ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<pevent_get_field_val>:
  endbr64
  test   %rsi,%rsi
  je     L3
  push   %r14
  mov    %r9d,%r14d
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  mov    %rdx,%rsi
  push   %rbx
  mov    %r8,%rbx
  call   <pevent_find_field@plt>
  mov    %r14d,%r9d
  mov    %rbx,%r8
  mov    %r13,%rcx
  pop    %rbx
  mov    %r12,%rdx
  mov    %rbp,%rdi
  mov    %rax,%rsi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <get_field_val@plt>
L3:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int pevent_get_field_val(struct trace_seq *s, struct event_format *event, const char *name, struct record *record, unsigned long long *val, int err) {
  struct format_field *field;
  if (!event)
    return -1;
  field = pevent_find_field(event, name);
  return get_field_val(s, field, name, record, val, err);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<skip_space>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L15
L29:
  addq   $0x1,-0x8(%rbp)
L15:
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jae    L11
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L29
L11:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *skip_space(char *p, char *e) {
  while (p < e && isspace(*p))
    p++;
  return p;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<skip_space>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %rdi,%rax
  cmp    %rsi,%rdi
  jae    L9
L24:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L1
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  jne    L24
  mov    %rbp,%rax
  jmp    L9
L1:
  mov    %rbx,%rax
L9:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *skip_space(char *p, char *e) {
  while (p < e && isspace(*p))
    p++;
  return p;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<skip_space>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cmp    %rsi,%rdi
  jb     L26
  jmp    L20
  nopw   0x0(%rax,%rax,1)
L24:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  je     L13
L26:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L24
L20:
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
L13:
  add    $0x8,%rsp
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *skip_space(char *p, char *e) {
  while (p < e && isspace(*p))
    p++;
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<skip_space>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cmp    %rsi,%rdi
  jb     L35
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L23:
  add    $0x1,%rbx
  cmp    %rbx,%rbp
  je     L13
L35:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L23
L1:
  add    $0x8,%rsp
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  ret
L13:
  add    $0x8,%rsp
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *skip_space(char *p, char *e) {
  while (p < e && isspace(*p))
    p++;
  return p;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<set_speed>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x14(%rbp),%ecx
  lea    -0xe(%rbp),%rax
  sub    $0x8,%rsp
  mov    -0x18(%rbp),%edx
  push   %rdx
  mov    %ecx,%r9d
  mov    $0x1,%r8d
  mov    $0x1,%ecx
  mov    $0x80,%edx
  lea    D15(%rip),%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  add    $0x10,%rsp
  lea    -0xe(%rbp),%rax
  mov    $0x5,%edx
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <write_tty@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D15
data_type: string
name: parse_data
arguments: data_label: D2
data_type: string

tool: D15: %c%c%c%c%c

tool: D2: /dev/ttyO1

assistant: ```c
void set_speed(int servo, int speed) {
  char cmd[6];
  sprintf(cmd, "%c%c%c%c%c", 0x80, 0x01, 0x01, servo, speed);
  write_tty("/dev/ttyO1", cmd, 5);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<set_speed>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%r9d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x2(%rsp),%rbx
  sub    $0x8,%rsp
  push   %rsi
  mov    $0x1,%r8d
  mov    $0x1,%ecx
  mov    $0x80,%edx
  lea    D17(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    $0x5,%edx
  mov    %rbx,%rsi
  lea    D2(%rip),%rdi
  call   <write_tty@plt>
  add    $0x10,%rsp
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x10,%rsp
  pop    %rbx
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D17
datatype: string
name: unpack_data
arguments: data_label: D2
datatype: string

