ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<strutil_insert>:
  endbr64
  push   %r15
  push   %r14
  mov    %rsi,%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmp    %rbx,%rax
  jbe    L2
  mov    %r14,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  mov    $0x1,%esi
  lea    (%r12,%rax,1),%rdi
  mov    %rax,%r15
  call   <calloc@plt>
  mov    %rax,%r13
  test   %rbx,%rbx
  jne    L14
  mov    %r15,%rdx
  mov    %r14,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strncat@plt>
L2:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    %r15,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <strncat@plt>
  mov    %r12,%rdx
  lea    0x0(%rbp,%rbx,1),%rsi
  mov    %r13,%rdi
  sub    %rbx,%rdx
  call   <strncat@plt>
  jmp    L2
```

assistant: ```c
char *strutil_insert(char *original, char *additional, size_t index) {
  size_t lo = strlen(original);
  size_t la = strlen(additional);
  if (index >= lo) {
    return NULL;
  }
  char *expanded = calloc(lo + la, sizeof(char));
  if (index == 0) {
    strncpy(expanded, additional, la);
    strncat(expanded, original, lo);
  } else {
    strncpy(expanded, original, index);
    strncat(expanded, additional, la);
    strncat(expanded, original + index, lo - index);
  }
  return expanded;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<SetIgorIntVar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x2c(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <StoreNumVar@plt>
  cmp    $0xffffffff,%eax
  jne    L42
  movl   $0x4,-0x20(%rbp)
  cmpl   $0x0,-0x30(%rbp)
  je     L31
  orl    $0x4000,-0x20(%rbp)
L31:
  mov    -0x20(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <Variable@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  je     L46
  mov    -0x1c(%rbp),%eax
  jmp    L49
L46:
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <StoreNumVar@plt>
L42:
  mov    -0x1c(%rbp),%eax
L49:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int SetIgorIntVar(const char *numVarName, int value, int forceGlobal) {
  int varType;
  double d1, d2;
  int result = 0;
  d1 = value;
  d2 = 0.0;
  if (StoreNumVar(numVarName, &d1, &d2) == -1) {
    varType = 4;
    if (forceGlobal)
      varType |= 0x4000;
    if (result = Variable(numVarName, varType))
      return result;
    StoreNumVar(numVarName, &d1, &d2);
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<SetIgorIntVar>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbx
  mov    %edx,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %esi,%xmm0
  movsd  %xmm0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <StoreNumVar@plt>
  mov    $0x0,%ebp
  cmp    $0xffffffff,%eax
  je     L37
L36:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  mov    %ebp,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L37:
  cmp    $0x1,%r12d
  sbb    %esi,%esi
  and    $0xffffc000,%esi
  add    $0x4004,%esi
  mov    %rbx,%rdi
  call   <Variable@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  jne    L36
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  call   <StoreNumVar@plt>
  jmp    L36
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int SetIgorIntVar(const char *numVarName, int value, int forceGlobal) {
  int varType;
  double d1, d2;
  int result = 0;
  d1 = value;
  d2 = 0.0;
  if (StoreNumVar(numVarName, &d1, &d2) == -1) {
    varType = 4;
    if (forceGlobal)
      varType |= 0x4000;
    if (result = Variable(numVarName, varType))
      return result;
    StoreNumVar(numVarName, &d1, &d2);
  }
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<SetIgorIntVar>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  push   %r13
  cvtsi2sd %esi,%xmm0
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r14
  lea    0x8(%rsp),%r13
  movq   $0x0,0x10(%rsp)
  mov    %r14,%rdx
  mov    %r13,%rsi
  movsd  %xmm0,0x8(%rsp)
  call   <StoreNumVar@plt>
  cmp    $0xffffffff,%eax
