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

```asm
<btff_memmove>:
  endbr64
  mov    %rsi,%rcx
  mov    %rdi,%rsi
  mov    %rdi,%rax
  or     %rcx,%rsi
  and    $0x3,%esi
  je     L63
L8:
  cmp    %rcx,%rax
  jae    L143
L115:
  test   %edx,%edx
  jle    L153
  lea    -0x1(%rdx),%edi
  xor    %edx,%edx
  lea    0x1(%rdi),%r8
  nopw   0x0(%rax,%rax,1)
L14:
  movzbl (%rcx,%rdx,1),%esi
  mov    %sil,(%rax,%rdx,1)
  mov    %rdx,%rsi
  add    $0x1,%rdx
  cmp    %rsi,%rdi
  jne    L14
  add    %r8,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L143:
  movslq %edx,%rsi
  add    %rcx,%rsi
  cmp    %rsi,%rax
  jae    L115
  cmp    %rcx,%rax
  ja     L105
L153:
  ret
  nopl   0x0(%rax)
L63:
  test   $0x3,%dl
  jne    L8
  cmp    %rcx,%rdi
  jb     L13
  movslq %edx,%rdi
  add    %rcx,%rdi
  cmp    %rdi,%rax
  jae    L13
  cmp    %rcx,%rax
  jbe    L153
  sub    $0x4,%edx
  js     L153
  movslq %edx,%rdx
  nopw   0x0(%rax,%rax,1)
L18:
  mov    (%rcx,%rdx,1),%rsi
  mov    %rsi,(%rax,%rdx,1)
  sub    $0x4,%rdx
  test   %edx,%edx
  jns    L18
  ret
  nopl   0x0(%rax)
L105:
  sub    $0x1,%edx
  js     L153
  movslq %edx,%rdx
L64:
  movzbl (%rcx,%rdx,1),%esi
  mov    %sil,(%rax,%rdx,1)
  sub    $0x1,%rdx
  test   %edx,%edx
  jns    L64
  ret
  nopl   0x0(%rax)
L13:
  cmp    $0x3,%edx
  jle    L153
  sub    $0x4,%edx
  shr    $0x2,%edx
  add    $0x1,%edx
  shl    $0x2,%rdx
  nopw   0x0(%rax,%rax,1)
L73:
  mov    (%rcx,%rsi,1),%rdi
  mov    %rdi,(%rax,%rsi,1)
  add    $0x4,%rsi
  cmp    %rdx,%rsi
  jne    L73
  add    %rsi,%rax
  ret
```

