ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码:

```asm
<writeSection>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  movzbl %sil,%edi
  push   %rbx
  mov    %rbp,%rsi
  mov    %rdx,%rbx
  call   <putc@plt>
  mov    %r12d,%eax
  mov    %rbp,%rsi
  movzbl %ah,%edi
  call   <putc@plt>
  mov    %r12,%rdi
  mov    %rbp,%rsi
  shr    $0x10,%rdi
  movzbl %dil,%edi
  call   <putc@plt>
  mov    %r12d,%edi
  mov    %rbp,%rsi
  shr    $0x18,%edi
  call   <putc@plt>
  movzbl %bl,%edi
  mov    %rbp,%rsi
  call   <putc@plt>
  movzbl %bh,%edi
  mov    %rbp,%rsi
  call   <putc@plt>
  mov    %rbx,%rdi
  mov    %rbp,%rsi
  shr    $0x10,%rdi
  movzbl %dil,%edi
  call   <putc@plt>
  mov    %ebx,%edi
  mov    %rbp,%rsi
  pop    %rbx
  shr    $0x18,%edi
  pop    %rbp
  pop    %r12
  jmp    1050 <putc@plt>
```

assistant: ```c
void writeSection(FILE *file, unsigned long num, unsigned long size) {
  putc((int)(num & 0xFF), file);
  putc((int)((num >> 8) & 0xFF), file);
  putc((int)((num >> 16) & 0xFF), file);
  putc((int)((num >> 24) & 0xFF), file);
  putc((int)(size & 0xFF), file);
  putc((int)((size >> 8) & 0xFF), file);
  putc((int)((size >> 16) & 0xFF), file);
  putc((int)((size >> 24) & 0xFF), file);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<writeSection>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  movzbl %sil,%edi
  push   %rbx
  mov    %rbp,%rsi
  mov    %rdx,%rbx
  call   <putc@plt>
  mov    %r12d,%eax
  mov    %rbp,%rsi
  movzbl %ah,%edi
  call   <putc@plt>
  mov    %r12,%rdi
  mov    %rbp,%rsi
  shr    $0x10,%rdi
  movzbl %dil,%edi
  call   <putc@plt>
  mov    %r12d,%edi
  mov    %rbp,%rsi
  shr    $0x18,%edi
  call   <putc@plt>
  movzbl %bl,%edi
  mov    %rbp,%rsi
  call   <putc@plt>
  movzbl %bh,%edi
  mov    %rbp,%rsi
  call   <putc@plt>
  mov    %rbx,%rdi
  mov    %rbp,%rsi
  shr    $0x10,%rdi
  movzbl %dil,%edi
  call   <putc@plt>
  mov    %ebx,%edi
  mov    %rbp,%rsi
  pop    %rbx
  shr    $0x18,%edi
  pop    %rbp
  pop    %r12
  jmp    1050 <putc@plt>
```

assistant: ```c
void writeSection(FILE *file, unsigned long num, unsigned long size) {
  putc((int)(num & 0xFF), file);
  putc((int)((num >> 8) & 0xFF), file);
  putc((int)((num >> 16) & 0xFF), file);
  putc((int)((num >> 24) & 0xFF), file);
  putc((int)(size & 0xFF), file);
  putc((int)((size >> 8) & 0xFF), file);
  putc((int)((size >> 16) & 0xFF), file);
  putc((int)((size >> 24) & 0xFF), file);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pi_pack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0xc(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <ceil@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L17
L10:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  shl    $0x4,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  add    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x2,-0x14(%rbp)
L17:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L10
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x30(%rbp),%rax
  mov    -0x18(%rbp),%edx
  mov    %edx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void pi_pack(char **output, int *output_size, char *input, int input_size) {
  int byte = 0, size = 0, i = 0, l = 0;
  char *out = NULL;
  out = malloc(sizeof(char) * ceil((double)(input_size / 2)) + 1);
  for (i = 0, l = input_size; i < l; i += 2) {
    byte = ((int)input[i] - 48) << 4;
    byte += (int)input[i + 1] - 48;
    out[size++] = (char)byte;
  }
  *output = out;
  *output_size = size;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<pi_pack>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %ecx,%r13d
  mov    %ecx,%eax
  shr    $0x1f,%eax
  add    %ecx,%eax
  sar    %eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  call   <ceil@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  test   %r13d,%r13d
  jle    L9
  sub    $0x1,%r13d
  shr    %r13d
  lea    0x1(%r13),%edi
  mov    $0x0,%edx
L14:
  movsbl (%rbx,%rdx,2),%ecx
  sub    $0x30,%ecx
  shl    $0x4,%ecx
  movsbl 0x1(%rbx,%rdx,2),%esi
  lea    -0x30(%rcx,%rsi,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L14
  add    $0x1,%r13d
L6:
  mov    %rax,(%r12)
  mov    %r13d,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  mov    $0x0,%r13d
  jmp    L6
```

