ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<mkdir>:
  endbr64
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D7(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int mkdir(const char *pathname, mode_t mode) {
  errno = ENOSYS;
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mkdir>:
  endbr64
  mov    D14(%rip),%rax
  mov    (%rax),%edx
  mov    D3(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int mkdir(const char *pathname, mode_t mode) {
  errno = ENOSYS;
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reflect_pos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L15
L16:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  neg    %edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L16
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void reflect_pos(int *pos, int n) {
  int i;
  for (i = 1; i <= n; i++)
    pos[i] = -pos[i];
}
```
user: What is the c source code of the assembly code below:

```asm
<reflect_pos>:
  endbr64
  test   %esi,%esi
  jle    L5
  lea    0x4(%rdi),%rax
  lea    -0x1(%rsi),%edx
  lea    0x8(%rdi,%rdx,4),%rdx
L9:
  negl   (%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L9
L5:
  ret
```

assistant: ```c
void reflect_pos(int *pos, int n) {
  int i;
  for (i = 1; i <= n; i++)
    pos[i] = -pos[i];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<reflect_pos>:
  endbr64
  test   %esi,%esi
  jle    L16
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi),%rax
  lea    0x8(%rdi,%rdx,4),%rdx
  nopl   0x0(%rax)
L14:
  negl   (%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L14
L16:
  ret
```

assistant: ```c
void reflect_pos(int *pos, int n) {
  int i;
  for (i = 1; i <= n; i++)
    pos[i] = -pos[i];
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<reflect_pos>:
  endbr64
  mov    %rdi,%rdx
  test   %esi,%esi
  jle    L51
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L30
  mov    %esi,%ecx
  lea    0x4(%rdi),%rax
  pxor   %xmm1,%xmm1
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x14(%rdi,%rcx,1),%rcx
  nopl   0x0(%rax)
L44:
  movdqu (%rax),%xmm2
  movdqa %xmm1,%xmm0
  add    $0x10,%rax
  psubd  %xmm2,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rcx,%rax
  jne    L44
  mov    %esi,%eax
  and    $0xfffffffc,%eax
  lea    0x1(%rax),%ecx
  cmp    %eax,%esi
  je     L45
L8:
  movslq %ecx,%rax
  lea    0x1(%rcx),%edi
  shl    $0x2,%rax
  negl   (%rdx,%rax,1)
  cmp    %edi,%esi
  jl     L51
  add    $0x2,%ecx
  negl   0x4(%rdx,%rax,1)
  cmp    %ecx,%esi
  jl     L51
  negl   0x8(%rdx,%rax,1)
L51:
  ret
  cs nopw 0x0(%rax,%rax,1)
L45:
  ret
L30:
  mov    $0x1,%ecx
  jmp    L8
```

assistant: ```c
void reflect_pos(int *pos, int n) {
  int i;
  for (i = 1; i <= n; i++)
    pos[i] = -pos[i];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bigint_grow_raw>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x1c(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    $0x4,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  nop
  leave
  ret
```

assistant: ```c
__attribute__((always_)) void bigint_grow_raw(const uint64_t *source0, unsigned int source_size, uint64_t *dest0, unsigned int dest_size) {
  unsigned int pad_size = (dest_size - source_size);
  memcpy(dest0, source0, sizeof(bigint_t(source_size)));
  memset((dest0 + source_size), 0, sizeof(bigint_t(pad_size)));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bigint_grow_raw>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %esi,%ebx
  mov    %rdx,%rbp
  mov    $0x4,%edx
  mov    %rdi,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  mov    %ebx,%ebx
  lea    0x0(%rbp,%rbx,4),%rdi
  mov    $0x4,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
__attribute__((always_)) void bigint_grow_raw(const uint64_t *source0, unsigned int source_size, uint64_t *dest0, unsigned int dest_size) {
  unsigned int pad_size = (dest_size - source_size);
  memcpy(dest0, source0, sizeof(bigint_t(source_size)));
  memset((dest0 + source_size), 0, sizeof(bigint_t(pad_size)));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<bigint_grow_raw>:
  endbr64
  push   %rbp
  mov    %rdx,%rbp
  mov    $0x4,%edx
  push   %rbx
  mov    %esi,%ebx
  mov    %rdi,%rsi
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <memcpy@plt>
  add    $0x8,%rsp
  lea    0x0(%rbp,%rbx,4),%rdi
  xor    %esi,%esi
  pop    %rbx
  mov    $0x4,%edx
  pop    %rbp
  jmp    1060 <memset@plt>
```

assistant: ```c
__attribute__((always_)) void bigint_grow_raw(const uint64_t *source0, unsigned int source_size, uint64_t *dest0, unsigned int dest_size) {
  unsigned int pad_size = (dest_size - source_size);
  memcpy(dest0, source0, sizeof(bigint_t(source_size)));
  memset((dest0 + source_size), 0, sizeof(bigint_t(pad_size)));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<bigint_grow_raw>:
  endbr64
  push   %rbp
  mov    %rdx,%rbp
  mov    $0x4,%edx
  push   %rbx
  mov    %esi,%ebx
  mov    %rdi,%rsi
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <memcpy@plt>
  add    $0x8,%rsp
  lea    0x0(%rbp,%rbx,4),%rdi
  xor    %esi,%esi
  pop    %rbx
  mov    $0x4,%edx
  pop    %rbp
  jmp    1060 <memset@plt>
```

