test   %ecx,%ecx
  je     L5
  mov    $0x0,%eax
L1:
  cmp    $0x3f,%ecx
  je     L8
  cmp    $0xa,%ecx
  je     L8
  mov    %rax,%rcx
  shl    $0x13,%rcx
  lea    (%rax,%rax,4),%rax
  movsbq %dl,%rdx
  sub    %rdx,%rax
  sub    %rcx,%rax
  add    $0x1,%rsi
  movzbl -0x1(%rsi),%edx
  movsbl %dl,%ecx
  test   %ecx,%ecx
  jne    L1
  ret
L5:
  mov    $0x0,%eax
L8:
  ret
```

assistant: ```c
unsigned long hash_gd5(char *uri) {
  unsigned long hash = 0;
  int c;
  while ((c = *uri++)) {
    if (c == '?' || c == '\n')
      break;
    hash = hash + (hash << 2) - (hash << 19) - c;
  }
  return hash;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<hash_gd5>:
  endbr64
  movsbq (%rdi),%rcx
  lea    0x1(%rdi),%rsi
  xor    %eax,%eax
  mov    %rcx,%rdx
  test   %ecx,%ecx
  je     L25
  nopl   (%rax)
L12:
  cmp    $0x3f,%ecx
  je     L25
  cmp    $0xa,%ecx
  je     L25
  mov    %rax,%rcx
  add    $0x1,%rsi
  lea    (%rax,%rax,4),%rax
  sub    %rdx,%rax
  movsbq -0x1(%rsi),%rdx
  shl    $0x13,%rcx
  sub    %rcx,%rax
  movsbl %dl,%ecx
  test   %ecx,%ecx
  jne    L12
  ret
  nopl   0x0(%rax)
L25:
  ret
```

assistant: ```c
unsigned long hash_gd5(char *uri) {
  unsigned long hash = 0;
  int c;
  while ((c = *uri++)) {
    if (c == '?' || c == '\n')
      break;
    hash = hash + (hash << 2) - (hash << 19) - c;
  }
  return hash;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hash_gd5>:
  endbr64
  movsbq (%rdi),%rcx
  lea    0x1(%rdi),%rsi
  xor    %eax,%eax
  mov    %rcx,%rdx
  test   %ecx,%ecx
  je     L32
  nopl   (%rax)
L27:
  cmp    $0x3f,%ecx
  je     L32
  cmp    $0xa,%ecx
  je     L32
  mov    %rax,%rcx
  add    $0x1,%rsi
  lea    (%rax,%rax,4),%rax
  sub    %rdx,%rax
  movsbq -0x1(%rsi),%rdx
  shl    $0x13,%rcx
  sub    %rcx,%rax
  movsbl %dl,%ecx
  test   %ecx,%ecx
  jne    L27
  ret
  nopl   0x0(%rax)
L32:
  ret
```

assistant: ```c
unsigned long hash_gd5(char *uri) {
  unsigned long hash = 0;
  int c;
  while ((c = *uri++)) {
    if (c == '?' || c == '\n')
      break;
    hash = hash + (hash << 2) - (hash << 19) - c;
  }
  return hash;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<Num2Ascii>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x0,-0x8(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L5
  negl   -0x14(%rbp)
  mov    -0x20(%rbp),%rax
  movb   $0x2d,(%rax)
  addq   $0x1,-0x20(%rbp)
  movl   $0x1,-0x8(%rbp)
L5:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <NumLength@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L6
L18:
  mov    -0x10(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L6:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L18
  movl   $0x0,-0xc(%rbp)
  jmp    L25
L15:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x10(%rbp)
  lea    0x30(%rax),%ecx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  addl   $0x1,-0xc(%rbp)
L25:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L15
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
int Num2Ascii(int Num, char *S) {
  int Len;
  int Scale = 1;
  int i;
  int NegFlag = 0;
  if (Num < 0) {
    Num = -Num;
    S[0] = '-';
    S++;
    NegFlag = 1;
  }
  Len = NumLength(Num);
  for (i = 0; i < Len - 1; i++)
    Scale = Scale * 10;
  for (i = 0; i < Len; i++) {
    S[i] = (Num / Scale) + '0';
    Num = Num % Scale;
    Scale = Scale / 10;
  }
  return (Len + NegFlag);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Num2Ascii>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %rsi,%r12
  mov    $0x0,%ebp
  test   %edi,%edi
  js     L54
L31:
  mov    %ebx,%edi
  call   <NumLength@plt>
  mov    %eax,%edi
  cmp    $0x1,%eax
  jle    L34
  lea    -0x1(%rax),%edx
  mov    $0x0,%eax
  mov    $0x1,%ecx
L33:
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
  add    $0x1,%eax
  cmp    %edx,%eax
  jne    L33
L35:
  test   %edi,%edi
  jle    L6
  mov    %r12,%rsi
  lea    -0x1(%rdi),%eax
  lea    0x1(%r12,%rax,1),%r8
L1:
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  add    $0x30,%eax
  mov    %al,(%rsi)
  mov    %edx,%ebx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ecx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%ecx
  add    $0x1,%rsi
  cmp    %r8,%rsi
  jne    L1
L6:
  lea    0x0(%rbp,%rdi,1),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L54:
  neg    %ebx
  movb   $0x2d,(%rsi)
  add    $0x1,%r12
  mov    $0x1,%ebp
  jmp    L31
L34:
  mov    $0x1,%ecx
  jmp    L35
```

