ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<eeprom_write>:
  endbr64
  push   %rbx
  mov    %esi,%eax
  lea    (%rdx,%r8,1),%esi
  cmp    %esi,%edx
  jae    L17
  mov    %edx,%esi
  shr    $0x8,%esi
  or     %eax,%esi
  movzbl %sil,%esi
  call   <i2c_write@plt>
  test   %rax,%rax
  setne  %bl
  movzbl %bl,%ebx
  mov    $0x2af8,%edi
  call   <usleep@plt>
L9:
  mov    %ebx,%eax
  pop    %rbx
  ret
L17:
  mov    $0x0,%ebx
  jmp    L9
```

assistant: ```c
int eeprom_write(int fd, unsigned dev_addr, unsigned offset, unsigned char *buffer, unsigned cnt) {
  unsigned end = offset + cnt;
  unsigned blk_off;
  int rcode = 0;
  while (offset < end) {
    unsigned alen, len, maxlen;
    unsigned char addr[2];
    blk_off = offset & 0xFF;
    addr[0] = offset >> 8;
    addr[1] = blk_off;
    alen = 2;
    addr[0] |= dev_addr;
    maxlen = (1 << 4) - ((blk_off) & ((1 << 4) - 1));
    len = end - offset;
    if (i2c_write(fd, addr[0], offset, buffer, len) != 0)
      rcode = 1;
    buffer += len;
    offset += len;
    usleep(11 * 1000);
  }
  return rcode;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<eeprom_write>:
  endbr64
  mov    %esi,%eax
  lea    (%rdx,%r8,1),%esi
  push   %r12
  cmp    %esi,%edx
  jae    L4
  mov    %edx,%esi
  xor    %r12d,%r12d
  shr    $0x8,%esi
  or     %eax,%esi
  movzbl %sil,%esi
  call   <i2c_write@plt>
  mov    $0x2af8,%edi
  test   %rax,%rax
  setne  %r12b
  call   <usleep@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %r12d,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int eeprom_write(int fd, unsigned dev_addr, unsigned offset, unsigned char *buffer, unsigned cnt) {
  unsigned end = offset + cnt;
  unsigned blk_off;
  int rcode = 0;
  while (offset < end) {
    unsigned alen, len, maxlen;
    unsigned char addr[2];
    blk_off = offset & 0xFF;
    addr[0] = offset >> 8;
    addr[1] = blk_off;
    alen = 2;
    addr[0] |= dev_addr;
    maxlen = (1 << 4) - ((blk_off) & ((1 << 4) - 1));
    len = end - offset;
    if (i2c_write(fd, addr[0], offset, buffer, len) != 0)
      rcode = 1;
    buffer += len;
    offset += len;
    usleep(11 * 1000);
  }
  return rcode;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<eeprom_write>:
  endbr64
  mov    %esi,%eax
  lea    (%rdx,%r8,1),%esi
  push   %r12
  cmp    %esi,%edx
  jae    L4
  mov    %edx,%esi
  xor    %r12d,%r12d
  shr    $0x8,%esi
  or     %eax,%esi
  movzbl %sil,%esi
  call   <i2c_write@plt>
  mov    $0x2af8,%edi
  test   %rax,%rax
  setne  %r12b
  call   <usleep@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %r12d,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int eeprom_write(int fd, unsigned dev_addr, unsigned offset, unsigned char *buffer, unsigned cnt) {
  unsigned end = offset + cnt;
  unsigned blk_off;
  int rcode = 0;
  while (offset < end) {
    unsigned alen, len, maxlen;
    unsigned char addr[2];
    blk_off = offset & 0xFF;
    addr[0] = offset >> 8;
    addr[1] = blk_off;
    alen = 2;
    addr[0] |= dev_addr;
    maxlen = (1 << 4) - ((blk_off) & ((1 << 4) - 1));
    len = end - offset;
    if (i2c_write(fd, addr[0], offset, buffer, len) != 0)
      rcode = 1;
    buffer += len;
    offset += len;
    usleep(11 * 1000);
  }
  return rcode;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<a_current_time_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <time@plt>
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <localtime_r@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rcx
  lea    D8(%rip),%rax
  mov    %rax,%rdx
  mov    $0x19,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <strftime@plt>
  test   %eax,%eax
  jne    L16
  movb   $0x0,D10(%rip)
L16:
  lea    D10(%rip),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D8
type_infered: string

tool: D8: %Y-%m-%d %T 

assistant: ```c
const char *a_current_time_str(void) {
  static char buffer[25];
  time_t now;
  struct tm tm;
  time(&now);
  localtime_r(&now, &tm);
  if (!strftime(buffer, sizeof(buffer), "%Y-%m-%d %T ", &tm))
