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

```asm
<isPalindrome>:
  endbr64
  mov    %edi,%r9d
  test   %edi,%edi
  je     L56
  js     L46
  mov    $0xcccccccd,%edi
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %r9d,%r8d
  mov    $0x1,%ecx
L59:
  mov    %r8d,%r8d
  mov    %ecx,%esi
  lea    (%rcx,%rcx,4),%ecx
  imul   %rdi,%r8
  add    %ecx,%ecx
  shr    $0x23,%r8
  test   %r8d,%r8d
  jne    L59
  mov    %r9d,%eax
  mov    %r9d,%ecx
  cltd
  imul   %rdi,%rcx
  idiv   %esi
  shr    $0x23,%rcx
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%r9d
  cmp    %eax,%r9d
  jne    L30
  mov    %edx,%edx
  imul   %rdi,%rdx
  shr    $0x23,%rdx
  mov    %rdx,%r9
  test   %edx,%edx
  jne    L28
L56:
  mov    $0x1,%r8d
L30:
  mov    %r8d,%eax
  ret
L46:
  xor    %r8d,%r8d
  jmp    L30
```

assistant: ```c
bool isPalindrome(int x) {
  if (x == 0)
    return true;
  if (x < 0)
    return false;
  int num_bit = 0, x0 = x, pow10 = 1;
  while (x0 != 0) {
    num_bit++;
    x0 /= 10;
    pow10 *= 10;
  }
  pow10 /= 10;
  int small = x % 10;
  int high = x / pow10;
  if (small != high)
    return false;
  x = x - high * pow10;
  x /= 10;
  return isPalindrome(x);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<isPalindrome>:
  endbr64
  mov    %edi,%r9d
  test   %edi,%edi
  je     L41
  js     L2
  mov    $0xcccccccd,%edi
  nopw   0x0(%rax,%rax,1)
L31:
  mov    %r9d,%r8d
  mov    $0x1,%ecx
L30:
  mov    %r8d,%r8d
  mov    %ecx,%esi
  lea    (%rcx,%rcx,4),%ecx
  imul   %rdi,%r8
  add    %ecx,%ecx
  shr    $0x23,%r8
  test   %r8d,%r8d
  jne    L30
  mov    %r9d,%eax
  mov    %r9d,%ecx
  cltd
  imul   %rdi,%rcx
  idiv   %esi
  shr    $0x23,%rcx
  lea    (%rcx,%rcx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%r9d
  cmp    %eax,%r9d
  jne    L11
  mov    %edx,%edx
  imul   %rdi,%rdx
  shr    $0x23,%rdx
  mov    %rdx,%r9
  test   %edx,%edx
  jne    L31
L41:
  mov    $0x1,%r8d
L11:
  mov    %r8d,%eax
  ret
L2:
  xor    %r8d,%r8d
  jmp    L11
```

assistant: ```c
bool isPalindrome(int x) {
  if (x == 0)
    return true;
  if (x < 0)
    return false;
  int num_bit = 0, x0 = x, pow10 = 1;
  while (x0 != 0) {
    num_bit++;
    x0 /= 10;
    pow10 *= 10;
  }
  pow10 /= 10;
  int small = x % 10;
  int high = x / pow10;
  if (small != high)
    return false;
  x = x - high * pow10;
  x /= 10;
  return isPalindrome(x);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  add    %edx,%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D7: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a = 1;
  int b = 1;
  printf("%d\n", a + ++b);
  printf("%d\n", a + b++);
  printf("%d\n", b);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x3,%esi
  lea    D4(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x3,%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D4
datatype: string

tool: D4: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a = 1;
  int b = 1;
  printf("%d\n", a + ++b);
  printf("%d\n", a + b++);
  printf("%d\n", b);
  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
  push   %rbp
  lea    D6(%rip),%rbp
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D6
datatype: string

tool: D6: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a = 1;
  int b = 1;
  printf("%d\n", a + ++b);
  printf("%d\n", a + b++);
  printf("%d\n", b);
  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
  push   %rbp
  lea    D5(%rip),%rbp
  mov    $0x3,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    $0x3,%esi
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type_guessed: string

tool: D5: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a = 1;
  int b = 1;
  printf("%d\n", a + ++b);
  printf("%d\n", a + b++);
  printf("%d\n", b);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<axp_set_pekinfo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int axp_set_pekinfo(struct platform_device *pdev) {
