ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x44434241,-0x3d(%rbp)
  movb   $0x0,-0x39(%rbp)
  movl   $0x5a5958,-0x41(%rbp)
  lea    -0x3d(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x60(%rbp)
  lea    -0x41(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x5c(%rbp)
  mov    -0x60(%rbp),%edx
  mov    -0x5c(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x58(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  movslq %eax,%rdx
  mov    %rdx,%r12
  mov    $0x0,%r13d
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L17:
  cmp    %rdx,%rsp
  je     L26
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L17
L26:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L36
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L36:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x50(%rbp)
  mov    -0x60(%rbp),%edx
  mov    -0x5c(%rbp),%eax
  add    %edx,%eax
  lea    0x1(%rax),%edx
  mov    -0x50(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x50(%rbp),%rdi
  mov    -0x5c(%rbp),%ecx
  mov    -0x60(%rbp),%edx
  lea    -0x41(%rbp),%rsi
  lea    -0x3d(%rbp),%rax
  mov    $0x0,%r9d
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <interleave@plt>
  mov    %rbx,%rsp
  nop
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void main() {
  char str1[] = "ABCD";
  char str2[] = "XYZ";
  int len1 = strlen(str1);
  int len2 = strlen(str2);
  char ils[len1 + len2 + 1];
  memset(ils, 0, len1 + len2 + 1);
  interleave(str1, str2, len1, len2, ils, 0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x44434241,-0x1d(%rbp)
  movb   $0x0,-0x19(%rbp)
  movl   $0x5a5958,-0x21(%rbp)
  lea    -0x1d(%rbp),%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  lea    -0x21(%rbp),%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  lea    0x1(%rbx,%rax,1),%edx
  movslq %edx,%rax
  add    $0xf,%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L18:
  cmp    %rcx,%rsp
  je     L29
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L18
L29:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L31
  orq    $0x0,-0x8(%rsp,%rax,1)
L31:
  mov    $0x0,%esi
  mov    %rsp,%rdi
  call   <memset@plt>
  lea    -0x21(%rbp),%rsi
  lea    -0x1d(%rbp),%rdi
  mov    $0x0,%r9d
  mov    %rsp,%r8
  mov    %r12d,%ecx
  mov    %ebx,%edx
  call   <interleave@plt>
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void main() {
  char str1[] = "ABCD";
  char str2[] = "XYZ";
  int len1 = strlen(str1);
  int len2 = strlen(str2);
  char ils[len1 + len2 + 1];
  memset(ils, 0, len1 + len2 + 1);
  interleave(str1, str2, len1, len2, ils, 0);
}
```
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
  push   %r15
  push   %r14
  lea    -0x2d(%rbp),%r14
  lea    -0x31(%rbp),%r15
  push   %r13
  mov    %r14,%rdi
  push   %r12
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movl   $0x44434241,-0x2d(%rbp)
  movb   $0x0,-0x29(%rbp)
  movl   $0x5a5958,-0x31(%rbp)
  call   <strlen@plt>
  mov    %r15,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  mov    %rsp,%rcx
  lea    0x1(%r12,%rax,1),%r8d
  mov    %eax,%r13d
  movslq %r8d,%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L48
L0:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L0
L48:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L2
L5:
  mov    %r8d,%edx
  xor    %esi,%esi
  mov    %rsp,%rdi
  call   <memset@plt>
  xor    %r9d,%r9d
  mov    %rsp,%r8
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <interleave@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  lea    -0x20(%rbp),%rsp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L2:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L5
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void main() {
  char str1[] = "ABCD";
  char str2[] = "XYZ";
  int len1 = strlen(str1);
  int len2 = strlen(str2);
  char ils[len1 + len2 + 1];
  memset(ils, 0, len1 + len2 + 1);
  interleave(str1, str2, len1, len2, ils, 0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  lea    -0x2d(%rbp),%r14
  lea    -0x31(%rbp),%r15
  push   %r13
  mov    %r14,%rdi
  push   %r12
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movl   $0x44434241,-0x2d(%rbp)
  movb   $0x0,-0x29(%rbp)
  movl   $0x5a5958,-0x31(%rbp)
  call   <strlen@plt>
  mov    %r15,%rdi
  mov    %eax,%r12d
  call   <strlen@plt>
  mov    %rsp,%rcx
  lea    0x1(%r12,%rax,1),%r8d
  mov    %eax,%r13d
  movslq %r8d,%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L11
L44:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L44
L11:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L47
L35:
  mov    %r8d,%edx
  xor    %esi,%esi
  mov    %rsp,%rdi
  call   <memset@plt>
  xor    %r9d,%r9d
  mov    %rsp,%r8
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <interleave@plt>
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  lea    -0x20(%rbp),%rsp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L47:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L35
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void main() {
