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

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  mov    $0x6,%edi
  call   <quotient@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0xd,%edi
  call   <quotient@plt>
  lea    D13(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x4,%esi
  mov    $0xc,%edi
  call   <modulus@plt>
  lea    D17(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    $0x2d9,%edi
  call   <modulus@plt>
  lea    D32(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D0
data_type: string
name: retrieve_data
arguments: data_label: D13
data_type: string
name: retrieve_data
arguments: data_label: D17
data_type: string
name: retrieve_data
arguments: data_label: D32
data_type: string

tool: D0: Quotient 6/3 = %d


tool: D13: Quotient 13/3: %d


tool: D17: Modulus 12 mod 4: %d


tool: D32: Modulus 729 mod 45: %d


assistant: ```c
int main(void) {
  printf("Quotient 6/3 = %d\n", quotient(6, 3));
  printf("Quotient 13/3: %d\n", quotient(13, 3));
  printf("Modulus 12 mod 4: %d\n", modulus(12, 4));
  printf("Modulus 729 mod 45: %d\n", modulus(729, 3));
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<is_dir>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  test   %rax,%rax
  jne    L4
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  je     L4
  mov    $0x1,%eax
  jmp    L5
L4:
  mov    $0x0,%eax
L5:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
int is_dir(const char *path) {
  struct stat st;
  return (stat(path, &st) == 0 && S_ISDIR(st.st_mode));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<is_dir>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L24
L22:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L16
  add    $0x18,%rsp
  ret
L24:
  mov    0x4(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L22
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_dir(const char *path) {
  struct stat st;
  return (stat(path, &st) == 0 && S_ISDIR(st.st_mode));
}
```
user: What is the c source code of the assembly code below:

```asm
<is_dir>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  je     L1
L11:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax)
L1:
  mov    0x4(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L11
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_dir(const char *path) {
  struct stat st;
  return (stat(path, &st) == 0 && S_ISDIR(st.st_mode));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<is_dir>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  je     L25
L8:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax)
L25:
  mov    0x4(%rsp),%edi
  call   <S_ISDIR@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L8
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int is_dir(const char *path) {
  struct stat st;
  return (stat(path, &st) == 0 && S_ISDIR(st.st_mode));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<my_qsort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L22
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <swap@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L24
L34:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  mov    -0x28(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   *%rcx
  test   %eax,%eax
  jns    L17
  addl   $0x1,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <swap@plt>
L17:
  addl   $0x1,-0x8(%rbp)
L24:
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jle    L34
  mov    -0x4(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <swap@plt>
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edi
  mov    -0x28(%rbp),%rdx
  mov    -0x1c(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rcx
  mov    %edi,%edx
  mov    %rax,%rdi
  call   <my_qsort>
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%esi
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <my_qsort>
  jmp    L5
L22:
  nop
L5:
  leave
  ret
```

assistant: ```c
void my_qsort(void *v[], int left, int right, int (*comp)(void *, void *, ...)) {
  int i, last;
  void swap(void *v[], int, int);
  if (left >= right)
    return;
  swap(v, left, (left + right) / 2);
  last = left;
  for (i = left + 1; i <= right; i++)
    if ((*comp)(v[i], v[left]) < 0)
      swap(v, ++last, i);
  swap(v, left, last);
  my_qsort(v, left, last - 1, comp);
  my_qsort(v, last + 1, right, comp);
}
```
user: Translate the below assembly code into C programming language:

```asm
<my_qsort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %esi,0xc(%rsp)
  cmp    %edx,%esi
  jl     L57
L24:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L57:
  mov    %rdi,%rbp
  mov    %edx,%r12d
  mov    %rcx,%r13
  mov    %esi,%ebx
  lea    (%rsi,%rdx,1),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  mov    %esi,%r14d
  call   <swap@plt>
  lea    0x1(%rbx),%ebx
  cmp    %ebx,%r12d
  jl     L67
  movslq %r14d,%rax
  lea    0x0(%rbp,%rax,8),%r15
  movslq %ebx,%rbx
  jmp    L65
L31:
  add    $0x1,%rbx
  cmp    %ebx,%r12d
  jl     L2
L65:
  mov    %ebx,0x8(%rsp)
  mov    0x0(%rbp,%rbx,8),%rdi
  mov    (%r15),%rsi
  mov    $0x0,%eax
  call   *%r13
  test   %eax,%eax
  jns    L31
  add    $0x1,%r14d
  mov    0x8(%rsp),%edx
  mov    %r14d,%esi
  mov    %rbp,%rdi
  call   <swap@plt>
  jmp    L31
L67:
  mov    0xc(%rsp),%r14d
L2:
  mov    %r14d,%edx
  mov    0xc(%rsp),%ebx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <swap@plt>
  lea    -0x1(%r14),%edx
  mov    %r13,%rcx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <my_qsort>
  lea    0x1(%r14),%esi
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %rbp,%rdi
  call   <my_qsort>
  jmp    L24
```

