ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<myHandler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <wait@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
data_type: string

tool: D8: Child %d terminated.


assistant: ```c
void myHandler(int signo) {
  int status;
  pid_t pid = wait(&status);
  printf("Child %d terminated.\n", pid);
  exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<myHandler>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <wait@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

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

tool: D3: Child %d terminated.


assistant: ```c
void myHandler(int signo) {
  int status;
  pid_t pid = wait(&status);
  printf("Child %d terminated.\n", pid);
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<myHandler>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <wait@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D6
data_type: string

tool: D6: Child %d terminated.


assistant: ```c
void myHandler(int signo) {
  int status;
  pid_t pid = wait(&status);
  printf("Child %d terminated.\n", pid);
  exit(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<myHandler>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <wait@plt>
  lea    D2(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

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

tool: D2: Child %d terminated.


assistant: ```c
void myHandler(int signo) {
  int status;
  pid_t pid = wait(&status);
  printf("Child %d terminated.\n", pid);
  exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<FilePrintBinary>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void FilePrintBinary(file, code, bits) void *file;
unsigned int code;
int bits;
{}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<FilePrintBinary>:
  endbr64
  ret
```

assistant: ```c
void FilePrintBinary(file, code, bits) void *file;
unsigned int code;
int bits;
{}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<FilePrintBinary>:
  endbr64
  ret
```

assistant: ```c
void FilePrintBinary(file, code, bits) void *file;
unsigned int code;
int bits;
{}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<FilePrintBinary>:
  endbr64
  ret
```

assistant: ```c
void FilePrintBinary(file, code, bits) void *file;
unsigned int code;
int bits;
{}
```
user: What would the C source look like for the assembly code provided here:

```asm
<utime_internal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <utime@plt>
  test   %eax,%eax
  jns    L1
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <rb_sys_fail@plt>
L1:
  nop
  leave
  ret
```

assistant: ```c
void utime_internal(path, arg) const char *path;
void *arg;
{
  struct utimbuf *utp = arg;
  if (utime(path, utp) < 0)
    rb_sys_fail(path);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<utime_internal>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x0,%eax
  call   <utime@plt>
  test   %eax,%eax
  js     L12
L7:
  pop    %rbx
  ret
L12:
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <rb_sys_fail@plt>
  jmp    L7
```

assistant: ```c
void utime_internal(path, arg) const char *path;
void *arg;
{
  struct utimbuf *utp = arg;
  if (utime(path, utp) < 0)
    rb_sys_fail(path);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<utime_internal>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  call   <utime@plt>
  test   %eax,%eax
  js     L5
  pop    %rbp
  ret
  nopl   (%rax)
L5:
  mov    %rbp,%rdi
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <rb_sys_fail@plt>
```

