ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_time>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,(%rsp),%eax
  cltq
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long get_time() {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return 1000000 * tv.tv_sec + tv.tv_usec;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_time>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  imul   $0xf4240,(%rsp),%eax
  cltq
  add    0x8(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long get_time() {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  return 1000000 * tv.tv_sec + tv.tv_usec;
}
```
user: What is the c source code of the assembly code below:

```asm
<mmx_copy_page>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <in_interrupt@plt>
  test   %rax,%rax
  je     L12
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <slow_copy_page@plt>
  jmp    L7
L12:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fast_copy_page@plt>
L7:
  nop
  leave
  ret
```

assistant: ```c
void mmx_copy_page(void *to, void *from) {
  if (in_interrupt())
    slow_copy_page(to, from);
  else
    fast_copy_page(to, from);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mmx_copy_page>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    $0x0,%eax
  call   <in_interrupt@plt>
  test   %rax,%rax
  je     L16
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <slow_copy_page@plt>
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <fast_copy_page@plt>
  jmp    L1
```

assistant: ```c
void mmx_copy_page(void *to, void *from) {
  if (in_interrupt())
    slow_copy_page(to, from);
  else
    fast_copy_page(to, from);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mmx_copy_page>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <in_interrupt@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  test   %rax,%rax
  je     L9
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1090 <slow_copy_page@plt>
  nop
L9:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1080 <fast_copy_page@plt>
```

assistant: ```c
void mmx_copy_page(void *to, void *from) {
  if (in_interrupt())
    slow_copy_page(to, from);
  else
    fast_copy_page(to, from);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mmx_copy_page>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <in_interrupt@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  test   %rax,%rax
  je     L2
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1090 <slow_copy_page@plt>
  nop
L2:
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  jmp    1080 <fast_copy_page@plt>
```

assistant: ```c
void mmx_copy_page(void *to, void *from) {
  if (in_interrupt())
    slow_copy_page(to, from);
  else
    fast_copy_page(to, from);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dvector>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  add    $0x2,%rax
  shl    $0x3,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L9
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <nrerror@plt>
L9:
  mov    -0x18(%rbp),%rdx
  mov    $0x1,%eax
  sub    %rdx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D8
data_type: string

tool: D8: Error(dvector): allocation failure


assistant: ```c
double *dvector(long nl, long nh) {
  double *v;
  v = (double *)malloc(((nh - nl + 1 + 1) * sizeof(double)));
  if (!v)
    nrerror("Error(dvector): allocation failure\n");
  return v - nl + 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dvector>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  sub    %rdi,%rsi
  lea    0x10(,%rsi,8),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L9
L12:
  mov    $0x1,%eax
  sub    %rbp,%rax
  lea    (%rbx,%rax,8),%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  lea    D4(%rip),%rdi
  call   <nrerror@plt>
  jmp    L12
```

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

tool: D4: Error(dvector): allocation failure


assistant: ```c
double *dvector(long nl, long nh) {
  double *v;
  v = (double *)malloc(((nh - nl + 1 + 1) * sizeof(double)));
  if (!v)
    nrerror("Error(dvector): allocation failure\n");
  return v - nl + 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dvector>:
  endbr64
  push   %rbp
  sub    %rdi,%rsi
  mov    %rdi,%rbp
  push   %rbx
  lea    0x10(,%rsi,8),%rdi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L1
L17:
  mov    $0x1,%eax
  add    $0x8,%rsp
  sub    %rbp,%rax
  lea    (%rbx,%rax,8),%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  lea    D6(%rip),%rdi
  call   <nrerror@plt>
  jmp    L17
```