  je     L15
L34:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x20,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L15:
  cmp    $0x1,%ebx
  mov    %rbp,%rdi
  sbb    %esi,%esi
  and    $0xffffc000,%esi
  add    $0x4004,%esi
  call   <Variable@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L34
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <StoreNumVar@plt>
  jmp    L34
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int SetIgorIntVar(const char *numVarName, int value, int forceGlobal) {
  int varType;
  double d1, d2;
  int result = 0;
  d1 = value;
  d2 = 0.0;
  if (StoreNumVar(numVarName, &d1, &d2) == -1) {
    varType = 4;
    if (forceGlobal)
      varType |= 0x4000;
    if (result = Variable(numVarName, varType))
      return result;
    StoreNumVar(numVarName, &d1, &d2);
  }
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<SetIgorIntVar>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  push   %r13
  cvtsi2sd %esi,%xmm0
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r14
  lea    0x8(%rsp),%r13
  movq   $0x0,0x10(%rsp)
  mov    %r14,%rdx
  mov    %r13,%rsi
  movsd  %xmm0,0x8(%rsp)
  call   <StoreNumVar@plt>
  cmp    $0xffffffff,%eax
  je     L5
L22:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x20,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L5:
  cmp    $0x1,%ebx
  mov    %rbp,%rdi
  sbb    %esi,%esi
  and    $0xffffc000,%esi
  add    $0x4004,%esi
  call   <Variable@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jne    L22
  mov    %r14,%rdx
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <StoreNumVar@plt>
  jmp    L22
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int SetIgorIntVar(const char *numVarName, int value, int forceGlobal) {
  int varType;
  double d1, d2;
  int result = 0;
  d1 = value;
  d2 = 0.0;
  if (StoreNumVar(numVarName, &d1, &d2) == -1) {
    varType = 4;
    if (forceGlobal)
      varType |= 0x4000;
    if (result = Variable(numVarName, varType))
      return result;
    StoreNumVar(numVarName, &d1, &d2);
  }
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <Set_Names@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <readfile@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main(int argc, char *argv[]) {
  Set_Names();
  readfile(argv);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    $0x0,%eax
  call   <Set_Names@plt>
  mov    %rbx,%rdi
  call   <readfile@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
main(int argc, char *argv[]) {
  Set_Names();
  readfile(argv);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rsi,%rbp
  call   <Set_Names@plt>
  mov    %rbp,%rdi
  call   <readfile@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
main(int argc, char *argv[]) {
  Set_Names();
  readfile(argv);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rsi,%rbp
  call   <Set_Names@plt>
  mov    %rbp,%rdi
  call   <readfile@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
main(int argc, char *argv[]) {
  Set_Names();
  readfile(argv);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L7
L29:
  mov    -0x18(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x18(%rbp)
L7:
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jne    L29
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L8
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L43
L8:
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L43:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L34
  call   <__stack_chk_fail@plt>
L34:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D18
type_infered: string
name: read_data
arguments: data_tag: D23
type_infered: string
name: read_data
arguments: data_tag: D24
type_infered: string
name: read_data
arguments: data_tag: D31
type_infered: string

tool: D18: Enter any integer to check whether it is palindrome or not: 

tool: D23: %d

tool: D24: 