  int ret = 0;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<axp_set_pekinfo>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int axp_set_pekinfo(struct platform_device *pdev) {
  int ret = 0;
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<axp_set_pekinfo>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int axp_set_pekinfo(struct platform_device *pdev) {
  int ret = 0;
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_file_permissions>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0xb,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %ebp,%edi
  call   <S_ISDIR@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xffffffc9,%eax
  add    $0x64,%eax
  mov    %al,(%rbx)
  mov    D29(%rip),%rax
  mov    %ebp,%edx
  and    (%rax),%edx
  cmp    $0x1,%edx
  sbb    %eax,%eax
  and    $0xffffffbb,%eax
  add    $0x72,%eax
  mov    %al,0x1(%rbx)
  mov    D2(%rip),%rax
  mov    %ebp,%ecx
  and    (%rax),%ecx
  cmp    $0x1,%ecx
  sbb    %eax,%eax
  and    $0xffffffb6,%eax
  add    $0x77,%eax
  mov    %al,0x2(%rbx)
  mov    D28(%rip),%rax
  mov    %ebp,%esi
  and    (%rax),%esi
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xffffffb5,%eax
  add    $0x78,%eax
  mov    %al,0x3(%rbx)
  mov    D79(%rip),%rax
  mov    %ebp,%edx
  and    (%rax),%edx
  cmp    $0x1,%edx
  sbb    %eax,%eax
  and    $0xffffffbb,%eax
  add    $0x72,%eax
  mov    %al,0x4(%rbx)
  mov    D48(%rip),%rax
  mov    %ebp,%ecx
  and    (%rax),%ecx
  cmp    $0x1,%ecx
  sbb    %eax,%eax
  and    $0xffffffb6,%eax
  add    $0x77,%eax
  mov    %al,0x5(%rbx)
  mov    D36(%rip),%rax
  mov    %ebp,%esi
  and    (%rax),%esi
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xffffffb5,%eax
  add    $0x78,%eax
  mov    %al,0x6(%rbx)
  mov    D13(%rip),%rax
  mov    %ebp,%edi
  and    (%rax),%edi
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0xffffffbb,%eax
  add    $0x72,%eax
  mov    %al,0x7(%rbx)
  mov    D30(%rip),%rax
  mov    %ebp,%edx
  and    (%rax),%edx
  cmp    $0x1,%edx
  sbb    %eax,%eax
  and    $0xffffffb6,%eax
  add    $0x77,%eax
  mov    %al,0x8(%rbx)
  mov    D22(%rip),%rax
  and    (%rax),%ebp
  cmp    $0x1,%ebp
  sbb    %eax,%eax
  and    $0xffffffb5,%eax
  add    $0x78,%eax
  mov    %al,0x9(%rbx)
  movb   $0x0,0xa(%rbx)
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *get_file_permissions(struct stat buf) {
  char *permissions = (char *)malloc(11 * sizeof(char));
  permissions[0] = S_ISDIR(buf.st_mode) ? 'd' : '-';
  permissions[1] = buf.st_mode & S_IRUSR ? 'r' : '-';
  permissions[2] = buf.st_mode & S_IWUSR ? 'w' : '-';
  permissions[3] = buf.st_mode & S_IXUSR ? 'x' : '-';
  permissions[4] = buf.st_mode & S_IRGRP ? 'r' : '-';
  permissions[5] = buf.st_mode & S_IWGRP ? 'w' : '-';
  permissions[6] = buf.st_mode & S_IXGRP ? 'x' : '-';
  permissions[7] = buf.st_mode & S_IROTH ? 'r' : '-';
  permissions[8] = buf.st_mode & S_IWOTH ? 'w' : '-';
  permissions[9] = buf.st_mode & S_IXOTH ? 'x' : '-';
  permissions[10] = '\0';
  return permissions;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_file_permissions>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  mov    $0xb,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %ebx,%edi
  mov    %rax,%r12
  call   <S_ISDIR@plt>
  movb   $0x0,0xa(%r12)
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xffffffc9,%eax
  add    $0x64,%eax
  mov    %al,(%r12)
  mov    D89(%rip),%rax
  mov    (%rax),%edx
  and    %ebx,%edx
  cmp    $0x1,%edx
  sbb    %eax,%eax
  and    $0xffffffbb,%eax
  add    $0x72,%eax
  mov    %al,0x1(%r12)
  mov    D38(%rip),%rax
  mov    (%rax),%ecx
  and    %ebx,%ecx
  cmp    $0x1,%ecx
  sbb    %eax,%eax
  and    $0xffffffb6,%eax
  add    $0x77,%eax
  mov    %al,0x2(%r12)
  mov    D73(%rip),%rax
  mov    (%rax),%esi
  and    %ebx,%esi
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xffffffb5,%eax