assistant: ```c
void pi_pack(char **output, int *output_size, char *input, int input_size) {
  int byte = 0, size = 0, i = 0, l = 0;
  char *out = NULL;
  out = malloc(sizeof(char) * ceil((double)(input_size / 2)) + 1);
  for (i = 0, l = input_size; i < l; i += 2) {
    byte = ((int)input[i] - 48) << 4;
    byte += (int)input[i + 1] - 48;
    out[size++] = (char)byte;
  }
  *output = out;
  *output_size = size;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pi_pack>:
  endbr64
  mov    %ecx,%eax
  pxor   %xmm0,%xmm0
  push   %r13
  mov    %rsi,%r13
  shr    $0x1f,%eax
  push   %r12
  mov    %rdi,%r12
  add    %ecx,%eax
  push   %rbp
  mov    %ecx,%ebp
  sar    %eax
  push   %rbx
  mov    %rdx,%rbx
  cvtsi2sd %eax,%xmm0
  sub    $0x8,%rsp
  call   <ceil@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rsi
  test   %ebp,%ebp
  jle    L11
  sub    $0x1,%ebp
  xor    %eax,%eax
  shr    %ebp
  lea    0x1(%rbp),%edi
  mov    %rdi,%rbp
  nopl   0x0(%rax,%rax,1)
L17:
  movsbl (%rbx,%rax,2),%edx
  movsbl 0x1(%rbx,%rax,2),%ecx
  sub    $0x30,%edx
  shl    $0x4,%edx
  lea    -0x30(%rdx,%rcx,1),%edx
  mov    %dl,(%rsi,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rdi
  jne    L17
L15:
  mov    %rsi,(%r12)
  mov    %ebp,0x0(%r13)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  xor    %ebp,%ebp
  jmp    L15
```

assistant: ```c
void pi_pack(char **output, int *output_size, char *input, int input_size) {
  int byte = 0, size = 0, i = 0, l = 0;
  char *out = NULL;
  out = malloc(sizeof(char) * ceil((double)(input_size / 2)) + 1);
  for (i = 0, l = input_size; i < l; i += 2) {
    byte = ((int)input[i] - 48) << 4;
    byte += (int)input[i + 1] - 48;
    out[size++] = (char)byte;
  }
  *output = out;
  *output_size = size;
}
```
user: Translate the below assembly code into C programming language:

```asm
<movehv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  %xmm0,-0x8(%rbp)
  movsd  %xmm1,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <hgoto@plt>
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <vgoto@plt>
  nop
  leave
  ret
```

assistant: ```c
void movehv(double h, double v) {
  hgoto(h);
  vgoto(v);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<movehv>:
  endbr64
  push   %rbx
  movq   %xmm1,%rbx
  call   <hgoto@plt>
  movq   %rbx,%xmm0
  call   <vgoto@plt>
  pop    %rbx
  ret
```

assistant: ```c
void movehv(double h, double v) {
  hgoto(h);
  vgoto(v);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<movehv>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm1,0x8(%rsp)
  call   <hgoto@plt>
  movsd  0x8(%rsp),%xmm1
  add    $0x18,%rsp
  movapd %xmm1,%xmm0
  jmp    1060 <vgoto@plt>
```

assistant: ```c
void movehv(double h, double v) {
  hgoto(h);
  vgoto(v);
}
```
user: Translate the below assembly code into C programming language:

