ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<buildNewBoard>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x40,%edi
  call   <malloc@plt>
  mov    $0x406,%edx
  pxor   %xmm0,%xmm0
  movabs $0x3333333333333333,%rcx
  movabs $0x3435363837363534,%rsi
  mov    %dx,0x4(%rax)
  movabs $0x101010101010102,%rdx
  movl   $0x5040302,(%rax)
  movb   $0x3,0x6(%rax)
  mov    %rdx,0x7(%rax)
  movb   $0x1,0xf(%rax)
  mov    %rcx,0x30(%rax)
  mov    %rsi,0x38(%rax)
  movups %xmm0,0x10(%rax)
  movups %xmm0,0x20(%rax)
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *buildNewBoard() {
  int i, idx;
  char *board = (char *)malloc(64);
  board[0] = 2;
  board[1] = 3;
  board[2] = 4;
  board[3] = 5;
  board[4] = 6;
  board[5] = 4;
  board[6] = 3;
  board[7] = 2;
  for (i = 0, idx = 8; i < 8; i++, idx++)
    board[idx] = 1;
  for (idx = 16; idx < 48; idx++)
    board[idx] = 0;
  for (i = 0, idx = 48; i < 8; i++, idx++)
    board[idx] = 1 + 50;
  board[56] = 2 + 50;
  board[57] = 3 + 50;
  board[58] = 4 + 50;
  board[59] = 5 + 50;
  board[60] = 6 + 50;
  board[61] = 4 + 50;
  board[62] = 3 + 50;
  board[63] = 2 + 50;
  return board;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<buildNewBoard>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x40,%edi
  call   <malloc@plt>
  mov    D3(%rip),%rdx
  pxor   %xmm0,%xmm0
  movabs $0x101010101010101,%rcx
  movabs $0x3333333333333333,%rsi
  mov    %rcx,0x8(%rax)
  mov    %rdx,(%rax)
  mov    D1(%rip),%rdx
  mov    %rsi,0x30(%rax)
  mov    %rdx,0x38(%rax)
  movups %xmm0,0x10(%rax)
  movups %xmm0,0x20(%rax)
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
data_type: byte[8]
name: retrieve_data
arguments: tag: D1
data_type: byte[8]

tool: {"D3": [0x02,0x03,0x04,0x05,0x06,0x04,0x03,0x02]}

tool: {"D1": [0x34,0x35,0x36,0x37,0x38,0x36,0x35,0x34]}

assistant: ```c
char *buildNewBoard() {
  int i, idx;
  char *board = (char *)malloc(64);
  board[0] = 2;
  board[1] = 3;
  board[2] = 4;
  board[3] = 5;
  board[4] = 6;
  board[5] = 4;
  board[6] = 3;
  board[7] = 2;
  for (i = 0, idx = 8; i < 8; i++, idx++)
    board[idx] = 1;
  for (idx = 16; idx < 48; idx++)
    board[idx] = 0;
  for (i = 0, idx = 48; i < 8; i++, idx++)
    board[idx] = 1 + 50;
  board[56] = 2 + 50;
  board[57] = 3 + 50;
  board[58] = 4 + 50;
  board[59] = 5 + 50;
  board[60] = 6 + 50;
  board[61] = 4 + 50;
  board[62] = 3 + 50;
  board[63] = 2 + 50;
  return board;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<phc_has_pps>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <phc_get_caps@plt>
  test   %rax,%rax
  je     L20
  mov    $0x0,%eax
  jmp    L7
L20:
  mov    -0xc(%rbp),%eax
L7:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int phc_has_pps(clockid_t clkid) {
  struct ptp_clock_caps caps;
  if (phc_get_caps(clkid, &caps))
    return 0;
  return caps.pps;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<phc_has_pps>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <phc_get_caps@plt>
  test   %rax,%rax
  mov    $0x0,%eax
  cmove  0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int phc_has_pps(clockid_t clkid) {
  struct ptp_clock_caps caps;
  if (phc_get_caps(clkid, &caps))
    return 0;
  return caps.pps;
}
```
user: Translate the below assembly code into C programming language:

