ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<_merge_sort>:
  endbr64
  cmp    %edx,%esi
  jl     L0
  ret
L0:
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebx
  mov    %edx,%ebp
  lea    (%rsi,%rdx,1),%eax
  mov    %eax,%r13d
  shr    $0x1f,%r13d
  add    %eax,%r13d
  sar    %r13d
  mov    %r13d,%edx
  call   <_merge_sort>
  lea    0x1(%r13),%esi
  mov    %ebp,%edx
  mov    %r12,%rdi
  call   <_merge_sort>
  mov    %ebp,%ecx
  mov    %r13d,%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <merge@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void _merge_sort(int array[], int low, int high) {
  int middle;
  if (low < high) {
    middle = (low + high) / 2;
    _merge_sort(array, low, middle);
    _merge_sort(array, middle + 1, high);
    merge(array, low, middle, high);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<my_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <my_strlen@plt>
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L29
  cmpq   $0x0,-0x18(%rbp)
  jne    L18
L29:
  mov    $0x0,%eax
  jmp    L37
L21:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x10(%rbp)
L18:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L21
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0xff,(%rax)
  mov    -0x8(%rbp),%rax
L37:
  leave
  ret
```

assistant: ```c
unsigned char *my_strdup(char *src) {
  int len;
  int lp;
  unsigned char *dest;
  len = my_strlen(src);
  lp = 0;
  dest = malloc(sizeof(unsigned char) * len + 1);
  if (dest == NULL || src == NULL)
    return (NULL);
  while (lp < len) {
    dest[lp] = src[lp];
    ++lp;
  }
  dest[lp] = 255;
  return (dest);
}
```
user: What is the c source code of the assembly code below:

```asm
<my_strdup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <my_strlen@plt>
  mov    %eax,%ebp
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L28
  test   %rbx,%rbx
  je     L28
  test   %ebp,%ebp
  jle    L37
  mov    %ebp,%esi
  mov    $0x0,%edx
L47:
  movzbl (%rbx,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L47
L8:
  movslq %ebp,%rbp
  movb   $0xff,(%rax,%rbp,1)
L46:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L37:
  mov    $0x0,%ebp
  jmp    L8
L28:
  mov    $0x0,%eax
  jmp    L46
```

