ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_mac_address>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  mov    $0x18,%edx
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <memset@plt>
  mov    D26(%rip),%rax
  mov    0x10(%rsp),%edi
  mov    %r12,%rsi
  mov    (%rax),%rdx
  sub    $0x1,%rdx
  call   <strncpy@plt>
  mov    D7(%rip),%rax
  mov    %r14,%rdx
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <ioctl@plt>
  test   %rax,%rax
  jne    L19
  mov    D9(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rsp)
  jne    L37
  mov    0x8(%rsp),%esi
  mov    $0x6,%edx
  mov    %r13,%rdi
  call   <memcpy@plt>
  xor    %eax,%eax
L46:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L48
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L37:
  mov    $0xfffffffe,%eax
  jmp    L46
  nopw   0x0(%rax,%rax,1)
L19:
  mov    $0xffffffff,%eax
  jmp    L46
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_mac_address(int fd, const char *dname, unsigned char *buffer) {
  struct ifreq dconf;
  memset(&dconf, 0, sizeof(dconf));
  strncpy(dconf.ifr_name, dname, IFNAMSIZ - 1);
  if (ioctl(fd, SIOCGIFHWADDR, &dconf) != 0)
    return -1;
  if (dconf.ifr_hwaddr.sa_family != ARPHRD_ETHER)
    return -2;
  memcpy(buffer, dconf.ifr_hwaddr.sa_data, 6);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_mac_address>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  mov    $0x18,%edx
  push   %r12
  mov    %rsi,%r12
  xor    %esi,%esi
  push   %rbp
  mov    %edi,%ebp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rdi
  call   <memset@plt>
  mov    D19(%rip),%rax
  mov    0x10(%rsp),%edi
  mov    %r12,%rsi
  mov    (%rax),%rdx
  sub    $0x1,%rdx
  call   <strncpy@plt>
  mov    D0(%rip),%rax
  mov    %r14,%rdx
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <ioctl@plt>
  test   %rax,%rax
  jne    L30
  mov    D15(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rsp)
  jne    L1
  mov    0x8(%rsp),%esi
  mov    $0x6,%edx
  mov    %r13,%rdi
  call   <memcpy@plt>
  xor    %eax,%eax
L34:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L46
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mov    $0xfffffffe,%eax
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L30:
  mov    $0xffffffff,%eax
  jmp    L34
L46:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int get_mac_address(int fd, const char *dname, unsigned char *buffer) {
  struct ifreq dconf;
  memset(&dconf, 0, sizeof(dconf));
  strncpy(dconf.ifr_name, dname, IFNAMSIZ - 1);
  if (ioctl(fd, SIOCGIFHWADDR, &dconf) != 0)
    return -1;
  if (dconf.ifr_hwaddr.sa_family != ARPHRD_ETHER)
    return -2;
  memcpy(buffer, dconf.ifr_hwaddr.sa_data, 6);
  return 0;
}
```
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
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L14
L17:
  mov    $0x0,%eax
  call   <random_int@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x8(%rbp)
L14:
  cmpl   $0x9,-0x8(%rbp)
  jle    L17
  mov    $0x0,%eax
  leave
  ret
```

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

tool: D4: %d


assistant: ```c
int main() {
  int integer = 0;
  for (int i = 0; i < 10; i++) {
    printf("%d\n", random_int());
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0xa,%ebx
  lea    D9(%rip),%rbp
L0:
  mov    $0x0,%eax
  call   <random_int@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L0
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D9
data_type: string

tool: D9: %d


assistant: ```c
int main() {
  int integer = 0;
  for (int i = 0; i < 10; i++) {
    printf("%d\n", random_int());
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L2:
  xor    %eax,%eax
  call   <random_int@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L2
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D6
datatype: string

tool: D6: %d


assistant: ```c
int main() {
  int integer = 0;
  for (int i = 0; i < 10; i++) {
    printf("%d\n", random_int());
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  mov    $0xa,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L1:
  xor    %eax,%eax
  call   <random_int@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  sub    $0x1,%ebx
  jne    L1
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
data_type: string

tool: D8: %d


assistant: ```c
int main() {
  int integer = 0;
  for (int i = 0; i < 10; i++) {
    printf("%d\n", random_int());
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tree_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <tree_clear@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void tree_free(struct tree *target) {
  tree_clear(target);
  free(target);
}
```
user: Translate the below assembly code into C programming language:

```asm
<tree_free>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <tree_clear@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  pop    %rbx
  ret
```

assistant: ```c
void tree_free(struct tree *target) {
  tree_clear(target);
  free(target);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<tree_free>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <tree_clear@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1060 <free@plt>
```

assistant: ```c
void tree_free(struct tree *target) {
  tree_clear(target);
  free(target);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<swap2cols>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x8(%rbp)
  mov    -0x8(%rbp),%ecx
  mov    %ecx,%edx
  shl    $0x2,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    %edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    %ecx,%edx
  shl    $0x2,%edx
  add    %ecx,%edx
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L4
L15:
  mov    -0xc(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  mov    -0xc(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0xc(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
  mov    -0xc(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L4:
  cmpl   $0x4,-0xc(%rbp)
  jle    L15
  nop
  nop
  leave
  ret
```

assistant: ```c
void swap2cols(char *key) {
  int i = rand() % 5;
  int j = rand() % 5;
  char temp;
  int k;
  for (k = 0; k < 5; k++) {
    temp = key[k * 5 + i];
    key[k * 5 + i] = key[k * 5 + j];
    key[k * 5 + j] = temp;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<swap2cols>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x21,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %ebx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%eax
  sub    %eax,%ebx
  movslq %ebx,%rsi
  lea    0x0(%rbp,%rsi,1),%rax
  lea    0x19(%rbp,%rsi,1),%r9
  movslq %edx,%rdx
L5:
  movzbl (%rax),%edi
  mov    %rax,%rcx
  sub    %rsi,%rcx
  movzbl (%rcx,%rdx,1),%r8d
  mov    %r8b,(%rax)
  mov    %dil,(%rcx,%rdx,1)
  add    $0x5,%rax
  cmp    %r9,%rax
  jne    L5
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void swap2cols(char *key) {
  int i = rand() % 5;
  int j = rand() % 5;
  char temp;
  int k;
  for (k = 0; k < 5; k++) {
    temp = key[k * 5 + i];
    key[k * 5 + i] = key[k * 5 + j];
    key[k * 5 + j] = temp;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<swap2cols>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <rand@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <rand@plt>
  mov    %ebp,%esi
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  sub    %ecx,%eax
  mov    %ebp,%ecx
  movslq %eax,%rdx
  movslq %ebp,%rax
  sar    $0x1f,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%eax
  sub    %eax,%esi
  movslq %esi,%rsi
  lea    (%rbx,%rsi,1),%rax
  lea    0x19(%rbx,%rsi,1),%r9
L5:
  mov    %rax,%rcx
  movzbl (%rax),%edi