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

tool: D6: Error(dvector): allocation failure


assistant: ```c
double *dvector(long nl, long nh) {
  double *v;
  v = (double *)malloc(((nh - nl + 1 + 1) * sizeof(double)));
  if (!v)
    nrerror("Error(dvector): allocation failure\n");
  return v - nl + 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dvector>:
  endbr64
  push   %rbp
  sub    %rdi,%rsi
  mov    %rdi,%rbp
  push   %rbx
  lea    0x10(,%rsi,8),%rdi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L17
L0:
  mov    $0x1,%eax
  add    $0x8,%rsp
  sub    %rbp,%rax
  lea    (%rbx,%rax,8),%rax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  lea    D4(%rip),%rdi
  call   <nrerror@plt>
  jmp    L0
```

tool_calls: name: disassemble_data
arguments: tag: D4
type: string

tool: D4: Error(dvector): allocation failure


assistant: ```c
double *dvector(long nl, long nh) {
  double *v;
  v = (double *)malloc(((nh - nl + 1 + 1) * sizeof(double)));
  if (!v)
    nrerror("Error(dvector): allocation failure\n");
  return v - nl + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sd_get_clock_divider>:
  endbr64
  push   %rbx
  mov    %esi,%r10d
  cmp    %edi,%esi
  jg     L101
  mov    %edi,%eax
  cltd
  idiv   %esi
  cmp    $0x1,%edx
  adc    $0xffffffff,%eax
  test   %eax,%eax
  jns    L26
  mov    $0x1f,%ecx
  test   $0x7fffffff,%eax
  je     L129
  jmp    L52
L101:
  mov    $0x1,%eax
L26:
  mov    $0x1f,%ecx
  mov    $0x1,%esi
L37:
  sub    $0x1,%ecx
  cmp    $0xffffffff,%ecx
  je     L20
  mov    %esi,%edx
  shl    %cl,%edx
  test   %eax,%edx
  je     L37
  not    %edx
  test   %eax,%edx
  je     L38
L52:
  add    $0x1,%ecx
L129:
  cmp    $0xffffffff,%ecx
  je     L5
L38:
  cmp    $0x1f,%ecx
  jg     L16
  test   %ecx,%ecx
  jne    L1
  mov    %ecx,%ebx
  mov    $0x1,%esi
  jmp    L121
L20:
  mov    $0x1f,%ecx
L1:
  sub    $0x1,%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  mov    %eax,%ecx
  mov    $0x3ff,%eax
  cmp    %eax,%ecx
  cmovg  %eax,%ecx
  mov    %ecx,%ebx
  shr    $0x2,%ebx
  and    $0xc0,%ebx
  mov    %ecx,%eax
  shl    $0x8,%eax
  movzwl %ax,%eax
  or     %eax,%ebx
  lea    (%rcx,%rcx,1),%esi
L121:
  mov    %edi,%eax
  cltd
  idiv   %esi
  mov    %eax,%r8d
  mov    %ebx,%r9d
  mov    %r10d,%edx
  mov    %edi,%esi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%eax
  pop    %rbx
  ret
L5:
  mov    $0x1f,%ecx
  jmp    L1
L16:
  mov    $0x1f,%ecx
  jmp    L1
```

tool_calls: name: disassemble_data
arguments: label: D3
datatype: string

tool: D3: EMMC: base_clock: %i, target_rate: %i, divisor: %08x, actual_clock: %i, ret: %08x


