ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<isPalindrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L10
L9:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L1
  mov    $0x0,%eax
  jmp    L34
L1:
  addl   $0x1,-0xc(%rbp)
  addl   $0x1,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
L10:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0xc(%rbp)
  jl     L9
  mov    $0x1,%eax
L34:
  leave
  ret
```

assistant: ```c
int isPalindrome(const char *string) {
  int i = 0, beg = 0, end = strlen(string) - 1;
  for (; i < (strlen(string) - 1) / 2; i++, beg++, end--) {
    if (string[beg] != string[end])
      return 0;
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<isPalindrome>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  call   <strlen@plt>
  lea    -0x1(%rax),%ebp
  movslq %ebp,%rbp
  add    %r12,%rbp
  mov    $0x0,%ebx
L0:
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  cmp    %ebx,%eax
  jle    L5
  movzbl (%r12,%rbx,1),%edx
  movzbl 0x0(%rbp),%eax
  add    $0x1,%rbx
  sub    $0x1,%rbp
  cmp    %al,%dl
  je     L0
  mov    $0x0,%eax
  jmp    L9
L5:
  mov    $0x1,%eax
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int isPalindrome(const char *string) {
  int i = 0, beg = 0, end = strlen(string) - 1;
  for (; i < (strlen(string) - 1) / 2; i++, beg++, end--) {
    if (string[beg] != string[end])
      return 0;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<isPalindrome>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    -0x1(%rax),%ebp
  movslq %ebp,%rbp
  add    %r12,%rbp
  jmp    L9
  nopl   (%rax)
L22:
  movzbl (%r12,%rbx,1),%edx
  movzbl 0x0(%rbp),%eax
  add    $0x1,%rbx
  sub    $0x1,%rbp
  cmp    %al,%dl
  jne    L11
L9:
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  cmp    %ebx,%eax
  jg     L22
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int isPalindrome(const char *string) {
  int i = 0, beg = 0, end = strlen(string) - 1;
  for (; i < (strlen(string) - 1) / 2; i++, beg++, end--) {
    if (string[beg] != string[end])
      return 0;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<isPalindrome>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  call   <strlen@plt>
  lea    -0x1(%rax),%ebp
  movslq %ebp,%rbp
  add    %r12,%rbp
  jmp    L3
  nopl   (%rax)
L11:
  movzbl (%r12,%rbx,1),%edx
  movzbl 0x0(%rbp),%eax
  add    $0x1,%rbx
  sub    $0x1,%rbp
  cmp    %al,%dl
  jne    L17
L3:
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  cmp    %ebx,%eax
  jg     L11
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int isPalindrome(const char *string) {
  int i = 0, beg = 0, end = strlen(string) - 1;
  for (; i < (strlen(string) - 1) / 2; i++, beg++, end--) {
    if (string[beg] != string[end])
      return 0;
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<HermiteInterpolate>:
  endbr64
  movapd %xmm1,%xmm11