  add    $0x5,%rax
  sub    %rsi,%rcx
  movzbl (%rcx,%rdx,1),%r8d
  mov    %r8b,-0x5(%rax)
  mov    %dil,(%rcx,%rdx,1)
  cmp    %r9,%rax
  jne    L5
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void swap2cols(char *key) {
  int i = rand() % 5;
  int j = rand() % 5;
  char temp;
  int k;
  for (k = 0; k < 5; k++) {
    temp = key[k * 5 + i];
    key[k * 5 + i] = key[k * 5 + j];
    key[k * 5 + j] = temp;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<swap2cols>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <rand@plt>
  movslq %eax,%rbx
  cltd
  imul   $0x66666667,%rbx,%rbx
  sar    $0x21,%rbx
  sub    %edx,%ebx
  lea    (%rbx,%rbx,4),%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  sub    %ecx,%eax
  movslq %ebx,%rcx
  mov    %eax,%edx
  cltq
  add    %rbp,%rcx
  add    %rbp,%rax
  movzbl (%rcx),%esi
  movzbl (%rax),%edi
  mov    %dil,(%rcx)
  lea    0x5(%rbx),%ecx
  mov    %sil,(%rax)
  lea    0x5(%rdx),%eax
  movslq %ecx,%rcx
  cltq
  add    %rbp,%rcx
  add    %rbp,%rax
  movzbl (%rcx),%esi
  movzbl (%rax),%edi
  mov    %dil,(%rcx)
  lea    0xa(%rbx),%ecx
  mov    %sil,(%rax)
  lea    0xa(%rdx),%eax
  movslq %ecx,%rcx
  cltq
  add    %rbp,%rcx
  add    %rbp,%rax
  movzbl (%rcx),%esi
  movzbl (%rax),%edi
  mov    %dil,(%rcx)
  lea    0xf(%rbx),%ecx
  add    $0x14,%ebx
  mov    %sil,(%rax)
  lea    0xf(%rdx),%eax
  movslq %ecx,%rcx
  add    $0x14,%edx
  cltq
  add    %rbp,%rcx
  movslq %edx,%rdx
  movslq %ebx,%rbx
  add    %rbp,%rax
  movzbl (%rcx),%esi
  add    %rbp,%rbx
  movzbl (%rax),%edi
  mov    %dil,(%rcx)
  mov    %sil,(%rax)
  lea    0x0(%rbp,%rdx,1),%rax
  movzbl (%rbx),%ecx
  movzbl (%rax),%edx
  mov    %dl,(%rbx)
  mov    %cl,(%rax)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void swap2cols(char *key) {
  int i = rand() % 5;
  int j = rand() % 5;
  char temp;
  int k;
  for (k = 0; k < 5; k++) {
    temp = key[k * 5 + i];
    key[k * 5 + i] = key[k * 5 + j];
    key[k * 5 + j] = temp;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rev_vals>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L18
L6:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L18:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0xc(%rbp)
  jl     L6
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rev_vals(double *vals, int n) {
  int i;
  double dum;
  for (i = 0; i < n / 2; ++i) {
    dum = vals[i];
    vals[i] = vals[n - 1 - i];
    vals[n - 1 - i] = dum;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rev_vals>:
  endbr64
  mov    %esi,%ecx
  shr    $0x1f,%ecx
  add    %esi,%ecx
  sar    %ecx
  cmp    $0x1,%esi
  jle    L16
  movslq %esi,%rsi
  lea    (%rdi,%rsi,8),%rdx
  mov    $0x0,%eax
L12:
  movsd  (%rdi,%rax,8),%xmm0
  movsd  -0x8(%rdx),%xmm1
  movsd  %xmm1,(%rdi,%rax,8)
  movsd  %xmm0,-0x8(%rdx)
  add    $0x1,%rax
  sub    $0x8,%rdx
  cmp    %eax,%ecx
  jg     L12
L16:
  ret
```

