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

```asm
<sanitize_image_name>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L40
  mov    %rsi,%rdi
  mov    %rsi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,(%r12)
  movsbl (%rbx),%edi
  mov    %rax,%rbp
  test   %dil,%dil
  je     L68
  xor    %r13d,%r13d
  xor    %r14d,%r14d
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L38:
  movb   $0x2d,0x0(%rbp)
L61:
  movsbl (%rbx),%edi
  lea    0x1(%r12),%rbp
  xor    %r13d,%r13d
  xor    %r14d,%r14d
  call   <tolower@plt>
  mov    %al,(%r12)
L98:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L68
L0:
  cmp    $0x2e,%dil
  je     L17
  cmp    $0x2b,%dil
  je     L14
  call   <isalnum@plt>
  test   %eax,%eax
  je     L88
L14:
  lea    0x1(%rbp),%r12
  test   %r13d,%r13d
  jne    L38
  test   %r14d,%r14d
  je     L60
  movb   $0x2e,0x0(%rbp)
  jmp    L61
  nopl   0x0(%rax)
L17:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  mov    $0x1,%r14d
  test   %dil,%dil
  jne    L0
L68:
  movb   $0x0,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L60:
  mov    %rbp,%r12
  jmp    L61
  nopl   (%rax)
L88:
  mov    $0x1,%r13d
  jmp    L98
L40:
  movq   $0x0,(%rdi)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
__attribute__((unused)) void sanitize_image_name(char **outp, char *in) {
  if (!in) {
    *outp = NULL;
    return;
  }
  char *out = malloc(strlen(in) + 1);
  *outp = out;
  bool dot = false, dash = false;
  for (; *in; in++) {
    if (*in == '.') {
      dot = true;
      continue;
    }
    if (*in != '+' && !isalnum(*in)) {
      dash = true;
      continue;
    }
    if (dash)
      *out++ = '-';
    else if (dot)
      *out++ = '.';
    dash = false;
    dot = false;
    *out++ = tolower(*in);
  }
  *out = 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
<sanitize_image_name>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  test   %rsi,%rsi
  je     L41
  mov    %rsi,%rdi
  mov    %rsi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <malloc@plt>
  mov    %rax,(%r12)
  movsbl (%rbx),%edi
  mov    %rax,%rbp
  test   %dil,%dil
  je     L58
  xor    %r13d,%r13d
  xor    %r14d,%r14d
  jmp    L77
  nopl   0x0(%rax,%rax,1)
L20:
  movb   $0x2d,0x0(%rbp)
L60:
  call   <tolower@plt>
  lea    0x1(%r12),%rbp
  xor    %r13d,%r13d
  xor    %r14d,%r14d
  mov    %al,(%r12)
L46:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L58
L77:
  cmp    $0x2e,%dil
  je     L57
  cmp    $0x2b,%dil
  je     L18
  call   <isalnum@plt>
  test   %eax,%eax
  je     L49
  movsbl (%rbx),%edi
L18:
  lea    0x1(%rbp),%r12
  test   %r13d,%r13d
  jne    L20
  test   %r14d,%r14d
  je     L109
  movb   $0x2e,0x0(%rbp)
  jmp    L60
  nopl   0x0(%rax)
L57:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  mov    $0x1,%r14d
  test   %dil,%dil
  jne    L77
L58:
  movb   $0x0,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L109:
  mov    %rbp,%r12
  jmp    L60
  nopl   (%rax)
L49:
  mov    $0x1,%r13d
  jmp    L46
L41:
  movq   $0x0,(%rdi)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
__attribute__((unused)) void sanitize_image_name(char **outp, char *in) {
  if (!in) {
    *outp = NULL;
    return;
  }
  char *out = malloc(strlen(in) + 1);
  *outp = out;
  bool dot = false, dash = false;
  for (; *in; in++) {
    if (*in == '.') {
      dot = true;
      continue;
    }
    if (*in != '+' && !isalnum(*in)) {
      dash = true;
      continue;
    }
    if (dash)
      *out++ = '-';
    else if (dot)
      *out++ = '.';
    dash = false;
    dot = false;
    *out++ = tolower(*in);
  }
  *out = 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<myabs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm1,%xmm0
  jb     L6
  movsd  -0x8(%rbp),%xmm0
  jmp    L16
L6:
  movsd  -0x8(%rbp),%xmm0
  movq   D8(%rip),%xmm1
  xorpd  %xmm1,%xmm0