  movapd %xmm4,%xmm10
  mulsd  %xmm4,%xmm10
  movapd %xmm4,%xmm8
  mulsd  %xmm10,%xmm8
  movsd  D20(%rip),%xmm15
  movapd %xmm6,%xmm1
  addsd  %xmm15,%xmm1
  movapd %xmm15,%xmm13
  subsd  %xmm5,%xmm13
  movapd %xmm11,%xmm5
  subsd  %xmm0,%xmm5
  movapd %xmm5,%xmm0
  mulsd  %xmm1,%xmm0
  mulsd  %xmm13,%xmm0
  movsd  D26(%rip),%xmm12
  mulsd  %xmm12,%xmm0
  movapd %xmm2,%xmm5
  subsd  %xmm11,%xmm5
  movapd %xmm15,%xmm14
  subsd  %xmm6,%xmm14
  movapd %xmm1,%xmm6
  mulsd  %xmm5,%xmm6
  mulsd  %xmm13,%xmm6
  mulsd  %xmm12,%xmm6
  movapd %xmm8,%xmm9
  addsd  %xmm8,%xmm9
  movapd %xmm10,%xmm7
  mulsd  D6(%rip),%xmm7
  mulsd  %xmm14,%xmm5
  mulsd  %xmm13,%xmm5
  mulsd  %xmm12,%xmm5
  movapd %xmm5,%xmm1
  addsd  %xmm0,%xmm1
  movapd %xmm10,%xmm5
  addsd  %xmm10,%xmm5
  movapd %xmm8,%xmm0
  subsd  %xmm5,%xmm0
  addsd  %xmm4,%xmm0
  mulsd  %xmm0,%xmm1
  movapd %xmm9,%xmm4
  subsd  %xmm7,%xmm4
  addsd  %xmm15,%xmm4
  mulsd  %xmm11,%xmm4
  addsd  %xmm4,%xmm1
  subsd  %xmm2,%xmm3
  mulsd  %xmm14,%xmm3
  mulsd  %xmm13,%xmm3
  mulsd  %xmm12,%xmm3
  addsd  %xmm6,%xmm3
  subsd  %xmm10,%xmm8
  mulsd  %xmm8,%xmm3
  addsd  %xmm3,%xmm1
  subsd  %xmm9,%xmm7
  mulsd  %xmm2,%xmm7
  addsd  %xmm7,%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: label: D20
type_infered: byte[8]
name: parse_data
arguments: label: D26
type_infered: byte[8]
name: parse_data
arguments: label: D6
type_infered: byte[8]

tool: {"D20": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D26": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

assistant: ```c
double HermiteInterpolate(double y0, double y1, double y2, double y3, double mu, double tension, double bias) {
  double m0, m1, mu2, mu3;
  double a0, a1, a2, a3;
  mu2 = mu * mu;
  mu3 = mu2 * mu;
  m0 = (y1 - y0) * (1 + bias) * (1 - tension) / 2;
  m0 += (y2 - y1) * (1 - bias) * (1 - tension) / 2;
  m1 = (y2 - y1) * (1 + bias) * (1 - tension) / 2;
  m1 += (y3 - y2) * (1 - bias) * (1 - tension) / 2;
  a0 = 2 * mu3 - 3 * mu2 + 1;
  a1 = mu3 - 2 * mu2 + mu;
  a2 = mu3 - mu2;
  a3 = -2 * mu3 + 3 * mu2;
  return (a0 * y1 + a1 * m0 + a2 * m1 + a3 * y2);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<HermiteInterpolate>:
  endbr64
  movapd %xmm1,%xmm11
  movapd %xmm6,%xmm1
  movsd  D26(%rip),%xmm15
  movapd %xmm4,%xmm10
  mulsd  %xmm4,%xmm10
  subsd  %xmm2,%xmm3
  movsd  D22(%rip),%xmm12
  movapd %xmm4,%xmm8
  movapd %xmm15,%xmm13
  addsd  %xmm15,%xmm1
  movapd %xmm15,%xmm14
  movsd  D1(%rip),%xmm7
  subsd  %xmm5,%xmm13
  movapd %xmm11,%xmm5
  subsd  %xmm6,%xmm14
  subsd  %xmm0,%xmm5
  movapd %xmm1,%xmm6
  mulsd  %xmm10,%xmm8
  mulsd  %xmm14,%xmm3
  movapd %xmm5,%xmm0
  movapd %xmm2,%xmm5
  subsd  %xmm11,%xmm5
  mulsd  %xmm1,%xmm0
  mulsd  %xmm10,%xmm7
  mulsd  %xmm5,%xmm6
  movapd %xmm8,%xmm9
  mulsd  %xmm14,%xmm5
  addsd  %xmm8,%xmm9
  mulsd  %xmm13,%xmm0
  mulsd  %xmm13,%xmm3