tool: D17: %c%c%c%c%c

tool: D2: /dev/ttyO1

assistant: ```c
void set_speed(int servo, int speed) {
  char cmd[6];
  sprintf(cmd, "%c%c%c%c%c", 0x80, 0x01, 0x01, servo, speed);
  write_tty("/dev/ttyO1", cmd, 5);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<set_speed>:
  endbr64
  push   %rbp
  mov    %edi,%r9d
  mov    $0x1,%r8d
  mov    $0x1,%ecx
  mov    $0x80,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x2(%rsp),%rbp
  sub    $0x8,%rsp
  push   %rsi
  mov    %rbp,%rdi
  lea    D9(%rip),%rsi
  call   <sprintf@plt>
  mov    $0x5,%edx
  mov    %rbp,%rsi
  lea    D12(%rip),%rdi
  call   <write_tty@plt>
  pop    %rax
  pop    %rdx
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x10,%rsp
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D9
type_infered: string
name: parse_data
arguments: data_tag: D12
type_infered: string

tool: D9: %c%c%c%c%c

tool: D12: /dev/ttyO1

assistant: ```c
void set_speed(int servo, int speed) {
  char cmd[6];
  sprintf(cmd, "%c%c%c%c%c", 0x80, 0x01, 0x01, servo, speed);
  write_tty("/dev/ttyO1", cmd, 5);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<set_speed>:
  endbr64
  push   %rbp
  mov    %edi,%r9d
  mov    $0x1,%r8d
  mov    $0x1,%ecx
  mov    $0x80,%edx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x2(%rsp),%rbp
  sub    $0x8,%rsp
  push   %rsi
  mov    %rbp,%rdi
  lea    D8(%rip),%rsi
  call   <sprintf@plt>
  mov    $0x5,%edx
  mov    %rbp,%rsi
  lea    D9(%rip),%rdi
  call   <write_tty@plt>
  pop    %rax
  pop    %rdx
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D8
data_type: string
name: disassemble_data
arguments: tag: D9
data_type: string

tool: D8: %c%c%c%c%c

tool: D9: /dev/ttyO1

assistant: ```c
void set_speed(int servo, int speed) {
  char cmd[6];
  sprintf(cmd, "%c%c%c%c%c", 0x80, 0x01, 0x01, servo, speed);
  write_tty("/dev/ttyO1", cmd, 5);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<arrangeCoins>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jg     L31
  mov    $0x0,%eax
  jmp    L16
L31:
  fildl  -0x24(%rbp)
  fldt   0xebf(%rip)        # 2000 <_fini+0xe3c>
  faddp  %st,%st(1)
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  call   <sqrt@plt>
  add    $0x10,%rsp
  fstpt  -0x40(%rbp)
  fldt   0xeb3(%rip)        # 2010 <_fini+0xe4c>
  lea    -0x10(%rsp),%rsp
  fstpt  (%rsp)
  call   <sqrt@plt>
  add    $0x10,%rsp
  fldt   -0x40(%rbp)
  fmulp  %st,%st(1)
  fldt   0xea7(%rip)        # 2020 <_fini+0xe5c>
  fsubrp %st,%st(1)
  fstpt  -0x10(%rbp)
  fldt   -0x10(%rbp)
  fnstcw -0x26(%rbp)
  movzwl -0x26(%rbp),%eax
  or     $0xc,%ah
  mov    %ax,-0x28(%rbp)
  fldcw  -0x28(%rbp)
  fistpl -0x14(%rbp)
  fldcw  -0x26(%rbp)
  fildl  -0x14(%rbp)
  fldt   -0x10(%rbp)
  fucomip %st(1),%st
  jp     L10
  fldt   -0x10(%rbp)
  fucomip %st(1),%st
  fstp   %st(0)
  je     L49
  jmp    L40
L10:
  fstp   %st(0)
L40:
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  jmp    L28
L49:
  mov    -0x14(%rbp),%eax
L28:
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
L16:
  leave
  ret
```

assistant: ```c
int arrangeCoins(int n) {
  if (n <= 0)
    return 0;
  long double f = sqrt((long double)n + 0.125) * sqrt((long double)2.0) - (long double)1.5;
  int i = (int)f;
  i = f == (long double)i ? i : (i + 1);
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<arrangeCoins>:
  endbr64
  mov    $0x0,%eax
  test   %edi,%edi
  jle    L25
  sub    $0x48,%rsp
  mov    %edi,0x10(%rsp)
  fildl  0x10(%rsp)
  fadds  0xec4(%rip)        # 2000 <_fini+0xe54>
  fstpt  (%rsp)
  call   <sqrt@plt>
  fstpt  0x10(%rsp)
  flds   0xeb6(%rip)        # 2004 <_fini+0xe58>
  fstpt  0x20(%rsp)
  mov    0x20(%rsp),%rax
  mov    0x28(%rsp),%edx
  push   %rdx
  push   %rax
  call   <sqrt@plt>
  fldt   0x20(%rsp)
  fmulp  %st,%st(1)
  fsubs  0xe9a(%rip)        # 2008 <_fini+0xe5c>
  fnstcw 0x4e(%rsp)
  movzwl 0x4e(%rsp),%eax
  or     $0xc,%ah
  mov    %ax,0x4c(%rsp)
  fld    %st(0)
  fldcw  0x4c(%rsp)
  fistpl 0x20(%rsp)
  fldcw  0x4e(%rsp)
  mov    0x20(%rsp),%eax
  fildl  0x20(%rsp)
  add    $0x20,%rsp
  fucomip %st(1),%st
  fstp   %st(0)
  jp     L15
  jne    L15
L23:
  add    $0x38,%rsp
  ret
L15:
  add    $0x1,%eax
  jmp    L23
L25:
  ret
```