assistant: ```c
void utime_internal(path, arg) const char *path;
void *arg;
{
  struct utimbuf *utp = arg;
  if (utime(path, utp) < 0)
    rb_sys_fail(path);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<utime_internal>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  call   <utime@plt>
  test   %eax,%eax
  js     L7
  pop    %rbp
  ret
  nopl   (%rax)
L7:
  mov    %rbp,%rdi
  xor    %eax,%eax
  pop    %rbp
  jmp    1070 <rb_sys_fail@plt>
```

assistant: ```c
void utime_internal(path, arg) const char *path;
void *arg;
{
  struct utimbuf *utp = arg;
  if (utime(path, utp) < 0)
    rb_sys_fail(path);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mapif_guild_message>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x230,%rsp
  mov    %edi,-0x214(%rbp)
  mov    %esi,-0x218(%rbp)
  mov    %rdx,-0x220(%rbp)
  mov    %rcx,-0x228(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x210(%rbp),%rax
  movw   $0x3837,(%rax)
  mov    -0x228(%rbp),%rax
  mov    %eax,%edx
  lea    -0x210(%rbp),%rax
  add    $0x2,%rax
  add    $0xc,%edx
  mov    %dx,(%rax)
  lea    -0x210(%rbp),%rax
  add    $0x4,%rax
  mov    -0x214(%rbp),%edx
  mov    %edx,(%rax)
  lea    -0x210(%rbp),%rax
  add    $0x8,%rax
  mov    -0x218(%rbp),%edx
  mov    %edx,(%rax)
  lea    -0x210(%rbp),%rax
  add    $0xc,%rax
  mov    -0x228(%rbp),%rdx
  mov    -0x220(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x228(%rbp),%rax
  lea    0xc(%rax),%rdx
  lea    -0x210(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <mapif_sendall@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int mapif_guild_message(int guild_id, int account_id, const char *mes, size_t len) {
  unsigned char buf[512];
  (*(unsigned short *)(((unsigned char *)(buf)) + (0))) = 0x3837;
  (*(unsigned short *)(((unsigned char *)(buf)) + (2))) = (unsigned short)(len + 12);
  (*(unsigned int *)(((unsigned char *)(buf)) + (4))) = guild_id;
  (*(unsigned int *)(((unsigned char *)(buf)) + (8))) = account_id;
  memcpy((((unsigned char *)(buf)) + (12)), mes, len);
  mapif_sendall(buf, len + 12);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mapif_guild_message>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x218,%rsp
  mov    %esi,%eax
  mov    %rdx,%rsi
  mov    %rcx,%rbx
  mov    %fs:0x28,%rdx
  mov    %rdx,0x208(%rsp)
  xor    %edx,%edx
  movw   $0x3837,(%rsp)
  lea    0xc(%rcx),%edx
  mov    %dx,0x2(%rsp)
  mov    %edi,0x4(%rsp)
  mov    %eax,0x8(%rsp)
  mov    %rsp,%rbp
  lea    0xc(%rsp),%rdi
  mov    %rcx,%rdx
  call   <memcpy@plt>
  lea    0xc(%rbx),%rsi
  mov    %rbp,%rdi
  call   <mapif_sendall@plt>
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x218,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int mapif_guild_message(int guild_id, int account_id, const char *mes, size_t len) {
  unsigned char buf[512];
  (*(unsigned short *)(((unsigned char *)(buf)) + (0))) = 0x3837;
  (*(unsigned short *)(((unsigned char *)(buf)) + (2))) = (unsigned short)(len + 12);
  (*(unsigned int *)(((unsigned char *)(buf)) + (4))) = guild_id;
  (*(unsigned int *)(((unsigned char *)(buf)) + (8))) = account_id;
  memcpy((((unsigned char *)(buf)) + (12)), mes, len);
  mapif_sendall(buf, len + 12);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mapif_guild_message>:
  endbr64
  push   %rbp
  mov    %esi,%r8d
  mov    %rdx,%rsi
  mov    %rcx,%rdx
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x218,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    $0x3837,%eax
  mov    %edi,0x4(%rsp)
  mov    %rsp,%rbp
  mov    %ax,(%rsp)
  lea    0xc(%rsp),%rdi
  lea    0xc(%rcx),%eax
  mov    %ax,0x2(%rsp)
  mov    %r8d,0x8(%rsp)
  call   <memcpy@plt>
  lea    0xc(%rbx),%rsi
  mov    %rbp,%rdi
  call   <mapif_sendall@plt>
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x218,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int mapif_guild_message(int guild_id, int account_id, const char *mes, size_t len) {
  unsigned char buf[512];
  (*(unsigned short *)(((unsigned char *)(buf)) + (0))) = 0x3837;
  (*(unsigned short *)(((unsigned char *)(buf)) + (2))) = (unsigned short)(len + 12);
  (*(unsigned int *)(((unsigned char *)(buf)) + (4))) = guild_id;
  (*(unsigned int *)(((unsigned char *)(buf)) + (8))) = account_id;
  memcpy((((unsigned char *)(buf)) + (12)), mes, len);
  mapif_sendall(buf, len + 12);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<mapif_guild_message>:
  endbr64
  push   %rbp
  mov    %esi,%r8d
  movd   %edi,%xmm0
  mov    %rdx,%rsi
  push   %rbx
  movd   %r8d,%xmm1
  mov    %rcx,%rbx
  mov    %rcx,%rdx
  punpckldq %xmm1,%xmm0
  sub    $0x218,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    $0x3837,%eax
  lea    0xc(%rsp),%rdi
  mov    %rsp,%rbp
  mov    %ax,(%rsp)
  lea    0xc(%rcx),%eax
  mov    %ax,0x2(%rsp)
  movq   %xmm0,0x4(%rsp)
  call   <memcpy@plt>
  lea    0xc(%rbx),%rsi
  mov    %rbp,%rdi
  call   <mapif_sendall@plt>
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x218,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int mapif_guild_message(int guild_id, int account_id, const char *mes, size_t len) {
  unsigned char buf[512];
  (*(unsigned short *)(((unsigned char *)(buf)) + (0))) = 0x3837;
  (*(unsigned short *)(((unsigned char *)(buf)) + (2))) = (unsigned short)(len + 12);
  (*(unsigned int *)(((unsigned char *)(buf)) + (4))) = guild_id;
  (*(unsigned int *)(((unsigned char *)(buf)) + (8))) = account_id;
  memcpy((((unsigned char *)(buf)) + (12)), mes, len);
  mapif_sendall(buf, len + 12);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D21(%rip),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x90(%rbp)
  lea    D48(%rip),%rax
  mov    %rax,%rsi
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x88(%rbp)
  jmp    L26
L3:
  lea    -0x80(%rbp),%rdx
  mov    -0x88(%rbp),%rax
  lea    D39(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fprintf@plt>
L26:
  mov    -0x90(%rbp),%rdx
  lea    -0x80(%rbp),%rax
  mov    $0x79,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L3
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D21
data_type: string
name: retrieve_data
arguments: data_tag: D16
data_type: string
name: retrieve_data
arguments: data_tag: D48
data_type: string
name: retrieve_data
arguments: data_tag: D40
data_type: string
name: retrieve_data
arguments: data_tag: D39
data_type: string

