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

```asm
<gds_write_endstr>:
  endbr64
  push   %r12
  lea    D7(%rip),%r12
  mov    $0x4,%eax
  push   %rbp
  mov    %edi,%ebp
  mov    %r12,%rdi
  sub    $0x8,%rsp
  mov    %ax,D22(%rip)
  call   <gds_swap2bytes@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    $0x2,%edx
  call   <write@plt>
  lea    D22(%rip),%r12
  mov    $0x700,%edx
  mov    %r12,%rdi
  mov    %dx,D15(%rip)
  call   <gds_swap2bytes@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    $0x2,%edx
  pop    %rbp
  pop    %r12
  jmp    1060 <write@plt>
```

assistant: ```c
void gds_write_endstr(int fd) {
  static short int count, token;
  count = 4;
  gds_swap2bytes((unsigned char *)&count);
  write(fd, &count, 2);
  token = 0x0700;
  gds_swap2bytes((unsigned char *)&token);
  write(fd, &token, 2);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<trdtoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
L18:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  test   %eax,%eax
  je     L80
  test   %eax,%eax
  jg     L0
  cmp    $0xffffffdc,%eax
  je     L58
  cmp    $0xffffffdc,%eax
  jg     L0
  cmp    $0xffffff8e,%eax
  je     L2
  cmp    $0xffffff99,%eax
  je     L10
  jmp    L0
L2:
  mov    -0x8(%rbp),%rax
  movb   $0xe0,(%rax)
  jmp    L0
L10:
  mov    -0x8(%rbp),%rax
  movb   $0xe4,(%rax)
  jmp    L0
L58:
  mov    -0x8(%rbp),%rax
  movb   $0xe6,(%rax)
  nop
L0:
  addq   $0x1,-0x8(%rbp)
  jmp    L18
L80:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void trdtoi(s) char *s;
{
  for (;; ++s)
    switch (*s) {
    case '\000':
      return;
    case '\216':
      *s = '\340';
      break;
    case '\231':
      *s = '\344';
      break;
    case '\334':
      *s = '\346';
      break;
    }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<trdtoi>:
  endbr64
  jmp    L58
L34:
  test   %al,%al
  jne    L44
  ret
L25:
  movb   $0xe0,(%rdi)
L44:
  add    $0x1,%rdi
L58:
  movzbl (%rdi),%eax
  cmp    $0xdc,%al
  je     L7
  cmp    $0xdd,%al
  jge    L34
  cmp    $0x8e,%al
  je     L25
  cmp    $0x99,%al
  jne    L44
  movb   $0xe4,(%rdi)
  jmp    L44
L7:
  movb   $0xe6,(%rdi)
  jmp    L44
```

assistant: ```c
void trdtoi(s) char *s;
{
  for (;; ++s)
    switch (*s) {
    case '\000':
      return;
    case '\216':
      *s = '\340';
      break;
    case '\231':
      *s = '\344';
      break;
    case '\334':
      *s = '\346';
      break;
    }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<trdtoi>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0xdc,%al
  je     L29
  nopl   0x0(%rax,%rax,1)
L46:
  cmp    $0xdd,%al
  jge    L11
  cmp    $0x8e,%al
  je     L10
  cmp    $0x99,%al
  jne    L77
  movb   $0xe4,(%rdi)
L77:
  add    $0x1,%rdi
L16:
  movzbl (%rdi),%eax
  cmp    $0xdc,%al
  jne    L46
L29:
  movb   $0xe6,(%rdi)
  add    $0x1,%rdi
  jmp    L16
  nopl   0x0(%rax,%rax,1)
L11:
  test   %al,%al
  jne    L77
  ret
  nopl   (%rax)
L10:
  movb   $0xe0,(%rdi)
  add    $0x1,%rdi
  jmp    L16
```