  char str1[] = "ABCD";
  char str2[] = "XYZ";
  int len1 = strlen(str1);
  int len2 = strlen(str2);
  char ils[len1 + len2 + 1];
  memset(ils, 0, len1 + len2 + 1);
  interleave(str1, str2, len1, len2, ils, 0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x64,-0x28(%rbp)
  movss  D19(%rip),%xmm0
  movss  %xmm0,-0x24(%rbp)
  movsd  D0(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movb   $0x41,-0x29(%rbp)
  movl   $0x64655a,-0x11(%rbp)
  movl   $0x77616853,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D75(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  cvtss2sd -0x24(%rbp),%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  movq   %rax,%xmm0
  lea    D81(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%eax
  mov    %eax,%esi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x11(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xd(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl -0x29(%rbp),%edx
  lea    -0xd(%rbp),%rcx
  lea    -0x11(%rbp),%rax
  mov    %rax,%rsi
  lea    D47(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D19
type_infered: byte[4]
name: parse_data
arguments: data_label: D0
type_infered: f64
name: parse_data
arguments: data_label: D75
type_infered: string
name: parse_data
arguments: data_label: D30
type_infered: string
name: parse_data
arguments: data_label: D81
type_infered: string
name: parse_data
arguments: data_label: D10
type_infered: string
name: parse_data
arguments: data_label: D5
type_infered: string
name: parse_data
arguments: data_label: D12
type_infered: string
name: parse_data
arguments: data_label: D47
type_infered: string

tool: {"D19": [0x7b,0x14,0x16,0x40]}

tool: D0: 56789.4532

tool: D75: you are %d miles away.


tool: D30: you have %f levels of power


tool: D81: you have %f awesome super powers.


tool: D10: I have an initial %c.


tool: D5: I have a first name %s.


tool: D12: I have a last name %s.


tool: D47: My whole name is %s %c %s.


assistant: ```c
int main(int argc, char *arg[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("you are %d miles away.\n", distance);
  printf("you have %f levels of power\n", power);
  printf("you have %f awesome super powers.\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  mov    $0x64,%esi
  lea    D46(%rip),%rdi
  call   <printf@plt>
  movsd  D17(%rip),%xmm0
  lea    D82(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  D40(%rip),%xmm0
  lea    D51(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xf(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x13(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D81(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rcx
  mov    $0x41,%edx
  mov    %rbx,%rsi
  lea    D74(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D46
data_type: string
name: read_data
arguments: label: D17
data_type: byte[8]
name: read_data
arguments: label: D82
data_type: string
name: read_data
arguments: label: D40
data_type: f64
name: read_data
arguments: label: D51
data_type: string
name: read_data
arguments: label: D15
data_type: string
name: read_data
arguments: label: D5
data_type: string
name: read_data
arguments: label: D81
data_type: string
name: read_data
arguments: label: D74
data_type: string

tool: D46: you are %d miles away.


tool: {"D17": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D82: you have %f levels of power


tool: D40: 56789.4532

tool: D51: you have %f awesome super powers.


tool: D15: I have an initial %c.


tool: D5: I have a first name %s.


tool: D81: I have a last name %s.


tool: D74: My whole name is %s %c %s.


assistant: ```c
int main(int argc, char *arg[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("you are %d miles away.\n", distance);
  printf("you have %f levels of power\n", power);
  printf("you have %f awesome super powers.\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x64,%esi
  lea    D32(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  lea    0xf(%rsp),%rbp
  lea    0x13(%rsp),%r12
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D73(%rip),%xmm0
  mov    $0x1,%eax
  lea    D67(%rip),%rdi
  call   <printf@plt>
  movsd  D78(%rip),%xmm0
  mov    $0x1,%eax
  lea    D82(%rip),%rdi
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D56(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x41,%edx
  mov    %rbp,%rsi
  lea    D80(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D32
data_type: string
name: extract_symbolic_data
arguments: label: D73
data_type: byte[8]
name: extract_symbolic_data
arguments: label: D67
data_type: string
name: extract_symbolic_data
arguments: label: D78
data_type: f64
name: extract_symbolic_data
arguments: label: D82
data_type: string
name: extract_symbolic_data
arguments: label: D56
data_type: string
name: extract_symbolic_data
arguments: label: D5
data_type: string
name: extract_symbolic_data
arguments: label: D40
data_type: string
name: extract_symbolic_data
arguments: label: D80
data_type: string

tool: D32: you are %d miles away.