assistant: ```c
void rev_vals(double *vals, int n) {
  int i;
  double dum;
  for (i = 0; i < n / 2; ++i) {
    dum = vals[i];
    vals[i] = vals[n - 1 - i];
    vals[n - 1 - i] = dum;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rev_vals>:
  endbr64
  mov    %esi,%ecx
  shr    $0x1f,%ecx
  add    %esi,%ecx
  sar    %ecx
  cmp    $0x1,%esi
  jle    L18
  sub    $0x1,%esi
  xor    %eax,%eax
  movslq %esi,%rsi
  lea    (%rdi,%rsi,8),%rdx
  xchg   %ax,%ax
L2:
  movsd  (%rdx),%xmm1
  movsd  (%rdi,%rax,8),%xmm0
  sub    $0x8,%rdx
  movsd  %xmm1,(%rdi,%rax,8)
  add    $0x1,%rax
  movsd  %xmm0,0x8(%rdx)
  cmp    %eax,%ecx
  jg     L2
L18:
  ret
```

assistant: ```c
void rev_vals(double *vals, int n) {
  int i;
  double dum;
  for (i = 0; i < n / 2; ++i) {
    dum = vals[i];
    vals[i] = vals[n - 1 - i];
    vals[n - 1 - i] = dum;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rev_vals>:
  endbr64
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  cmp    $0x1,%esi
  jle    L18
  movslq %esi,%rdx
  mov    %r8d,%eax
  lea    -0x1(%rsi),%ecx
  shl    $0x3,%rdx
  shl    $0x3,%rax
  mov    %rdx,%r9
  sub    %rax,%r9
  cmp    %r9,%rax
  jg     L57
  cmp    $0x5,%esi
  jle    L57
  mov    %r8d,%esi
  mov    %rdi,%rax
  lea    -0x10(%rdi,%rdx,1),%rdx
  shr    %esi
  sub    $0x1,%esi
  shl    $0x4,%rsi
  lea    0x10(%rdi,%rsi,1),%rsi
  nopl   0x0(%rax)
L10:
  movupd (%rdx),%xmm1
  movupd (%rax),%xmm0
  add    $0x10,%rax
  sub    $0x10,%rdx
  shufpd $0x1,%xmm1,%xmm1
  shufpd $0x1,%xmm0,%xmm0
  movups %xmm1,-0x10(%rax)
  movups %xmm0,0x10(%rdx)
  cmp    %rax,%rsi
  jne    L10
  mov    %r8d,%esi
  and    $0xfffffffe,%esi
  and    $0x1,%r8d
  je     L18
  movslq %esi,%rax
  lea    (%rdi,%rax,8),%rdx
  mov    %ecx,%eax
  sub    %esi,%eax
  movsd  (%rdx),%xmm0
  cltq
  lea    (%rdi,%rax,8),%rax
  movsd  (%rax),%xmm1
  movsd  %xmm1,(%rdx)
  movsd  %xmm0,(%rax)
  ret
  nopl   (%rax)
L57:
  movslq %ecx,%rax
  lea    (%rdi,%rax,8),%rdx
  xor    %eax,%eax
  nopl   0x0(%rax)
L32:
  movsd  (%rdx),%xmm1
  movsd  (%rdi,%rax,8),%xmm0
  sub    $0x8,%rdx
  movsd  %xmm1,(%rdi,%rax,8)
  add    $0x1,%rax
  movsd  %xmm0,0x8(%rdx)
  cmp    %eax,%r8d
  jg     L32
L18:
  ret
```

