ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<ix_max>:
  endbr64
  mov    %rdi,%rdx
  cmp    %rsi,%rdi
  jae    L51
  lea    -0x1(%rsi),%rcx
  mov    %rdi,%rax
  sub    %rdi,%rcx
  mov    %rcx,%rdi
  shr    $0x3,%rdi
  add    $0x1,%rdi
  cmp    $0x17,%rcx
  jbe    L35
  mov    %rdi,%rcx
  pxor   %xmm3,%xmm3
  shr    $0x2,%rcx
  shl    $0x5,%rcx
  add    %rdx,%rcx
  nopl   0x0(%rax)
L29:
  movdqu (%rax),%xmm1
  movdqu 0x10(%rax),%xmm4
  add    $0x20,%rax
  movdqa %xmm1,%xmm2
  shufps $0xdd,%xmm4,%xmm1
  shufps $0x88,%xmm4,%xmm2
  movdqa %xmm2,%xmm0
  pcmpgtd %xmm1,%xmm0
  pand   %xmm0,%xmm2
  pandn  %xmm1,%xmm0
  por    %xmm2,%xmm0
  movdqa %xmm0,%xmm1
  pcmpgtd %xmm3,%xmm1
  pand   %xmm1,%xmm0
  pandn  %xmm3,%xmm1
  movdqa %xmm1,%xmm3
  por    %xmm0,%xmm3
  cmp    %rax,%rcx
  jne    L29
  movdqa %xmm3,%xmm1
  mov    %rdi,%rcx
  psrldq $0x8,%xmm1
  and    $0xfffffffffffffffc,%rcx
  movdqa %xmm1,%xmm0
  lea    (%rdx,%rcx,8),%rdx
  pcmpgtd %xmm3,%xmm0
  pand   %xmm0,%xmm1
  pandn  %xmm3,%xmm0
  por    %xmm1,%xmm0
  movdqa %xmm0,%xmm2
  psrldq $0x4,%xmm2
  movdqa %xmm2,%xmm1
  pcmpgtd %xmm0,%xmm1
  pand   %xmm1,%xmm2
  pandn  %xmm0,%xmm1
  por    %xmm2,%xmm1
  movd   %xmm1,%eax
  cmp    %rcx,%rdi
  je     L34
L44:
  mov    (%rdx),%ecx
  mov    0x4(%rdx),%edi
  cmp    %edi,%ecx
  cmovl  %edi,%ecx
  cmp    %ecx,%eax
  cmovl  %ecx,%eax
  lea    0x8(%rdx),%rcx
  cmp    %rcx,%rsi
  jbe    L22
  mov    0x8(%rdx),%ecx
  mov    0xc(%rdx),%edi
  cmp    %edi,%ecx
  cmovl  %edi,%ecx
  cmp    %ecx,%eax
  cmovl  %ecx,%eax
  lea    0x10(%rdx),%rcx
  cmp    %rcx,%rsi
  jbe    L22
  mov    0x10(%rdx),%ecx
  mov    0x14(%rdx),%edx
  cmp    %edx,%ecx
  cmovge %ecx,%edx
  cmp    %edx,%eax
  cmovl  %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L51:
  xor    %eax,%eax
L22:
  ret
  nopl   0x0(%rax,%rax,1)
L34:
  ret
L35:
  xor    %eax,%eax
  jmp    L44
```

assistant: ```c
int ix_max(int *ix, int *end) {
  int max = 0;
  while (ix < end) {
    int x = *ix++;
    if (max < x)
      max = x;
    x = *ix++;
    if (max < x)
      max = x;
  }
  return max;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<mul_digit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %esi,%eax
  mov    %edx,-0x1c(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %edi,%edx
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x18(%rbp)
  movsbl -0x14(%rbp),%eax
  lea    -0x30(%rax),%edx
  movsbl -0x18(%rbp),%eax
  sub    $0x30,%eax
  imul   %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  mov    %ecx,%eax
  add    $0x30,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x1c(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int mul_digit(char a, char b, int carry, char *c) {
  int res;
  res = (a - '0') * (b - '0') + carry;
  carry = res / 10;
  *c = res % 10 + '0';
  return carry;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<mul_digit>:
  endbr64
  mov    %edx,%eax
  movsbl %dil,%edi
  lea    -0x30(%rdi),%edx
  movsbl %sil,%esi
  sub    $0x30,%esi
  imul   %esi,%edx
  add    %eax,%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,(%rcx)
  ret
```