assistant: ```c
uint32_t sd_get_clock_divider(uint32_t base_clock, uint32_t target_rate) {
  uint32_t targetted_divisor = 0;
  if (target_rate > base_clock)
    targetted_divisor = 1;
  else {
    targetted_divisor = base_clock / target_rate;
    uint32_t mod = base_clock % target_rate;
    if (mod)
      targetted_divisor--;
  }
  if (1) {
    int divisor = -1;
    for (int first_bit = 31; first_bit >= 0; first_bit--) {
      uint32_t bit_test = (1 << first_bit);
      if (targetted_divisor & bit_test) {
        divisor = first_bit;
        targetted_divisor &= ~bit_test;
        if (targetted_divisor) {
          divisor++;
        }
        break;
      }
    }
    if (divisor == -1)
      divisor = 31;
    if (divisor >= 32)
      divisor = 31;
    if (divisor != 0)
      divisor = (1 << (divisor - 1));
    if (divisor >= 0x400)
      divisor = 0x3ff;
    uint32_t freq_select = divisor & 0xff;
    uint32_t upper_bits = (divisor >> 8) & 0x3;
    uint32_t ret = (freq_select << 8) | (upper_bits << 6) | (0 << 5);
    int denominator = 1;
    if (divisor != 0)
      denominator = divisor * 2;
    int actual_clock = base_clock / denominator;
    printf("EMMC: base_clock: %i, target_rate: %i, divisor: %08x, "
           "actual_clock: %i, ret: %08x\n",
           base_clock, target_rate, divisor, actual_clock, ret);
    return ret;
  } else {
    printf("EMMC: unsupported host version\n");
    return 0xffffffff;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sd_get_clock_divider>:
  endbr64
  push   %r12
  mov    %edi,%r10d
  mov    %esi,%r11d
  mov    $0x1,%eax
  cmp    %edi,%esi
  jg     L1
  mov    %edi,%eax
  cltd
  idiv   %esi
  cmp    $0x1,%edx
  adc    $0xffffffff,%eax
L1:
  mov    $0x1f,%ecx
  mov    $0x1,%edi
  jmp    L33
  nopl   0x0(%rax)
L39:
  sub    $0x1,%ecx
  jb     L10
L33:
  mov    %edi,%edx
  shl    %cl,%edx
  test   %edx,%eax
  je     L39
  not    %edx
  test   %eax,%edx
  jne    L47
  test   %ecx,%ecx
  jne    L40
  mov    %r10d,%r8d
  xor    %r12d,%r12d
  jmp    L65
  nop
L47:
  add    $0x1,%ecx
  cmp    $0x20,%ecx
  jne    L40
L10:
  mov    $0xffc0,%r12d
  mov    $0x7fe,%esi
  mov    $0x3ff,%ecx
L69:
  mov    %r10d,%eax
  cltd
  idiv   %esi
  mov    %eax,%r8d
L65:
  mov    %r12d,%r9d
  mov    %r11d,%edx
  mov    %r10d,%esi
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
L40:
  sub    $0x1,%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  mov    %eax,%ecx
  mov    $0x3ff,%eax
  cmp    %eax,%ecx
  cmovg  %eax,%ecx
  mov    %ecx,%r12d
  mov    %ecx,%eax
  lea    (%rcx,%rcx,1),%esi
  sar    $0x8,%r12d
  shl    $0x8,%eax
  shl    $0x6,%r12d
  movzwl %ax,%eax
  or     %eax,%r12d
  jmp    L69
```

tool_calls: name: read_data
arguments: tag: D6
type_infered: string

tool: D6: EMMC: base_clock: %i, target_rate: %i, divisor: %08x, actual_clock: %i, ret: %08x


assistant: ```c
uint32_t sd_get_clock_divider(uint32_t base_clock, uint32_t target_rate) {
  uint32_t targetted_divisor = 0;
  if (target_rate > base_clock)
    targetted_divisor = 1;
  else {
    targetted_divisor = base_clock / target_rate;
    uint32_t mod = base_clock % target_rate;
    if (mod)
      targetted_divisor--;
  }
  if (1) {
    int divisor = -1;
    for (int first_bit = 31; first_bit >= 0; first_bit--) {
      uint32_t bit_test = (1 << first_bit);
      if (targetted_divisor & bit_test) {
        divisor = first_bit;
        targetted_divisor &= ~bit_test;
        if (targetted_divisor) {
          divisor++;
        }
        break;
      }
    }
    if (divisor == -1)
      divisor = 31;
    if (divisor >= 32)
      divisor = 31;
    if (divisor != 0)
      divisor = (1 << (divisor - 1));
    if (divisor >= 0x400)
      divisor = 0x3ff;