L16:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
data_type: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
double myabs(double x) {
  if (x >= 0) {
    return x;
  } else {
    return -x;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
data_type: string
name: extract_symbolic_data
arguments: label: D5
data_type: string

tool: D0: Content-Type:text/html;



tool: D5: hello CGI~~


assistant: ```c
int main() {
  printf("Content-Type:text/html;\n\n");
  printf("hello CGI~~\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D13
data_type: string
name: unpack_data
arguments: label: D4
data_type: string

tool: D13: Content-Type:text/html;



tool: D4: hello CGI~~


assistant: ```c
int main() {
  printf("Content-Type:text/html;\n\n");
  printf("hello CGI~~\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
data_type: string
name: disassemble_data
arguments: tag: D5
data_type: string

tool: D4: Content-Type:text/html;



tool: D5: hello CGI~~


assistant: ```c
int main() {
  printf("Content-Type:text/html;\n\n");
  printf("hello CGI~~\n");
  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
  sub    $0x8,%rsp
  lea    D14(%rip),%rdi
  call   <printf@plt>
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D14
type_infered: string
name: retrieve_data
arguments: data_label: D9
type_infered: string

tool: D14: Content-Type:text/html;



tool: D9: hello CGI~~


assistant: ```c
int main() {
  printf("Content-Type:text/html;\n\n");
  printf("hello CGI~~\n");
  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
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x68db8bad,%rax,%rax
  shr    $0x20,%rax
  sar    $0xc,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x2710,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <Section1@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <Section2@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <Section3@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <Section4@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D19
type_guessed: string

tool: D2: Insert a number between 1000 and 9999: 

tool: D19: %d

assistant: ```c
int main() {
  int number;
  printf("Insert a number between 1000 and 9999: ");
  scanf("%d", &number);
  number %= 10000;
  Section1(number);
  Section2(number);
  Section3(number);
  Section4(number);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  movslq %eax,%rdi
  imul   $0x68db8bad,%rdi,%rdi
  sar    $0x2c,%rdi
  cltd
  sub    %edx,%edi
  imul   $0x2710,%edi,%edx
  sub    %edx,%eax
  mov    %eax,%edi
  mov    %eax,0x4(%rsp)
  call   <Section1@plt>
  mov    0x4(%rsp),%edi
  call   <Section2@plt>
  mov    0x4(%rsp),%edi
  call   <Section3@plt>
  mov    0x4(%rsp),%edi
  call   <Section4@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D9
data_type: string
name: parse_data
arguments: data_tag: D15
data_type: string

tool: D9: Insert a number between 1000 and 9999: 

tool: D15: %d

assistant: ```c
int main() {
  int number;
  printf("Insert a number between 1000 and 9999: ");
  scanf("%d", &number);
  number %= 10000;
  Section1(number);
  Section2(number);
  Section3(number);
  Section4(number);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D17(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rdi
  mov    %rdi,%rax
  imul   $0x68db8bad,%rdi,%rdi
  cltd
  sar    $0x2c,%rdi
  sub    %edx,%edi
  imul   $0x2710,%edi,%edx
  sub    %edx,%eax
  mov    %eax,%edi
  mov    %eax,0x4(%rsp)
  call   <Section1@plt>
  mov    0x4(%rsp),%edi
  call   <Section2@plt>
  mov    0x4(%rsp),%edi
  call   <Section3@plt>
  mov    0x4(%rsp),%edi
  call   <Section4@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D17
type_guessed: string
name: unpack_data
arguments: data_tag: D2
type_guessed: string

tool: D17: Insert a number between 1000 and 9999: 

tool: D2: %d