    buffer[0] = '\0';
  return buffer;
}
```
user: Translate the below assembly code into C programming language:

```asm
<a_current_time_str>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <time@plt>
  lea    0x4(%rsp),%rbx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <localtime_r@plt>
  mov    %rbx,%rcx
  lea    D35(%rip),%rdx
  mov    $0x19,%esi
  lea    D26(%rip),%rdi
  call   <strftime@plt>
  test   %eax,%eax
  jne    L6
  movb   $0x0,D26(%rip)
L6:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  lea    D26(%rip),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D35
type: string

tool: D35: %Y-%m-%d %T 

assistant: ```c
const char *a_current_time_str(void) {
  static char buffer[25];
  time_t now;
  struct tm tm;
  time(&now);
  localtime_r(&now, &tm);
  if (!strftime(buffer, sizeof(buffer), "%Y-%m-%d %T ", &tm))
    buffer[0] = '\0';
  return buffer;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<a_current_time_str>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x4(%rsp),%r12
  mov    %rbp,%rdi
  call   <time@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <localtime_r@plt>
  mov    %r12,%rcx
  mov    $0x19,%esi
  lea    D11(%rip),%r12
  lea    D19(%rip),%rdx
  mov    %r12,%rdi
  call   <strftime@plt>
  test   %eax,%eax
  jne    L14
  movb   $0x0,D11(%rip)
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D19
type_guessed: string

tool: D19: %Y-%m-%d %T 

assistant: ```c
const char *a_current_time_str(void) {
  static char buffer[25];
  time_t now;
  struct tm tm;
  time(&now);
  localtime_r(&now, &tm);
  if (!strftime(buffer, sizeof(buffer), "%Y-%m-%d %T ", &tm))
    buffer[0] = '\0';
  return buffer;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<a_current_time_str>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x4(%rsp),%r12
  mov    %rbp,%rdi
  call   <time@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <localtime_r@plt>
  mov    %r12,%rcx
  mov    $0x19,%esi
  lea    D18(%rip),%r12
  lea    D27(%rip),%rdx
  mov    %r12,%rdi
  call   <strftime@plt>
  test   %eax,%eax
  jne    L4
  movb   $0x0,D18(%rip)
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D27
type: string

tool: D27: %Y-%m-%d %T 

assistant: ```c
const char *a_current_time_str(void) {
  static char buffer[25];
  time_t now;
  struct tm tm;
  time(&now);
  localtime_r(&now, &tm);
  if (!strftime(buffer, sizeof(buffer), "%Y-%m-%d %T ", &tm))
    buffer[0] = '\0';
  return buffer;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<irc_voice>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x220,%rsp
  mov    %rdi,-0x218(%rbp)
  mov    %rsi,-0x220(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x220(%rbp),%rcx
  mov    -0x218(%rbp),%rdx
  lea    -0x210(%rbp),%rax
  mov    %rcx,%r9
  mov    %rdx,%r8
  lea    D13(%rip),%rdx
  mov    %rdx,%rcx
  lea    D10(%rip),%rdx
  mov    $0x1fd,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  lea    -0x210(%rbp),%rax
  mov    %rax,%rdi
  call   <net_send@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D13
datatype: string
name: parse_data
arguments: data_tag: D10
datatype: string

tool: D13: MODE

tool: D10: %s %s +v %s

assistant: ```c
void irc_voice(char *channel, char *nick) {
  char msg[(512 - 3)];
  snprintf(msg, (512 - 3), "%s %s +v %s", "MODE", channel, nick);
  net_send(msg);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<irc_voice>:
  endbr64
  push   %rbx
  sub    $0x210,%rsp
  mov    %rdi,%r8
  mov    %rsi,%r9
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    D16(%rip),%rcx
  lea    D14(%rip),%rdx
  mov    $0x1fd,%esi
  mov    %rbx,%rdi
  call   <snprintf@plt>
  mov    %rbx,%rdi
  call   <net_send@plt>
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x210,%rsp
  pop    %rbx
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D16
data_type: string
name: disassemble_data
arguments: label: D14
data_type: string

tool: D16: MODE

tool: D14: %s %s +v %s

assistant: ```c