assistant: ```c
unsigned char *my_strdup(char *src) {
  int len;
  int lp;
  unsigned char *dest;
  len = my_strlen(src);
  lp = 0;
  dest = malloc(sizeof(unsigned char) * len + 1);
  if (dest == NULL || src == NULL)
    return (NULL);
  while (lp < len) {
    dest[lp] = src[lp];
    ++lp;
  }
  dest[lp] = 255;
  return (dest);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_strdup>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <my_strlen@plt>
  movslq %eax,%rbp
  lea    0x1(%rbp),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L5
  test   %rbx,%rbx
  je     L5
  test   %ebp,%ebp
  jle    L19
  lea    -0x1(%rbp),%esi
  xor    %edx,%edx
L49:
  movzbl (%rbx,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  mov    %rdx,%rcx
  add    $0x1,%rdx
  cmp    %rcx,%rsi
  jne    L49
  add    %rax,%rbp
L26:
  movb   $0xff,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L5:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L19:
  mov    %rax,%rbp
  jmp    L26
```

assistant: ```c
unsigned char *my_strdup(char *src) {
  int len;
  int lp;
  unsigned char *dest;
  len = my_strlen(src);
  lp = 0;
  dest = malloc(sizeof(unsigned char) * len + 1);
  if (dest == NULL || src == NULL)
    return (NULL);
  while (lp < len) {
    dest[lp] = src[lp];
    ++lp;
  }
  dest[lp] = 255;
  return (dest);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<my_strdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <my_strlen@plt>
  movslq %eax,%rbx
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L73
  test   %rbp,%rbp
  je     L73
  mov    %rax,%r8
  test   %ebx,%ebx
  jle    L130
  lea    0x1(%rbp),%rcx
  mov    %r8,%rdx
  lea    -0x1(%rbx),%eax
  sub    %rcx,%rdx
  cmp    $0xe,%rdx
  jbe    L105
  cmp    $0x6,%eax
  jbe    L105
  cmp    $0xe,%eax
  jbe    L39
  mov    %ebx,%edx
  xor    %eax,%eax
  shr    $0x4,%edx
  shl    $0x4,%rdx
  nopl   0x0(%rax)
L0:
  movdqu 0x0(%rbp,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L0
  mov    %ebx,%edx
  and    $0xfffffff0,%edx
  mov    %edx,%eax
  cmp    %edx,%ebx
  je     L92
  mov    %ebx,%r12d
  sub    %edx,%r12d
  lea    -0x1(%r12),%ecx
  cmp    $0x6,%ecx
  jbe    L139
L6:
  mov    0x0(%rbp,%rdx,1),%rcx
  mov    %rcx,(%r8,%rdx,1)
  mov    %r12d,%edx
  and    $0xfffffff8,%edx
  add    %edx,%eax
  cmp    %edx,%r12d
  je     L92
L139:
  movslq %eax,%rdx
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x1(%rax),%edx
  cmp    %edx,%ebx
  jle    L92
  movslq %edx,%rdx
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x2(%rax),%edx
  cmp    %edx,%ebx
  jle    L92
  movslq %edx,%rdx
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x3(%rax),%edx
  cmp    %edx,%ebx
  jle    L92
  movslq %edx,%rdx
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x4(%rax),%edx
  cmp    %edx,%ebx
  jle    L92
  movslq %edx,%rdx
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  lea    0x5(%rax),%edx
  cmp    %edx,%ebx
  jle    L92
  movslq %edx,%rdx
  add    $0x6,%eax
  movzbl 0x0(%rbp,%rdx,1),%ecx
  mov    %cl,(%r8,%rdx,1)
  cmp    %eax,%ebx
  jle    L92
  cltq
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %dl,(%r8,%rax,1)
L92:
  add    %r8,%rbx
L164:
  movb   $0xff,(%rbx)
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L73:
  xor    %r8d,%r8d
  pop    %rbx
  pop    %rbp
  mov    %r8,%rax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L105:
  mov    %eax,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L143:
  movzbl 0x0(%rbp,%rax,1),%ecx
  mov    %cl,(%r8,%rax,1)
  mov    %rax,%rcx
  add    $0x1,%rax
  cmp    %rdx,%rcx
  jne    L143
  jmp    L92
  nopw   0x0(%rax,%rax,1)
L130:
  mov    %rax,%rbx
  jmp    L164
L39:
  mov    %ebx,%r12d
  xor    %edx,%edx
  xor    %eax,%eax
  jmp    L6
```

assistant: ```c
unsigned char *my_strdup(char *src) {
  int len;
  int lp;
  unsigned char *dest;
  len = my_strlen(src);
  lp = 0;
  dest = malloc(sizeof(unsigned char) * len + 1);
  if (dest == NULL || src == NULL)
    return (NULL);
  while (lp < len) {
    dest[lp] = src[lp];
    ++lp;
  }
  dest[lp] = 255;
  return (dest);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <make_atan2_table@plt>
  mov    $0x0,%eax
  call   <init_fec@plt>
  mov    $0x0,%eax
  call   <read_from_stdin@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  make_atan2_table();
  init_fec();
  read_from_stdin();
  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
  mov    $0x0,%eax
  call   <make_atan2_table@plt>
  mov    $0x0,%eax
  call   <init_fec@plt>
  mov    $0x0,%eax
  call   <read_from_stdin@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  make_atan2_table();
  init_fec();
  read_from_stdin();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <make_atan2_table@plt>
  xor    %eax,%eax
  call   <init_fec@plt>
  xor    %eax,%eax
  call   <read_from_stdin@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  make_atan2_table();
  init_fec();
  read_from_stdin();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <make_atan2_table@plt>
  xor    %eax,%eax
  call   <init_fec@plt>
  xor    %eax,%eax
  call   <read_from_stdin@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  make_atan2_table();
  init_fec();
  read_from_stdin();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<cmd_zhome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x2,-0x14(%rbp)
  je     L26
  lea    D2(%rip),%rax
  mov    %rax,%rsi
  mov    $0xfffffffe,%edi
  call   <error@plt>
  jmp    L24
L26:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  lea    -0x9(%rbp),%rdx
  lea    D16(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sscanf@plt>
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <zaber_home@plt>
  mov    $0x0,%eax
L24:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D2
type_guessed: string
name: read_data
arguments: tag: D16
type_guessed: string