tool: {"D73": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D67: you have %f levels of power


tool: D78: 56789.4532

tool: D82: you have %f awesome super powers.


tool: D56: I have an initial %c.


tool: D5: I have a first name %s.


tool: D40: I have a last name %s.


tool: D80: My whole name is %s %c %s.


assistant: ```c
int main(int argc, char *arg[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("you are %d miles away.\n", distance);
  printf("you have %f levels of power\n", power);
  printf("you have %f awesome super powers.\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x64,%esi
  lea    D2(%rip),%rdi
  push   %rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x64655a,0xf(%rsp)
  lea    0xf(%rsp),%rbp
  lea    0x13(%rsp),%r12
  movl   $0x77616853,0x13(%rsp)
  movb   $0x0,0x17(%rsp)
  call   <printf@plt>
  movsd  D48(%rip),%xmm0
  mov    $0x1,%eax
  lea    D86(%rip),%rdi
  call   <printf@plt>
  movsd  D21(%rip),%xmm0
  mov    $0x1,%eax
  lea    D64(%rip),%rdi
  call   <printf@plt>
  mov    $0x41,%esi
  lea    D8(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D87(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rcx
  mov    $0x41,%edx
  mov    %rbp,%rsi
  lea    D35(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
data_type: string
name: extract_symbolic_data
arguments: tag: D48
data_type: byte[8]
name: extract_symbolic_data
arguments: tag: D86
data_type: string
name: extract_symbolic_data
arguments: tag: D21
data_type: f64
name: extract_symbolic_data
arguments: tag: D64
data_type: string
name: extract_symbolic_data
arguments: tag: D8
data_type: string
name: extract_symbolic_data
arguments: tag: D87
data_type: string
name: extract_symbolic_data
arguments: tag: D42
data_type: string
name: extract_symbolic_data
arguments: tag: D35
data_type: string

tool: D2: you are %d miles away.


tool: {"D48": [0x00,0x00,0x00,0x60,0x8f,0xc2,0x02,0x40]}

tool: D86: you have %f levels of power


tool: D21: 56789.4532

tool: D64: you have %f awesome super powers.


tool: D8: I have an initial %c.


tool: D87: I have a first name %s.


tool: D42: I have a last name %s.


tool: D35: My whole name is %s %c %s.


assistant: ```c
int main(int argc, char *arg[]) {
  int distance = 100;
  float power = 2.345f;
  double super_power = 56789.4532;
  char initial = 'A';
  char first_name[] = "Zed";
  char last_name[] = "Shaw";
  printf("you are %d miles away.\n", distance);
  printf("you have %f levels of power\n", power);
  printf("you have %f awesome super powers.\n", super_power);
  printf("I have an initial %c.\n", initial);
  printf("I have a first name %s.\n", first_name);
  printf("I have a last name %s.\n", last_name);
  printf("My whole name is %s %c %s.\n", first_name, initial, last_name);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<i2d_RSA_PUBKEY>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L30
  mov    $0x0,%eax
  jmp    L0
L30:
  mov    $0x0,%eax
  call   <EVP_PKEY_new@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L26
  mov    D2(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <ASN1error@plt>
  mov    $0x0,%eax
  jmp    L0
L26:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <EVP_PKEY_set1_RSA@plt>
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <i2d_PUBKEY@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <EVP_PKEY_free@plt>
  mov    -0xc(%rbp),%eax
L0:
  leave
  ret
```

assistant: ```c
int i2d_RSA_PUBKEY(RSA *a, unsigned char **pp) {
  EVP_PKEY *pktmp;
  int ret;
  if (!a)
    return 0;
  pktmp = EVP_PKEY_new();
  if (!pktmp) {
    ASN1error(ERR_R_MALLOC_FAILURE);
    return 0;
  }
  EVP_PKEY_set1_RSA(pktmp, a);
  ret = i2d_PUBKEY(pktmp, pp);
  EVP_PKEY_free(pktmp);
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<i2d_RSA_PUBKEY>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%r12d
  test   %rdi,%rdi
  je     L27
  mov    %rdi,%rbx
  mov    %rsi,%r13
  mov    $0x0,%eax
  call   <EVP_PKEY_new@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L19