void irc_voice(char *channel, char *nick) {
  char msg[(512 - 3)];
  snprintf(msg, (512 - 3), "%s %s +v %s", "MODE", channel, nick);
  net_send(msg);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<irc_voice>:
  endbr64
  push   %rbp
  mov    %rdi,%r8
  mov    %rsi,%r9
  lea    D10(%rip),%rcx
  lea    D5(%rip),%rdx
  mov    $0x1fd,%esi
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <snprintf@plt>
  mov    %rbp,%rdi
  call   <net_send@plt>
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x210,%rsp
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D10
data_type: string
name: unpack_data
arguments: label: D5
data_type: string

tool: D10: MODE

tool: D5: %s %s +v %s

assistant: ```c
void irc_voice(char *channel, char *nick) {
  char msg[(512 - 3)];
  snprintf(msg, (512 - 3), "%s %s +v %s", "MODE", channel, nick);
  net_send(msg);
}
```
user: What is the c source code of the assembly code below:

```asm
<irc_voice>:
  endbr64
  push   %rbp
  mov    %rdi,%r8
  mov    %rsi,%r9
  lea    D6(%rip),%rcx
  lea    D16(%rip),%rdx
  mov    $0x1fd,%esi
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <snprintf@plt>
  mov    %rbp,%rdi
  call   <net_send@plt>
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x210,%rsp
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D6
type_guessed: string
name: read_data
arguments: label: D16
type_guessed: string

tool: D6: MODE

tool: D16: %s %s +v %s

assistant: ```c
void irc_voice(char *channel, char *nick) {
  char msg[(512 - 3)];
  snprintf(msg, (512 - 3), "%s %s +v %s", "MODE", channel, nick);
  net_send(msg);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<be32dec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  shl    $0x8,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  shl    $0x10,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  shl    $0x18,%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t be32dec(const void *pp) {
  const uint8_t *p = (uint8_t const *)pp;
  return ((uint32_t)(p[3]) + ((uint32_t)(p[2]) << 8) + ((uint32_t)(p[1]) << 16) + ((uint32_t)(p[0]) << 24));
}
```
user: Translate the below assembly code into C programming language:

```asm
<be32dec>:
  endbr64
  mov    0x10(%rdi),%eax
  shl    $0x8,%eax
  add    0x18(%rdi),%eax
  mov    0x8(%rdi),%edx
  shl    $0x10,%edx
  add    %edx,%eax
  mov    (%rdi),%rdx
  shl    $0x18,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint32_t be32dec(const void *pp) {
  const uint8_t *p = (uint8_t const *)pp;
  return ((uint32_t)(p[3]) + ((uint32_t)(p[2]) << 8) + ((uint32_t)(p[1]) << 16) + ((uint32_t)(p[0]) << 24));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<be32dec>:
  endbr64
  mov    0x10(%rdi),%eax
  mov    0x8(%rdi),%edx
  shl    $0x8,%eax
  shl    $0x10,%edx
  add    0x18(%rdi),%eax
  add    %edx,%eax
  mov    (%rdi),%rdx
  shl    $0x18,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint32_t be32dec(const void *pp) {
  const uint8_t *p = (uint8_t const *)pp;
  return ((uint32_t)(p[3]) + ((uint32_t)(p[2]) << 8) + ((uint32_t)(p[1]) << 16) + ((uint32_t)(p[0]) << 24));
}
```
user: What is the c source code of the assembly code below:

```asm
<be32dec>:
  endbr64
  mov    0x10(%rdi),%eax
  mov    0x8(%rdi),%edx
  shl    $0x8,%eax
  shl    $0x10,%edx
  add    0x18(%rdi),%eax
  add    %edx,%eax
  mov    (%rdi),%rdx
  shl    $0x18,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint32_t be32dec(const void *pp) {
  const uint8_t *p = (uint8_t const *)pp;
  return ((uint32_t)(p[3]) + ((uint32_t)(p[2]) << 8) + ((uint32_t)(p[1]) << 16) + ((uint32_t)(p[0]) << 24));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<plusOne>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L53
L35:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    $0x1,%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%ecx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,(%rsi)
  cmpl   $0x0,-0xc(%rbp)
  jne    L17
  mov    -0x28(%rbp),%rax
  mov    -0x1c(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  jmp    L51