The number %d is palindrome

tool: D31: 
The number %d is not a palindrome

assistant: ```c
int main(void) {
  int n, rev = 0, d, num;
  printf("Enter any integer to check whether it is palindrome or not: ");
  scanf("%d", &n);
  num = n;
  while (n != 0) {
    d = n % 10;
    rev = rev * 10 + d;
    n = n / 10;
  }
  if (num == rev)
    printf("\nThe number %d is palindrome", num);
  else
    printf("\nThe number %d is not a palindrome", num);
  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
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D37(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  je     L0
  mov    %edi,%edx
  mov    $0x0,%ecx
L16:
  lea    (%rcx,%rcx,4),%esi
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rsi,2),%ecx
  mov    %eax,%edx
  test   %eax,%eax
  jne    L16
  movl   $0x0,0x4(%rsp)
  cmp    %ecx,%edi
  je     L0
  mov    %edi,%esi
  lea    D12(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L13:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L0:
  mov    %edi,%esi
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L13
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D37
type_guessed: string
name: disassemble_data
arguments: tag: D14
type_guessed: string
name: disassemble_data
arguments: tag: D12
type_guessed: string
name: disassemble_data
arguments: tag: D13
type_guessed: string

tool: D37: Enter any integer to check whether it is palindrome or not: 

tool: D14: %d

tool: D12: 
The number %d is not a palindrome

tool: D13: 
The number %d is palindrome

assistant: ```c
int main(void) {
  int n, rev = 0, d, num;
  printf("Enter any integer to check whether it is palindrome or not: ");
  scanf("%d", &n);
  num = n;
  while (n != 0) {
    d = n % 10;
    rev = rev * 10 + d;
    n = n / 10;
  }
  if (num == rev)
    printf("\nThe number %d is palindrome", num);
  else
    printf("\nThe number %d is not a palindrome", num);
  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    D4(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  xor    %ecx,%ecx
  mov    %r8d,%edx
  test   %r8d,%r8d
  je     L25
  nopl   0x0(%rax)
L17:
  movslq %edx,%rax
  lea    (%rcx,%rcx,4),%esi
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rsi,2),%ecx
  mov    %eax,%edx
  test   %eax,%eax
  jne    L17
  movl   $0x0,0x4(%rsp)
  cmp    %ecx,%r8d
  je     L25
  mov    %r8d,%esi
  lea    D32(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L14:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L25:
  mov    %r8d,%esi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L14
L45:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D4
type_infered: string
name: unpack_data
arguments: data_tag: D18
type_infered: string
name: unpack_data
arguments: data_tag: D32
type_infered: string
name: unpack_data
arguments: data_tag: D22
type_infered: string

tool: D4: Enter any integer to check whether it is palindrome or not: 

tool: D18: %d

tool: D32: 
The number %d is not a palindrome

tool: D22: 
The number %d is palindrome

assistant: ```c
int main(void) {
  int n, rev = 0, d, num;
  printf("Enter any integer to check whether it is palindrome or not: ");
  scanf("%d", &n);
  num = n;
  while (n != 0) {
    d = n % 10;
    rev = rev * 10 + d;
    n = n / 10;
  }
  if (num == rev)
    printf("\nThe number %d is palindrome", num);
  else
    printf("\nThe number %d is not a palindrome", num);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D38(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%r8d
  xor    %ecx,%ecx
  mov    %r8d,%edx
  test   %r8d,%r8d
  je     L23
  nopl   0x0(%rax)
L44:
  movslq %edx,%rax
  lea    (%rcx,%rcx,4),%esi
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rsi,2),%ecx
  mov    %eax,%edx
  test   %eax,%eax
  jne    L44
  movl   $0x0,0x4(%rsp)
  cmp    %ecx,%r8d
  je     L23
  mov    %r8d,%esi