tool: D21: r

tool: D16: file.in

tool: D48: w

tool: D40: p118file.out

tool: D39: %s

assistant: ```c
int main() {
  char line[120];
  FILE *instream, *outstream;
  instream = fopen("file.in", "r");
  outstream = fopen("p118file.out", "w");
  while (fgets(line, 121, instream))
    fprintf(outstream, "%s", line);
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    D29(%rip),%rsi
  lea    D17(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  lea    D45(%rip),%rsi
  lea    D49(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  lea    D37(%rip),%r12
  jmp    L10
L12:
  mov    %rsp,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <fprintf@plt>
L10:
  mov    %rsp,%rdi
  mov    %rbx,%rdx
  mov    $0x79,%esi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L12
  mov    $0x0,%edi
  call   <exit@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    $0x0,%eax
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D29
type_infered: string
name: parse_data
arguments: data_label: D17
type_infered: string
name: parse_data
arguments: data_label: D45
type_infered: string
name: parse_data
arguments: data_label: D49
type_infered: string
name: parse_data
arguments: data_label: D37
type_infered: string

tool: D29: r

tool: D17: file.in

tool: D45: w

tool: D49: p118file.out

tool: D37: %s

assistant: ```c
int main() {
  char line[120];
  FILE *instream, *outstream;
  instream = fopen("file.in", "r");
  outstream = fopen("p118file.out", "w");
  while (fgets(line, 121, instream))
    fprintf(outstream, "%s", line);
  exit(0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D10(%rip),%rsi
  lea    D18(%rip),%rdi
  push   %r12
  lea    D27(%rip),%r13
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <fopen@plt>
  lea    D7(%rip),%rsi
  lea    D42(%rip),%rdi
  mov    %rax,%rbp
  call   <fopen@plt>
  mov    %rax,%r12
  jmp    L4
  nopl   0x0(%rax)
L9:
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <fprintf@plt>
L4:
  mov    %rbp,%rdx
  mov    $0x79,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L9
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D10
type_infered: string
name: extract_symbolic_data
arguments: tag: D18
type_infered: string
name: extract_symbolic_data
arguments: tag: D27
type_infered: string
name: extract_symbolic_data
arguments: tag: D7
type_infered: string
name: extract_symbolic_data
arguments: tag: D42
type_infered: string

tool: D10: r

tool: D18: file.in

tool: D27: %s

tool: D7: w

tool: D42: p118file.out

assistant: ```c
int main() {
  char line[120];
  FILE *instream, *outstream;
  instream = fopen("file.in", "r");
  outstream = fopen("p118file.out", "w");
  while (fgets(line, 121, instream))
    fprintf(outstream, "%s", line);
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  lea    D12(%rip),%rsi
  lea    D43(%rip),%rdi
  push   %r12
  lea    D31(%rip),%r13
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  call   <fopen@plt>
  lea    D22(%rip),%rsi
  lea    D32(%rip),%rdi
  mov    %rax,%rbp
  call   <fopen@plt>
  mov    %rax,%r12
  jmp    L29
  nopl   0x0(%rax)
L28:
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <fprintf@plt>
L29:
  mov    %rbp,%rdx
  mov    $0x79,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L28
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D12
type_infered: string
name: read_data
arguments: data_tag: D43
type_infered: string
name: read_data
arguments: data_tag: D31
type_infered: string
name: read_data
arguments: data_tag: D22
type_infered: string
name: read_data
arguments: data_tag: D32
type_infered: string

tool: D12: r

tool: D43: file.in

tool: D31: %s

tool: D22: w

tool: D32: p118file.out

assistant: ```c
int main() {
  char line[120];
  FILE *instream, *outstream;
  instream = fopen("file.in", "r");
  outstream = fopen("p118file.out", "w");
  while (fgets(line, 121, instream))
    fprintf(outstream, "%s", line);
  exit(0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<z_now_ms>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <z_ms@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
time_t z_now_ms() {
  struct timeval t;
  gettimeofday(&t, NULL);
  return z_ms(&t);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<z_now_ms>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <gettimeofday@plt>
  mov    %rbx,%rdi
  call   <z_ms@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x10,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
time_t z_now_ms() {
  struct timeval t;
  gettimeofday(&t, NULL);
  return z_ms(&t);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<z_now_ms>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rdi
  call   <gettimeofday@plt>
  mov    %rbp,%rdi
  call   <z_ms@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x10,%rsp
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
time_t z_now_ms() {
  struct timeval t;
  gettimeofday(&t, NULL);
  return z_ms(&t);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<z_now_ms>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  mov    %rbp,%rdi
  call   <gettimeofday@plt>
  mov    %rbp,%rdi
  call   <z_ms@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x10,%rsp
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
time_t z_now_ms() {
  struct timeval t;
  gettimeofday(&t, NULL);
  return z_ms(&t);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<symbol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1f,-0x4(%rbp)
  jle    L20
  cmpl   $0x7f,-0x4(%rbp)
  jne    L25
L20:
  movb   $0x5e,D11(%rip)
  mov    -0x4(%rbp),%eax
  and    $0x1f,%eax
  add    $0x40,%eax
  mov    %al,D57(%rip)
  movb   $0x0,D23(%rip)
  jmp    L28
L25:
  mov    -0x4(%rbp),%eax
  mov    %al,D3(%rip)
  movb   $0x0,D57(%rip)
L28:
  lea    D3(%rip),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *symbol(int c) {
  static char str[3];
  if (c < ' ' || c == 127) {
    str[0] = '^';
    str[1] = (c & 037) + '@';
    str[2] = 0;
  } else {
    str[0] = c;
    str[1] = 0;
  }
  return (str);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<symbol>:
  endbr64
  cmp    $0x1f,%edi
  jle    L5
  cmp    $0x7f,%edi
  je     L5
  mov    %dil,D27(%rip)
  movb   $0x0,D26(%rip)
  jmp    L23
L5:
  movb   $0x5e,D27(%rip)
  and    $0x1f,%edi
  add    $0x40,%edi
  mov    %dil,D32(%rip)
  movb   $0x0,D32(%rip)
L23:
  lea    D27(%rip),%rax
  ret
```