  mulsd  %xmm13,%xmm6
  mulsd  %xmm13,%xmm5
  mulsd  %xmm12,%xmm0
  mulsd  %xmm12,%xmm3
  mulsd  %xmm12,%xmm6
  mulsd  %xmm12,%xmm5
  addsd  %xmm6,%xmm3
  movapd %xmm5,%xmm1
  movapd %xmm10,%xmm5
  addsd  %xmm10,%xmm5
  addsd  %xmm0,%xmm1
  movapd %xmm8,%xmm0
  subsd  %xmm10,%xmm8
  subsd  %xmm5,%xmm0
  mulsd  %xmm8,%xmm3
  addsd  %xmm4,%xmm0
  movapd %xmm9,%xmm4
  subsd  %xmm7,%xmm4
  subsd  %xmm9,%xmm7
  mulsd  %xmm0,%xmm1
  addsd  %xmm15,%xmm4
  mulsd  %xmm2,%xmm7
  mulsd  %xmm11,%xmm4
  addsd  %xmm4,%xmm1
  addsd  %xmm3,%xmm1
  addsd  %xmm7,%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D26
type_guessed: byte[8]
name: unpack_data
arguments: data_label: D22
type_guessed: byte[8]
name: unpack_data
arguments: data_label: D1
type_guessed: byte[8]

tool: {"D26": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D22": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D1": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

assistant: ```c
double HermiteInterpolate(double y0, double y1, double y2, double y3, double mu, double tension, double bias) {
  double m0, m1, mu2, mu3;
  double a0, a1, a2, a3;
  mu2 = mu * mu;
  mu3 = mu2 * mu;
  m0 = (y1 - y0) * (1 + bias) * (1 - tension) / 2;
  m0 += (y2 - y1) * (1 - bias) * (1 - tension) / 2;
  m1 = (y2 - y1) * (1 + bias) * (1 - tension) / 2;
  m1 += (y3 - y2) * (1 - bias) * (1 - tension) / 2;
  a0 = 2 * mu3 - 3 * mu2 + 1;
  a1 = mu3 - 2 * mu2 + mu;
  a2 = mu3 - mu2;
  a3 = -2 * mu3 + 3 * mu2;
  return (a0 * y1 + a1 * m0 + a2 * m1 + a3 * y2);
}
```
user: Translate the below assembly code into C programming language:

```asm
<HermiteInterpolate>:
  endbr64
  movapd %xmm1,%xmm11
  movapd %xmm6,%xmm1
  movsd  D14(%rip),%xmm15
  movapd %xmm4,%xmm10
  mulsd  %xmm4,%xmm10
  subsd  %xmm2,%xmm3
  movsd  D8(%rip),%xmm12
  movapd %xmm4,%xmm8
  movapd %xmm15,%xmm13
  addsd  %xmm15,%xmm1
  movapd %xmm15,%xmm14
  movsd  D0(%rip),%xmm7
  subsd  %xmm5,%xmm13
  movapd %xmm11,%xmm5
  subsd  %xmm6,%xmm14
  subsd  %xmm0,%xmm5
  movapd %xmm1,%xmm6
  mulsd  %xmm10,%xmm8
  mulsd  %xmm14,%xmm3
  movapd %xmm5,%xmm0
  movapd %xmm2,%xmm5
  subsd  %xmm11,%xmm5
  mulsd  %xmm1,%xmm0
  mulsd  %xmm10,%xmm7
  mulsd  %xmm5,%xmm6
  movapd %xmm8,%xmm9
  mulsd  %xmm14,%xmm5
  addsd  %xmm8,%xmm9
  mulsd  %xmm13,%xmm0
  mulsd  %xmm13,%xmm3
  mulsd  %xmm13,%xmm6
  mulsd  %xmm13,%xmm5
  mulsd  %xmm12,%xmm0
  mulsd  %xmm12,%xmm3
  mulsd  %xmm12,%xmm6
  mulsd  %xmm12,%xmm5
  addsd  %xmm6,%xmm3
  movapd %xmm5,%xmm1
  movapd %xmm10,%xmm5
  addsd  %xmm10,%xmm5
  addsd  %xmm0,%xmm1
  movapd %xmm8,%xmm0
  subsd  %xmm10,%xmm8
  subsd  %xmm5,%xmm0
  mulsd  %xmm8,%xmm3
  addsd  %xmm4,%xmm0
  movapd %xmm9,%xmm4
  subsd  %xmm7,%xmm4
  subsd  %xmm9,%xmm7
  mulsd  %xmm0,%xmm1
  addsd  %xmm15,%xmm4
  mulsd  %xmm2,%xmm7
  mulsd  %xmm11,%xmm4
  addsd  %xmm4,%xmm1
  addsd  %xmm3,%xmm1
  addsd  %xmm7,%xmm1