  add    $0x78,%eax
  mov    %al,0x3(%r12)
  mov    D26(%rip),%rax
  mov    (%rax),%edx
  and    %ebx,%edx
  cmp    $0x1,%edx
  sbb    %eax,%eax
  and    $0xffffffbb,%eax
  add    $0x72,%eax
  mov    %al,0x4(%r12)
  mov    D37(%rip),%rax
  mov    (%rax),%ecx
  and    %ebx,%ecx
  cmp    $0x1,%ecx
  sbb    %eax,%eax
  and    $0xffffffb6,%eax
  add    $0x77,%eax
  mov    %al,0x5(%r12)
  mov    D79(%rip),%rax
  mov    (%rax),%esi
  and    %ebx,%esi
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xffffffb5,%eax
  add    $0x78,%eax
  mov    %al,0x6(%r12)
  mov    D51(%rip),%rax
  mov    (%rax),%edi
  and    %ebx,%edi
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0xffffffbb,%eax
  add    $0x72,%eax
  mov    %al,0x7(%r12)
  mov    D35(%rip),%rax
  mov    (%rax),%edx
  and    %ebx,%edx
  cmp    $0x1,%edx
  sbb    %eax,%eax
  and    $0xffffffb6,%eax
  add    $0x77,%eax
  mov    %al,0x8(%r12)
  mov    D48(%rip),%rax
  and    (%rax),%ebx
  cmp    $0x1,%ebx
  sbb    %eax,%eax
  and    $0xffffffb5,%eax
  add    $0x78,%eax
  mov    %al,0x9(%r12)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *get_file_permissions(struct stat buf) {
  char *permissions = (char *)malloc(11 * sizeof(char));
  permissions[0] = S_ISDIR(buf.st_mode) ? 'd' : '-';
  permissions[1] = buf.st_mode & S_IRUSR ? 'r' : '-';
  permissions[2] = buf.st_mode & S_IWUSR ? 'w' : '-';
  permissions[3] = buf.st_mode & S_IXUSR ? 'x' : '-';
  permissions[4] = buf.st_mode & S_IRGRP ? 'r' : '-';
  permissions[5] = buf.st_mode & S_IWGRP ? 'w' : '-';
  permissions[6] = buf.st_mode & S_IXGRP ? 'x' : '-';
  permissions[7] = buf.st_mode & S_IROTH ? 'r' : '-';
  permissions[8] = buf.st_mode & S_IWOTH ? 'w' : '-';
  permissions[9] = buf.st_mode & S_IXOTH ? 'x' : '-';
  permissions[10] = '\0';
  return permissions;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_file_permissions>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  mov    $0xb,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %ebx,%edi
  mov    %rax,%r12
  call   <S_ISDIR@plt>
  movb   $0x0,0xa(%r12)
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xffffffc9,%eax
  add    $0x64,%eax
  mov    %al,(%r12)
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  and    %ebx,%edx
  cmp    $0x1,%edx
  sbb    %eax,%eax
  and    $0xffffffbb,%eax
  add    $0x72,%eax
  mov    %al,0x1(%r12)
  mov    D23(%rip),%rax
  mov    (%rax),%ecx
  and    %ebx,%ecx
  cmp    $0x1,%ecx
  sbb    %eax,%eax
  and    $0xffffffb6,%eax
  add    $0x77,%eax
  mov    %al,0x2(%r12)
  mov    D88(%rip),%rax
  mov    (%rax),%esi
  and    %ebx,%esi
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xffffffb5,%eax
  add    $0x78,%eax
  mov    %al,0x3(%r12)
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  and    %ebx,%edx
  cmp    $0x1,%edx
  sbb    %eax,%eax
  and    $0xffffffbb,%eax
  add    $0x72,%eax
  mov    %al,0x4(%r12)
  mov    D30(%rip),%rax
  mov    (%rax),%ecx
  and    %ebx,%ecx
  cmp    $0x1,%ecx
  sbb    %eax,%eax
  and    $0xffffffb6,%eax
  add    $0x77,%eax
  mov    %al,0x5(%r12)
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  and    %ebx,%esi
  cmp    $0x1,%esi
  sbb    %eax,%eax
  and    $0xffffffb5,%eax
  add    $0x78,%eax
  mov    %al,0x6(%r12)
  mov    D50(%rip),%rax
  mov    (%rax),%edi
  and    %ebx,%edi
  cmp    $0x1,%edi
  sbb    %eax,%eax
  and    $0xffffffbb,%eax
  add    $0x72,%eax
  mov    %al,0x7(%r12)
  mov    D54(%rip),%rax
  mov    (%rax),%edx
  and    %ebx,%edx
  cmp    $0x1,%edx
  sbb    %eax,%eax
  and    $0xffffffb6,%eax
  add    $0x77,%eax
  mov    %al,0x8(%r12)
  mov    D20(%rip),%rax
  and    (%rax),%ebx
  cmp    $0x1,%ebx
  sbb    %eax,%eax
  and    $0xffffffb5,%eax
  add    $0x78,%eax
  mov    %al,0x9(%r12)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *get_file_permissions(struct stat buf) {
  char *permissions = (char *)malloc(11 * sizeof(char));