    uint32_t freq_select = divisor & 0xff;
    uint32_t upper_bits = (divisor >> 8) & 0x3;
    uint32_t ret = (freq_select << 8) | (upper_bits << 6) | (0 << 5);
    int denominator = 1;
    if (divisor != 0)
      denominator = divisor * 2;
    int actual_clock = base_clock / denominator;
    printf("EMMC: base_clock: %i, target_rate: %i, divisor: %08x, "
           "actual_clock: %i, ret: %08x\n",
           base_clock, target_rate, divisor, actual_clock, ret);
    return ret;
  } else {
    printf("EMMC: unsupported host version\n");
    return 0xffffffff;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sd_get_clock_divider>:
  endbr64
  push   %r12
  mov    %edi,%r10d
  mov    %esi,%r11d
  mov    $0x1,%eax
  cmp    %edi,%esi
  jg     L45
  mov    %edi,%eax
  cltd
  idiv   %esi
  cmp    $0x1,%edx
  adc    $0xffffffff,%eax
L45:
  mov    $0x1f,%ecx
  mov    $0x1,%edi
  jmp    L62
  nopl   0x0(%rax)
L31:
  sub    $0x1,%ecx
  jb     L54
L62:
  mov    %edi,%edx
  shl    %cl,%edx
  test   %edx,%eax
  je     L31
  not    %edx
  test   %eax,%edx
  jne    L15
  test   %ecx,%ecx
  jne    L59
  mov    %r10d,%r8d
  xor    %r12d,%r12d
  jmp    L40
  nop
L15:
  add    $0x1,%ecx
  cmp    $0x20,%ecx
  jne    L59
L54:
  mov    $0xffc0,%r12d
  mov    $0x7fe,%esi
  mov    $0x3ff,%ecx
L58:
  mov    %r10d,%eax
  cltd
  idiv   %esi
  mov    %eax,%r8d
L40:
  mov    %r12d,%r9d
  mov    %r11d,%edx
  mov    %r10d,%esi
  xor    %eax,%eax
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
L59:
  sub    $0x1,%ecx
  mov    $0x1,%eax
  shl    %cl,%eax
  mov    %eax,%ecx
  mov    $0x3ff,%eax
  cmp    %eax,%ecx
  cmovg  %eax,%ecx
  mov    %ecx,%r12d
  mov    %ecx,%eax
  lea    (%rcx,%rcx,1),%esi
  shl    $0x8,%r12d
  sar    $0x8,%eax
  movzwl %r12w,%r12d
  shl    $0x6,%eax
  or     %eax,%r12d
  jmp    L58
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type_guessed: string

tool: D9: EMMC: base_clock: %i, target_rate: %i, divisor: %08x, actual_clock: %i, ret: %08x


assistant: ```c
uint32_t sd_get_clock_divider(uint32_t base_clock, uint32_t target_rate) {
  uint32_t targetted_divisor = 0;
  if (target_rate > base_clock)
    targetted_divisor = 1;
  else {
    targetted_divisor = base_clock / target_rate;
    uint32_t mod = base_clock % target_rate;
    if (mod)
      targetted_divisor--;
  }
  if (1) {
    int divisor = -1;
    for (int first_bit = 31; first_bit >= 0; first_bit--) {
      uint32_t bit_test = (1 << first_bit);
      if (targetted_divisor & bit_test) {
        divisor = first_bit;
        targetted_divisor &= ~bit_test;
        if (targetted_divisor) {
          divisor++;
        }
        break;
      }
    }
    if (divisor == -1)
      divisor = 31;
    if (divisor >= 32)
      divisor = 31;
    if (divisor != 0)
      divisor = (1 << (divisor - 1));
    if (divisor >= 0x400)
      divisor = 0x3ff;
    uint32_t freq_select = divisor & 0xff;
    uint32_t upper_bits = (divisor >> 8) & 0x3;
    uint32_t ret = (freq_select << 8) | (upper_bits << 6) | (0 << 5);
    int denominator = 1;
    if (divisor != 0)
      denominator = divisor * 2;
    int actual_clock = base_clock / denominator;
    printf("EMMC: base_clock: %i, target_rate: %i, divisor: %08x, "
           "actual_clock: %i, ret: %08x\n",
           base_clock, target_rate, divisor, actual_clock, ret);
    return ret;
  } else {
    printf("EMMC: unsupported host version\n");
    return 0xffffffff;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hexDigit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  js     L24
  cmpl   $0x9,-0x4(%rbp)
  jg     L24
  mov    -0x4(%rbp),%eax
  add    $0x30,%eax
  jmp    L9
L24:
  mov    -0x4(%rbp),%eax
  add    $0x37,%eax
L9:
  pop    %rbp
  ret
```