assistant: ```c
__attribute__((always_)) void bigint_grow_raw(const uint64_t *source0, unsigned int source_size, uint64_t *dest0, unsigned int dest_size) {
  unsigned int pad_size = (dest_size - source_size);
  memcpy(dest0, source0, sizeof(bigint_t(source_size)));
  memset((dest0 + source_size), 0, sizeof(bigint_t(pad_size)));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hc_fftw_to_mx>:
  endbr64
  test   $0x1,%cl
  je     L35
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdi)
  movq   $0x0,(%rsi)
  lea    0x1(%rcx),%rax
  mov    %rax,%r8
  shr    $0x3f,%r8
  add    %rax,%r8
  sar    %r8
  cmp    $0x2,%rcx
  jle    L31
  lea    -0x8(,%rcx,8),%rcx
  mov    $0x1,%eax
  movq   D14(%rip),%xmm2
L17:
  movsd  (%rdx,%rax,8),%xmm1
  movsd  (%rdx,%rcx,1),%xmm0
  movsd  %xmm1,(%rdi,%rax,8)
  movsd  %xmm0,(%rsi,%rax,8)
  movsd  %xmm1,(%rdi,%rcx,1)
  xorpd  %xmm2,%xmm0
  movsd  %xmm0,(%rsi,%rcx,1)
  add    $0x1,%rax
  sub    $0x8,%rcx
  cmp    %r8,%rax
  jl     L17
  ret
L35:
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdi)
  movq   $0x0,(%rsi)
  mov    %rcx,%r8
  shr    $0x3f,%r8
  add    %rcx,%r8
  sar    %r8
  cmp    $0x3,%rcx
  jle    L47
  lea    -0x8(,%rcx,8),%rcx
  mov    $0x1,%eax
  movq   D14(%rip),%xmm2
L36:
  movsd  (%rdx,%rax,8),%xmm1
  movsd  (%rdx,%rcx,1),%xmm0
  movsd  %xmm1,(%rdi,%rax,8)
  movsd  %xmm0,(%rsi,%rax,8)
  movsd  %xmm1,(%rdi,%rcx,1)
  xorpd  %xmm2,%xmm0
  movsd  %xmm0,(%rsi,%rcx,1)
  add    $0x1,%rax
  sub    $0x8,%rcx
  cmp    %r8,%rax
  jl     L36
L47:
  movsd  (%rdx,%r8,8),%xmm0
  movsd  %xmm0,(%rdi,%r8,8)
  movq   $0x0,(%rsi,%r8,8)
L31:
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D14
data_type: f32[2]