tool: D2: Usage: zhome [unit]

tool: D16: %hhu

assistant: ```c
int cmd_zhome(int argc, char *argv[]) {
  unsigned char unit;
  if (argc != 2)
    return error((-2), "Usage: zhome [unit]");
  sscanf(argv[1], "%hhu", &unit);
  zaber_home(unit);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cmd_zhome>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L21
  lea    D9(%rip),%rsi
  mov    $0xfffffffe,%edi
  call   <error@plt>
L8:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0x18,%rsp
  ret
L21:
  lea    0x7(%rsp),%rdx
  mov    0x8(%rsi),%rdi
  lea    D0(%rip),%rsi
  call   <sscanf@plt>
  movzbl 0x7(%rsp),%edi
  call   <zaber_home@plt>
  mov    $0x0,%eax
  jmp    L8
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D9
type_guessed: string
name: retrieve_data
arguments: data_label: D0
type_guessed: string

tool: D9: Usage: zhome [unit]

tool: D0: %hhu

assistant: ```c
int cmd_zhome(int argc, char *argv[]) {
  unsigned char unit;
  if (argc != 2)
    return error((-2), "Usage: zhome [unit]");
  sscanf(argv[1], "%hhu", &unit);
  zaber_home(unit);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cmd_zhome>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L28
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  lea    D12(%rip),%rsi
  mov    $0xfffffffe,%edi
  add    $0x18,%rsp
  jmp    10a0 <error@plt>
  nopw   0x0(%rax,%rax,1)
L28:
  mov    0x8(%rsi),%rdi
  lea    0x7(%rsp),%rdx
  lea    D19(%rip),%rsi
  call   <sscanf@plt>
  movzbl 0x7(%rsp),%edi
  call   <zaber_home@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D12
type: string
name: retrieve_data
arguments: label: D19
type: string

tool: D12: Usage: zhome [unit]

tool: D19: %hhu

assistant: ```c
int cmd_zhome(int argc, char *argv[]) {
  unsigned char unit;
  if (argc != 2)
    return error((-2), "Usage: zhome [unit]");
  sscanf(argv[1], "%hhu", &unit);
  zaber_home(unit);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<cmd_zhome>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L2
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  lea    D13(%rip),%rsi
  mov    $0xfffffffe,%edi
  add    $0x18,%rsp
  jmp    10a0 <error@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  mov    0x8(%rsi),%rdi
  lea    0x7(%rsp),%rdx
  lea    D3(%rip),%rsi
  call   <sscanf@plt>
  movzbl 0x7(%rsp),%edi
  call   <zaber_home@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
datatype: string
name: retrieve_data
arguments: data_tag: D3
datatype: string

tool: D13: Usage: zhome [unit]

tool: D3: %hhu

assistant: ```c
int cmd_zhome(int argc, char *argv[]) {
  unsigned char unit;
  if (argc != 2)
    return error((-2), "Usage: zhome [unit]");
  sscanf(argv[1], "%hhu", &unit);
  zaber_home(unit);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<svcraw_stat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  pop    %rbp
  ret
```

assistant: ```c
enum xprt_stat svcraw_stat(SVCXPRT *__xprt) {
  (void)__xprt;
  return (XPRT_IDLE);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<svcraw_stat>:
  endbr64
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
enum xprt_stat svcraw_stat(SVCXPRT *__xprt) {
  (void)__xprt;
  return (XPRT_IDLE);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<svcraw_stat>:
  endbr64
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  ret
```