assistant: ```c
void my_qsort(void *v[], int left, int right, int (*comp)(void *, void *, ...)) {
  int i, last;
  void swap(void *v[], int, int);
  if (left >= right)
    return;
  swap(v, left, (left + right) / 2);
  last = left;
  for (i = left + 1; i <= right; i++)
    if ((*comp)(v[i], v[left]) < 0)
      swap(v, ++last, i);
  swap(v, left, last);
  my_qsort(v, left, last - 1, comp);
  my_qsort(v, last + 1, right, comp);
}
```
user: What is the c source code of the assembly code below?

```asm
<my_qsort>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %rcx,(%rsp)
L25:
  cmp    %r15d,%ebp
  jle    L18
L76:
  lea    0x0(%rbp,%r15,1),%eax
  mov    %r15d,%esi
  mov    %rbx,%rdi
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  call   <swap@plt>
  lea    0x1(%r15),%ecx
  cmp    %ecx,%ebp
  jl     L49
  movslq %r15d,%rax
  movslq %ecx,%r14
  mov    %r15d,%r13d
  lea    (%rbx,%rax,8),%rax
  mov    %rax,0x8(%rsp)
  jmp    L47
L19:
  add    $0x1,%r14
  cmp    %r14d,%ebp
  jl     L27
L47:
  mov    0x8(%rsp),%rax
  mov    (%rbx,%r14,8),%rdi
  mov    (%rsp),%rcx
  mov    (%rax),%rsi
  xor    %eax,%eax
  call   *%rcx
  test   %eax,%eax
  jns    L19
  add    $0x1,%r13d
  mov    %r14d,%edx
  mov    %rbx,%rdi
  add    $0x1,%r14
  mov    %r13d,%esi
  call   <swap@plt>
  cmp    %r14d,%ebp
  jge    L47
L27:
  mov    %r15d,%esi
  mov    %r13d,%edx
  mov    %rbx,%rdi
  call   <swap@plt>
  mov    (%rsp),%rcx
  mov    %r15d,%esi
  lea    -0x1(%r13),%edx
  mov    %rbx,%rdi
  lea    0x1(%r13),%r15d
  call   <my_qsort>
  cmp    %ebp,%r15d
  jl     L76
L18:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L49:
  mov    %r15d,%edx
  mov    %r15d,%esi
  lea    -0x1(%r15),%ebp
  mov    %rbx,%rdi
  call   <swap@plt>
  jmp    L25
```

assistant: ```c
void my_qsort(void *v[], int left, int right, int (*comp)(void *, void *, ...)) {
  int i, last;
  void swap(void *v[], int, int);
  if (left >= right)
    return;
  swap(v, left, (left + right) / 2);
  last = left;
  for (i = left + 1; i <= right; i++)
    if ((*comp)(v[i], v[left]) < 0)
      swap(v, ++last, i);
  swap(v, left, last);
  my_qsort(v, left, last - 1, comp);
  my_qsort(v, last + 1, right, comp);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<my_qsort>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %rcx,(%rsp)
L5:
  cmp    %r15d,%ebp
  jle    L23
L6:
  lea    0x0(%rbp,%r15,1),%eax
  mov    %r15d,%esi
  mov    %rbx,%rdi
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %eax,%edx
  sar    %edx
  call   <swap@plt>
  lea    0x1(%r15),%ecx
  cmp    %ecx,%ebp
  jl     L52
  movslq %r15d,%rax
  movslq %ecx,%r14
  mov    %r15d,%r13d
  lea    (%rbx,%rax,8),%rax
  mov    %rax,0x8(%rsp)
  jmp    L17
L13:
  add    $0x1,%r14
  cmp    %r14d,%ebp
  jl     L73
L17:
  mov    0x8(%rsp),%rax
  mov    (%rbx,%r14,8),%rdi
  mov    (%rsp),%rcx
  mov    (%rax),%rsi
  xor    %eax,%eax
  call   *%rcx
  test   %eax,%eax
  jns    L13
  add    $0x1,%r13d
  mov    %r14d,%edx
  mov    %rbx,%rdi
  add    $0x1,%r14
  mov    %r13d,%esi
  call   <swap@plt>
  cmp    %r14d,%ebp
  jge    L17
L73:
  mov    %r15d,%esi
  mov    %r13d,%edx
  mov    %rbx,%rdi
  call   <swap@plt>
  mov    (%rsp),%rcx
  mov    %r15d,%esi
  lea    -0x1(%r13),%edx
  mov    %rbx,%rdi
  lea    0x1(%r13),%r15d
  call   <my_qsort>
  cmp    %ebp,%r15d
  jl     L6
L23:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L52:
  mov    %r15d,%edx
  mov    %r15d,%esi
  lea    -0x1(%r15),%ebp
  mov    %rbx,%rdi
  call   <swap@plt>
  jmp    L5
```