tool: {"D14": [0.0,-0.0]}

assistant: ```c
void hc_fftw_to_mx(register double *Xr, register double *Xi, register double *X, long N) {
  register long k;
  register double re, im;
  if (N % 2) {
    Xr[0] = X[0];
    Xi[0] = 0.0;
    for (k = 1; k < ((N + 1) / 2); ++k) {
      re = X[k];
      im = X[N - k];
      Xr[k] = re;
      Xi[k] = im;
      Xr[N - k] = re;
      Xi[N - k] = -im;
    }
  } else {
    Xr[0] = X[0];
    Xi[0] = 0.0;
    for (k = 1; k < (N / 2); ++k) {
      re = X[k];
      im = X[N - k];
      Xr[k] = re;
      Xi[k] = im;
      Xr[N - k] = re;
      Xi[N - k] = -im;
    }
    Xr[N / 2] = X[N / 2];
    Xi[N / 2] = 0.0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hc_fftw_to_mx>:
  endbr64
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdi)
  movq   $0x0,(%rsi)
  test   $0x1,%cl
  je     L15
  lea    0x1(%rcx),%rax
  mov    %rax,%r8
  shr    $0x3f,%r8
  add    %rax,%r8
  sar    %r8
  cmp    $0x2,%rcx
  jle    L3
  movq   D4(%rip),%xmm2
  lea    -0x8(,%rcx,8),%rcx
  mov    $0x1,%eax
  nopl   0x0(%rax)
L6:
  movsd  (%rdx,%rcx,1),%xmm0
  movsd  (%rdx,%rax,8),%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
  movsd  %xmm0,(%rsi,%rax,8)
  xorpd  %xmm2,%xmm0
  add    $0x1,%rax
  movsd  %xmm1,(%rdi,%rcx,1)
  movsd  %xmm0,(%rsi,%rcx,1)
  sub    $0x8,%rcx
  cmp    %r8,%rax
  jl     L6
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  ret
  nopl   0x0(%rax)
L15:
  mov    %rcx,%r8
  shr    $0x3f,%r8
  add    %rcx,%r8
  sar    %r8
  cmp    $0x3,%rcx
  jle    L18
  movq   D4(%rip),%xmm2
  lea    -0x8(,%rcx,8),%rcx
  mov    $0x1,%eax
L36:
  movsd  (%rdx,%rcx,1),%xmm0
  movsd  (%rdx,%rax,8),%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
  movsd  %xmm0,(%rsi,%rax,8)
  xorpd  %xmm2,%xmm0
  add    $0x1,%rax
  movsd  %xmm1,(%rdi,%rcx,1)
  movsd  %xmm0,(%rsi,%rcx,1)
  sub    $0x8,%rcx
  cmp    %r8,%rax
  jl     L36
L18:
  movsd  (%rdx,%r8,8),%xmm0
  movsd  %xmm0,(%rdi,%r8,8)
  movq   $0x0,(%rsi,%r8,8)
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D4
data_type: f32[2]