```asm
<phc_has_pps>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <phc_get_caps@plt>
  test   %rax,%rax
  mov    $0x0,%eax
  cmove  0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int phc_has_pps(clockid_t clkid) {
  struct ptp_clock_caps caps;
  if (phc_get_caps(clkid, &caps))
    return 0;
  return caps.pps;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<phc_has_pps>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <phc_get_caps@plt>
  test   %rax,%rax
  mov    $0x0,%eax
  cmove  0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int phc_has_pps(clockid_t clkid) {
  struct ptp_clock_caps caps;
  if (phc_get_caps(clkid, &caps))
    return 0;
  return caps.pps;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rtp_set_ssrc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x20(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  mov    -0x8(%rbp),%rdx
  add    $0x24,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%rax
  add    $0x8,%rax
  mov    -0x8(%rbp),%rdx
  add    $0x28,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%rax
  add    $0xc,%rax
  mov    -0x8(%rbp),%rdx
  add    $0x2c,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rtp_set_ssrc(uint8_t *p_rtp, const uint8_t pi_ssrc[4]) {
  p_rtp[8] = pi_ssrc[0];
  p_rtp[9] = pi_ssrc[1];
  p_rtp[10] = pi_ssrc[2];
  p_rtp[11] = pi_ssrc[3];
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<rtp_set_ssrc>:
  endbr64
  mov    (%rsi),%eax
  mov    %eax,0x20(%rdi)
  mov    0x4(%rsi),%eax
  mov    %eax,0x24(%rdi)
  mov    0x8(%rsi),%eax
  mov    %eax,0x28(%rdi)
  mov    0xc(%rsi),%eax
  mov    %eax,0x2c(%rdi)
  ret
```

assistant: ```c
void rtp_set_ssrc(uint8_t *p_rtp, const uint8_t pi_ssrc[4]) {
  p_rtp[8] = pi_ssrc[0];
  p_rtp[9] = pi_ssrc[1];
  p_rtp[10] = pi_ssrc[2];
  p_rtp[11] = pi_ssrc[3];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rtp_set_ssrc>:
  endbr64
  mov    (%rsi),%eax
  mov    %eax,0x20(%rdi)
  mov    0x4(%rsi),%eax
  mov    %eax,0x24(%rdi)
  mov    0x8(%rsi),%eax
  mov    %eax,0x28(%rdi)
  mov    0xc(%rsi),%eax
  mov    %eax,0x2c(%rdi)
  ret
```

assistant: ```c
void rtp_set_ssrc(uint8_t *p_rtp, const uint8_t pi_ssrc[4]) {
  p_rtp[8] = pi_ssrc[0];
  p_rtp[9] = pi_ssrc[1];
  p_rtp[10] = pi_ssrc[2];
  p_rtp[11] = pi_ssrc[3];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<rtp_set_ssrc>:
  endbr64
  mov    (%rsi),%eax
  mov    %eax,0x20(%rdi)
  mov    0x4(%rsi),%eax
  mov    %eax,0x24(%rdi)
  mov    0x8(%rsi),%eax
  mov    %eax,0x28(%rdi)
  mov    0xc(%rsi),%eax
  mov    %eax,0x2c(%rdi)
  ret
```

assistant: ```c
void rtp_set_ssrc(uint8_t *p_rtp, const uint8_t pi_ssrc[4]) {
  p_rtp[8] = pi_ssrc[0];
  p_rtp[9] = pi_ssrc[1];
  p_rtp[10] = pi_ssrc[2];
  p_rtp[11] = pi_ssrc[3];
}
```
user: Translate the below assembly code into C programming language:

```asm
<leftbig>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  jne    L54
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L47
L54:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L16
L63:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%ecx
  mov    -0x1c(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x4(%rbp)
L16:
  cmpl   $0x0,-0x4(%rbp)
  jg     L63
  movl   $0x1,-0x4(%rbp)
  jmp    L45
L12:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  addl   $0x1,-0x4(%rbp)
L45:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L12
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L67
L47:
  nop
L67:
  pop    %rbp
  ret
```