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L28:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L23:
  mov    %r8d,%esi
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L28
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D38
type_guessed: string
name: disassemble_data
arguments: label: D11
type_guessed: string
name: disassemble_data
arguments: label: D3
type_guessed: string
name: disassemble_data
arguments: label: D26
type_guessed: string

tool: D38: Enter any integer to check whether it is palindrome or not: 

tool: D11: %d

tool: D3: 
The number %d is not a palindrome

tool: D26: 
The number %d is palindrome

assistant: ```c
int main(void) {
  int n, rev = 0, d, num;
  printf("Enter any integer to check whether it is palindrome or not: ");
  scanf("%d", &n);
  num = n;
  while (n != 0) {
    d = n % 10;
    rev = rev * 10 + d;
    n = n / 10;
  }
  if (num == rev)
    printf("\nThe number %d is palindrome", num);
  else
    printf("\nThe number %d is not a palindrome", num);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<eot>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x28(%rbp)
  movsd  %xmm1,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  call   <eqc@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  call   <al_sun@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  movq   %rax,%xmm0
  call   <cos_to_earth@plt>
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%edi
  movq   %rax,%xmm0
  call   <ra_sun@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  movsd  -0x10(%rbp),%xmm0
  subsd  -0x18(%rbp),%xmm0
  subsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double eot(double t, double to_earth) {
  double ec, al, ra;
  ec = eqc(t);
  al = al_sun(t);
  ra = ra_sun(t, cos_to_earth(to_earth));
  return -ec + al - ra;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<eot>:
  endbr64
  push   %r14
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  movq   %xmm0,%rbp
  movq   %xmm1,%r12
  call   <eqc@plt>
  movq   %xmm0,%r14
  movq   %rbp,%xmm0
  call   <al_sun@plt>
  movq   %xmm0,%rbx
  movq   %r12,%xmm0
  call   <cos_to_earth@plt>
  mov    %eax,%edi
  movq   %rbp,%xmm0
  call   <ra_sun@plt>
  movapd %xmm0,%xmm1
  movq   %rbx,%xmm0
  movq   %r14,%xmm2
  subsd  %xmm2,%xmm0
  subsd  %xmm1,%xmm0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
```

assistant: ```c
double eot(double t, double to_earth) {
  double ec, al, ra;
  ec = eqc(t);
  al = al_sun(t);
  ra = ra_sun(t, cos_to_earth(to_earth));
  return -ec + al - ra;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<eot>:
  endbr64
  sub    $0x28,%rsp
  movsd  %xmm1,0x18(%rsp)
  movsd  %xmm0,0x8(%rsp)
  call   <eqc@plt>
  movsd  0x8(%rsp),%xmm3
  movsd  %xmm0,(%rsp)
  movapd %xmm3,%xmm0
  movsd  %xmm3,0x10(%rsp)
  call   <al_sun@plt>
  movsd  0x18(%rsp),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm1,%xmm0
  call   <cos_to_earth@plt>
  movsd  0x10(%rsp),%xmm3
  mov    %eax,%edi
  movapd %xmm3,%xmm0
  call   <ra_sun@plt>
  movsd  0x8(%rsp),%xmm2
  subsd  (%rsp),%xmm2
  add    $0x28,%rsp
  subsd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
  ret
```

assistant: ```c
double eot(double t, double to_earth) {
  double ec, al, ra;
  ec = eqc(t);
  al = al_sun(t);
  ra = ra_sun(t, cos_to_earth(to_earth));
  return -ec + al - ra;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<eot>:
  endbr64
  sub    $0x28,%rsp
  movsd  %xmm1,0x18(%rsp)
  movsd  %xmm0,0x8(%rsp)
  call   <eqc@plt>
  movsd  0x8(%rsp),%xmm3
  movsd  %xmm0,(%rsp)
  movapd %xmm3,%xmm0
  movsd  %xmm3,0x10(%rsp)
  call   <al_sun@plt>
  movsd  0x18(%rsp),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movapd %xmm1,%xmm0
  call   <cos_to_earth@plt>
  movsd  0x10(%rsp),%xmm3
  mov    %eax,%edi
  movapd %xmm3,%xmm0
  call   <ra_sun@plt>
  movsd  0x8(%rsp),%xmm2
  subsd  (%rsp),%xmm2
  add    $0x28,%rsp
  subsd  %xmm0,%xmm2
  movapd %xmm2,%xmm0
  ret
```