L17:
  subl   $0x1,-0x10(%rbp)
L53:
  cmpl   $0x0,-0x10(%rbp)
  jns    L35
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x1,(%rax)
  movl   $0x0,-0x10(%rbp)
  jmp    L2
L14:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  add    $0x1,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x10(%rbp)
L2:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L14
  mov    -0x8(%rbp),%rax
L51:
  leave
  ret
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int carry = 0;
  int *res;
  int i;
  i = digitsSize - 1;
  while (i >= 0) {
    carry = (digits[i] + 1) / 10;
    digits[i] = (digits[i] + 1) % 10;
    if (carry == 0) {
      returnSize[0] = digitsSize;
      return digits;
    }
    i--;
  }
  returnSize[0] = digitsSize + 1;
  res = (int *)malloc(sizeof(int) * (digitsSize + 1));
  res[0] = 1;
  for (i = 0; i < digitsSize; i++) {
    res[i + 1] = digits[i];
  }
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<plusOne>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %rdx,%r8
  mov    %esi,%ecx
  sub    $0x1,%ecx
  js     L46
  movslq %ecx,%rcx
L39:
  mov    (%rbx,%rcx,4),%esi
  lea    0x1(%rsi),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,(%rbx,%rcx,4)
  add    $0xa,%esi
  cmp    $0x12,%esi
  jbe    L54
  sub    $0x1,%rcx
  test   %ecx,%ecx
  jns    L39
L46:
  lea    0x1(%rbp),%edi
  mov    %edi,(%r8)
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%rsi
  movl   $0x1,(%rax)
  test   %ebp,%ebp
  jle    L38
  lea    0x4(%rax),%rdx
  lea    -0x1(%rbp),%ecx
  lea    0x8(%rax,%rcx,4),%rcx
  sub    %rax,%rbx
L30:
  mov    -0x4(%rbx,%rdx,1),%eax
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  cmp    %rcx,%rdx
  jne    L30
  jmp    L38
L54:
  mov    %ebp,(%r8)
  mov    %rbx,%rsi
L38:
  mov    %rsi,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int carry = 0;
  int *res;
  int i;
  i = digitsSize - 1;
  while (i >= 0) {
    carry = (digits[i] + 1) / 10;
    digits[i] = (digits[i] + 1) % 10;
    if (carry == 0) {
      returnSize[0] = digitsSize;
      return digits;
    }
    i--;
  }
  returnSize[0] = digitsSize + 1;
  res = (int *)malloc(sizeof(int) * (digitsSize + 1));
  res[0] = 1;
  for (i = 0; i < digitsSize; i++) {
    res[i + 1] = digits[i];
  }
  return res;
}
```
user: What is the c source code of the assembly code below?

```asm
<plusOne>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  mov    %rdx,%r8
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x1,%r12d
  js     L8
  movslq %r12d,%rsi
  jmp    L2
  xchg   %ax,%ax
L35:
  sub    $0x1,%rsi
  test   %esi,%esi
  js     L8
L2:
  mov    (%rbx,%rsi,4),%ecx
  lea    0x1(%rcx),%edx
  add    $0xa,%ecx
  movslq %edx,%rax
  mov    %edx,%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edi
  sar    $0x22,%rax
  sub    %edi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,(%rbx,%rsi,4)
  cmp    $0x12,%ecx
  ja     L35
  mov    %ebp,(%r8)
  mov    %rbx,%rax
L57:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  lea    0x1(%rbp),%edi
  mov    %edi,(%r8)
  shl    $0x2,%edi
  call   <malloc@plt>
  movl   $0x1,(%rax)
  test   %ebp,%ebp
  jle    L57
  mov    %r12d,%r12d
  lea    0x4(%rax),%rdx
  sub    %rax,%rbx
  lea    0x8(%rax,%r12,4),%rsi
  nopw   0x0(%rax,%rax,1)
L4:
  mov    -0x4(%rbx,%rdx,1),%ecx
  add    $0x4,%rdx
  mov    %ecx,-0x4(%rdx)
  cmp    %rdx,%rsi
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int carry = 0;
  int *res;
  int i;
  i = digitsSize - 1;
  while (i >= 0) {
    carry = (digits[i] + 1) / 10;
    digits[i] = (digits[i] + 1) % 10;
    if (carry == 0) {
      returnSize[0] = digitsSize;
      return digits;
    }
    i--;
  }
  returnSize[0] = digitsSize + 1;
  res = (int *)malloc(sizeof(int) * (digitsSize + 1));
  res[0] = 1;
  for (i = 0; i < digitsSize; i++) {
    res[i + 1] = digits[i];
  }
  return res;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<plusOne>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  mov    %rdx,%r8