tool: {"D4": [0.0,-0.0]}

assistant: ```c
void hc_fftw_to_mx(register double *Xr, register double *Xi, register double *X, long N) {
  register long k;
  register double re, im;
  if (N % 2) {
    Xr[0] = X[0];
    Xi[0] = 0.0;
    for (k = 1; k < ((N + 1) / 2); ++k) {
      re = X[k];
      im = X[N - k];
      Xr[k] = re;
      Xi[k] = im;
      Xr[N - k] = re;
      Xi[N - k] = -im;
    }
  } else {
    Xr[0] = X[0];
    Xi[0] = 0.0;
    for (k = 1; k < (N / 2); ++k) {
      re = X[k];
      im = X[N - k];
      Xr[k] = re;
      Xi[k] = im;
      Xr[N - k] = re;
      Xi[N - k] = -im;
    }
    Xr[N / 2] = X[N / 2];
    Xi[N / 2] = 0.0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<hc_fftw_to_mx>:
  endbr64
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rdi)
  movq   $0x0,(%rsi)
  test   $0x1,%cl
  je     L46
  lea    0x1(%rcx),%rax
  mov    %rax,%r8
  shr    $0x3f,%r8
  add    %rax,%r8
  sar    %r8
  cmp    $0x2,%rcx
  jle    L39
  movq   D16(%rip),%xmm2
  lea    -0x8(,%rcx,8),%rcx
  mov    $0x1,%eax
  nopl   0x0(%rax)
L48:
  movsd  (%rdx,%rcx,1),%xmm0
  movsd  (%rdx,%rax,8),%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
  movsd  %xmm0,(%rsi,%rax,8)
  xorpd  %xmm2,%xmm0
  add    $0x1,%rax
  movsd  %xmm1,(%rdi,%rcx,1)
  movsd  %xmm0,(%rsi,%rcx,1)
  sub    $0x8,%rcx
  cmp    %r8,%rax
  jl     L48
  ret
  nopl   0x0(%rax,%rax,1)
L39:
  ret
  nopl   0x0(%rax)
L46:
  mov    %rcx,%r8
  shr    $0x3f,%r8
  add    %rcx,%r8
  sar    %r8
  cmp    $0x3,%rcx
  jle    L25
  movq   D16(%rip),%xmm2
  lea    -0x8(,%rcx,8),%rcx
  mov    $0x1,%eax
L22:
  movsd  (%rdx,%rcx,1),%xmm0
  movsd  (%rdx,%rax,8),%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
  movsd  %xmm0,(%rsi,%rax,8)
  xorpd  %xmm2,%xmm0
  add    $0x1,%rax
  movsd  %xmm1,(%rdi,%rcx,1)
  movsd  %xmm0,(%rsi,%rcx,1)
  sub    $0x8,%rcx
  cmp    %r8,%rax
  jl     L22
L25:
  movsd  (%rdx,%r8,8),%xmm0
  movsd  %xmm0,(%rdi,%r8,8)
  movq   $0x0,(%rsi,%r8,8)
  ret
```

tool_calls: name: read_data
arguments: data_label: D16
datatype: f32[2]

tool: {"D16": [0.0,-0.0]}

assistant: ```c
void hc_fftw_to_mx(register double *Xr, register double *Xi, register double *X, long N) {
  register long k;
  register double re, im;
  if (N % 2) {
    Xr[0] = X[0];
    Xi[0] = 0.0;
    for (k = 1; k < ((N + 1) / 2); ++k) {
      re = X[k];
      im = X[N - k];
      Xr[k] = re;
      Xi[k] = im;
      Xr[N - k] = re;
      Xi[N - k] = -im;
    }
  } else {
    Xr[0] = X[0];
    Xi[0] = 0.0;
    for (k = 1; k < (N / 2); ++k) {
      re = X[k];
      im = X[N - k];
      Xr[k] = re;
      Xi[k] = im;
      Xr[N - k] = re;
      Xi[N - k] = -im;
    }
    Xr[N / 2] = X[N / 2];
    Xi[N / 2] = 0.0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<ReverseNumber>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L11
L5:
  mov    -0x4(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%esi
  mov    -0x14(%rbp),%ecx
  mov    %ecx,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  mov    %eax,%edx
  shr    $0x3,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  lea    (%rsi,%rdx,1),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    $0xcccccccd,%eax
  imul   %rdx,%rax
  shr    $0x20,%rax
  shr    $0x3,%eax
  mov    %eax,-0x14(%rbp)
L11:
  cmpl   $0x0,-0x14(%rbp)
  jne    L5
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int ReverseNumber(unsigned int num) {
  unsigned int rev_num = 0;
  while (num != 0) {
    rev_num = rev_num * 10 + num % 10;
    num /= 10;
  }
  return rev_num;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ReverseNumber>:
  endbr64
  test   %edi,%edi
  je     L23
  mov    $0x0,%edx
  mov    $0xcccccccd,%r8d
L1:
  lea    (%rdx,%rdx,4),%esi
  mov    %edi,%eax
  imul   %r8,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %edi,%ecx
  sub    %edx,%ecx
  lea    (%rcx,%rsi,2),%edx
  mov    %edi,%ecx
  mov    %eax,%edi
  cmp    $0x9,%ecx
  ja     L1
L29:
  mov    %edx,%eax
  ret
L23:
  mov    %edi,%edx
  jmp    L29
```