assistant: ```c
void my_qsort(void *v[], int left, int right, int (*comp)(void *, void *, ...)) {
  int i, last;
  void swap(void *v[], int, int);
  if (left >= right)
    return;
  swap(v, left, (left + right) / 2);
  last = left;
  for (i = left + 1; i <= right; i++)
    if ((*comp)(v[i], v[left]) < 0)
      swap(v, ++last, i);
  swap(v, left, last);
  my_qsort(v, left, last - 1, comp);
  my_qsort(v, last + 1, right, comp);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sysfs_devname_dev_to_sys>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L27
  jmp    L22
L3:
  mov    -0x8(%rbp),%rax
  movb   $0x21,(%rax)
L22:
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L3
L27:
  nop
  leave
  ret
```

assistant: ```c
void sysfs_devname_dev_to_sys(char *name) {
  char *c;
  if (name)
    while ((c = strchr(name, '/')))
      c[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
<sysfs_devname_dev_to_sys>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rdi,%rdi
  jne    L12
  jmp    L23
L22:
  movb   $0x21,(%rax)
L12:
  mov    $0x2f,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L22
L23:
  pop    %rbx
  ret
```

assistant: ```c
void sysfs_devname_dev_to_sys(char *name) {
  char *c;
  if (name)
    while ((c = strchr(name, '/')))
      c[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
<sysfs_devname_dev_to_sys>:
  endbr64
  test   %rdi,%rdi
  je     L12
  push   %rbx
  mov    %rdi,%rbx
  jmp    L15
  nop
L7:
  movb   $0x21,(%rax)
L15:
  mov    $0x2f,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L7
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  ret
```

assistant: ```c
void sysfs_devname_dev_to_sys(char *name) {
  char *c;
  if (name)
    while ((c = strchr(name, '/')))
      c[0] = '!';
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sysfs_devname_dev_to_sys>:
  endbr64
  test   %rdi,%rdi
  je     L10
  push   %rbx
  mov    %rdi,%rbx
  jmp    L7
  nop
L3:
  movb   $0x21,(%rax)
L7:
  mov    $0x2f,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L3
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  ret
```

assistant: ```c
void sysfs_devname_dev_to_sys(char *name) {
  char *c;
  if (name)
    while ((c = strchr(name, '/')))
      c[0] = '!';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<logger_get_rotatelogger_filename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    0x8(%rax),%ecx
  mov    -0x20(%rbp),%rax
  mov    0x4(%rax),%eax
  lea    0x1(%rax),%r8d
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  lea    0x76c(%rax),%edi
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rsp
  push   %rcx
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %rdx,%rcx
  lea    D7(%rip),%rdx
  mov    %rax,%rdi
  call   <snprintf@plt>
  add    $0x10,%rsp
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D7
type: string