assistant: ```c
void rev_vals(double *vals, int n) {
  int i;
  double dum;
  for (i = 0; i < n / 2; ++i) {
    dum = vals[i];
    vals[i] = vals[n - 1 - i];
    vals[n - 1 - i] = dum;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  movb   $0x0,-0x15(%rbp)
  movq   $0x1,-0x10(%rbp)
  lea    -0x15(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L37
L1:
  mov    -0x14(%rbp),%edx
  mov    -0x30(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x15(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x14(%rbp)
  lea    -0x15(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %rax,-0x10(%rbp)
L37:
  movzbl -0x15(%rbp),%eax
  cmp    $0xa,%al
  je     L32
  cmpq   $0x0,-0x10(%rbp)
  jg     L1
L32:
  mov    -0x14(%rbp),%edx
  mov    -0x38(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

assistant: ```c
int read_line(int fd, char *buf, int *count) {
  unsigned int i = 0;
  char a = 0;
  ssize_t ret = 1;
  ret = read(fd, &a, 1);
  while (a != '\n' && ret > 0) {
    buf[i] = a;
    i++;
    ret = read(fd, &a, 1);
  }
  *count = i;
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_line>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    %rdx,%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x7(%rsp)
  lea    0x7(%rsp),%rsi
  mov    $0x1,%edx
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  test   %rax,%rax
  jle    L17
  cmp    $0xa,%dl
  je     L17
  mov    $0x0,%ebx
  lea    0x7(%rsp),%r13
L24:
  mov    %ebx,%eax
  mov    %dl,(%r12,%rax,1)
  add    $0x1,%ebx
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  test   %rax,%rax
  jle    L7
  cmp    $0xa,%dl
  jne    L24
L7:
  mov    %ebx,(%r14)
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L50
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  mov    $0x0,%ebx
  jmp    L7
L50:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_line(int fd, char *buf, int *count) {
  unsigned int i = 0;
  char a = 0;
  ssize_t ret = 1;
  ret = read(fd, &a, 1);
  while (a != '\n' && ret > 0) {
    buf[i] = a;
    i++;
    ret = read(fd, &a, 1);
  }
  *count = i;
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<read_line>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  mov    $0x1,%edx
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%r12
  movb   $0x0,0x7(%rsp)
  mov    %r12,%rsi
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  test   %rax,%rax
  jg     L32
  jmp    L39
  xchg   %ax,%ax
L41:
  mov    %r14d,%eax
  mov    %r12,%rsi
  mov    %ebx,%edi
  add    $0x1,%r14d
  mov    %dl,0x0(%rbp,%rax,1)
  mov    $0x1,%edx
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  test   %rax,%rax
  jle    L39
L32:
  cmp    $0xa,%dl
  jne    L41
L39:
  mov    %r14d,0x0(%r13)
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L13
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_line(int fd, char *buf, int *count) {
  unsigned int i = 0;
  char a = 0;
  ssize_t ret = 1;
  ret = read(fd, &a, 1);
  while (a != '\n' && ret > 0) {
    buf[i] = a;
    i++;
    ret = read(fd, &a, 1);
  }
  *count = i;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_line>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  mov    $0x1,%edx
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x7(%rsp),%r12
  movb   $0x0,0x7(%rsp)
  mov    %r12,%rsi
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  test   %rax,%rax
  jg     L22
  jmp    L20
  xchg   %ax,%ax
L25:
  mov    %r14d,%eax
  mov    %r12,%rsi
  mov    %ebx,%edi
  add    $0x1,%r14d
  mov    %dl,0x0(%rbp,%rax,1)
  mov    $0x1,%edx
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  test   %rax,%rax
  jle    L20
L22:
  cmp    $0xa,%dl
  jne    L25
L20:
  mov    %r14d,0x0(%r13)
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_line(int fd, char *buf, int *count) {
  unsigned int i = 0;
  char a = 0;
  ssize_t ret = 1;
  ret = read(fd, &a, 1);
  while (a != '\n' && ret > 0) {
    buf[i] = a;
    i++;
    ret = read(fd, &a, 1);
  }
  *count = i;
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <environ_exit_application_name@plt>
  mov    %rax,-0x8(%rbp)
  cmpl   $0x3,-0x14(%rbp)
  jne    L40
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  lea    D13(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L35
  mov    -0x20(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <attribute_list_dictionary@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L35
L40:
  cmpl   $0x2,-0x14(%rbp)
  jne    L35
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  lea    D18(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L33
  mov    $0x0,%eax
  call   <attribute_list_folder_attribute@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L35
L33:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <attribute_list_attribute_name@plt>
  mov    $0x0,%edi
  call   <exit@plt>
L35:
  mov    $0x0,%eax
  call   <attribute_list_without_folder@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D13
type: string
name: disassemble_data
arguments: tag: D18
type: string

tool: D13: $dictionary

tool: D18: with_table

assistant: ```c
int main(int argc, char **argv) {
  char *application_name;
  if ((application_name = environ_exit_application_name(argv[0]))) {
  };
  if (argc == 3) {
    if (strcmp(argv[2], "$dictionary") != 0) {
      attribute_list_dictionary(argv[2]);
      exit(0);
    }
  } else if (argc == 2) {
    if (strcmp(argv[1], "with_table") == 0) {
      attribute_list_folder_attribute();
      exit(0);
    } else {
      attribute_list_attribute_name(argv[1]);
      exit(0);
    }
  }
  attribute_list_without_folder();
  return 0;
}
```