assistant: ```c
int mul_digit(char a, char b, int carry, char *c) {
  int res;
  res = (a - '0') * (b - '0') + carry;
  carry = res / 10;
  *c = res % 10 + '0';
  return carry;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mul_digit>:
  endbr64
  movsbl %dil,%edi
  movsbl %sil,%esi
  mov    %edx,%r8d
  sub    $0x30,%esi
  lea    -0x30(%rdi),%edx
  imul   %esi,%edx
  add    %r8d,%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,(%rcx)
  ret
```

assistant: ```c
int mul_digit(char a, char b, int carry, char *c) {
  int res;
  res = (a - '0') * (b - '0') + carry;
  carry = res / 10;
  *c = res % 10 + '0';
  return carry;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mul_digit>:
  endbr64
  movsbl %dil,%edi
  movsbl %sil,%esi
  mov    %edx,%r8d
  sub    $0x30,%esi
  lea    -0x30(%rdi),%edx
  imul   %esi,%edx
  add    %r8d,%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,(%rcx)
  ret
```

assistant: ```c
int mul_digit(char a, char b, int carry, char *c) {
  int res;
  res = (a - '0') * (b - '0') + carry;
  carry = res / 10;
  *c = res % 10 + '0';
  return carry;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<slaDrange>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <sla_drange_@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double slaDrange(double angle) {
  double ANGLE;
  double result;
  ANGLE = angle;
  result = sla_drange_(&ANGLE);
  return result;
}
```
user: What is the c source code of the assembly code below?

```asm
<slaDrange>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movsd  %xmm0,(%rsp)
  mov    %rsp,%rdi
  call   <sla_drange_@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double slaDrange(double angle) {
  double ANGLE;
  double result;
  ANGLE = angle;
  result = sla_drange_(&ANGLE);
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<slaDrange>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movsd  %xmm0,(%rsp)
  call   <sla_drange_@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double slaDrange(double angle) {
  double ANGLE;
  double result;
  ANGLE = angle;
  result = sla_drange_(&ANGLE);
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<slaDrange>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movsd  %xmm0,(%rsp)
  call   <sla_drange_@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
double slaDrange(double angle) {
  double ANGLE;
  double result;
  ANGLE = angle;
  result = sla_drange_(&ANGLE);
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Q_strcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rax,%rcx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <Q_strncpyz@plt>
  nop
  leave
  ret
```

assistant: ```c
void Q_strcat(char *dest, int size, const char *src) {
  int l1;
  l1 = strlen(dest);
  if (l1 >= size) {
    ;
  }
  Q_strncpyz(dest + l1, src, size - l1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Q_strcat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%ebp
  mov    %rdx,%r12
  call   <strlen@plt>
  mov    %ebp,%edx
  sub    %eax,%edx
  cltq
  lea    (%rbx,%rax,1),%rdi
  mov    %r12,%rsi
  call   <Q_strncpyz@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void Q_strcat(char *dest, int size, const char *src) {
  int l1;
  l1 = strlen(dest);
  if (l1 >= size) {
    ;
  }
  Q_strncpyz(dest + l1, src, size - l1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Q_strcat>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %ebp,%edx
  mov    %r12,%rsi
  sub    %eax,%edx
  cltq
  lea    (%rbx,%rax,1),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <Q_strncpyz@plt>
```

assistant: ```c
void Q_strcat(char *dest, int size, const char *src) {
  int l1;
  l1 = strlen(dest);
  if (l1 >= size) {
    ;
  }
  Q_strncpyz(dest + l1, src, size - l1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<Q_strcat>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %ebp,%edx
  mov    %r12,%rsi
  sub    %eax,%edx
  cltq
  lea    (%rbx,%rax,1),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1070 <Q_strncpyz@plt>
```

assistant: ```c
void Q_strcat(char *dest, int size, const char *src) {
  int l1;
  l1 = strlen(dest);
  if (l1 >= size) {
    ;
  }
  Q_strncpyz(dest + l1, src, size - l1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<br_find_bin_dir>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x0,%edi
  call   <br_find_prefix@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L27
  cmpq   $0x0,-0x18(%rbp)
  je     L33
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  jmp    L18
L33:
  mov    $0x0,%eax
  jmp    L18
L27:
  mov    -0x10(%rbp),%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <br_build_path@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
L18:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D7
type: string

tool: D7: bin

assistant: ```c
char *br_find_bin_dir(const char *default_bin_dir) {
  char *prefix, *dir;
  prefix = br_find_prefix((const char *)((char *)0));
  if (prefix == (char *)((char *)0)) {
    if (default_bin_dir != (const char *)((char *)0))
      return strdup(default_bin_dir);
    else
      return (char *)((char *)0);
  }
  dir = br_build_path(prefix, "bin");
  free(prefix);
  return dir;
}
```
user: What is the c source code of the assembly code below?

```asm
<br_find_bin_dir>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    $0x0,%edi
  call   <br_find_prefix@plt>
  test   %rax,%rax
  je     L6
  mov    %rax,%rbx
  lea    D5(%rip),%rsi
  mov    %rax,%rdi
  call   <br_build_path@plt>
  mov    %rax,%r12
  mov    %rbx,%rdi
  call   <free@plt>
L8:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  mov    %rbp,%r12
  test   %rbp,%rbp
  je     L8
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,%r12
  jmp    L8
```