assistant: ```c
int main() {
  int number;
  printf("Insert a number between 1000 and 9999: ");
  scanf("%d", &number);
  number %= 10000;
  Section1(number);
  Section2(number);
  Section3(number);
  Section4(number);
  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
  sub    $0x18,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rdi
  mov    %rdi,%rax
  imul   $0x68db8bad,%rdi,%rdi
  cltd
  sar    $0x2c,%rdi
  sub    %edx,%edi
  imul   $0x2710,%edi,%edx
  sub    %edx,%eax
  mov    %eax,%edi
  mov    %eax,0x4(%rsp)
  call   <Section1@plt>
  mov    0x4(%rsp),%edi
  call   <Section2@plt>
  mov    0x4(%rsp),%edi
  call   <Section3@plt>
  mov    0x4(%rsp),%edi
  call   <Section4@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
datatype: string
name: extract_symbolic_data
arguments: data_tag: D13
datatype: string

tool: D5: Insert a number between 1000 and 9999: 

tool: D13: %d

assistant: ```c
int main() {
  int number;
  printf("Insert a number between 1000 and 9999: ");
  scanf("%d", &number);
  number %= 10000;
  Section1(number);
  Section2(number);
  Section3(number);
  Section4(number);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<test_shrt_cst>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movw   $0x8000,-0x4(%rbp)
  movw   $0x7fff,-0x2(%rbp)
  movswl -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0xffffffff,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0x0,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0x1,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  movswl -0x2(%rbp),%eax
  mov    %eax,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  nop
  leave
  ret
```

assistant: ```c
void test_shrt_cst(void) {
  const short min = (-0x7fff - 1);
  const short max = 0x7fff;
  sink(f_shrt_1(min));
  sink(f_shrt_1(-1));
  sink(f_shrt_1(0));
  sink(f_shrt_1(1));
  sink(f_shrt_1(max));
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_shrt_cst>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffff8000,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0xffffffff,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0x0,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0x1,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0x7fff,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void test_shrt_cst(void) {
  const short min = (-0x7fff - 1);
  const short max = 0x7fff;
  sink(f_shrt_1(min));
  sink(f_shrt_1(-1));
  sink(f_shrt_1(0));
  sink(f_shrt_1(1));
  sink(f_shrt_1(max));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<test_shrt_cst>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffff8000,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0xffffffff,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  xor    %edi,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0x1,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0x7fff,%edi
  call   <f_shrt_1@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    1070 <sink@plt>
```

assistant: ```c
void test_shrt_cst(void) {
  const short min = (-0x7fff - 1);
  const short max = 0x7fff;
  sink(f_shrt_1(min));
  sink(f_shrt_1(-1));
  sink(f_shrt_1(0));
  sink(f_shrt_1(1));
  sink(f_shrt_1(max));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test_shrt_cst>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffff8000,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0xffffffff,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  xor    %edi,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0x1,%edi
  call   <f_shrt_1@plt>
  mov    %eax,%edi
  call   <sink@plt>
  mov    $0x7fff,%edi
  call   <f_shrt_1@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    1070 <sink@plt>
```

assistant: ```c
void test_shrt_cst(void) {
  const short min = (-0x7fff - 1);
  const short max = 0x7fff;
  sink(f_shrt_1(min));
  sink(f_shrt_1(-1));