assistant: ```c
unsigned int ReverseNumber(unsigned int num) {
  unsigned int rev_num = 0;
  while (num != 0) {
    rev_num = rev_num * 10 + num % 10;
    num /= 10;
  }
  return rev_num;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ReverseNumber>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L4
  mov    $0xcccccccd,%r9d
  nopl   0x0(%rax)
L7:
  mov    %edi,%eax
  mov    %edi,%ecx
  lea    (%r8,%r8,4),%esi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%ecx
  mov    %edi,%edx
  mov    %eax,%edi
  lea    (%rcx,%rsi,2),%r8d
  cmp    $0x9,%edx
  ja     L7
L4:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int ReverseNumber(unsigned int num) {
  unsigned int rev_num = 0;
  while (num != 0) {
    rev_num = rev_num * 10 + num % 10;
    num /= 10;
  }
  return rev_num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ReverseNumber>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L6
  mov    $0xcccccccd,%r9d
  nopl   0x0(%rax)
L2:
  mov    %edi,%eax
  mov    %edi,%ecx
  lea    (%r8,%r8,4),%esi
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%ecx
  mov    %edi,%edx
  mov    %eax,%edi
  lea    (%rcx,%rsi,2),%r8d
  cmp    $0x9,%edx
  ja     L2
L6:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int ReverseNumber(unsigned int num) {
  unsigned int rev_num = 0;
  while (num != 0) {
    rev_num = rev_num * 10 + num % 10;
    num /= 10;
  }
  return rev_num;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<_a32be>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  sar    $0x18,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    -0xc(%rbp),%edx
  sar    $0x10,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    -0xc(%rbp),%edx
  sar    $0x8,%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0xc(%rax),%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void _a32be(uint8_t *d, uint_fast32_t val) {
  d[0] = val >> 24;
  d[1] = val >> 16;
  d[2] = val >> 8;
  d[3] = val;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<_a32be>:
  endbr64
  mov    %esi,%eax
  sar    $0x18,%eax
  mov    %eax,(%rdi)
  mov    %esi,%eax
  sar    $0x10,%eax
  mov    %eax,0x4(%rdi)
  mov    %esi,%eax
  sar    $0x8,%eax
  mov    %eax,0x8(%rdi)
  mov    %esi,0xc(%rdi)
  ret
```

assistant: ```c
void _a32be(uint8_t *d, uint_fast32_t val) {
  d[0] = val >> 24;
  d[1] = val >> 16;
  d[2] = val >> 8;
  d[3] = val;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<_a32be>:
  endbr64
  mov    %esi,%eax
  mov    %esi,0xc(%rdi)
  sar    $0x18,%eax
  mov    %eax,(%rdi)
  mov    %esi,%eax
  sar    $0x10,%eax
  mov    %eax,0x4(%rdi)
  mov    %esi,%eax
  sar    $0x8,%eax
  mov    %eax,0x8(%rdi)
  ret
```

assistant: ```c
void _a32be(uint8_t *d, uint_fast32_t val) {
  d[0] = val >> 24;
  d[1] = val >> 16;
  d[2] = val >> 8;
  d[3] = val;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_a32be>:
  endbr64
  mov    %esi,%eax
  mov    %esi,%edx
  movd   %esi,%xmm2
  sar    $0x18,%eax
  sar    $0x8,%edx
  movd   %eax,%xmm0
  mov    %esi,%eax
  movd   %edx,%xmm1
  sar    $0x10,%eax
  punpckldq %xmm2,%xmm1
  movd   %eax,%xmm3
  punpckldq %xmm3,%xmm0
  punpcklqdq %xmm1,%xmm0
  movups %xmm0,(%rdi)
  ret
```

assistant: ```c
void _a32be(uint8_t *d, uint_fast32_t val) {
  d[0] = val >> 24;
  d[1] = val >> 16;
  d[2] = val >> 8;
  d[3] = val;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<copy_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x230,%rsp
  mov    %rdi,-0x228(%rbp)
  mov    %rsi,-0x230(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x218(%rbp)
  jmp    L55
L9:
  mov    -0x218(%rbp),%edx
  mov    %edx,%eax
  shl    $0x3,%eax
  add    %edx,%eax
  shl    $0xb,%eax
  mov    %eax,%ecx
  mov    -0x228(%rbp),%rax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <fseek@plt>
  movl   $0x0,-0x214(%rbp)
  jmp    L59
L50:
  mov    -0x230(%rbp),%rdx
  lea    -0x210(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x200,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fread@plt>
  test   %rax,%rax
  je     L61
  mov    -0x228(%rbp),%rdx
  lea    -0x210(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x200,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fwrite@plt>
  addl   $0x1,-0x214(%rbp)
L59:
  cmpl   $0x11,-0x214(%rbp)
  jle    L50
  addl   $0x1,-0x218(%rbp)
L55:
  cmpl   $0x4f,-0x218(%rbp)
  jle    L9
  jmp    L47
L61:
  nop
L47:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L15
  call   <__stack_chk_fail@plt>
L15:
  leave
  ret
```