assistant: ```c
double eot(double t, double to_earth) {
  double ec, al, ra;
  ec = eqc(t);
  al = al_sun(t);
  ra = ra_sun(t, cos_to_earth(to_earth));
  return -ec + al - ra;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rtc_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x44e00804,%eax
  movl   $0x2,(%rax)
  mov    $0x1,%edi
  call   <leds_set@plt>
  mov    $0x44e00800,%eax
  movl   $0x2,(%rax)
  mov    $0x2,%edi
  call   <leds_set@plt>
  mov    $0x44e3e06c,%eax
  movl   $0x83e70b13,(%rax)
  mov    $0x44e3e070,%eax
  movl   $0x95a4f1e0,(%rax)
  mov    $0x44e3e040,%eax
  movl   $0x1,(%rax)
  mov    $0x44e3e054,%eax
  movl   $0x48,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rtc_init() {
  (*(volatile unsigned int *)(0x44E00804) = (unsigned int)(0x2));
  leds_set(1);
  (*(volatile unsigned int *)(0x44E00800) = (unsigned int)(0x2));
  leds_set(2);
  (*(volatile unsigned int *)(0x44E3E000 + 0x6C) = (unsigned int)(0x83E70B13));
  (*(volatile unsigned int *)(0x44E3E000 + 0x70) = (unsigned int)(0x95A4F1E0));
  (*(volatile unsigned int *)(0x44E3E000 + 0x40) = (unsigned int)(0x1));
  (*(volatile unsigned int *)(0x44E3E000 + 0x54) = (unsigned int)((1 << 3) | (1 << 6)));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<rtc_init>:
  endbr64
  sub    $0x8,%rsp
  movl   $0x2,0x44e00804
  mov    $0x1,%edi
  call   <leds_set@plt>
  movl   $0x2,0x44e00800
  mov    $0x2,%edi
  call   <leds_set@plt>
  movl   $0x83e70b13,0x44e3e06c
  movl   $0x95a4f1e0,0x44e3e070
  movl   $0x1,0x44e3e040
  movl   $0x48,0x44e3e054
  add    $0x8,%rsp
  ret
```

assistant: ```c
void rtc_init() {
  (*(volatile unsigned int *)(0x44E00804) = (unsigned int)(0x2));
  leds_set(1);
  (*(volatile unsigned int *)(0x44E00800) = (unsigned int)(0x2));
  leds_set(2);
  (*(volatile unsigned int *)(0x44E3E000 + 0x6C) = (unsigned int)(0x83E70B13));
  (*(volatile unsigned int *)(0x44E3E000 + 0x70) = (unsigned int)(0x95A4F1E0));
  (*(volatile unsigned int *)(0x44E3E000 + 0x40) = (unsigned int)(0x1));
  (*(volatile unsigned int *)(0x44E3E000 + 0x54) = (unsigned int)((1 << 3) | (1 << 6)));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rtc_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  movl   $0x2,0x44e00804
  call   <leds_set@plt>
  mov    $0x2,%edi
  movl   $0x2,0x44e00800
  call   <leds_set@plt>
  movl   $0x83e70b13,0x44e3e06c
  movl   $0x95a4f1e0,0x44e3e070
  movl   $0x1,0x44e3e040
  movl   $0x48,0x44e3e054
  add    $0x8,%rsp
  ret
```

assistant: ```c
void rtc_init() {
  (*(volatile unsigned int *)(0x44E00804) = (unsigned int)(0x2));
  leds_set(1);
  (*(volatile unsigned int *)(0x44E00800) = (unsigned int)(0x2));
  leds_set(2);
  (*(volatile unsigned int *)(0x44E3E000 + 0x6C) = (unsigned int)(0x83E70B13));
  (*(volatile unsigned int *)(0x44E3E000 + 0x70) = (unsigned int)(0x95A4F1E0));
  (*(volatile unsigned int *)(0x44E3E000 + 0x40) = (unsigned int)(0x1));
  (*(volatile unsigned int *)(0x44E3E000 + 0x54) = (unsigned int)((1 << 3) | (1 << 6)));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<rtc_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  movl   $0x2,0x44e00804
  call   <leds_set@plt>
  mov    $0x2,%edi
  movl   $0x2,0x44e00800
  call   <leds_set@plt>
  movl   $0x83e70b13,0x44e3e06c
  movl   $0x95a4f1e0,0x44e3e070
  movl   $0x1,0x44e3e040
  movl   $0x48,0x44e3e054
  add    $0x8,%rsp
  ret
```

assistant: ```c
void rtc_init() {
  (*(volatile unsigned int *)(0x44E00804) = (unsigned int)(0x2));
  leds_set(1);
  (*(volatile unsigned int *)(0x44E00800) = (unsigned int)(0x2));
  leds_set(2);
  (*(volatile unsigned int *)(0x44E3E000 + 0x6C) = (unsigned int)(0x83E70B13));
  (*(volatile unsigned int *)(0x44E3E000 + 0x70) = (unsigned int)(0x95A4F1E0));
  (*(volatile unsigned int *)(0x44E3E000 + 0x40) = (unsigned int)(0x1));
  (*(volatile unsigned int *)(0x44E3E000 + 0x54) = (unsigned int)((1 << 3) | (1 << 6)));
}
```