  sink(f_shrt_1(0));
  sink(f_shrt_1(1));
  sink(f_shrt_1(max));
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<pack_one>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  movq   $0x2,-0x10(%rbp)
  jmp    L18
L21:
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  add    %rax,-0x8(%rbp)
  cmpq   $0x4,-0x10(%rbp)
  jg     L27
  shlq   $0x8,-0x8(%rbp)
L27:
  addq   $0x1,-0x10(%rbp)
L18:
  cmpq   $0x5,-0x10(%rbp)
  jle    L21
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long pack_one(long nfrequency, long digitized[]) {
  long i;
  unsigned long packed;
  packed = 0;
  for (i = 2; i <= 5; i++) {
    packed += digitized[i];
    if (i < 5)
      packed *= 256;
  }
  return (packed);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<pack_one>:
  endbr64
  mov    $0x3,%eax
  mov    $0x0,%edx
  jmp    L38
L28:
  cmp    $0x5,%rax
  jle    L39
  mov    %rcx,%rax
  ret
L39:
  add    $0x1,%rax
L38:
  mov    %rdx,%rcx
  add    -0x8(%rsi,%rax,8),%rcx
  mov    %rcx,%rdx
  lea    -0x1(%rax),%rdi
  cmp    $0x4,%rdi
  jg     L28
  shl    $0x8,%rdx
  jmp    L39
```

assistant: ```c
unsigned long pack_one(long nfrequency, long digitized[]) {
  long i;
  unsigned long packed;
  packed = 0;
  for (i = 2; i <= 5; i++) {
    packed += digitized[i];
    if (i < 5)
      packed *= 256;
  }
  return (packed);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<pack_one>:
  endbr64
  mov    0x10(%rsi),%rax
  shl    $0x8,%rax
  add    0x18(%rsi),%rax
  shl    $0x8,%rax
  add    0x20(%rsi),%rax
  shl    $0x8,%rax
  add    0x28(%rsi),%rax
  ret
```

assistant: ```c
unsigned long pack_one(long nfrequency, long digitized[]) {
  long i;
  unsigned long packed;
  packed = 0;
  for (i = 2; i <= 5; i++) {
    packed += digitized[i];
    if (i < 5)
      packed *= 256;
  }
  return (packed);
}
```
user: Translate the below assembly code into C programming language:

```asm
<pack_one>:
  endbr64
  mov    0x10(%rsi),%rax
  shl    $0x8,%rax
  add    0x18(%rsi),%rax
  shl    $0x8,%rax
  add    0x20(%rsi),%rax
  shl    $0x8,%rax
  add    0x28(%rsi),%rax
  ret
```

assistant: ```c
unsigned long pack_one(long nfrequency, long digitized[]) {
  long i;
  unsigned long packed;
  packed = 0;
  for (i = 2; i <= 5; i++) {
    packed += digitized[i];
    if (i < 5)
      packed *= 256;
  }
  return (packed);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<apply_mods>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    $0x1000,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <copy_str@plt>
  jmp    L27
L18:
  mov    -0x3c(%rbp),%esi
  lea    -0x18(%rbp),%rcx
  mov    -0x38(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %esi,%r8d
  mov    %rax,%rsi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <apply_mod@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  je     L36
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    $0x1000,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  call   <copy_str@plt>
  mov    -0x18(%rbp),%eax
  cltq
  add    %rax,-0x38(%rbp)
  addl   $0x1,-0x14(%rbp)
L27:
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L18
  jmp    L8
L36:
  nop
L8:
  lea    D31(%rip),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L43
  call   <__stack_chk_fail@plt>
L43:
  leave
  ret
```

assistant: ```c
const char *apply_mods(const char path[], const char parent[], const char mod[], int for_shell) {
  static char buf[4096];
  int napplied = 0;
  copy_str(buf, sizeof(buf), path);
  while (*mod != '\0') {
    int mod_len;
    const char *const p = apply_mod(buf, parent, mod, &mod_len, for_shell);
    if (p == NULL) {
      break;
    }
    copy_str(buf, sizeof(buf), p);
    mod += mod_len;
    napplied++;
  }
  return buf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<apply_mods>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%r12