assistant: ```c
int arrangeCoins(int n) {
  if (n <= 0)
    return 0;
  long double f = sqrt((long double)n + 0.125) * sqrt((long double)2.0) - (long double)1.5;
  int i = (int)f;
  i = f == (long double)i ? i : (i + 1);
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<arrangeCoins>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  jle    L23
  sub    $0x48,%rsp
  mov    %edi,0x10(%rsp)
  fildl  0x10(%rsp)
  fadds  0xec0(%rip)        # 2000 <_fini+0xe3c>
  fstpt  (%rsp)
  call   <sqrt@plt>
  fstpt  0x20(%rsp)
  flds   0xeb2(%rip)        # 2004 <_fini+0xe40>
  fstpt  0x10(%rsp)
  mov    0x10(%rsp),%rax
  mov    0x18(%rsp),%edx
  push   %rdx
  push   %rax
  call   <sqrt@plt>
  fldt   0x30(%rsp)
  fnstcw 0x4e(%rsp)
  movzwl 0x4e(%rsp),%eax
  fmulp  %st,%st(1)
  fsubs  0xe8d(%rip)        # 2008 <_fini+0xe44>
  or     $0xc,%ah
  mov    %ax,0x4c(%rsp)
  fld    %st(0)
  fldcw  0x4c(%rsp)
  fistpl 0x20(%rsp)
  fldcw  0x4e(%rsp)
  fildl  0x20(%rsp)
  mov    0x20(%rsp),%eax
  add    $0x20,%rsp
  fucomip %st(1),%st
  fstp   %st(0)
  jp     L0
  jne    L0
  add    $0x38,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  add    $0x1,%eax
  add    $0x38,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L23:
  ret
```

assistant: ```c
int arrangeCoins(int n) {
  if (n <= 0)
    return 0;
  long double f = sqrt((long double)n + 0.125) * sqrt((long double)2.0) - (long double)1.5;
  int i = (int)f;
  i = f == (long double)i ? i : (i + 1);
  return i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<arrangeCoins>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  jle    L1
  sub    $0x48,%rsp
  mov    %edi,0x10(%rsp)
  fildl  0x10(%rsp)
  fadds  0xec0(%rip)        # 2000 <_fini+0xe3c>
  fstpt  (%rsp)
  call   <sqrt@plt>
  fstpt  0x20(%rsp)
  flds   0xeb2(%rip)        # 2004 <_fini+0xe40>
  fstpt  0x10(%rsp)
  mov    0x10(%rsp),%rax
  mov    0x18(%rsp),%edx
  push   %rdx
  push   %rax
  call   <sqrt@plt>
  fldt   0x30(%rsp)
  fnstcw 0x4e(%rsp)
  movzwl 0x4e(%rsp),%eax
  fmulp  %st,%st(1)
  fsubs  0xe8d(%rip)        # 2008 <_fini+0xe44>
  or     $0xc,%ah
  mov    %ax,0x4c(%rsp)
  fld    %st(0)
  fldcw  0x4c(%rsp)
  fistpl 0x20(%rsp)
  fldcw  0x4e(%rsp)
  fildl  0x20(%rsp)
  mov    0x20(%rsp),%eax
  add    $0x20,%rsp
  fucomip %st(1),%st
  fstp   %st(0)
  jp     L29
  jne    L29
  add    $0x38,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L29:
  add    $0x1,%eax
  add    $0x38,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  ret
```