assistant: ```c
void leftbig(int *x, int l) {
  int i;
  if (x[0] == 1 && x[1] == 0)
    return;
  for (i = x[0]; i >= 1; i--)
    x[i + l] = x[i];
  for (i = 1; i <= l; i++)
    x[i] = 0;
  x[0] += l;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<leftbig>:
  endbr64
  mov    (%rdi),%eax
  cmp    $0x1,%eax
  je     L7
  test   %eax,%eax
  jle    L15
L14:
  cltq
  movslq %esi,%rdx
  lea    (%rdi,%rdx,4),%rcx
L2:
  mov    (%rdi,%rax,4),%edx
  mov    %edx,(%rcx,%rax,4)
  sub    $0x1,%rax
  test   %eax,%eax
  jg     L2
L15:
  test   %esi,%esi
  jle    L56
  lea    0x4(%rdi),%rax
  lea    -0x1(%rsi),%edx
  lea    0x8(%rdi,%rdx,4),%rdx
L35:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L35
L56:
  add    %esi,(%rdi)
  ret
L7:
  cmpl   $0x0,0x4(%rdi)
  jne    L14
  ret
```

assistant: ```c
void leftbig(int *x, int l) {
  int i;
  if (x[0] == 1 && x[1] == 0)
    return;
  for (i = x[0]; i >= 1; i--)
    x[i + l] = x[i];
  for (i = 1; i <= l; i++)
    x[i] = 0;
  x[0] += l;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<leftbig>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  movslq (%rdi),%rax
  cmp    $0x1,%eax
  je     L32
  test   %eax,%eax
  jle    L36
  movslq %esi,%rdx
  lea    (%rdi,%rdx,4),%rcx
  nopw   0x0(%rax,%rax,1)
L38:
  mov    (%rbx,%rax,4),%edx
  mov    %edx,(%rcx,%rax,4)
  sub    $0x1,%rax
  test   %eax,%eax
  jg     L38
L36:
  test   %ebp,%ebp
  jle    L43
  movslq %ebp,%rdx
  lea    0x4(%rbx),%rdi
  xor    %esi,%esi
  shl    $0x2,%rdx
  call   <memset@plt>
L43:
  add    %ebp,(%rbx)
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L32:
  mov    0x4(%rdi),%eax
  test   %eax,%eax
  je     L8
  lea    0x1(%rsi),%edx
  movslq %edx,%rdx
  mov    %eax,(%rdi,%rdx,4)
  jmp    L36
```