assistant: ```c
enum xprt_stat svcraw_stat(SVCXPRT *__xprt) {
  (void)__xprt;
  return (XPRT_IDLE);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<precedence>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  jne    L1
  mov    $0xa,%eax
  jmp    L11
L1:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L78
  mov    $0xa,%eax
  jmp    L11
L78:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2b,%al
  jne    L37
  mov    $0x1,%eax
  jmp    L11
L37:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L32
  mov    $0x1,%eax
  jmp    L11
L32:
  mov    $0x0,%eax
L11:
  pop    %rbp
  ret
```

assistant: ```c
int precedence(char *operator) {
  if (*operator== '*')
    return 10;
  if (*operator== '/')
    return 10;
  if (*operator== '+')
    return 1;
  if (*operator== '-')
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<precedence>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x2f,%al
  je     L10
  cmp    $0x2a,%al
  je     L10
  sub    $0x2b,%eax
  test   $0xfd,%al
  sete   %al
  movzbl %al,%eax
  ret
L10:
  mov    $0xa,%eax
  ret
```

assistant: ```c
int precedence(char *operator) {
  if (*operator== '*')
    return 10;
  if (*operator== '/')
    return 10;
  if (*operator== '+')
    return 1;
  if (*operator== '-')
    return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<precedence>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x2a,%al
  je     L10
  cmp    $0x2f,%al
  je     L10
  sub    $0x2b,%eax
  test   $0xfd,%al
  sete   %al
  movzbl %al,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  mov    $0xa,%eax
  ret
```

assistant: ```c
int precedence(char *operator) {
  if (*operator== '*')
    return 10;
  if (*operator== '/')
    return 10;
  if (*operator== '+')
    return 1;
  if (*operator== '-')
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<precedence>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0x2a,%al
  je     L7
  cmp    $0x2f,%al
  je     L7
  sub    $0x2b,%eax
  test   $0xfd,%al
  sete   %al
  movzbl %al,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  mov    $0xa,%eax
  ret
```

assistant: ```c
int precedence(char *operator) {
  if (*operator== '*')
    return 10;
  if (*operator== '/')
    return 10;
  if (*operator== '+')
    return 1;
  if (*operator== '-')
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_G_CopyString>:
  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,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rcx
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <trap_MemAlloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *_G_CopyString(const char *in, const char *filename, int fileline) {
  char *out;
  out = trap_MemAlloc(strlen(in) + 1, filename, fileline);
  strcpy(out, in);
  return out;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<_G_CopyString>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %edx,%r12d
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  mov    %r12d,%edx
  mov    %rbx,%rsi
  call   <trap_MemAlloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *_G_CopyString(const char *in, const char *filename, int fileline) {
  char *out;
  out = trap_MemAlloc(strlen(in) + 1, filename, fileline);
  strcpy(out, in);
  return out;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_G_CopyString>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r13d,%edx
  mov    %r12,%rsi
  lea    0x1(%rax),%rdi
  call   <trap_MemAlloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *_G_CopyString(const char *in, const char *filename, int fileline) {
  char *out;
  out = trap_MemAlloc(strlen(in) + 1, filename, fileline);
  strcpy(out, in);
  return out;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<_G_CopyString>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r13d,%edx
  mov    %r12,%rsi
  lea    0x1(%rax),%rdi
  call   <trap_MemAlloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *_G_CopyString(const char *in, const char *filename, int fileline) {
  char *out;
  out = trap_MemAlloc(strlen(in) + 1, filename, fileline);
  strcpy(out, in);
  return out;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xa,%esi
  mov    $0x2,%edi
  call   <power@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
type_guessed: string

tool: D4: 2^10 = %d


assistant: ```c
int main() {
  printf("2^10 = %d\n", power(2, 10));
  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
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  mov    $0x2,%edi
  call   <power@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D9
datatype: string