assistant: ```c
int arrangeCoins(int n) {
  if (n <= 0)
    return 0;
  long double f = sqrt((long double)n + 0.125) * sqrt((long double)2.0) - (long double)1.5;
  int i = (int)f;
  i = f == (long double)i ? i : (i + 1);
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  cltd
  idivl  -0x8(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L14
  mov    -0xc(%rbp),%eax
  jmp    L15
L14:
  mov    $0x0,%eax
L15:
  pop    %rbp
  ret
```

assistant: ```c
int sum(int k, int blockSize, int distance) {
  if ((k + distance + 1) % blockSize == 0)
    return distance;
  else
    return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum>:
  endbr64
  mov    %edx,%ecx
  lea    0x1(%rdx,%rdi,1),%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  mov    $0x0,%eax
  cmovne %eax,%ecx
  mov    %ecx,%eax
  ret
```

assistant: ```c
int sum(int k, int blockSize, int distance) {
  if ((k + distance + 1) % blockSize == 0)
    return distance;
  else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<sum>:
  endbr64
  lea    0x1(%rdx,%rdi,1),%eax
  mov    %edx,%r8d
  cltd
  idiv   %esi
  xor    %eax,%eax
  test   %edx,%edx
  cmovne %eax,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int sum(int k, int blockSize, int distance) {
  if ((k + distance + 1) % blockSize == 0)
    return distance;
  else
    return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sum>:
  endbr64
  lea    0x1(%rdx,%rdi,1),%eax
  mov    %edx,%r8d
  cltd
  idiv   %esi
  xor    %eax,%eax
  test   %edx,%edx
  cmovne %eax,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int sum(int k, int blockSize, int distance) {
  if ((k + distance + 1) % blockSize == 0)
    return distance;
  else
    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
<test_hello2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movq   $0x0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x2a,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void test_hello2() {
  int *s = NULL;
  *s = 42;
}
```
user: What is the c source code of the assembly code below:

```asm
<test_hello2>:
  endbr64
  movl   $0x2a,0x0
  ret
```

assistant: ```c
void test_hello2() {
  int *s = NULL;
  *s = 42;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test_hello2>:
  endbr64
  movl   $0x0,0x0
  ud2
```

assistant: ```c
void test_hello2() {
  int *s = NULL;
  *s = 42;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_hello2>:
  endbr64
  movl   $0x0,0x0
  ud2
```

assistant: ```c
void test_hello2() {
  int *s = NULL;
  *s = 42;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<rab>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0xc,%rax
  movl   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
rab(char *pt, int *pti) {
  pt[2] = 0;
  pti[3] = 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<rab>:
  endbr64
  movb   $0x0,0x2(%rdi)
  movl   $0x0,0xc(%rsi)
  ret
```