assistant: ```c
void *btff_memmove(register void *dest, register void *src, register int n) {
  if ((0x03 & (unsigned long)dest) || (0x03 & (unsigned long)src) || (0x03 & n)) {
    if (dest < src || src + n <= dest)
      for (n--; n >= 0; n--, dest++, src++)
        (*(unsigned char *)dest) = (*(unsigned char *)src);
    else if (src < dest)
      for (n--; n >= 0; n--)
        (*(unsigned char *)(dest + n)) = (*(unsigned char *)(src + n));
  } else {
    if (dest < src || src + n <= dest)
      for (n -= 4; n >= 0; n -= 4, dest += 4, src += 4)
        (*(unsigned long *)dest) = (*(unsigned long *)src);
    else if (src < dest)
      for (n -= 4; n >= 0; n -= 4)
        (*(unsigned long *)(dest + n)) = (*(unsigned long *)(src + n));
  }
  return dest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<running_stairs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L11
  mov    $0x0,%eax
  jmp    L2
L11:
  cmpl   $0x0,-0x14(%rbp)
  jne    L31
  mov    $0x1,%eax
  jmp    L2
L31:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <running_stairs>
  mov    %rax,%rbx
  mov    -0x14(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,%edi
  call   <running_stairs>
  add    %rax,%rbx
  mov    -0x14(%rbp),%eax
  sub    $0x3,%eax
  mov    %eax,%edi
  call   <running_stairs>
  add    %rbx,%rax
L2:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long running_stairs(int s) {
  if (s < 0) {
    return 0;
  } else if (s == 0) {
    return 1;
  } else {
    return running_stairs(s - 1) + running_stairs(s - 2) + running_stairs(s - 3);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<running_stairs>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  test   %edi,%edi
  js     L15
  mov    %edi,%ebx
  mov    $0x1,%eax
  jne    L10
L15:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  lea    -0x1(%rdi),%edi
  call   <running_stairs>
  mov    %rax,%rbp
  lea    -0x2(%rbx),%edi
  call   <running_stairs>
  add    %rax,%rbp
  lea    -0x3(%rbx),%edi
  call   <running_stairs>
  add    %rbp,%rax
  jmp    L15
```

assistant: ```c
long running_stairs(int s) {
  if (s < 0) {
    return 0;
  } else if (s == 0) {
    return 1;
  } else {
    return running_stairs(s - 1) + running_stairs(s - 2) + running_stairs(s - 3);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<wait_msec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x10624dd3,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x3e8,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  imul   $0xf4240,%edx,%eax
  mov    %eax,-0xc(%rbp)
  lea    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <nanosleep@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
int wait_msec(int msec) {
  struct timespec req;
  req.tv_sec = msec / 1000;
  req.tv_nsec = (msec % 1000) * 1000 * 1000;
  return nanosleep(&req, NULL);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<wait_msec>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movslq %edi,%rax
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  mov    %edi,0x4(%rsp)
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <nanosleep@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int wait_msec(int msec) {
  struct timespec req;
  req.tv_sec = msec / 1000;
  req.tv_nsec = (msec % 1000) * 1000 * 1000;
  return nanosleep(&req, NULL);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<wait_msec>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movslq %edi,%rax
  sar    $0x1f,%edx
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  sub    %edx,%eax
  mov    %eax,(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  mov    %edi,0x4(%rsp)
  mov    %rsp,%rdi
  call   <nanosleep@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int wait_msec(int msec) {
  struct timespec req;
  req.tv_sec = msec / 1000;
  req.tv_nsec = (msec % 1000) * 1000 * 1000;
  return nanosleep(&req, NULL);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<wait_msec>:
  endbr64
  sub    $0x18,%rsp
  mov    %edi,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movslq %edi,%rax
  sar    $0x1f,%edx
  imul   $0x10624dd3,%rax,%rax
  sar    $0x26,%rax
  sub    %edx,%eax
  imul   $0x3e8,%eax,%edx
  movd   %eax,%xmm0
  sub    %edx,%edi
  imul   $0xf4240,%edi,%edi
  movd   %edi,%xmm1
  mov    %rsp,%rdi
  punpckldq %xmm1,%xmm0
  movq   %xmm0,(%rsp)
  call   <nanosleep@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int wait_msec(int msec) {
  struct timespec req;
  req.tv_sec = msec / 1000;
  req.tv_nsec = (msec % 1000) * 1000 * 1000;
  return nanosleep(&req, NULL);
}
```
user: Translate the below assembly code into C programming language:

```asm
<mkrecord>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L18
  mov    $0x0,%eax
  jmp    L4
L18:
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%rax
L4:
  leave
  ret
```

assistant: ```c
char *mkrecord(uint32_t size, int pattern) {
  char *record;
  if ((record = malloc(size)) == NULL)
    return NULL;
  memset(record, pattern, size);
  return record;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mkrecord>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %esi,%r12d
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L4
  mov    %ebp,%edx
  mov    %r12d,%esi
  mov    %rax,%rdi
  call   <memset@plt>
L4:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *mkrecord(uint32_t size, int pattern) {
  char *record;
  if ((record = malloc(size)) == NULL)
    return NULL;
  memset(record, pattern, size);
  return record;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<mkrecord>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L4
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <memset@plt>
L4:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *mkrecord(uint32_t size, int pattern) {
  char *record;
  if ((record = malloc(size)) == NULL)
    return NULL;
  memset(record, pattern, size);
  return record;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mkrecord>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L0
  mov    %r13d,%edx
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <memset@plt>
L0:
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *mkrecord(uint32_t size, int pattern) {
  char *record;
  if ((record = malloc(size)) == NULL)
    return NULL;
  memset(record, pattern, size);
  return record;
}
```
user: What is the c source code of the assembly code below?