assistant: ```c
void trdtoi(s) char *s;
{
  for (;; ++s)
    switch (*s) {
    case '\000':
      return;
    case '\216':
      *s = '\340';
      break;
    case '\231':
      *s = '\344';
      break;
    case '\334':
      *s = '\346';
      break;
    }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<trdtoi>:
  endbr64
  movzbl (%rdi),%eax
  cmp    $0xdc,%al
  je     L41
  nopl   0x0(%rax,%rax,1)
L1:
  cmp    $0xdd,%al
  jge    L55
  cmp    $0x8e,%al
  je     L22
  cmp    $0x99,%al
  jne    L4
  movb   $0xe4,(%rdi)
L4:
  add    $0x1,%rdi
L58:
  movzbl (%rdi),%eax
  cmp    $0xdc,%al
  jne    L1
L41:
  movb   $0xe6,(%rdi)
  add    $0x1,%rdi
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L55:
  test   %al,%al
  jne    L4
  ret
  nopl   (%rax)
L22:
  movb   $0xe0,(%rdi)
  add    $0x1,%rdi
  jmp    L58
```

assistant: ```c
void trdtoi(s) char *s;
{
  for (;; ++s)
    switch (*s) {
    case '\000':
      return;
    case '\216':
      *s = '\340';
      break;
    case '\231':
      *s = '\344';
      break;
    case '\334':
      *s = '\346';
      break;
    }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<f_abs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    (%rax),%eax
  and    $0x7fffffff,%eax
  mov    %eax,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  movss  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
float f_abs(float f) {
  int i = ((*(int *)&f) & 0x7fffffff);
  return (*(float *)&i);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<f_abs>:
  endbr64
  movd   %xmm0,%eax
  and    $0x7fffffff,%eax
  movd   %eax,%xmm0
  ret
```

assistant: ```c
float f_abs(float f) {
  int i = ((*(int *)&f) & 0x7fffffff);
  return (*(float *)&i);
}
```
user: What is the c source code of the assembly code below:

```asm
<f_abs>:
  endbr64
  movd   %xmm0,%eax
  and    $0x7fffffff,%eax
  movd   %eax,%xmm0
  ret
```

assistant: ```c
float f_abs(float f) {
  int i = ((*(int *)&f) & 0x7fffffff);
  return (*(float *)&i);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<f_abs>:
  endbr64
  movd   %xmm0,%eax
  and    $0x7fffffff,%eax
  movd   %eax,%xmm0
  ret
```

assistant: ```c
float f_abs(float f) {
  int i = ((*(int *)&f) & 0x7fffffff);
  return (*(float *)&i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<tickOneSecond>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x3e8,-0xc(%rbp)
  mov    $0x0,%eax
  call   <clock@plt>
  mov    %rax,-0x8(%rbp)
  nop
L7:
  mov    $0x0,%eax
  call   <clock@plt>
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  cmp    %rdx,%rax
  jl     L7
  nop
  nop
  leave
  ret
```

assistant: ```c
void tickOneSecond() {
  clock_t start;
  int pauseLength = 1000;
  start = clock();
  while (clock() < start + pauseLength)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<tickOneSecond>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <clock@plt>
  lea    0x3e8(%rax),%rbx
L2:
  mov    $0x0,%eax
  call   <clock@plt>
  cmp    %rbx,%rax
  jl     L2
  pop    %rbx
  ret
```

assistant: ```c
void tickOneSecond() {
  clock_t start;
  int pauseLength = 1000;
  start = clock();
  while (clock() < start + pauseLength)
    ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<tickOneSecond>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <clock@plt>
  lea    0x3e8(%rax),%rbx
  nopl   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  call   <clock@plt>
  cmp    %rbx,%rax
  jl     L7
  pop    %rbx
  ret
```

assistant: ```c
void tickOneSecond() {
  clock_t start;
  int pauseLength = 1000;
  start = clock();
  while (clock() < start + pauseLength)
    ;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<tickOneSecond>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <clock@plt>
  lea    0x3e8(%rax),%rbx
  nopl   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  call   <clock@plt>
  cmp    %rbx,%rax
  jl     L7
  pop    %rbx
  ret
```

assistant: ```c
void tickOneSecond() {
  clock_t start;
  int pauseLength = 1000;
  start = clock();
  while (clock() < start + pauseLength)
    ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<scalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L13
  cmpq   $0x0,-0x18(%rbp)
  je     L13
  mov    -0x20(%rbp),%rax
  cltq
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  imul   %edx,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <fail@plt>
L13:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
data_type: string