assistant: ```c
rab(char *pt, int *pti) {
  pt[2] = 0;
  pti[3] = 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<rab>:
  endbr64
  movb   $0x0,0x2(%rdi)
  movl   $0x0,0xc(%rsi)
  ret
```

assistant: ```c
rab(char *pt, int *pti) {
  pt[2] = 0;
  pti[3] = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<atohx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L54
L81:
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x1c(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  cmp    %rax,%rdx
  jbe    L129
  mov    -0x30(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jne    L59
L129:
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L73
  mov    $0x30,%eax
  jmp    L72
L73:
  mov    $0x57,%eax
L72:
  sub    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  je     L40
  mov    $0x30,%eax
  jmp    L83
L40:
  mov    $0x57,%eax
L83:
  sub    %eax,-0x18(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  js     L84
  cmpl   $0xf,-0x14(%rbp)
  jg     L84
  cmpl   $0x0,-0x18(%rbp)
  js     L84
  cmpl   $0xf,-0x18(%rbp)
  jle    L123
L84:
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
  jmp    L126
L123:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    -0x14(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%ecx
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  or     %ecx,%esi
  mov    %esi,%edx
  mov    %dl,(%rax)
  mov    -0x34(%rbp),%eax
  add    $0x2,%eax
  mov    %eax,%eax
  add    %rax,-0x30(%rbp)
L54:
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L81
  jmp    L98
L59:
  nop
L98:
  mov    -0x28(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
L126:
  leave
  ret
```