assistant: ```c
void copy_file(FILE *img, FILE *sys) {
  char buf[512];
  for (int i = 1; i < 80; i++) {
    fseek(img, 18 * 2 * 512 * i, 0);
    for (int j = 0; j < 18; j++) {
      if (fread(buf, 1, 512, sys))
        fwrite(buf, 1, 512, img);
      else
        return;
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<copy_file>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    $0x4800,%r14d
  mov    %rsp,%rbp
L28:
  mov    $0x0,%edx
  mov    %r14d,%esi
  mov    %r12,%rdi
  call   <fseek@plt>
  mov    $0x12,%ebx
L12:
  mov    %r13,%rcx
  mov    $0x200,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  test   %rax,%rax
  je     L31
  mov    %r12,%rcx
  mov    $0x200,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fwrite@plt>
  sub    $0x1,%ebx
  jne    L12
  add    $0x4800,%r14d
  cmp    $0x168000,%r14d
  jne    L28
L31:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void copy_file(FILE *img, FILE *sys) {
  char buf[512];
  for (int i = 1; i < 80; i++) {
    fseek(img, 18 * 2 * 512 * i, 0);
    for (int j = 0; j < 18; j++) {
      if (fread(buf, 1, 512, sys))
        fwrite(buf, 1, 512, img);
      else
        return;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<copy_file>:
  endbr64
  push   %r14
  mov    $0x4800,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
L7:
  xor    %edx,%edx
  mov    %r14d,%esi
  mov    %r12,%rdi
  mov    $0x12,%ebx
  call   <fseek@plt>
  nopw   0x0(%rax,%rax,1)
L22:
  mov    %r13,%rcx
  mov    $0x200,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  test   %rax,%rax
  je     L24
  mov    %r12,%rcx
  mov    $0x200,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fwrite@plt>
  sub    $0x1,%ebx
  jne    L22
  add    $0x4800,%r14d
  cmp    $0x168000,%r14d
  jne    L7
  nopl   0x0(%rax)
L24:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void copy_file(FILE *img, FILE *sys) {
  char buf[512];
  for (int i = 1; i < 80; i++) {
    fseek(img, 18 * 2 * 512 * i, 0);
    for (int j = 0; j < 18; j++) {
      if (fread(buf, 1, 512, sys))
        fwrite(buf, 1, 512, img);
      else
        return;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<copy_file>:
  endbr64
  push   %r14
  mov    $0x4800,%r14d
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
L27:
  xor    %edx,%edx
  mov    %r14d,%esi
  mov    %r12,%rdi
  mov    $0x12,%ebx
  call   <fseek@plt>
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %r13,%rcx
  mov    $0x200,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  test   %rax,%rax
  je     L38
  mov    %r12,%rcx
  mov    $0x200,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fwrite@plt>
  sub    $0x1,%ebx
  jne    L26
  add    $0x4800,%r14d
  cmp    $0x168000,%r14d
  jne    L27
  nopl   0x0(%rax)
L38:
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L35:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void copy_file(FILE *img, FILE *sys) {
  char buf[512];
  for (int i = 1; i < 80; i++) {
    fseek(img, 18 * 2 * 512 * i, 0);
    for (int j = 0; j < 18; j++) {
      if (fread(buf, 1, 512, sys))
        fwrite(buf, 1, 512, img);
      else
        return;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strtolower>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L13
  mov    $0x0,%eax
  jmp    L42
L13:
  cmpq   $0x0,-0x30(%rbp)
  jne    L8
  mov    -0x28(%rbp),%rax
  movb   $0x0,(%rax)
  jmp    L10
L8:
  movq   $0x0,-0x18(%rbp)
  jmp    L41
L2:
  mov    -0x18(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    -0x18(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %al,(%rbx)
  addq   $0x1,-0x18(%rbp)
L41:
  mov    -0x18(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L2
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L10:
  mov    -0x28(%rbp),%rax
L42:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *strtolower(char *target, char *source) {
  long i;
  if (target == NULL)
    return NULL;
  if (source == NULL)
    target[0] = '\0';
  else {
    for (i = 0; source[i] != '\0'; i++)
      target[i] = tolower(source[i]);
    target[i] = '\0';
  }
  return target;
}
```