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <EVP_PKEY_set1_RSA@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <i2d_PUBKEY@plt>
  mov    %eax,%r12d
  mov    %rbp,%rdi
  call   <EVP_PKEY_free@plt>
L27:
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L19:
  mov    D0(%rip),%rax
  mov    (%rax),%edi
  call   <ASN1error@plt>
  jmp    L27
```

assistant: ```c
int i2d_RSA_PUBKEY(RSA *a, unsigned char **pp) {
  EVP_PKEY *pktmp;
  int ret;
  if (!a)
    return 0;
  pktmp = EVP_PKEY_new();
  if (!pktmp) {
    ASN1error(ERR_R_MALLOC_FAILURE);
    return 0;
  }
  EVP_PKEY_set1_RSA(pktmp, a);
  ret = i2d_PUBKEY(pktmp, pp);
  EVP_PKEY_free(pktmp);
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<i2d_RSA_PUBKEY>:
  endbr64
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L0
  xor    %eax,%eax
  mov    %rdi,%rbp
  mov    %rsi,%r14
  call   <EVP_PKEY_new@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L18
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <EVP_PKEY_set1_RSA@plt>
  mov    %r12,%rdi
  mov    %r14,%rsi
  call   <i2d_PUBKEY@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <EVP_PKEY_free@plt>
L0:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L18:
  mov    D2(%rip),%rax
  mov    (%rax),%edi
  call   <ASN1error@plt>
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int i2d_RSA_PUBKEY(RSA *a, unsigned char **pp) {
  EVP_PKEY *pktmp;
  int ret;
  if (!a)
    return 0;
  pktmp = EVP_PKEY_new();
  if (!pktmp) {
    ASN1error(ERR_R_MALLOC_FAILURE);
    return 0;
  }
  EVP_PKEY_set1_RSA(pktmp, a);
  ret = i2d_PUBKEY(pktmp, pp);
  EVP_PKEY_free(pktmp);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<i2d_RSA_PUBKEY>:
  endbr64
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L14
  xor    %eax,%eax
  mov    %rdi,%rbp
  mov    %rsi,%r14
  call   <EVP_PKEY_new@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L1
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <EVP_PKEY_set1_RSA@plt>
  mov    %r12,%rdi
  mov    %r14,%rsi
  call   <i2d_PUBKEY@plt>
  mov    %r12,%rdi
  mov    %eax,%r13d
  call   <EVP_PKEY_free@plt>
L14:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L1:
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <ASN1error@plt>
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int i2d_RSA_PUBKEY(RSA *a, unsigned char **pp) {
  EVP_PKEY *pktmp;
  int ret;
  if (!a)
    return 0;
  pktmp = EVP_PKEY_new();
  if (!pktmp) {
    ASN1error(ERR_R_MALLOC_FAILURE);
    return 0;
  }
  EVP_PKEY_set1_RSA(pktmp, a);
  ret = i2d_PUBKEY(pktmp, pp);
  EVP_PKEY_free(pktmp);
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<writeriffhead>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x52,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movb   $0x49,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movb   $0x46,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movb   $0x46,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    -0xc(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0xc(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x5,%rax
  mov    %dl,(%rax)
  mov    -0xc(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x6,%rax
  mov    %dl,(%rax)
  mov    -0xc(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    $0x7,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  movb   $0x57,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x9,%rax
  movb   $0x41,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0xa,%rax
  movb   $0x56,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0xb,%rax
  movb   $0x45,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void writeriffhead(unsigned char *hdr, int flength) {
  *(hdr) = 'R';
  *(hdr + 1) = 'I';
  *(hdr + 2) = 'F';
  *(hdr + 3) = 'F';
  *(hdr + 4) = (unsigned char)(flength & 0xff);
  *(hdr + 5) = (unsigned char)((flength & 0xff00) >> 8);
  *(hdr + 6) = (unsigned char)((flength & 0xff0000) >> 16);
  *(hdr + 7) = (unsigned char)((flength & 0xff000000) >> 24);
  *(hdr + 8) = 'W';
  *(hdr + 9) = 'A';
  *(hdr + 10) = 'V';
  *(hdr + 11) = 'E';
}
```