```asm
<movehv>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm1,0x8(%rsp)
  call   <hgoto@plt>
  movsd  0x8(%rsp),%xmm1
  add    $0x18,%rsp
  movapd %xmm1,%xmm0
  jmp    1060 <vgoto@plt>
```

assistant: ```c
void movehv(double h, double v) {
  hgoto(h);
  vgoto(v);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<apply_h_mod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L8
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  lea    D4(%rip),%rdx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L20
L8:
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rsi
  mov    -0x20(%rbp),%rax
  mov    %rdx,%rcx
  lea    D2(%rip),%rdx
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <is_root_dir@plt>
  test   %eax,%eax
  jne    L20
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <is_root_dir@plt>
  test   %eax,%eax
  jne    L20
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L20:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
type: string
name: unpack_data
arguments: tag: D2
type: string

tool: D4: .

tool: D2: %s

assistant: ```c
int apply_h_mod(const char *path, char *buf, size_t buf_len) {
  char *p = strrchr(path, '/');
  if (p == NULL) {
    snprintf(buf, buf_len, ".");
  } else {
    snprintf(buf, buf_len, "%s", path);
    if (!is_root_dir(path)) {
      buf[p - path + 1] = '\0';
      if (!is_root_dir(buf))
        buf[p - path] = '\0';
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<apply_h_mod>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    $0x2f,%esi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L42
  mov    %rax,%rbp
  mov    %rbx,%rcx
  lea    D17(%rip),%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  mov    %rbx,%rdi
  call   <is_root_dir@plt>
  test   %eax,%eax
  je     L45
L2:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L42:
  lea    D6(%rip),%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <snprintf@plt>
  jmp    L2
L45:
  sub    %rbx,%rbp
  movb   $0x0,0x1(%r12,%rbp,1)
  mov    %r12,%rdi
  call   <is_root_dir@plt>
  test   %eax,%eax
  jne    L2
  movb   $0x0,(%r12,%rbp,1)
  jmp    L2
```

tool_calls: name: parse_data
arguments: label: D17
type_guessed: string
name: parse_data
arguments: label: D6
type_guessed: string

tool: D17: %s

tool: D6: .

assistant: ```c
int apply_h_mod(const char *path, char *buf, size_t buf_len) {
  char *p = strrchr(path, '/');
  if (p == NULL) {
    snprintf(buf, buf_len, ".");
  } else {
    snprintf(buf, buf_len, "%s", path);
    if (!is_root_dir(path)) {
      buf[p - path + 1] = '\0';
      if (!is_root_dir(buf))
        buf[p - path] = '\0';
    }
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<apply_h_mod>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  mov    $0x2f,%esi
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strrchr@plt>
  test   %rax,%rax
  je     L26
  mov    %r12,%rdi
  mov    %rax,%rbp
  mov    %rbx,%rcx
  xor    %eax,%eax
  lea    D19(%rip),%rdx
  mov    %r13,%rsi
  call   <snprintf@plt>
  mov    %rbx,%rdi
  call   <is_root_dir@plt>
  test   %eax,%eax
  je     L27
L23:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L27:
  sub    %rbx,%rbp
  mov    %r12,%rdi
  movb   $0x0,0x1(%r12,%rbp,1)
  call   <is_root_dir@plt>
  test   %eax,%eax
  jne    L23
  movb   $0x0,(%r12,%rbp,1)
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  mov    %r13,%rsi
  mov    %r12,%rdi
  lea    D16(%rip),%rdx
  xor    %eax,%eax
  call   <snprintf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: data_label: D19
type: string
name: parse_data
arguments: data_label: D16
type: string

tool: D19: %s

tool: D16: .