assistant: ```c
char *atohx(char *dst, const char *src, const unsigned int spacing) {
  char *ret = dst;
  const char *srcPtr = src;
  const unsigned int length = strlen(srcPtr);
  int lsb, msb;
  for (lsb = 0, msb = 0; *src; src += 2 + spacing) {
    if ((src + 1 + spacing) > srcPtr + length && src - srcPtr) {
      break;
    }
    msb = tolower(*src);
    lsb = tolower(*(src + 1));
    msb -= isdigit(msb) ? 0x30 : 0x57;
    lsb -= isdigit(lsb) ? 0x30 : 0x57;
    if ((msb < 0x0 || msb > 0xf) || (lsb < 0x0 || lsb > 0xf)) {
      *ret = 0;
      return NULL;
    }
    *dst++ = (char)(lsb | (msb << 4));
  }
  *dst = 0;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<atohx>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%r13
  mov    %rdi,0x18(%rsp)
  mov    %rsi,%r15
  mov    %edx,%ebx
  mov    %rsi,%rdi
  call   <strlen@plt>
  movzbl (%r15),%edi
  test   %dil,%dil
  je     L53
  mov    %ebx,%edx
  lea    0x1(%rdx),%rcx
  mov    %rcx,0x10(%rsp)
  mov    %eax,%eax
  lea    (%r15,%rax,1),%r14
  mov    %r15,%r12
  lea    0x2(%rbx),%eax
  mov    %eax,0xc(%rsp)
L63:
  movsbl %dil,%edi
  call   <tolower@plt>
  mov    %eax,%ebx
  movsbl 0x1(%r12),%edi
  call   <tolower@plt>
  mov    %eax,%ebp
  mov    %ebx,%edi
  call   <isdigit@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0x27,%eax
  add    $0x30,%eax
  sub    %eax,%ebx
  mov    %ebp,%edi
  call   <isdigit@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0x27,%eax
  add    $0x30,%eax
  sub    %eax,%ebp
  mov    %ebx,%eax
  or     %ebp,%eax
  cmp    $0xf,%eax
  ja     L31
  add    $0x1,%r13
  shl    $0x4,%ebx
  or     %ebp,%ebx
  mov    %bl,-0x1(%r13)
  mov    0xc(%rsp),%eax
  add    %rax,%r12
  movzbl (%r12),%edi
  test   %dil,%dil
  je     L10
  mov    0x10(%rsp),%rax
  add    %r12,%rax
  cmp    %rax,%r14
  jae    L63
  cmp    %r12,%r15
  je     L63
L10:
  movb   $0x0,0x0(%r13)
  mov    0x18(%rsp),%rax
L59:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L31:
  mov    0x18(%rsp),%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
  jmp    L59
L53:
  mov    0x18(%rsp),%r13
  jmp    L10
```

assistant: ```c
char *atohx(char *dst, const char *src, const unsigned int spacing) {
  char *ret = dst;
  const char *srcPtr = src;
  const unsigned int length = strlen(srcPtr);
  int lsb, msb;
  for (lsb = 0, msb = 0; *src; src += 2 + spacing) {
    if ((src + 1 + spacing) > srcPtr + length && src - srcPtr) {
      break;
    }
    msb = tolower(*src);
    lsb = tolower(*(src + 1));
    msb -= isdigit(msb) ? 0x30 : 0x57;
    lsb -= isdigit(lsb) ? 0x30 : 0x57;
    if ((msb < 0x0 || msb > 0xf) || (lsb < 0x0 || lsb > 0xf)) {
      *ret = 0;
      return NULL;
    }
    *dst++ = (char)(lsb | (msb << 4));
  }
  *dst = 0;
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<atohx>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    %rdi,0x18(%rsp)
  mov    %rsi,%rdi
  mov    %rsi,0x10(%rsp)
  call   <strlen@plt>
  movsbl (%rbx),%edi
  test   %dil,%dil
  je     L51
  mov    %eax,%eax
  mov    %r13d,%r14d
  mov    %rbx,%r15
  add    $0x2,%r13d
  add    %rbx,%rax
  add    $0x1,%r14
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax)
L0:
  call   <tolower@plt>
  movsbl 0x1(%r15),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  mov    %ebx,%edi
  mov    %eax,%ebp
  call   <isdigit@plt>
  mov    %ebp,%edi
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0x27,%eax
  add    $0x30,%eax
  sub    %eax,%ebx
  call   <isdigit@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0x27,%eax
  add    $0x30,%eax
  sub    %eax,%ebp
  mov    %ebx,%eax
  or     %ebp,%eax
  cmp    $0xf,%eax
  ja     L1
  shl    $0x4,%ebx
  mov    %r13d,%eax
  add    $0x1,%r12
  or     %ebp,%ebx
  add    %rax,%r15
  mov    %bl,-0x1(%r12)
  movsbl (%r15),%edi
  test   %dil,%dil
  je     L46
  lea    (%r15,%r14,1),%rax
  cmp    %rax,0x8(%rsp)
  jae    L0
  cmp    %r15,0x10(%rsp)
  je     L0
L46:
  movb   $0x0,(%r12)
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    0x18(%rsp),%rax
  movb   $0x0,(%rax)
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L51:
  mov    0x18(%rsp),%r12
  jmp    L46
```

assistant: ```c
char *atohx(char *dst, const char *src, const unsigned int spacing) {
  char *ret = dst;
  const char *srcPtr = src;
  const unsigned int length = strlen(srcPtr);
  int lsb, msb;
  for (lsb = 0, msb = 0; *src; src += 2 + spacing) {
    if ((src + 1 + spacing) > srcPtr + length && src - srcPtr) {
      break;
    }
    msb = tolower(*src);
    lsb = tolower(*(src + 1));
    msb -= isdigit(msb) ? 0x30 : 0x57;
    lsb -= isdigit(lsb) ? 0x30 : 0x57;
    if ((msb < 0x0 || msb > 0xf) || (lsb < 0x0 || lsb > 0xf)) {
      *ret = 0;
      return NULL;
    }
    *dst++ = (char)(lsb | (msb << 4));
  }
  *dst = 0;
  return ret;
}
```