tool: D7: %s.%04d-%02d-%02d

assistant: ```c
int logger_get_rotatelogger_filename(char *filename, size_t filename_len, const char *prefix, struct tm *m_tm) {
  snprintf(filename, filename_len, "%s.%04d-%02d-%02d", prefix, (m_tm->tm_year + 1900), (m_tm->tm_mon + 1), (m_tm->tm_mday));
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<logger_get_rotatelogger_filename>:
  endbr64
  sub    $0x10,%rsp
  mov    0x4(%rcx),%r8d
  mov    (%rcx),%eax
  mov    0x8(%rcx),%ecx
  push   %rcx
  lea    0x1(%r8),%r9d
  lea    0x76c(%rax),%r8d
  mov    %rdx,%rcx
  lea    D4(%rip),%rdx
  call   <snprintf@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
type_infered: string

tool: D4: %s.%04d-%02d-%02d

assistant: ```c
int logger_get_rotatelogger_filename(char *filename, size_t filename_len, const char *prefix, struct tm *m_tm) {
  snprintf(filename, filename_len, "%s.%04d-%02d-%02d", prefix, (m_tm->tm_year + 1900), (m_tm->tm_mon + 1), (m_tm->tm_mday));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<logger_get_rotatelogger_filename>:
  endbr64
  sub    $0x10,%rsp
  mov    %rdx,%r10
  mov    (%rcx),%eax
  mov    0x4(%rcx),%edx
  mov    0x8(%rcx),%ecx
  lea    0x1(%rdx),%r9d
  lea    0x76c(%rax),%r8d
  push   %rcx
  lea    D8(%rip),%rdx
  mov    %r10,%rcx
  call   <snprintf@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
type_infered: string

tool: D8: %s.%04d-%02d-%02d

assistant: ```c
int logger_get_rotatelogger_filename(char *filename, size_t filename_len, const char *prefix, struct tm *m_tm) {
  snprintf(filename, filename_len, "%s.%04d-%02d-%02d", prefix, (m_tm->tm_year + 1900), (m_tm->tm_mon + 1), (m_tm->tm_mday));
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<logger_get_rotatelogger_filename>:
  endbr64
  sub    $0x10,%rsp
  mov    %rdx,%r10
  mov    (%rcx),%eax
  mov    0x4(%rcx),%edx
  mov    0x8(%rcx),%ecx
  lea    0x1(%rdx),%r9d
  lea    0x76c(%rax),%r8d
  push   %rcx
  lea    D3(%rip),%rdx
  mov    %r10,%rcx
  call   <snprintf@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

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

tool: D3: %s.%04d-%02d-%02d

assistant: ```c
int logger_get_rotatelogger_filename(char *filename, size_t filename_len, const char *prefix, struct tm *m_tm) {
  snprintf(filename, filename_len, "%s.%04d-%02d-%02d", prefix, (m_tm->tm_year + 1900), (m_tm->tm_mon + 1), (m_tm->tm_mday));
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x20(%rbp)
  movl   $0x2,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  movl   $0x4,-0x14(%rbp)
  movl   $0x5,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  mov    $0x7,%edx
  mov    $0x5,%esi
  mov    %rax,%rdi
  call   <twoSum@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <free@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: disassemble_data
arguments: label: D6
type: string

tool: D6: %d, %d


assistant: ```c
int main() {
  int nums[5] = {1, 2, 3, 4, 5};
  int *indices = twoSum(nums, 5, 7);
  printf("%d, %d\n", indices[0], indices[1]);
  free(indices);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x4,0xc(%rsp)
  movl   $0x5,0x10(%rsp)
  mov    %rsp,%rdi
  mov    $0x7,%edx
  mov    $0x5,%esi
  call   <twoSum@plt>
  mov    %rax,%rbx
  mov    0x4(%rax),%edx
  mov    (%rax),%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D7
type_infered: string