assistant: ```c
int apply_h_mod(const char *path, char *buf, size_t buf_len) {
  char *p = strrchr(path, '/');
  if (p == NULL) {
    snprintf(buf, buf_len, ".");
  } else {
    snprintf(buf, buf_len, "%s", path);
    if (!is_root_dir(path)) {
      buf[p - path + 1] = '\0';
      if (!is_root_dir(buf))
        buf[p - path] = '\0';
    }
  }
  return 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
<apply_h_mod>:
  endbr64
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rsi,%r12
  mov    $0x2f,%esi
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strrchr@plt>
  test   %rax,%rax
  je     L24
  mov    %r12,%rdi
  mov    %rax,%rbp
  mov    %rbx,%rcx
  xor    %eax,%eax
  lea    D10(%rip),%rdx
  mov    %r13,%rsi
  call   <snprintf@plt>
  mov    %rbx,%rdi
  call   <is_root_dir@plt>
  test   %eax,%eax
  je     L27
L5:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L27:
  sub    %rbx,%rbp
  mov    %r12,%rdi
  movb   $0x0,0x1(%r12,%rbp,1)
  call   <is_root_dir@plt>
  test   %eax,%eax
  jne    L5
  movb   $0x0,(%r12,%rbp,1)
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L24:
  mov    %r13,%rsi
  mov    %r12,%rdi
  lea    D0(%rip),%rdx
  xor    %eax,%eax
  call   <snprintf@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D10
type_guessed: string
name: disassemble_data
arguments: data_label: D0
type_guessed: string

tool: D10: %s

tool: D0: .

assistant: ```c
int apply_h_mod(const char *path, char *buf, size_t buf_len) {
  char *p = strrchr(path, '/');
  if (p == NULL) {
    snprintf(buf, buf_len, ".");
  } else {
    snprintf(buf, buf_len, "%s", path);
    if (!is_root_dir(path)) {
      buf[p - path + 1] = '\0';
      if (!is_root_dir(buf))
        buf[p - path] = '\0';
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L8
L16:
  lea    -0x18(%rbp),%rcx
  lea    -0x1c(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x20(%rbp),%eax
  mov    -0x1c(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x20(%rbp),%eax
  lea    (%rcx,%rax,1),%edx
  mov    -0x18(%rbp),%eax
  imul   %edx,%eax
  mov    -0x20(%rbp),%ecx
  mov    -0x1c(%rbp),%edx
  lea    (%rdx,%rcx,1),%esi
  cltd
  idiv   %esi
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L8:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L16
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D16
datatype: string
name: unpack_data
arguments: data_label: D2
datatype: string
name: unpack_data
arguments: data_label: D1
datatype: string

tool: D16: %d

tool: D2: %d %d %d

tool: D1: %d


assistant: ```c
int main() {
  int a, b, c, n, i;
  int total;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d %d", &a, &b, &c);
    total = c * (a + (a - b)) / (a + b);
    printf("%d\n", total);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x14(%rsp)
  jle    L5
  mov    $0x1,%ebx
  lea    0x10(%rsp),%r14
  lea    0xc(%rsp),%r13
  lea    D27(%rip),%r12
  lea    D4(%rip),%rbp
L15:
  lea    0x8(%rsp),%rsi
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%edx
  mov    %edx,%eax
  sub    0xc(%rsp),%eax
  add    %edx,%eax
  imul   0x10(%rsp),%eax
  mov    %edx,%ecx
  add    0xc(%rsp),%ecx
  cltd
  idiv   %ecx
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x14(%rsp)
  jge    L15
L5:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D1
type_infered: string
name: disassemble_data
arguments: data_label: D27
type_infered: string
name: disassemble_data
arguments: data_label: D4
type_infered: string