```asm
<string_append>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x30(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *string_append(const char *prefix, const char *suffix) {
  char *buf = (char *)malloc(strlen(prefix) + strlen(suffix));
  buf[0] = '\0';
  strcat(buf, prefix);
  strcat(buf, suffix);
  return buf;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<string_append>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    (%rbx,%rax,1),%rdi
  call   <malloc@plt>
  mov    %rax,%rbx
  movb   $0x0,(%rax)
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strcat@plt>
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *string_append(const char *prefix, const char *suffix) {
  char *buf = (char *)malloc(strlen(prefix) + strlen(suffix));
  buf[0] = '\0';
  strcat(buf, prefix);
  strcat(buf, suffix);
  return buf;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<string_append>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    (%r12,%rax,1),%rdi
  call   <malloc@plt>
  mov    %r13,%rsi
  movb   $0x0,(%rax)
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcat@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *string_append(const char *prefix, const char *suffix) {
  char *buf = (char *)malloc(strlen(prefix) + strlen(suffix));
  buf[0] = '\0';
  strcat(buf, prefix);
  strcat(buf, suffix);
  return buf;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<string_append>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <strlen@plt>
  lea    (%r12,%rax,1),%rdi
  call   <malloc@plt>
  mov    %r13,%rsi
  movb   $0x0,(%rax)
  mov    %rax,%r12
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcat@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *string_append(const char *prefix, const char *suffix) {
  char *buf = (char *)malloc(strlen(prefix) + strlen(suffix));
  buf[0] = '\0';
  strcat(buf, prefix);
  strcat(buf, suffix);
  return buf;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<image_char2double>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L2
L13:
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L24:
  mov    -0x8(%rbp),%eax
  cltq
  imul   $0x310,%rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  movzbl %al,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  imul   $0x1880,%rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  movsd  D6(%rip),%xmm1
  divsd  %xmm1,%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  movsd  %xmm0,(%rdx,%rax,8)
  addl   $0x1,-0x4(%rbp)
L15:
  cmpl   $0x30f,-0x4(%rbp)
  jle    L24
  addl   $0x1,-0x8(%rbp)
L2:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L13
  nop
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D6
type_infered: f64

tool: D6: 255

assistant: ```c
void image_char2double(int num_data, unsigned char data_image_char[][784], double data_image[][784]) {
  int i, j;
  for (i = 0; i < num_data; i++)
    for (j = 0; j < 784; j++)
      data_image[i][j] = (double)data_image_char[i][j] / 255.0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<image_char2double>:
  endbr64
  mov    %edi,%ecx
  mov    %rdx,%r8
  test   %edi,%edi
  jle    L8
  mov    %rsi,%rdi
  lea    -0x1(%rcx),%edx
  imul   $0x310,%rdx,%rdx
  lea    0x310(%rsi,%rdx,1),%r9
  movsd  D0(%rip),%xmm1
L20:
  mov    %rdi,%rsi
  mov    %r8,%rcx
  mov    $0x0,%eax
L2:
  movzbl (%rsi,%rax,1),%edx
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  add    $0x1,%rax
  cmp    $0x310,%rax
  jne    L2
  add    $0x310,%rdi
  add    $0x1880,%r8
  cmp    %r9,%rdi
  jne    L20
L8:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
type_infered: f64