tool: D2: %s: allocation of %d bytes failed


assistant: ```c
void *scalloc(size_t nmemb, size_t size, const char *file) {
  void *res = calloc(nmemb, size);
  if (!res && nmemb)
    fail("%s: allocation of %d bytes failed\n", file, (int)size * nmemb);
  return res;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<scalloc>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%r13
  call   <calloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L17
  test   %rbp,%rbp
  jne    L8
L17:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  mov    %r12d,%edx
  imul   %ebp,%edx
  mov    %r13,%rsi
  lea    D4(%rip),%rdi
  call   <fail@plt>
  jmp    L17
```

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

tool: D4: %s: allocation of %d bytes failed


assistant: ```c
void *scalloc(size_t nmemb, size_t size, const char *file) {
  void *res = calloc(nmemb, size);
  if (!res && nmemb)
    fail("%s: allocation of %d bytes failed\n", file, (int)size * nmemb);
  return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<scalloc>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L7
  test   %rbx,%rbx
  jne    L9
L7:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %ebp,%edx
  mov    %r13,%rsi
  lea    D7(%rip),%rdi
  imul   %ebx,%edx
  call   <fail@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: label: D7
type: string

tool: D7: %s: allocation of %d bytes failed


assistant: ```c
void *scalloc(size_t nmemb, size_t size, const char *file) {
  void *res = calloc(nmemb, size);
  if (!res && nmemb)
    fail("%s: allocation of %d bytes failed\n", file, (int)size * nmemb);
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<scalloc>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L10
  test   %rbx,%rbx
  jne    L11
L10:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %ebp,%edx
  mov    %r13,%rsi
  lea    D7(%rip),%rdi
  imul   %ebx,%edx
  call   <fail@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
datatype: string

tool: D7: %s: allocation of %d bytes failed


assistant: ```c
void *scalloc(size_t nmemb, size_t size, const char *file) {
  void *res = calloc(nmemb, size);
  if (!res && nmemb)
    fail("%s: allocation of %d bytes failed\n", file, (int)size * nmemb);
  return res;
}
```
user: What is the c source code of the assembly code below:

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x41,-0x3c(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %al,-0x12(%rbp)
  movl   $0x1,-0x50(%rbp)
  movl   $0x3,-0x4c(%rbp)
  movl   $0x5,-0x48(%rbp)
  movl   $0xc,-0x44(%rbp)
  mov    -0x50(%rbp),%eax
  cmp    $0x1,%eax
  je     L84
  mov    $0x1,%eax
  jmp    L61
L84:
  mov    -0x4c(%rbp),%eax
  cmp    $0x3,%eax
  je     L18
  mov    $0x1,%eax
  jmp    L61
L18:
  mov    -0x48(%rbp),%eax
  cmp    $0x5,%eax
  je     L48
  mov    $0x1,%eax
  jmp    L61
L48:
  mov    -0x44(%rbp),%eax
  cmp    $0xc,%eax
  je     L87
  mov    $0x1,%eax
  jmp    L61
L87:
  mov    $0x0,%eax
L61:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