tool: D7: %d, %d


assistant: ```c
int main() {
  int nums[5] = {1, 2, 3, 4, 5};
  int *indices = twoSum(nums, 5, 7);
  printf("%d, %d\n", indices[0], indices[1]);
  free(indices);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x7,%edx
  mov    $0x5,%esi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x200000001,%rax
  mov    %rsp,%rdi
  movl   $0x5,0x10(%rsp)
  mov    %rax,(%rsp)
  movabs $0x400000003,%rax
  mov    %rax,0x8(%rsp)
  call   <twoSum@plt>
  lea    D9(%rip),%rdi
  mov    0x4(%rax),%edx
  mov    (%rax),%esi
  mov    %rax,%rbp
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D9
type: string

tool: D9: %d, %d


assistant: ```c
int main() {
  int nums[5] = {1, 2, 3, 4, 5};
  int *indices = twoSum(nums, 5, 7);
  printf("%d, %d\n", indices[0], indices[1]);
  free(indices);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x7,%edx
  mov    $0x5,%esi
  sub    $0x20,%rsp
  movdqa D10(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movl   $0x5,0x10(%rsp)
  movaps %xmm0,(%rsp)
  call   <twoSum@plt>
  lea    D9(%rip),%rdi
  mov    0x4(%rax),%edx
  mov    (%rax),%esi
  mov    %rax,%rbp
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D10
type_guessed: byte[16]
name: unpack_data
arguments: data_label: D9
type_guessed: string

tool: {"D10": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D9: %d, %d


assistant: ```c
int main() {
  int nums[5] = {1, 2, 3, 4, 5};
  int *indices = twoSum(nums, 5, 7);
  printf("%d, %d\n", indices[0], indices[1]);
  free(indices);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<unixDlOpen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%edx
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  or     %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <dlopen@plt>
  leave
  ret
```

assistant: ```c
void *unixDlOpen(sqlite3_vfs *NotUsed, const char *zFilename) {
  (void)(NotUsed);
  return dlopen(zFilename, RTLD_NOW | RTLD_GLOBAL);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unixDlOpen>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdi
  mov    D4(%rip),%rax
  mov    (%rax),%esi
  mov    D16(%rip),%rax
  or     (%rax),%esi
  call   <dlopen@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *unixDlOpen(sqlite3_vfs *NotUsed, const char *zFilename) {
  (void)(NotUsed);
  return dlopen(zFilename, RTLD_NOW | RTLD_GLOBAL);
}
```
user: What is the c source code of the assembly code below?

```asm
<unixDlOpen>:
  endbr64
  mov    D8(%rip),%rax
  mov    %rsi,%rdi
  mov    (%rax),%esi
  mov    D11(%rip),%rax
  or     (%rax),%esi
  jmp    1050 <dlopen@plt>
```

assistant: ```c
void *unixDlOpen(sqlite3_vfs *NotUsed, const char *zFilename) {
  (void)(NotUsed);
  return dlopen(zFilename, RTLD_NOW | RTLD_GLOBAL);
}
```
user: What is the c source code of the assembly code below:

```asm
<unixDlOpen>:
  endbr64
  mov    D0(%rip),%rax
  mov    %rsi,%rdi
  mov    (%rax),%esi
  mov    D3(%rip),%rax
  or     (%rax),%esi
  jmp    1050 <dlopen@plt>
```

assistant: ```c
void *unixDlOpen(sqlite3_vfs *NotUsed, const char *zFilename) {
  (void)(NotUsed);
  return dlopen(zFilename, RTLD_NOW | RTLD_GLOBAL);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<pumas_version>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L9
  mov    -0x8(%rbp),%rax
  movl   $0x1,(%rax)
L9:
  cmpq   $0x0,-0x10(%rbp)
  je     L14
  mov    -0x10(%rbp),%rax
  movl   $0x1,(%rax)
L14:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void pumas_version(int *major, int *minor) {
  if (major != NULL)
    *major = 1;
  if (minor != NULL)
    *minor = 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<pumas_version>:
  endbr64
  test   %rdi,%rdi
  je     L2
  movl   $0x1,(%rdi)
L2:
  test   %rsi,%rsi
  je     L9
  movl   $0x1,(%rsi)
L9:
  ret
```

assistant: ```c
void pumas_version(int *major, int *minor) {
  if (major != NULL)
    *major = 1;
  if (minor != NULL)
    *minor = 1;
}
```