assistant: ```c
void leftbig(int *x, int l) {
  int i;
  if (x[0] == 1 && x[1] == 0)
    return;
  for (i = x[0]; i >= 1; i--)
    x[i + l] = x[i];
  for (i = 1; i <= l; i++)
    x[i] = 0;
  x[0] += l;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<leftbig>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  mov    (%rdi),%edx
  cmp    $0x1,%edx
  je     L70
  test   %edx,%edx
  jle    L103
  movslq %esi,%rax
  lea    -0x1(%rdx),%esi
  lea    0xc(,%rax,4),%rcx
  cmp    $0x2,%esi
  jbe    L66
  cmp    $0x8,%rcx
  jbe    L66
  movslq %edx,%rcx
  add    %rcx,%rax
  lea    -0xc(%rdi,%rcx,4),%rdi
  mov    %edx,%ecx
  shr    $0x2,%ecx
  lea    -0xc(%rbx,%rax,4),%rsi
  xor    %eax,%eax
  neg    %rcx
  shl    $0x4,%rcx
  nopl   0x0(%rax)
L14:
  movdqu (%rdi,%rax,1),%xmm0
  movups %xmm0,(%rsi,%rax,1)
  sub    $0x10,%rax
  cmp    %rcx,%rax
  jne    L14
  mov    %edx,%eax
  and    $0x3,%eax
  and    $0x3,%edx
  je     L103
  movslq %eax,%rdx
  mov    (%rbx,%rdx,4),%ecx
  lea    0x0(%rbp,%rax,1),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rbx,%rdx,4)
  mov    %eax,%edx
  sub    $0x1,%edx
  je     L103
  movslq %edx,%rcx
  add    %ebp,%edx
  lea    0x0(,%rcx,4),%rsi
  mov    (%rbx,%rcx,4),%ecx
  movslq %edx,%rdx
  mov    %ecx,(%rbx,%rdx,4)
  sub    $0x2,%eax
  je     L103
  mov    -0x4(%rbx,%rsi,1),%edx
  add    %ebp,%eax
  cltq
  mov    %edx,(%rbx,%rax,4)
L103:
  test   %ebp,%ebp
  jle    L88
  movslq %ebp,%rdx
  lea    0x4(%rbx),%rdi
  xor    %esi,%esi
  shl    $0x2,%rdx
  call   <memset@plt>
L88:
  add    %ebp,(%rbx)
L82:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L70:
  mov    0x4(%rdi),%eax
  test   %eax,%eax
  je     L82
  lea    0x1(%rsi),%edx
  movslq %edx,%rdx
  mov    %eax,(%rdi,%rdx,4)
  jmp    L103
  nopw   0x0(%rax,%rax,1)
L66:
  movslq %edx,%rax
  lea    -0xc(%rbx,%rcx,1),%rcx
L25:
  mov    (%rbx,%rax,4),%edx
  mov    %edx,(%rcx,%rax,4)
  sub    $0x1,%rax
  test   %eax,%eax
  jg     L25
  jmp    L103
```

assistant: ```c
void leftbig(int *x, int l) {
  int i;
  if (x[0] == 1 && x[1] == 0)
    return;
  for (i = x[0]; i >= 1; i--)
    x[i + l] = x[i];
  for (i = 1; i <= l; i++)
    x[i] = 0;
  x[0] += l;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<readline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  movl   $0x50,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L42
  lea    D12(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <die@plt>
L42:
  lea    -0x1d(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x14(%rbp)
  jmp    L35
L94:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L10
  addl   $0x28,-0x18(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L10
  lea    D12(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <die@plt>
L10:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x1d(%rbp),%eax
  mov    %al,(%rdx)
  lea    -0x1d(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x14(%rbp)
L35:
  movzbl -0x1d(%rbp),%eax
  cmp    $0xa,%al
  je     L18
  movzbl -0x1d(%rbp),%eax
  cmp    $0xd,%al
  je     L18
  cmpl   $0x0,-0x14(%rbp)
  jne    L94
L18:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  cmpl   $0x0,-0x14(%rbp)
  je     L5
  mov    -0x10(%rbp),%rax
  jmp    L29
L5:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
L29:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D12
datatype: string

tool: D12: memory full.

assistant: ```c
char *readline(int fd) {
  ssize_t readno, allocno, readok;
  char *line, thechar;
  readno = 0;
  allocno = 80;
  line = malloc(allocno);
  if (!line)
    die(1, "memory full.");
  readok = read(fd, &thechar, 1);
  while (thechar != '\n' && thechar != '\r' && readok) {
    readno++;
    if (readno >= allocno) {
      allocno = allocno + 40;
      line = realloc(line, allocno);
      if (!line)
        die(1, "memory full.");
    }
    line[readno - 1] = thechar;
    readok = read(fd, &thechar, 1);
  }
  line[readno] = '\0';
  if (readok)
    return line;
  else {
    free(line);
    return NULL;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<readline>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r14d
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x50,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L63
L74:
  lea    0x7(%rsp),%rsi
  mov    $0x1,%edx
  mov    %r14d,%edi
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  cmp    $0xa,%dl
  setne  %cl
  cmp    $0xd,%dl
  setne  %dl
  test   %dl,%cl
  je     L85
  mov    $0x0,%ebx
  mov    $0x50,%r13d
  test   %eax,%eax
  jne    L100
L85:
  mov    $0x0,%r15d
L64:
  movslq %r15d,%r15
  movb   $0x0,(%r12,%r15,1)
  test   %eax,%eax
  je     L86
L41:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L70
  mov    %r12,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L63:
  lea    D6(%rip),%rsi
  mov    $0x1,%edi
  call   <die@plt>
  jmp    L74
L20:
  add    $0x1,%rbx
L100:
  mov    %ebx,%ebp
  lea    0x1(%rbx),%r15d
  movzbl 0x7(%rsp),%eax
  mov    %al,(%r12,%rbx,1)
  lea    0x7(%rsp),%rsi
  mov    $0x1,%edx
  mov    %r14d,%edi
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  cmp    $0xa,%dl
  setne  %cl
  cmp    $0xd,%dl
  setne  %dl
  test   %dl,%cl
  je     L64
  test   %eax,%eax
  je     L64
  add    $0x2,%ebp
  cmp    %ebp,%r13d
  jg     L20
  add    $0x28,%r13d
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <realloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L20
  lea    D6(%rip),%rsi
  mov    $0x1,%edi
  call   <die@plt>
  jmp    L20
L86:
  mov    %r12,%rdi
  call   <free@plt>
  mov    $0x0,%r12d
  jmp    L41
L70:
  call   <__stack_chk_fail@plt>
```