assistant: ```c
char hexDigit(uint8_t d) {
  if (d >= 0 && d <= 9)
    return d + '0';
  else
    return d + 'A' - 10;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hexDigit>:
  endbr64
  lea    0x30(%rdi),%edx
  lea    0x37(%rdi),%eax
  cmp    $0x9,%edi
  cmovbe %edx,%eax
  ret
```

assistant: ```c
char hexDigit(uint8_t d) {
  if (d >= 0 && d <= 9)
    return d + '0';
  else
    return d + 'A' - 10;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<hexDigit>:
  endbr64
  lea    0x30(%rdi),%edx
  lea    0x37(%rdi),%eax
  cmp    $0x9,%edi
  cmovbe %edx,%eax
  ret
```

assistant: ```c
char hexDigit(uint8_t d) {
  if (d >= 0 && d <= 9)
    return d + '0';
  else
    return d + 'A' - 10;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<hexDigit>:
  endbr64
  lea    0x30(%rdi),%edx
  lea    0x37(%rdi),%eax
  cmp    $0x9,%edi
  cmovbe %edx,%eax
  ret
```

assistant: ```c
char hexDigit(uint8_t d) {
  if (d >= 0 && d <= 9)
    return d + '0';
  else
    return d + 'A' - 10;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<XTeaEncrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  mov    %eax,-0x24(%rbp)
  movl   $0x0,-0x20(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L8
L16:
  mov    -0x24(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  sar    $0x5,%eax
  xor    %eax,%edx
  mov    -0x24(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x20(%rbp),%eax
  cltq
  and    $0x3,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  xor    %ecx,%eax
  add    %eax,-0x28(%rbp)
  mov    -0x20(%rbp),%eax
  sub    $0x61c88647,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x28(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%eax
  sar    $0x5,%eax
  xor    %eax,%edx
  mov    -0x28(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x20(%rbp),%eax
  sar    $0xb,%eax
  cltq
  and    $0x3,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  xor    %ecx,%eax
  add    %eax,-0x24(%rbp)
  addl   $0x1,-0x1c(%rbp)
L8:
  cmpl   $0x1f,-0x1c(%rbp)
  jbe    L16
  mov    -0x8(%rbp),%rax
  mov    -0x28(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0x24(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void XTeaEncrypt(const unsigned char src[8], unsigned char dst[8], const unsigned char ckey[16]) {
  const uint32_t *s = (const uint32_t *)src;
  uint32_t s0 = s[0];
  uint32_t s1 = s[1];
  uint32_t sum = 0;
  uint32_t *key = (uint32_t *)ckey;
  uint32_t *d = (uint32_t *)dst;
  unsigned int i;
  for (i = 0; i < 32; i++) {
    s0 += (((s1 << 4) ^ (s1 >> 5)) + s1) ^ (sum + key[sum & 3]);
    sum += 0x9E3779B9;
    s1 += (((s0 << 4) ^ (s0 >> 5)) + s0) ^ (sum + key[(sum >> 11) & 3]);
  }
  d[0] = s0;
  d[1] = s1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<XTeaEncrypt>:
  endbr64
  mov    %rsi,%r8
  mov    %rdx,%r9
  mov    (%rdi),%esi
  mov    0x4(%rdi),%ecx
  mov    $0x0,%eax
L5:
  mov    %eax,%edx
  and    $0x3,%edx
  mov    %eax,%edi
  add    (%r9,%rdx,4),%edi
  mov    %ecx,%edx
  shl    $0x4,%edx
  mov    %ecx,%r10d
  sar    $0x5,%r10d
  xor    %r10d,%edx
  add    %ecx,%edx
  xor    %edi,%edx
  add    %edx,%esi
  sub    $0x61c88647,%eax
  mov    %eax,%edx
  sar    $0xb,%edx
  and    $0x3,%edx
  mov    %eax,%edi
  add    (%r9,%rdx,4),%edi
  mov    %esi,%edx
  shl    $0x4,%edx
  mov    %esi,%r10d
  sar    $0x5,%r10d
  xor    %r10d,%edx
  add    %esi,%edx
  xor    %edi,%edx
  add    %edx,%ecx
  cmp    $0xc6ef3720,%eax
  jne    L5
  mov    %esi,(%r8)
  mov    %ecx,0x4(%r8)
  ret
```