tool: D0: 255

assistant: ```c
void image_char2double(int num_data, unsigned char data_image_char[][784], double data_image[][784]) {
  int i, j;
  for (i = 0; i < num_data; i++)
    for (j = 0; j < 784; j++)
      data_image[i][j] = (double)data_image_char[i][j] / 255.0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<image_char2double>:
  endbr64
  mov    %rsi,%rcx
  mov    %edi,%eax
  mov    %rdx,%rsi
  test   %edi,%edi
  jle    L22
  sub    $0x1,%eax
  movsd  D5(%rip),%xmm1
  imul   $0x310,%rax,%rax
  lea    0x310(%rcx,%rax,1),%rdi
L10:
  xor    %eax,%eax
  nopl   0x0(%rax)
L19:
  movzbl (%rcx,%rax,1),%edx
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    $0x310,%rax
  jne    L19
  add    $0x310,%rcx
  add    $0x1880,%rsi
  cmp    %rdi,%rcx
  jne    L10
L22:
  ret
```

tool_calls: name: unpack_data
arguments: label: D5
type_guessed: f64

tool: D5: 255

assistant: ```c
void image_char2double(int num_data, unsigned char data_image_char[][784], double data_image[][784]) {
  int i, j;
  for (i = 0; i < num_data; i++)
    for (j = 0; j < 784; j++)
      data_image[i][j] = (double)data_image_char[i][j] / 255.0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<image_char2double>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%rcx
  mov    %rdx,%r8
  test   %edi,%edi
  jle    L14
  sub    $0x1,%eax
  pxor   %xmm6,%xmm6
  pxor   %xmm4,%xmm4
  movapd D4(%rip),%xmm3
  imul   $0x310,%rax,%rax
  movsd  D4(%rip),%xmm5
  lea    0x310(%rsi),%rdi
  lea    0x620(%rsi,%rax,1),%r9
L23:
  mov    %r8,%rsi
  mov    %r8,%rax
  add    $0x1880,%r8
  cmp    %rdi,%rsi
  jae    L10
  cmp    %r8,%rcx
  jb     L29
L10:
  mov    %rcx,%rdx
  nopl   0x0(%rax,%rax,1)
L49:
  movdqu (%rdx),%xmm0
  add    $0x10,%rdx
  sub    $0xffffffffffffff80,%rax
  movdqa %xmm0,%xmm1
  punpckhbw %xmm6,%xmm0
  punpcklbw %xmm6,%xmm1
  movdqa %xmm0,%xmm2
  punpckhwd %xmm4,%xmm0
  movdqa %xmm1,%xmm7
  punpckhwd %xmm4,%xmm1
  punpcklwd %xmm4,%xmm2
  punpcklwd %xmm4,%xmm7
  cvtdq2pd %xmm7,%xmm8
  divpd  %xmm3,%xmm8
  pshufd $0xee,%xmm7,%xmm7
  cvtdq2pd %xmm7,%xmm7
  divpd  %xmm3,%xmm7
  movups %xmm8,-0x80(%rax)
  movups %xmm7,-0x70(%rax)
  cvtdq2pd %xmm1,%xmm7
  pshufd $0xee,%xmm1,%xmm1
  cvtdq2pd %xmm1,%xmm1
  divpd  %xmm3,%xmm1
  divpd  %xmm3,%xmm7
  movups %xmm1,-0x50(%rax)
  cvtdq2pd %xmm2,%xmm1
  pshufd $0xee,%xmm2,%xmm2
  cvtdq2pd %xmm2,%xmm2
  divpd  %xmm3,%xmm1
  movups %xmm7,-0x60(%rax)
  divpd  %xmm3,%xmm2
  movups %xmm1,-0x40(%rax)
  cvtdq2pd %xmm0,%xmm1
  pshufd $0xee,%xmm0,%xmm0
  cvtdq2pd %xmm0,%xmm0
  divpd  %xmm3,%xmm1
  movups %xmm2,-0x30(%rax)
  divpd  %xmm3,%xmm0
  movups %xmm1,-0x20(%rax)
  movups %xmm0,-0x10(%rax)
  cmp    %rdi,%rdx
  jne    L49
L9:
  add    $0x310,%rdi
  add    $0x310,%rcx
  cmp    %r9,%rdi
  jne    L23
L14:
  ret
L29:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L45:
  movzbl (%rcx,%rax,1),%edx
  pxor   %xmm0,%xmm0
  cvtsi2sd %edx,%xmm0
  divsd  %xmm5,%xmm0
  movsd  %xmm0,(%rsi,%rax,8)
  add    $0x1,%rax
  cmp    $0x310,%rax
  jne    L45
  jmp    L9
```

tool_calls: name: parse_data
arguments: label: D4
type_guessed: f64[2]