tool: D1: %d

tool: D27: %d %d %d

tool: D4: %d


assistant: ```c
int main() {
  int a, b, c, n, i;
  int total;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d %d", &a, &b, &c);
    total = c * (a + (a - b)) / (a + b);
    printf("%d\n", total);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  lea    D13(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L23
  mov    $0x1,%r15d
  lea    0x10(%rsp),%r14
  lea    0xc(%rsp),%r13
  lea    0x8(%rsp),%r12
  lea    D27(%rip),%rbp
  lea    D21(%rip),%rbx
  xchg   %ax,%ax
L12:
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <scanf@plt>
  mov    0x8(%rsp),%edx
  mov    0xc(%rsp),%ecx
  mov    %rbx,%rdi
  mov    %edx,%eax
  sub    0xc(%rsp),%eax
  add    %edx,%ecx
  add    %edx,%eax
  imul   0x10(%rsp),%eax
  cltd
  idiv   %ecx
  mov    %eax,%esi
  call   <printf@plt>
  cmp    %r15d,0x14(%rsp)
  jge    L12
L23:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D13
type_guessed: string
name: read_data
arguments: data_tag: D27
type_guessed: string
name: read_data
arguments: data_tag: D21
type_guessed: string

tool: D13: %d

tool: D27: %d %d %d

tool: D21: %d


assistant: ```c
int main() {
  int a, b, c, n, i;
  int total;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d %d", &a, &b, &c);
    total = c * (a + (a - b)) / (a + b);
    printf("%d\n", total);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D17(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rsi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jle    L7
  mov    $0x1,%r15d
  lea    0x10(%rsp),%r14
  lea    0xc(%rsp),%r13
  lea    0x8(%rsp),%r12
  lea    D10(%rip),%rbp
  lea    D29(%rip),%rbx
  xchg   %ax,%ax
L13:
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r15d
  call   <scanf@plt>
  mov    0x8(%rsp),%edx
  mov    0xc(%rsp),%ecx
  mov    %rbx,%rdi
  mov    %edx,%eax
  sub    0xc(%rsp),%eax
  add    %edx,%ecx
  add    %edx,%eax
  imul   0x10(%rsp),%eax
  cltd
  idiv   %ecx
  mov    %eax,%esi
  call   <printf@plt>
  cmp    %r15d,0x14(%rsp)
  jge    L13
L7:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D17
type_infered: string
name: parse_data
arguments: label: D10
type_infered: string
name: parse_data
arguments: label: D29
type_infered: string

tool: D17: %d

tool: D10: %d %d %d

tool: D29: %d


assistant: ```c
int main() {
  int a, b, c, n, i;
  int total;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d %d", &a, &b, &c);
    total = c * (a + (a - b)) / (a + b);
    printf("%d\n", total);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ins_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  sub    -0x28(%rbp),%rax
  sar    $0x2,%rax
  add    $0x1,%eax
  mov    %eax,-0x18(%rbp)
  cmpl   $0x0,-0x18(%rbp)
  jle    L93
  cmpl   $0x2,-0x18(%rbp)
  jne    L77
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sort2@plt>
  jmp    L86
L77:
  cmpl   $0x3,-0x18(%rbp)
  jne    L46
  mov    -0x28(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sort3@plt>
  jmp    L86
L46:
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L60
L43:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  sub    $0x4,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L8
L33:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jge    L13
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subq   $0x4,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jae    L33
  jmp    L99
L13:
  nop
L99:
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,(%rdx)
  addq   $0x4,-0x10(%rbp)
L60:
  mov    -0x10(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jbe    L43
  jmp    L86
L93:
  nop
L86:
  leave
  ret
```

assistant: ```c
void ins_sort(int *start, int *end) {
  int *left;
  int *right;
  int n;
  int value;
  n = end - start + 1;
  if (n < 1)
    return;
  if (n == 2) {
    sort2(start, end);
    return;
  }
  if (n == 3) {
    sort3(start, start + 1, end);
    return;
  }
  left = start;
  for (left = start; left <= end; left++) {
    value = *left;
    for (right = left - 1; right >= start; right--) {
      if (*right <= value)
        break;
      *(right + 1) = *right;
    }
    *(right + 1) = value;
  }
}
```