assistant: ```c
char *symbol(int c) {
  static char str[3];
  if (c < ' ' || c == 127) {
    str[0] = '^';
    str[1] = (c & 037) + '@';
    str[2] = 0;
  } else {
    str[0] = c;
    str[1] = 0;
  }
  return (str);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<symbol>:
  endbr64
  cmp    $0x1f,%edi
  jle    L19
  cmp    $0x7f,%edi
  je     L19
  mov    %dil,D22(%rip)
  lea    D22(%rip),%rax
  movb   $0x0,D28(%rip)
  ret
  nopl   0x0(%rax)
L19:
  and    $0x1f,%edi
  movb   $0x5e,D22(%rip)
  lea    D22(%rip),%rax
  add    $0x40,%edi
  movb   $0x0,D56(%rip)
  mov    %dil,D33(%rip)
  ret
```

assistant: ```c
char *symbol(int c) {
  static char str[3];
  if (c < ' ' || c == 127) {
    str[0] = '^';
    str[1] = (c & 037) + '@';
    str[2] = 0;
  } else {
    str[0] = c;
    str[1] = 0;
  }
  return (str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<symbol>:
  endbr64
  cmp    $0x1f,%edi
  jle    L7
  mov    %edi,%ecx
  xor    %edx,%edx
  cmp    $0x7f,%edi
  je     L7
L25:
  xor    %eax,%eax
  mov    %cl,%al
  mov    %dl,%ah
  mov    %ax,D21(%rip)
  lea    D21(%rip),%rax
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  and    $0x1f,%edi
  movb   $0x0,D5(%rip)
  mov    $0x5e,%ecx
  lea    0x40(%rdi),%edx
  jmp    L25
```