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

tool: D6: memory full.

assistant: ```c
char *readline(int fd) {
  ssize_t readno, allocno, readok;
  char *line, thechar;
  readno = 0;
  allocno = 80;
  line = malloc(allocno);
  if (!line)
    die(1, "memory full.");
  readok = read(fd, &thechar, 1);
  while (thechar != '\n' && thechar != '\r' && readok) {
    readno++;
    if (readno >= allocno) {
      allocno = allocno + 40;
      line = realloc(line, allocno);
      if (!line)
        die(1, "memory full.");
    }
    line[readno - 1] = thechar;
    readok = read(fd, &thechar, 1);
  }
  line[readno] = '\0';
  if (readok)
    return line;
  else {
    free(line);
    return NULL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<readline>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edi,%r13d
  mov    $0x50,%edi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L149
L108:
  lea    0x7(%rsp),%r14
  mov    $0x1,%edx
  mov    %r13d,%edi
  mov    %r14,%rsi
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  cmp    $0xd,%dl
  je     L140
  cmp    $0xa,%dl
  je     L140
  xor    %ebx,%ebx
  mov    $0x50,%r12d
  mov    %r15,%rbp
  test   %eax,%eax
  jne    L72
  jmp    L54
  nopl   0x0(%rax,%rax,1)
L89:
  mov    %rbp,%rbx
L72:
  mov    %dl,(%r15,%rbx,1)
  mov    %r14,%rsi
  mov    $0x1,%edx
  mov    %r13d,%edi
  lea    0x1(%rbx),%rbp
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  cmp    $0xa,%dl
  je     L153
  cmp    $0xd,%dl
  je     L153
  test   %eax,%eax
  je     L79
  add    $0x2,%ebx
  cmp    %ebx,%r12d
  jg     L89
  add    $0x28,%r12d
  mov    %r15,%rdi
  mov    %r12d,%esi
  call   <realloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L51
L159:
  movzbl 0x7(%rsp),%edx
  jmp    L89
  nopl   0x0(%rax)
L51:
  lea    D7(%rip),%rsi
  mov    $0x1,%edi
  call   <die@plt>
  jmp    L159
  nopl   0x0(%rax,%rax,1)
L153:
  add    %r15,%rbp
L48:
  movb   $0x0,0x0(%rbp)
  test   %eax,%eax
  je     L147
L26:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L79:
  add    %r15,%rbp
L54:
  movb   $0x0,0x0(%rbp)
L147:
  mov    %r15,%rdi
  xor    %r15d,%r15d
  call   <free@plt>
  jmp    L26
  nopl   0x0(%rax)
L149:
  lea    D7(%rip),%rsi
  mov    $0x1,%edi
  call   <die@plt>
  jmp    L108
L140:
  mov    %r15,%rbp
  jmp    L48
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D7
type_infered: string

tool: D7: memory full.