  permissions[0] = S_ISDIR(buf.st_mode) ? 'd' : '-';
  permissions[1] = buf.st_mode & S_IRUSR ? 'r' : '-';
  permissions[2] = buf.st_mode & S_IWUSR ? 'w' : '-';
  permissions[3] = buf.st_mode & S_IXUSR ? 'x' : '-';
  permissions[4] = buf.st_mode & S_IRGRP ? 'r' : '-';
  permissions[5] = buf.st_mode & S_IWGRP ? 'w' : '-';
  permissions[6] = buf.st_mode & S_IXGRP ? 'x' : '-';
  permissions[7] = buf.st_mode & S_IROTH ? 'r' : '-';
  permissions[8] = buf.st_mode & S_IWOTH ? 'w' : '-';
  permissions[9] = buf.st_mode & S_IXOTH ? 'x' : '-';
  permissions[10] = '\0';
  return permissions;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<le2uint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x18,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  or     %edx,%eax
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t le2uint(const unsigned char *buf) {
  uint32_t res = (buf[3] << 24) | (buf[2] << 16) | (buf[1] << 8) | buf[0];
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<le2uint>:
  endbr64
  movzbl (%rdi),%eax
  ret
```

assistant: ```c
uint32_t le2uint(const unsigned char *buf) {
  uint32_t res = (buf[3] << 24) | (buf[2] << 16) | (buf[1] << 8) | buf[0];
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<le2uint>:
  endbr64
  movzbl (%rdi),%eax
  ret
```

assistant: ```c
uint32_t le2uint(const unsigned char *buf) {
  uint32_t res = (buf[3] << 24) | (buf[2] << 16) | (buf[1] << 8) | buf[0];
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<le2uint>:
  endbr64
  movzbl (%rdi),%eax
  ret
```

assistant: ```c
uint32_t le2uint(const unsigned char *buf) {
  uint32_t res = (buf[3] << 24) | (buf[2] << 16) | (buf[1] << 8) | buf[0];
  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
<hexcdiff_str_auto>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <hexcdiff_msg_size@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdi
  mov    -0x18(%rbp),%rcx
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <hexcdiff@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *hexcdiff_str_auto(const char *expect_str, const char *actual_str) {
  size_t expect_len = strlen(expect_str);
  size_t actual_len = strlen(actual_str);
  size_t msg_len = hexcdiff_msg_size(expect_len, actual_len);
  char *msg = malloc(msg_len);
  hexcdiff((void *)expect_str, expect_len, (void *)actual_str, actual_len, msg);
  return msg;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<hexcdiff_str_auto>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r13
  call   <strlen@plt>
  mov    %rax,%r12
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,%r14
  mov    %rax,%rsi
  mov    %r12,%rdi
  call   <hexcdiff_msg_size@plt>
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%r8
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <hexcdiff@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *hexcdiff_str_auto(const char *expect_str, const char *actual_str) {
  size_t expect_len = strlen(expect_str);
  size_t actual_len = strlen(actual_str);
  size_t msg_len = hexcdiff_msg_size(expect_len, actual_len);
  char *msg = malloc(msg_len);
  hexcdiff((void *)expect_str, expect_len, (void *)actual_str, actual_len, msg);
  return msg;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hexcdiff_str_auto>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %rax,%r13
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  mov    %rax,%r15
  call   <hexcdiff_msg_size@plt>
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rax,%r12
  mov    %rbp,%rdi
  mov    %rax,%r8
  call   <hexcdiff@plt>
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *hexcdiff_str_auto(const char *expect_str, const char *actual_str) {
  size_t expect_len = strlen(expect_str);
  size_t actual_len = strlen(actual_str);
  size_t msg_len = hexcdiff_msg_size(expect_len, actual_len);
  char *msg = malloc(msg_len);
  hexcdiff((void *)expect_str, expect_len, (void *)actual_str, actual_len, msg);