tool_calls: name: parse_data
arguments: label: D5
type_guessed: string

tool: D5: bin

assistant: ```c
char *br_find_bin_dir(const char *default_bin_dir) {
  char *prefix, *dir;
  prefix = br_find_prefix((const char *)((char *)0));
  if (prefix == (char *)((char *)0)) {
    if (default_bin_dir != (const char *)((char *)0))
      return strdup(default_bin_dir);
    else
      return (char *)((char *)0);
  }
  dir = br_build_path(prefix, "bin");
  free(prefix);
  return dir;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<br_find_bin_dir>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <br_find_prefix@plt>
  test   %rax,%rax
  je     L19
  mov    %rax,%rbp
  lea    D9(%rip),%rsi
  mov    %rax,%rdi
  call   <br_build_path@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  test   %r12,%r12
  je     L3
  add    $0x8,%rsp
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    10b0 <strdup@plt>
  nopl   0x0(%rax)
L3:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_infered: string

tool: D9: bin

assistant: ```c
char *br_find_bin_dir(const char *default_bin_dir) {
  char *prefix, *dir;
  prefix = br_find_prefix((const char *)((char *)0));
  if (prefix == (char *)((char *)0)) {
    if (default_bin_dir != (const char *)((char *)0))
      return strdup(default_bin_dir);
    else
      return (char *)((char *)0);
  }
  dir = br_build_path(prefix, "bin");
  free(prefix);
  return dir;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<br_find_bin_dir>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  push   %rbp
  sub    $0x8,%rsp
  call   <br_find_prefix@plt>
  test   %rax,%rax
  je     L8
  mov    %rax,%rbp
  lea    D8(%rip),%rsi
  mov    %rax,%rdi
  call   <br_build_path@plt>
  mov    %rbp,%rdi
  mov    %rax,%r12
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  test   %r12,%r12
  je     L16
  add    $0x8,%rsp
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    10b0 <strdup@plt>
  nopl   0x0(%rax)
L16:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
datatype: string