assistant: ```c
void XTeaEncrypt(const unsigned char src[8], unsigned char dst[8], const unsigned char ckey[16]) {
  const uint32_t *s = (const uint32_t *)src;
  uint32_t s0 = s[0];
  uint32_t s1 = s[1];
  uint32_t sum = 0;
  uint32_t *key = (uint32_t *)ckey;
  uint32_t *d = (uint32_t *)dst;
  unsigned int i;
  for (i = 0; i < 32; i++) {
    s0 += (((s1 << 4) ^ (s1 >> 5)) + s1) ^ (sum + key[sum & 3]);
    sum += 0x9E3779B9;
    s1 += (((s0 << 4) ^ (s0 >> 5)) + s0) ^ (sum + key[(sum >> 11) & 3]);
  }
  d[0] = s0;
  d[1] = s1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<XTeaEncrypt>:
  endbr64
  mov    %rsi,%r8
  mov    0x4(%rdi),%ecx
  mov    (%rdi),%esi
  mov    %rdx,%r9
  xor    %eax,%eax
  nopl   0x0(%rax)
L3:
  mov    %eax,%edx
  mov    %ecx,%r10d
  and    $0x3,%edx
  sar    $0x5,%r10d
  mov    (%r9,%rdx,4),%edi
  mov    %ecx,%edx
  shl    $0x4,%edx
  xor    %r10d,%edx
  add    %eax,%edi
  sub    $0x61c88647,%eax
  add    %ecx,%edx
  xor    %edi,%edx
  add    %edx,%esi
  mov    %eax,%edx
  sar    $0xb,%edx
  mov    %esi,%r10d
  and    $0x3,%edx
  sar    $0x5,%r10d
  mov    (%r9,%rdx,4),%edi
  mov    %esi,%edx
  shl    $0x4,%edx
  xor    %r10d,%edx
  add    %eax,%edi
  add    %esi,%edx
  xor    %edi,%edx
  add    %edx,%ecx
  cmp    $0xc6ef3720,%eax
  jne    L3
  mov    %esi,(%r8)
  mov    %ecx,0x4(%r8)
  ret
```

assistant: ```c
void XTeaEncrypt(const unsigned char src[8], unsigned char dst[8], const unsigned char ckey[16]) {
  const uint32_t *s = (const uint32_t *)src;
  uint32_t s0 = s[0];
  uint32_t s1 = s[1];
  uint32_t sum = 0;
  uint32_t *key = (uint32_t *)ckey;
  uint32_t *d = (uint32_t *)dst;
  unsigned int i;
  for (i = 0; i < 32; i++) {
    s0 += (((s1 << 4) ^ (s1 >> 5)) + s1) ^ (sum + key[sum & 3]);
    sum += 0x9E3779B9;
    s1 += (((s0 << 4) ^ (s0 >> 5)) + s0) ^ (sum + key[(sum >> 11) & 3]);
  }
  d[0] = s0;
  d[1] = s1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<XTeaEncrypt>:
  endbr64
  mov    %rsi,%r8
  mov    0x4(%rdi),%ecx
  mov    (%rdi),%esi
  mov    %rdx,%r9
  xor    %eax,%eax
  nopl   0x0(%rax)
L9:
  mov    %eax,%edx
  mov    %ecx,%r10d
  and    $0x3,%edx
  sar    $0x5,%r10d
  mov    (%r9,%rdx,4),%edi
  mov    %ecx,%edx
  shl    $0x4,%edx
  xor    %r10d,%edx
  add    %eax,%edi
  sub    $0x61c88647,%eax
  add    %ecx,%edx
  xor    %edi,%edx
  add    %edx,%esi
  mov    %eax,%edx
  sar    $0xb,%edx
  mov    %esi,%r10d
  and    $0x3,%edx
  sar    $0x5,%r10d
  mov    (%r9,%rdx,4),%edi
  mov    %esi,%edx
  shl    $0x4,%edx
  xor    %r10d,%edx
  add    %eax,%edi
  add    %esi,%edx
  xor    %edi,%edx
  add    %edx,%ecx
  cmp    $0xc6ef3720,%eax
  jne    L9
  mov    %esi,(%r8)
  mov    %ecx,0x4(%r8)
  ret
```

assistant: ```c
void XTeaEncrypt(const unsigned char src[8], unsigned char dst[8], const unsigned char ckey[16]) {
  const uint32_t *s = (const uint32_t *)src;
  uint32_t s0 = s[0];
  uint32_t s1 = s[1];
  uint32_t sum = 0;
  uint32_t *key = (uint32_t *)ckey;
  uint32_t *d = (uint32_t *)dst;
  unsigned int i;
  for (i = 0; i < 32; i++) {
    s0 += (((s1 << 4) ^ (s1 >> 5)) + s1) ^ (sum + key[sum & 3]);
    sum += 0x9E3779B9;
    s1 += (((s0 << 4) ^ (s0 >> 5)) + s0) ^ (sum + key[(sum >> 11) & 3]);
  }
  d[0] = s0;
  d[1] = s1;
}
```