assistant: ```c
int test() {
  int a[10];
  a[1] = 65;
  char s[10];
  s[0] = a[1];
  int b[4] = {1, 3, 5, 12};
  if (b[0] != 1)
    return 1;
  if (b[1] != 3)
    return 1;
  if (b[2] != 5)
    return 1;
  if (b[3] != 12)
    return 1;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int test() {
  int a[10];
  a[1] = 65;
  char s[10];
  s[0] = a[1];
  int b[4] = {1, 3, 5, 12};
  if (b[0] != 1)
    return 1;
  if (b[1] != 3)
    return 1;
  if (b[2] != 5)
    return 1;
  if (b[3] != 12)
    return 1;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int test() {
  int a[10];
  a[1] = 65;
  char s[10];
  s[0] = a[1];
  int b[4] = {1, 3, 5, 12};
  if (b[0] != 1)
    return 1;
  if (b[1] != 3)
    return 1;
  if (b[2] != 5)
    return 1;
  if (b[3] != 12)
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<test>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int test() {
  int a[10];
  a[1] = 65;
  char s[10];
  s[0] = a[1];
  int b[4] = {1, 3, 5, 12};
  if (b[0] != 1)
    return 1;
  if (b[1] != 3)
    return 1;
  if (b[2] != 5)
    return 1;
  if (b[3] != 12)
    return 1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<copy_result>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x18(%rbp)
  mov    %r8,-0x30(%rbp)
  mov    %r9,-0x38(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L16
L15:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L15
  movl   $0x0,-0x4(%rbp)
  jmp    L33
L17:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L33:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L17
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void copy_result(int in, int *const isrc, int *const itgt, int dn, double *const dsrc, double *const dtgt) {
  int i;
  for (i = 0; i < in; i++)
    itgt[i] = isrc[i];
  for (i = 0; i < dn; i++)
    dtgt[i] = dsrc[i];
}
```
user: Translate the below assembly code into C programming language:

```asm
<copy_result>:
  endbr64
  mov    %rsi,%r10
  test   %edi,%edi
  jle    L7
  mov    %edi,%edi
  mov    $0x0,%eax
L39:
  mov    (%r10,%rax,4),%esi
  mov    %esi,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L39
L7:
  test   %ecx,%ecx
  jle    L1
  mov    %ecx,%ecx
  mov    $0x0,%eax
L30:
  movsd  (%r8,%rax,8),%xmm0
  movsd  %xmm0,(%r9,%rax,8)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L30
L1:
  ret
```

assistant: ```c
void copy_result(int in, int *const isrc, int *const itgt, int dn, double *const dsrc, double *const dtgt) {
  int i;
  for (i = 0; i < in; i++)
    itgt[i] = isrc[i];
  for (i = 0; i < dn; i++)
    dtgt[i] = dsrc[i];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<copy_result>:
  endbr64
  mov    %rsi,%r10
  test   %edi,%edi
  jle    L21
  movslq %edi,%rdi
  xor    %eax,%eax
L1:
  mov    (%r10,%rax,4),%esi
  mov    %esi,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L1
L21:
  test   %ecx,%ecx
  jle    L18
  movslq %ecx,%rcx
  xor    %eax,%eax
  nopl   0x0(%rax)
L5:
  movsd  (%r8,%rax,8),%xmm0
  movsd  %xmm0,(%r9,%rax,8)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L5
L18:
  ret
```