tool: D8: bin

assistant: ```c
char *br_find_bin_dir(const char *default_bin_dir) {
  char *prefix, *dir;
  prefix = br_find_prefix((const char *)((char *)0));
  if (prefix == (char *)((char *)0)) {
    if (default_bin_dir != (const char *)((char *)0))
      return strdup(default_bin_dir);
    else
      return (char *)((char *)0);
  }
  dir = br_build_path(prefix, "bin");
  free(prefix);
  return dir;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cd_track_length_frames>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <raw_track_length@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,%rdi
  call   <raw_track_length@plt>
  mov    %rax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <cd_present@plt>
  test   %eax,%eax
  jne    L15
  mov    $0x0,%eax
  jmp    L4
L15:
  mov    -0x8(%rbp),%rax
  sub    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
L4:
  leave
  ret
```

assistant: ```c
unsigned long cd_track_length_frames(u_char trknum) {
  long frames = raw_track_length(trknum);
  long prevframes = raw_track_length(trknum - 1);
  if (!cd_present())
    return 0;
  frames -= prevframes;
  return frames;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<cd_track_length_frames>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <raw_track_length@plt>
  mov    %rax,%rbx
  lea    -0x1(%rbp),%rdi
  call   <raw_track_length@plt>
  mov    %rax,%rbp
  mov    $0x0,%eax
  call   <cd_present@plt>
  sub    %rbp,%rbx
  test   %eax,%eax
  mov    $0x0,%eax
  cmovne %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned long cd_track_length_frames(u_char trknum) {
  long frames = raw_track_length(trknum);
  long prevframes = raw_track_length(trknum - 1);
  if (!cd_present())
    return 0;
  frames -= prevframes;
  return frames;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<cd_track_length_frames>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <raw_track_length@plt>
  lea    -0x1(%rbp),%rdi
  mov    %rax,%rbx
  call   <raw_track_length@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <cd_present@plt>
  sub    %rbp,%rbx
  test   %eax,%eax
  mov    $0x0,%eax
  cmovne %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned long cd_track_length_frames(u_char trknum) {
  long frames = raw_track_length(trknum);
  long prevframes = raw_track_length(trknum - 1);
  if (!cd_present())
    return 0;
  frames -= prevframes;
  return frames;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cd_track_length_frames>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <raw_track_length@plt>
  lea    -0x1(%rbp),%rdi
  mov    %rax,%rbx
  call   <raw_track_length@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <cd_present@plt>
  sub    %rbp,%rbx
  test   %eax,%eax
  mov    $0x0,%eax
  cmovne %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned long cd_track_length_frames(u_char trknum) {
  long frames = raw_track_length(trknum);
  long prevframes = raw_track_length(trknum - 1);
  if (!cd_present())
    return 0;
  frames -= prevframes;
  return frames;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ERR>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L2
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L2:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %edx,%ecx
  mov    %rax,%rdx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <write_message@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

assistant: ```c
void ERR(char *format, ...) {
  va_list args;
  va_start(args, format);
  write_message(1, 1, format, args);
  va_end(args);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ERR>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %rbx,%rdx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <write_message@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void ERR(char *format, ...) {
  va_list args;
  va_start(args, format);
  write_message(1, 1, format, args);
  va_end(args);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ERR>:
  endbr64
  push   %r12
  mov    %rdi,%rsi
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <write_message@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void ERR(char *format, ...) {
  va_list args;
  va_start(args, format);
  write_message(1, 1, format, args);
  va_end(args);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ERR>:
  endbr64
  push   %r12
  mov    %rdi,%rsi
  mov    %rdi,%r12
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r12,%rdx
  mov    $0x1,%esi
  mov    $0x1,%edi
  call   <write_message@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void ERR(char *format, ...) {
  va_list args;
  va_start(args, format);
  write_message(1, 1, format, args);
  va_end(args);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<endswith>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jbe    L31
  mov    $0x0,%eax
  jmp    L38
L31:
  mov    -0x18(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  je     L14
  mov    $0x0,%eax
  jmp    L38
L14:
  mov    $0x1,%eax
L38:
  leave
  ret
```

assistant: ```c
int endswith(char *haystack, char *needle) {
  size_t len = strlen(haystack);
  size_t nlen = strlen(needle);
  if (nlen > len)
    return (0);
  size_t pos = len - nlen;
  if (memcmp(&haystack[pos], needle, nlen) != 0)
    return (0);
  return (1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<endswith>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  call   <strlen@plt>
  mov    %rax,%rbp
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  cmp    %rdx,%rbp
  jae    L11
L14:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  sub    %rdx,%rbp
  lea    (%rbx,%rbp,1),%rdi
  mov    %r12,%rsi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L14
```

assistant: ```c
int endswith(char *haystack, char *needle) {
  size_t len = strlen(haystack);
  size_t nlen = strlen(needle);
  if (nlen > len)
    return (0);
  size_t pos = len - nlen;
  if (memcmp(&haystack[pos], needle, nlen) != 0)
    return (0);
  return (1);
}
```
user: What is the c source code of the assembly code below?

```asm
<endswith>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    %rax,%rdx
  xor    %eax,%eax
  cmp    %rdx,%rbp
  jb     L3
  sub    %rdx,%rbp
  mov    %r12,%rsi
  lea    (%rbx,%rbp,1),%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L3:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int endswith(char *haystack, char *needle) {
  size_t len = strlen(haystack);
  size_t nlen = strlen(needle);
  if (nlen > len)
    return (0);
  size_t pos = len - nlen;
  if (memcmp(&haystack[pos], needle, nlen) != 0)
    return (0);
  return (1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<endswith>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    %rax,%rdx
  xor    %eax,%eax
  cmp    %rdx,%rbp
  jb     L5
  sub    %rdx,%rbp
  mov    %r12,%rsi
  lea    (%rbx,%rbp,1),%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int endswith(char *haystack, char *needle) {
  size_t len = strlen(haystack);
  size_t nlen = strlen(needle);
  if (nlen > len)
    return (0);
  size_t pos = len - nlen;
  if (memcmp(&haystack[pos], needle, nlen) != 0)
    return (0);
  return (1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<checkizero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L4
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <xlFmtError@plt>
L4:
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D5
datatype: string

tool: D5: division by zero

assistant: ```c
void checkizero(long iarg) {
  if (iarg == 0)
    xlFmtError("division by zero");
}
```