assistant: ```c
char *readline(int fd) {
  ssize_t readno, allocno, readok;
  char *line, thechar;
  readno = 0;
  allocno = 80;
  line = malloc(allocno);
  if (!line)
    die(1, "memory full.");
  readok = read(fd, &thechar, 1);
  while (thechar != '\n' && thechar != '\r' && readok) {
    readno++;
    if (readno >= allocno) {
      allocno = allocno + 40;
      line = realloc(line, allocno);
      if (!line)
        die(1, "memory full.");
    }
    line[readno - 1] = thechar;
    readok = read(fd, &thechar, 1);
  }
  line[readno] = '\0';
  if (readok)
    return line;
  else {
    free(line);
    return NULL;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<readline>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %edi,%r13d
  mov    $0x50,%edi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L10
L7:
  lea    0x7(%rsp),%r14
  mov    $0x1,%edx
  mov    %r13d,%edi
  mov    %r14,%rsi
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  cmp    $0xd,%dl
  je     L95
  cmp    $0xa,%dl
  je     L95
  xor    %ebx,%ebx
  mov    $0x50,%r12d
  mov    %r15,%rbp
  test   %eax,%eax
  jne    L106
  jmp    L60
  nopl   0x0(%rax,%rax,1)
L134:
  mov    %rbp,%rbx
L106:
  mov    %dl,(%r15,%rbx,1)
  mov    %r14,%rsi
  mov    $0x1,%edx
  mov    %r13d,%edi
  lea    0x1(%rbx),%rbp
  call   <read@plt>
  movzbl 0x7(%rsp),%edx
  cmp    $0xa,%dl
  je     L11
  cmp    $0xd,%dl
  je     L11
  test   %eax,%eax
  je     L40
  add    $0x2,%ebx
  cmp    %ebx,%r12d
  jg     L134
  add    $0x28,%r12d
  mov    %r15,%rdi
  mov    %r12d,%esi
  call   <realloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L120
L72:
  movzbl 0x7(%rsp),%edx
  jmp    L134
  nopl   0x0(%rax)
L120:
  lea    D15(%rip),%rsi
  mov    $0x1,%edi
  call   <die@plt>
  jmp    L72
  nopl   0x0(%rax,%rax,1)
L11:
  add    %r15,%rbp
L65:
  movb   $0x0,0x0(%rbp)
  test   %eax,%eax
  je     L24
L36:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L40:
  add    %r15,%rbp
L60:
  movb   $0x0,0x0(%rbp)
L24:
  mov    %r15,%rdi
  xor    %r15d,%r15d
  call   <free@plt>
  jmp    L36
  nopl   0x0(%rax)
L10:
  lea    D15(%rip),%rsi
  mov    $0x1,%edi
  call   <die@plt>
  jmp    L7
L95:
  mov    %r15,%rbp
  jmp    L65
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D15
data_type: string

tool: D15: memory full.

assistant: ```c
char *readline(int fd) {
  ssize_t readno, allocno, readok;
  char *line, thechar;
  readno = 0;
  allocno = 80;
  line = malloc(allocno);
  if (!line)
    die(1, "memory full.");
  readok = read(fd, &thechar, 1);
  while (thechar != '\n' && thechar != '\r' && readok) {
    readno++;
    if (readno >= allocno) {
      allocno = allocno + 40;
      line = realloc(line, allocno);
      if (!line)
        die(1, "memory full.");
    }
    line[readno - 1] = thechar;
    readok = read(fd, &thechar, 1);
  }
  line[readno] = '\0';
  if (readok)
    return line;
  else {
    free(line);
    return NULL;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<InitCTX>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <SSL_library_init@plt>
  mov    $0x0,%eax
  call   <OpenSSL_add_all_algorithms@plt>
  mov    $0x0,%eax
  call   <SSL_load_error_strings@plt>
  mov    $0x0,%eax
  call   <TLSv1_client_method@plt>
  mov    %eax,%edi
  call   <SSL_CTX_new@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L8
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <ERR_print_errors_fp@plt>
  mov    $0x0,%eax
  call   <abort@plt>
L8:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
SSL_CTX *InitCTX(void) {
  SSL_CTX *ctx;
  SSL_library_init();
  OpenSSL_add_all_algorithms();
  SSL_load_error_strings();
  ctx = SSL_CTX_new(TLSv1_client_method());
  if (ctx == NULL) {
    ERR_print_errors_fp(stderr);
    abort();
  }
  return ctx;
}
```