tool: {"D4": [255.0,255.0]}

assistant: ```c
void image_char2double(int num_data, unsigned char data_image_char[][784], double data_image[][784]) {
  int i, j;
  for (i = 0; i < num_data; i++)
    for (j = 0; j < 784; j++)
      data_image[i][j] = (double)data_image_char[i][j] / 255.0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<encode_nonprint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x5c,-0xe(%rbp)
  movb   $0x75,-0xd(%rbp)
  movb   $0x30,-0xc(%rbp)
  movb   $0x30,-0xb(%rbp)
  mov    $0x0,%edi
  call   <get_hexdigit@plt>
  mov    %al,-0xa(%rbp)
  movsbl -0x1c(%rbp),%eax
  and    $0xf,%eax
  mov    %eax,%edi
  call   <get_hexdigit@plt>
  mov    %al,-0x9(%rbp)
  lea    -0xe(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x6,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <buffer_append_str@plt>
  test   %rax,%rax
  jns    L15
  mov    $0x1,%eax
  jmp    L6
L15:
  mov    $0x1,%eax
L6:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
int encode_nonprint(struct buffer *output, char c) {
  char str[6] = {'\\', 'u', '0', '0', get_hexdigit((c & 0xf0) >> 8), get_hexdigit(c & 0xf)};
  if (buffer_append_str(output, str, sizeof(str)) < 0)
    return 1;
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<encode_nonprint>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x5c,0x2(%rsp)
  movb   $0x75,0x3(%rsp)
  movb   $0x30,0x4(%rsp)
  movb   $0x30,0x5(%rsp)
  mov    $0x0,%edi
  call   <get_hexdigit@plt>
  mov    %al,0x6(%rsp)
  and    $0xf,%ebx
  mov    %ebx,%edi
  call   <get_hexdigit@plt>
  mov    %al,0x7(%rsp)
  lea    0x2(%rsp),%rsi
  mov    $0x6,%edx
  mov    %rbp,%rdi
  call   <buffer_append_str@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x1,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int encode_nonprint(struct buffer *output, char c) {
  char str[6] = {'\\', 'u', '0', '0', get_hexdigit((c & 0xf0) >> 8), get_hexdigit(c & 0xf)};
  if (buffer_append_str(output, str, sizeof(str)) < 0)
    return 1;
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<encode_nonprint>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %esi,%ebx
  and    $0xf,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x3030755c,0x2(%rsp)
  call   <get_hexdigit@plt>
  mov    %ebx,%edi
  mov    %al,0x6(%rsp)
  call   <get_hexdigit@plt>
  lea    0x2(%rsp),%rsi
  mov    $0x6,%edx
  mov    %rbp,%rdi
  mov    %al,0x7(%rsp)
  call   <buffer_append_str@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int encode_nonprint(struct buffer *output, char c) {
  char str[6] = {'\\', 'u', '0', '0', get_hexdigit((c & 0xf0) >> 8), get_hexdigit(c & 0xf)};
  if (buffer_append_str(output, str, sizeof(str)) < 0)
    return 1;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<encode_nonprint>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %esi,%ebx
  and    $0xf,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x3030755c,(%rsp)
  call   <get_hexdigit@plt>
  mov    %ebx,%edi
  mov    %al,0x4(%rsp)
  call   <get_hexdigit@plt>
  mov    %rsp,%rsi
  mov    $0x6,%edx
  mov    %rbp,%rdi
  mov    %al,0x5(%rsp)
  call   <buffer_append_str@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int encode_nonprint(struct buffer *output, char c) {
  char str[6] = {'\\', 'u', '0', '0', get_hexdigit((c & 0xf0) >> 8), get_hexdigit(c & 0xf)};
  if (buffer_append_str(output, str, sizeof(str)) < 0)
    return 1;
  return 1;
}
```
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
  mov    $0x0,%eax
  call   <show_info_a@plt>
  mov    $0x0,%eax
  call   <show_info_b@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  show_info_a();
  show_info_b();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <show_info_a@plt>
  mov    $0x0,%eax
  call   <show_info_b@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  show_info_a();
  show_info_b();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <show_info_a@plt>
  xor    %eax,%eax
  call   <show_info_b@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  show_info_a();
  show_info_b();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <show_info_a@plt>
  xor    %eax,%eax
  call   <show_info_b@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  show_info_a();
  show_info_b();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<intSize>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x2,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
int intSize() {
  int intArray[10];
  int *intPtr1;
  int *intPtr2;
  return 2;
}
```