  mov    %rdx,%rbx
  mov    %ecx,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rdx
  mov    $0x1000,%esi
  lea    D11(%rip),%rdi
  call   <copy_str@plt>
  cmpb   $0x0,(%rbx)
  je     L0
  lea    D11(%rip),%rbp
L23:
  lea    0x4(%rsp),%rcx
  mov    %r13d,%r8d
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <apply_mod@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  je     L0
  mov    $0x1000,%esi
  mov    %rbp,%rdi
  call   <copy_str@plt>
  movslq 0x4(%rsp),%rax
  add    %rax,%rbx
  cmpb   $0x0,(%rbx)
  jne    L23
L0:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  lea    D11(%rip),%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
const char *apply_mods(const char path[], const char parent[], const char mod[], int for_shell) {
  static char buf[4096];
  int napplied = 0;
  copy_str(buf, sizeof(buf), path);
  while (*mod != '\0') {
    int mod_len;
    const char *const p = apply_mod(buf, parent, mod, &mod_len, for_shell);
    if (p == NULL) {
      break;
    }
    copy_str(buf, sizeof(buf), p);
    mod += mod_len;
    napplied++;
  }
  return buf;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<apply_mods>:
  endbr64
  push   %r14
  lea    D1(%rip),%r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  mov    $0x1000,%esi
  push   %rbp
  mov    %ecx,%ebp
  push   %rbx
  mov    %rdx,%rbx
  mov    %rdi,%rdx
  mov    %r14,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <copy_str@plt>
  cmpb   $0x0,(%rbx)
  je     L17
  lea    0x4(%rsp),%r13
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L1:
  mov    $0x1000,%esi
  mov    %r14,%rdi
  call   <copy_str@plt>
  movslq 0x4(%rsp),%rax
  add    %rax,%rbx
  cmpb   $0x0,(%rbx)
  je     L17
L39:
  mov    %rbx,%rdx
  mov    %ebp,%r8d
  mov    %r13,%rcx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <apply_mod@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L1
L17:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0x10,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L41:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
const char *apply_mods(const char path[], const char parent[], const char mod[], int for_shell) {
  static char buf[4096];
  int napplied = 0;
  copy_str(buf, sizeof(buf), path);
  while (*mod != '\0') {
    int mod_len;
    const char *const p = apply_mod(buf, parent, mod, &mod_len, for_shell);
    if (p == NULL) {
      break;
    }
    copy_str(buf, sizeof(buf), p);
    mod += mod_len;
    napplied++;
  }
  return buf;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<apply_mods>:
  endbr64
  push   %r14
  lea    D2(%rip),%r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  mov    $0x1000,%esi
  push   %rbp
  mov    %ecx,%ebp
  push   %rbx
  mov    %rdx,%rbx
  mov    %rdi,%rdx
  mov    %r14,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <copy_str@plt>
  cmpb   $0x0,(%rbx)
  je     L18
  lea    0x4(%rsp),%r13
  jmp    L35
  nopl   0x0(%rax,%rax,1)
L28:
  mov    $0x1000,%esi
  mov    %r14,%rdi
  call   <copy_str@plt>
  movslq 0x4(%rsp),%rax
  add    %rax,%rbx
  cmpb   $0x0,(%rbx)
  je     L18
L35:
  mov    %rbx,%rdx
  mov    %ebp,%r8d
  mov    %r13,%rcx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <apply_mod@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  jne    L28
L18:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x10,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
const char *apply_mods(const char path[], const char parent[], const char mod[], int for_shell) {
  static char buf[4096];
  int napplied = 0;
  copy_str(buf, sizeof(buf), path);
  while (*mod != '\0') {
    int mod_len;
    const char *const p = apply_mod(buf, parent, mod, &mod_len, for_shell);
    if (p == NULL) {
      break;
    }
    copy_str(buf, sizeof(buf), p);
    mod += mod_len;
    napplied++;
  }
  return buf;
}
```