  return msg;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<hexcdiff_str_auto>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %rax,%r13
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  mov    %rax,%r15
  call   <hexcdiff_msg_size@plt>
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %r15,%rcx
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rax,%r12
  mov    %rbp,%rdi
  mov    %rax,%r8
  call   <hexcdiff@plt>
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *hexcdiff_str_auto(const char *expect_str, const char *actual_str) {
  size_t expect_len = strlen(expect_str);
  size_t actual_len = strlen(actual_str);
  size_t msg_len = hexcdiff_msg_size(expect_len, actual_len);
  char *msg = malloc(msg_len);
  hexcdiff((void *)expect_str, expect_len, (void *)actual_str, actual_len, msg);
  return msg;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3e8,-0x28(%rbp)
  movl   $0x4b0,-0x24(%rbp)
  mov    $0x8,%esi
  mov    $0xf4240,%edi
  call   <polybench_alloc_data@plt>
  mov    %rax,-0x18(%rbp)
  mov    $0x8,%esi
  mov    $0x124f80,%edi
  call   <polybench_alloc_data@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdi
  mov    -0x18(%rbp),%rcx
  lea    -0x20(%rbp),%rdx
  mov    -0x24(%rbp),%esi
  mov    -0x28(%rbp),%eax
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <init_array@plt>
  mov    -0x20(%rbp),%rdi
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x24(%rbp),%esi
  mov    -0x28(%rbp),%eax
  movq   %rdi,%xmm0
  mov    %eax,%edi
  call   <kernel_trmm@plt>
  cmpl   $0x2a,-0x34(%rbp)
  jle    L11
  mov    -0x40(%rbp),%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  test   %eax,%eax
  jne    L11
  mov    -0x10(%rbp),%rdx
  mov    -0x24(%rbp),%ecx
  mov    -0x28(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <print_array@plt>
L11:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int m = 1000;
  int n = 1200;
  double alpha;
  double(*A)[1000 + 0][1000 + 0];
  A = (double(*)[1000 + 0][1000 + 0]) polybench_alloc_data((1000 + 0) * (1000 + 0), sizeof(double));
  ;
  double(*B)[1000 + 0][1200 + 0];
  B = (double(*)[1000 + 0][1200 + 0]) polybench_alloc_data((1000 + 0) * (1200 + 0), sizeof(double));
  ;
  init_array(m, n, &alpha, *A, *B);
  ;
  kernel_trmm(m, n, alpha, *A, *B);
  ;
  ;
  if (argc > 42 && !strcmp(argv[0], ""))
    print_array(m, n, *B);
  free((void *)A);
  ;
  free((void *)B);
  ;
  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   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r12d
  mov    %rsi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x8,%esi
  mov    $0xf4240,%edi
  call   <polybench_alloc_data@plt>
  mov    %rax,%rbp
  mov    $0x8,%esi
  mov    $0x124f80,%edi
  call   <polybench_alloc_data@plt>
  mov    %rax,%rbx
  mov    %rsp,%rdx
  mov    %rax,%r8
  mov    %rbp,%rcx
  mov    $0x4b0,%esi
  mov    $0x3e8,%edi
  call   <init_array@plt>
  mov    %rbx,%rcx
  mov    %rbp,%rdx
  movsd  (%rsp),%xmm0
  mov    $0x4b0,%esi
  mov    $0x3e8,%edi
  call   <kernel_trmm@plt>
  cmp    $0x2a,%r12d
  jle    L29
  mov    0x0(%r13),%rax
  cmpb   $0x0,(%rax)
  je     L37
L29:
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L37:
  mov    %rbx,%rdx
  mov    $0x4b0,%esi
  mov    $0x3e8,%edi
  call   <print_array@plt>
  jmp    L29
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  int m = 1000;
  int n = 1200;
  double alpha;
  double(*A)[1000 + 0][1000 + 0];
  A = (double(*)[1000 + 0][1000 + 0]) polybench_alloc_data((1000 + 0) * (1000 + 0), sizeof(double));
  ;
  double(*B)[1000 + 0][1200 + 0];
  B = (double(*)[1000 + 0][1200 + 0]) polybench_alloc_data((1000 + 0) * (1200 + 0), sizeof(double));
  ;
  init_array(m, n, &alpha, *A, *B);
  ;
  kernel_trmm(m, n, alpha, *A, *B);
  ;
  ;
  if (argc > 42 && !strcmp(argv[0], ""))
    print_array(m, n, *B);
  free((void *)A);
  ;
  free((void *)B);
  ;
  return 0;
}
```