assistant: ```c
int Num2Ascii(int Num, char *S) {
  int Len;
  int Scale = 1;
  int i;
  int NegFlag = 0;
  if (Num < 0) {
    Num = -Num;
    S[0] = '-';
    S++;
    NegFlag = 1;
  }
  Len = NumLength(Num);
  for (i = 0; i < Len - 1; i++)
    Scale = Scale * 10;
  for (i = 0; i < Len; i++) {
    S[i] = (Num / Scale) + '0';
    Num = Num % Scale;
    Scale = Scale / 10;
  }
  return (Len + NegFlag);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Num2Ascii>:
  endbr64
  push   %rbp
  mov    %edi,%edx
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  test   %edi,%edi
  jns    L1
  movb   $0x2d,(%rsi)
  neg    %edx
  add    $0x1,%rbp
  mov    $0x1,%ebx
L1:
  mov    %edx,%edi
  mov    %edx,0xc(%rsp)
  call   <NumLength@plt>
  mov    0xc(%rsp),%edx
  mov    %eax,%r8d
  lea    -0x1(%rax),%eax
  test   %eax,%eax
  jle    L8
  xor    %edi,%edi
  mov    $0x1,%ecx
  nopl   0x0(%rax,%rax,1)
L46:
  lea    (%rcx,%rcx,4),%ecx
  add    $0x1,%edi
  add    %ecx,%ecx
  cmp    %eax,%edi
  jne    L46
L4:
  test   %r8d,%r8d
  jle    L27
  mov    %rbp,%rsi
  lea    0x1(%rbp,%rax,1),%r9
  mov    $0xcccccccd,%edi
  cs nopw 0x0(%rax,%rax,1)
L26:
  mov    %edx,%eax
  add    $0x1,%rsi
  cltd
  idiv   %ecx
  mov    %ecx,%ecx
  imul   %rdi,%rcx
  shr    $0x23,%rcx
  add    $0x30,%eax
  mov    %al,-0x1(%rsi)
  cmp    %r9,%rsi
  jne    L26
L27:
  add    $0x18,%rsp
  lea    (%rbx,%r8,1),%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L8:
  mov    $0x1,%ecx
  jmp    L4
```

assistant: ```c
int Num2Ascii(int Num, char *S) {
  int Len;
  int Scale = 1;
  int i;
  int NegFlag = 0;
  if (Num < 0) {
    Num = -Num;
    S[0] = '-';
    S++;
    NegFlag = 1;
  }
  Len = NumLength(Num);
  for (i = 0; i < Len - 1; i++)
    Scale = Scale * 10;
  for (i = 0; i < Len; i++) {
    S[i] = (Num / Scale) + '0';
    Num = Num % Scale;
    Scale = Scale / 10;
  }
  return (Len + NegFlag);
}
```
user: Translate the below assembly code into C programming language:

```asm
<Num2Ascii>:
  endbr64
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rsi,%rbx
  test   %edi,%edi
  jns    L75
  movb   $0x2d,(%rsi)
  neg    %r12d
  add    $0x1,%rbx
  mov    $0x1,%ebp
L75:
  mov    %r12d,%edi
  call   <NumLength@plt>
  mov    %eax,%edi
  cmp    $0x1,%eax
  jle    L46
  lea    -0x1(%rax),%esi
  cmp    $0x2,%eax
  je     L63
  movq   D3(%rip),%xmm0
  mov    %esi,%edx
  xor    %eax,%eax
  shr    %edx
L69:
  movdqa %xmm0,%xmm1
  add    $0x1,%eax
  pslld  $0x2,%xmm1
  paddd  %xmm1,%xmm0
  pslld  $0x1,%xmm0
  cmp    %eax,%edx
  jne    L69
  movd   %xmm0,%ecx
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm2,%eax
  imul   %eax,%ecx
  and    $0x1,%esi
  je     L24
L54:
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
L24:
  test   %edi,%edi
  jle    L73
  lea    -0x1(%rdi),%eax
  mov    %rbx,%rsi
  mov    $0xcccccccd,%r8d
  lea    0x1(%rbx,%rax,1),%r9
  nopl   0x0(%rax,%rax,1)
L40:
  mov    %r12d,%eax
  add    $0x1,%rsi
  cltd
  idiv   %ecx
  mov    %ecx,%ecx
  imul   %r8,%rcx
  shr    $0x23,%rcx
  add    $0x30,%eax
  mov    %edx,%r12d
  mov    %al,-0x1(%rsi)
  cmp    %r9,%rsi
  jne    L40
L73:
  lea    0x0(%rbp,%rdi,1),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L46:
  mov    $0x1,%ecx
  jmp    L24
L63:
  mov    $0x1,%ecx
  jmp    L54
```

tool_calls: name: unpack_data
arguments: label: D3
type_infered: byte[8]

tool: {"D3": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int Num2Ascii(int Num, char *S) {
  int Len;
  int Scale = 1;
  int i;
  int NegFlag = 0;
  if (Num < 0) {
    Num = -Num;
    S[0] = '-';
    S++;
    NegFlag = 1;
  }
  Len = NumLength(Num);
  for (i = 0; i < Len - 1; i++)
    Scale = Scale * 10;
  for (i = 0; i < Len; i++) {
    S[i] = (Num / Scale) + '0';
    Num = Num % Scale;
    Scale = Scale / 10;
  }
  return (Len + NegFlag);
}
```