assistant: ```c
char *symbol(int c) {
  static char str[3];
  if (c < ' ' || c == 127) {
    str[0] = '^';
    str[1] = (c & 037) + '@';
    str[2] = 0;
  } else {
    str[0] = c;
    str[1] = 0;
  }
  return (str);
}
```
user: What is the c source code of the assembly code below?

```asm
<dmi_string_cook>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  jmp    L28
L40:
  mov    -0x10(%rbp),%rax
  movb   $0x25,(%rax)
L28:
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L40
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L59
L19:
  addq   $0x1,-0x10(%rbp)
L59:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L106
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L19
  jmp    L106
L90:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L106:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L90
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L70
L75:
  addq   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L75
  subq   $0x1,-0x10(%rbp)
  jmp    L89
L4:
  subq   $0x1,-0x10(%rbp)
L89:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L4
  addq   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
L70:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *dmi_string_cook(char *s) {
  char *p, *q;
  while ((p = strchr(s, '/')) != NULL)
    *p = '%';
  p = s, q = s;
  while ((*p != '\0') && isspace(*p))
    p++;
  while (*p != '\0')
    *q = *p, q++, p++;
  *q = '\0';
  p = s;
  if (*p != '\0') {
    do
      p++;
    while (*p != '\0');
    p--;
    while (isspace(*p))
      p--;
    p++;
    *p = '\0';
  }
  return s;
}
```