tool: D9: 2^10 = %d


assistant: ```c
int main() {
  printf("2^10 = %d\n", power(2, 10));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  mov    $0x2,%edi
  call   <power@plt>
  lea    D3(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D3: 2^10 = %d


assistant: ```c
int main() {
  printf("2^10 = %d\n", power(2, 10));
  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
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xa,%esi
  mov    $0x2,%edi
  call   <power@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
data_type: string

tool: D4: 2^10 = %d


assistant: ```c
int main() {
  printf("2^10 = %d\n", power(2, 10));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mult_const_i_3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8d,-0x2c(%rbp)
  mov    %r9d,-0x30(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L44
L42:
  movl   $0x0,-0x8(%rbp)
  jmp    L33
L26:
  movl   $0x0,-0x4(%rbp)
  jmp    L54
L0:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%rcx
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x3,%rdx
  add    %rcx,%rdx
  mov    (%rdx),%rcx
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  add    %rcx,%rdx
  imul   -0x1c(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L54:
  mov    0x20(%rbp),%eax
  sub    0x18(%rbp),%eax
  add    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jle    L0
  addl   $0x1,-0x8(%rbp)
L33:
  mov    0x10(%rbp),%eax
  sub    -0x30(%rbp),%eax
  cmp    %eax,-0x8(%rbp)
  jle    L26
  addl   $0x1,-0xc(%rbp)
L44:
  mov    -0x2c(%rbp),%eax
  sub    -0x20(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jle    L42
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int ***mult_const_i_3(int ***res, int cn, int ***vec, int lo1, int hi1, int lo2, int hi2, int lo3, int hi3) {
  int i, j, k;
  for (i = 0; i <= hi1 - lo1; i++)
    for (j = 0; j <= hi2 - lo2; j++)
      for (k = 0; k <= hi3 - lo3 + 1; k++)
        res[i][j][k] = cn * vec[i][j][k];
  return res;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<mult_const_i_3>:
  endbr64
  mov    %rdi,%r11
  mov    %r8d,%eax
  sub    %ecx,%eax
  js     L17
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r10d
  mov    %rdx,%rdi
  mov    %r11,%r8
  mov    %eax,%eax
  lea    0x8(%rdx,%rax,8),%r13
  mov    0x30(%rsp),%r12d
  sub    %r9d,%r12d
  mov    %r12d,%eax
  lea    0x8(,%rax,8),%rbp
  mov    0x40(%rsp),%ebx
  sub    0x38(%rsp),%ebx
  lea    0x1(%rbx),%eax
  lea    0x4(,%rax,4),%r9
  jmp    L14
L74:
  mov    (%r8),%rcx
  mov    (%rcx,%rdx,1),%rcx
  mov    (%rdi),%rsi
  mov    (%rsi,%rdx,1),%rsi
  mov    %r10d,%r14d
  imul   (%rsi,%rax,1),%r14d
  mov    %r14d,(%rcx,%rax,1)
  add    $0x4,%rax
  cmp    %r9,%rax
  jne    L74
L8:
  add    $0x8,%rdx
  cmp    %rbp,%rdx
  je     L27
L83:
  mov    $0x0,%eax
  cmp    $0xffffffff,%ebx
  jge    L74
  jmp    L8
L27:
  add    $0x8,%rdi
  add    $0x8,%r8
  cmp    %r13,%rdi
  je     L38
L14:
  mov    $0x0,%edx
  test   %r12d,%r12d
  jns    L83
  jmp    L27
L38:
  mov    %r11,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  mov    %rdi,%rax
  ret
```

assistant: ```c
int ***mult_const_i_3(int ***res, int cn, int ***vec, int lo1, int hi1, int lo2, int hi2, int lo3, int hi3) {
  int i, j, k;
  for (i = 0; i <= hi1 - lo1; i++)
    for (j = 0; j <= hi2 - lo2; j++)
      for (k = 0; k <= hi3 - lo3 + 1; k++)
        res[i][j][k] = cn * vec[i][j][k];
  return res;
}
```