assistant: ```c
void copy_result(int in, int *const isrc, int *const itgt, int dn, double *const dsrc, double *const dtgt) {
  int i;
  for (i = 0; i < in; i++)
    itgt[i] = isrc[i];
  for (i = 0; i < dn; i++)
    dtgt[i] = dsrc[i];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copy_result>:
  endbr64
  push   %rbx
  test   %edi,%edi
  jle    L18
  lea    0x4(%rsi),%r10
  mov    %rdx,%rax
  sub    %r10,%rax
  cmp    $0x8,%rax
  jbe    L47
  lea    -0x1(%rdi),%eax
  cmp    $0x2,%eax
  jbe    L47
  mov    %edi,%r10d
  xor    %eax,%eax
  shr    $0x2,%r10d
  shl    $0x4,%r10
  nopw   0x0(%rax,%rax,1)
L105:
  movdqu (%rsi,%rax,1),%xmm1
  movups %xmm1,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %r10,%rax
  jne    L105
  mov    %edi,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%dil
  je     L18
  mov    %eax,%r11d
  mov    (%rsi,%r11,4),%ebx
  lea    0x0(,%r11,4),%r10
  mov    %ebx,(%rdx,%r11,4)
  lea    0x1(%rax),%r11d
  cmp    %r11d,%edi
  jle    L18
  mov    0x4(%rsi,%r10,1),%r11d
  add    $0x2,%eax
  mov    %r11d,0x4(%rdx,%r10,1)
  cmp    %eax,%edi
  jle    L18
  mov    0x8(%rsi,%r10,1),%eax
  mov    %eax,0x8(%rdx,%r10,1)
L18:
  test   %ecx,%ecx
  jle    L3
  lea    0x8(%r8),%rax
  cmp    %rax,%r9
  je     L140
  lea    -0x1(%rcx),%eax
  cmp    $0x2,%eax
  jbe    L140
  mov    %ecx,%edx
  xor    %eax,%eax
  shr    %edx
  shl    $0x4,%rdx
  nopl   0x0(%rax,%rax,1)
L54:
  movupd (%r8,%rax,1),%xmm2
  movups %xmm2,(%r9,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L54
  mov    %ecx,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%ecx
  je     L3
  movsd  (%r8,%rax,8),%xmm0
  movsd  %xmm0,(%r9,%rax,8)
L3:
  pop    %rbx
  ret
  nopl   0x0(%rax)
L47:
  mov    %edi,%edi
  xor    %eax,%eax
  nopl   0x0(%rax)
L66:
  mov    (%rsi,%rax,4),%r10d
  mov    %r10d,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L66
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L140:
  mov    %ecx,%ecx
  xor    %eax,%eax
  nopl   0x0(%rax)
L89:
  movsd  (%r8,%rax,8),%xmm0
  movsd  %xmm0,(%r9,%rax,8)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L89
  pop    %rbx
  ret
```

assistant: ```c
void copy_result(int in, int *const isrc, int *const itgt, int dn, double *const dsrc, double *const dtgt) {
  int i;
  for (i = 0; i < in; i++)
    itgt[i] = isrc[i];
  for (i = 0; i < dn; i++)
    dtgt[i] = dsrc[i];
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<scalar_triple>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <cross_product@plt>
  lea    -0x20(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <dot_product@plt>
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double scalar_triple(const double *u, const double *v, const double *w) {
  double tmp[3];
  cross_product(w, u, tmp);
  return dot_product(v, tmp);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<scalar_triple>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rax
  mov    %rsi,%rbx
  mov    %rdx,%rdi
  mov    %fs:0x28,%rdx
  mov    %rdx,0x18(%rsp)
  xor    %edx,%edx
  mov    %rsp,%rbp
  mov    %rbp,%rdx
  mov    %rax,%rsi
  call   <cross_product@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <dot_product@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double scalar_triple(const double *u, const double *v, const double *w) {
  double tmp[3];
  cross_product(w, u, tmp);
  return dot_product(v, tmp);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<scalar_triple>:
  endbr64
  push   %r12
  mov    %rdi,%r8
  mov    %rdx,%rdi
  push   %rbp
  mov    %rsi,%rbp
  mov    %r8,%rsi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdx
  call   <cross_product@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <dot_product@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double scalar_triple(const double *u, const double *v, const double *w) {
  double tmp[3];
  cross_product(w, u, tmp);
  return dot_product(v, tmp);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<scalar_triple>:
  endbr64
  push   %r12
  mov    %rdi,%r8
  mov    %rdx,%rdi
  push   %rbp
  mov    %rsi,%rbp
  mov    %r8,%rsi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdx
  call   <cross_product@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <dot_product@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double scalar_triple(const double *u, const double *v, const double *w) {
  double tmp[3];
  cross_product(w, u, tmp);
  return dot_product(v, tmp);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<systemInitPower>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <__init_main_power@plt>
  mov    $0x0,%eax
  call   <__init_usb_host@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void systemInitPower(void) {
  __init_main_power();
  __init_usb_host();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<systemInitPower>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <__init_main_power@plt>
  mov    $0x0,%eax
  call   <__init_usb_host@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void systemInitPower(void) {
  __init_main_power();
  __init_usb_host();
}
```
user: Translate the below assembly code into C programming language:

```asm
<systemInitPower>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <__init_main_power@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <__init_usb_host@plt>
```

assistant: ```c
void systemInitPower(void) {
  __init_main_power();
  __init_usb_host();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<systemInitPower>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <__init_main_power@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <__init_usb_host@plt>
```

assistant: ```c
void systemInitPower(void) {
  __init_main_power();
  __init_usb_host();
}
```