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x1,%r12d
  js     L22
  movslq %r12d,%rsi
  jmp    L17
  xchg   %ax,%ax
L122:
  sub    $0x1,%rsi
  test   %esi,%esi
  js     L22
L17:
  mov    (%rbx,%rsi,4),%ecx
  lea    0x1(%rcx),%edx
  add    $0xa,%ecx
  movslq %edx,%rax
  mov    %edx,%edi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edi
  sar    $0x22,%rax
  sub    %edi,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,(%rbx,%rsi,4)
  cmp    $0x12,%ecx
  ja     L122
  mov    %ebp,(%r8)
  mov    %rbx,%r8
L102:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L22:
  lea    0x1(%rbp),%edi
  mov    %edi,(%r8)
  shl    $0x2,%edi
  call   <malloc@plt>
  movl   $0x1,(%rax)
  mov    %rax,%r8
  test   %ebp,%ebp
  jle    L102
  lea    0x4(%rax),%rdx
  lea    0x4(%rbx),%rsi
  mov    %rdx,%rcx
  mov    %rdx,%rax
  sub    %rsi,%rcx
  cmp    $0x8,%rcx
  jbe    L128
  cmp    $0x2,%r12d
  jbe    L128
  mov    %ebp,%edx
  shr    $0x2,%edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x14(%r8,%rdx,1),%rcx
  mov    %rbx,%rdx
  sub    %r8,%rdx
  nop
L4:
  movdqu -0x4(%rax,%rdx,1),%xmm0
  add    $0x10,%rax
  movups %xmm0,-0x10(%rax)
  cmp    %rcx,%rax
  jne    L4
  mov    %ebp,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%bpl
  je     L102
  mov    %eax,%ecx
  lea    0x0(,%rcx,4),%rdx
  mov    (%rbx,%rcx,4),%ecx
  mov    %ecx,0x4(%r8,%rdx,1)
  lea    0x1(%rax),%ecx
  cmp    %ecx,%ebp
  jle    L102
  mov    0x4(%rbx,%rdx,1),%ecx
  add    $0x2,%eax
  mov    %ecx,0x8(%r8,%rdx,1)
  cmp    %eax,%ebp
  jle    L102
  mov    0x8(%rbx,%rdx,1),%eax
  mov    %eax,0xc(%r8,%rdx,1)
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L128:
  mov    %r12d,%r12d
  sub    %r8,%rbx
  lea    0x8(%r8,%r12,4),%rcx
  nop
L5:
  mov    -0x4(%rbx,%rdx,1),%eax
  add    $0x4,%rdx
  mov    %eax,-0x4(%rdx)
  cmp    %rcx,%rdx
  jne    L5
  jmp    L102
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int carry = 0;
  int *res;
  int i;
  i = digitsSize - 1;
  while (i >= 0) {
    carry = (digits[i] + 1) / 10;
    digits[i] = (digits[i] + 1) % 10;
    if (carry == 0) {
      returnSize[0] = digitsSize;
      return digits;
    }
    i--;
  }
  returnSize[0] = digitsSize + 1;
  res = (int *)malloc(sizeof(int) * (digitsSize + 1));
  res[0] = 1;
  for (i = 0; i < digitsSize; i++) {
    res[i + 1] = digits[i];
  }
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<clearDecodeTable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L6:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1ffffff,(%rax)
  addl   $0x1,-0x4(%rbp)
L17:
  cmpl   $0xff,-0x4(%rbp)
  jle    L6
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void clearDecodeTable(uint32_t *ary) {
  int i = 0;
  for (i = 0; i < 256; ++i) {
    ary[i] = 0x01FFFFFF;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<clearDecodeTable>:
  endbr64
  lea    0x400(%rdi),%rax
L7:
  movl   $0x1ffffff,(%rdi)
  add    $0x4,%rdi
  cmp    %rax,%rdi
  jne    L7
  ret
```

assistant: ```c
void clearDecodeTable(uint32_t *ary) {
  int i = 0;
  for (i = 0; i < 256; ++i) {
    ary[i] = 0x01FFFFFF;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<clearDecodeTable>:
  endbr64
  lea    0x400(%rdi),%rax
  nopl   0x0(%rax,%rax,1)
L6:
  movl   $0x1ffffff,(%rdi)
  add    $0x4,%rdi
  cmp    %rax,%rdi
  jne    L6
  ret
```

assistant: ```c
void clearDecodeTable(uint32_t *ary) {
  int i = 0;
  for (i = 0; i < 256; ++i) {
    ary[i] = 0x01FFFFFF;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<clearDecodeTable>:
  endbr64
  movdqa D2(%rip),%xmm0
  lea    0x400(%rdi),%rax
  nopl   0x0(%rax,%rax,1)
L2:
  movups %xmm0,(%rdi)
  add    $0x10,%rdi
  cmp    %rdi,%rax
  jne    L2
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D2
type_guessed: byte[16]

tool: {"D2": [0xff,0xff,0xff,0x01,0xff,0xff,0xff,0x01,0xff,0xff,0xff,0x01,0xff,0xff,0xff,0x01]}

assistant: ```c
void clearDecodeTable(uint32_t *ary) {
  int i = 0;
  for (i = 0; i < 256; ++i) {
    ary[i] = 0x01FFFFFF;
  }
}
```