  movapd %xmm1,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D14
data_type: byte[8]
name: retrieve_data
arguments: tag: D8
data_type: byte[8]
name: retrieve_data
arguments: tag: D0
data_type: byte[8]

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f]}

tool: {"D8": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

assistant: ```c
double HermiteInterpolate(double y0, double y1, double y2, double y3, double mu, double tension, double bias) {
  double m0, m1, mu2, mu3;
  double a0, a1, a2, a3;
  mu2 = mu * mu;
  mu3 = mu2 * mu;
  m0 = (y1 - y0) * (1 + bias) * (1 - tension) / 2;
  m0 += (y2 - y1) * (1 - bias) * (1 - tension) / 2;
  m1 = (y2 - y1) * (1 + bias) * (1 - tension) / 2;
  m1 += (y3 - y2) * (1 - bias) * (1 - tension) / 2;
  a0 = 2 * mu3 - 3 * mu2 + 1;
  a1 = mu3 - 2 * mu2 + mu;
  a2 = mu3 - mu2;
  a3 = -2 * mu3 + 3 * mu2;
  return (a0 * y1 + a1 * m0 + a2 * m1 + a3 * y2);
}
```
user: What is the c source code of the assembly code below?

```asm
<sprint_mov_bit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  and    $0x80,%eax
  test   %eax,%eax
  je     L7
  mov    -0xc(%rbp),%eax
  and    $0x7,%eax
  mov    %eax,%ecx
  mov    -0xc(%rbp),%eax
  and    $0xf8,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  lea    D9(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
  jmp    L9
L7:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  lea    D8(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <sprintf@plt>
L9:
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
data_type: string
name: extract_symbolic_data
arguments: label: D8
data_type: string

tool: D9: MOV   C, 0x%02X.%X                 

tool: D8: MOV   C, 0x%02X                   

assistant: ```c
void sprint_mov_bit(char *str, uint8_t bit) {
  if (bit & 0x80)
    sprintf(str,
            "MOV   C, "
            "0x%02X"
            ".%X                 ",
            bit & 0xF8, bit & 0x07);
  else
    sprintf(str,
            "MOV   C, "
            "0x%02X"
            "                   ",
            bit);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sprint_mov_bit>:
  endbr64
  sub    $0x8,%rsp
  mov    %esi,%edx
  test   $0x80,%sil
  je     L12
  mov    %esi,%ecx
  and    $0x7,%ecx
  and    $0xf8,%edx
  lea    D13(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
L4:
  add    $0x8,%rsp
  ret
L12:
  lea    D17(%rip),%rsi
  mov    $0x0,%eax
  call   <sprintf@plt>
  jmp    L4
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
datatype: string
name: extract_symbolic_data
arguments: tag: D17
datatype: string

tool: D13: MOV   C, 0x%02X.%X                 

tool: D17: MOV   C, 0x%02X                   

assistant: ```c
void sprint_mov_bit(char *str, uint8_t bit) {
  if (bit & 0x80)
    sprintf(str,
            "MOV   C, "
            "0x%02X"
            ".%X                 ",
            bit & 0xF8, bit & 0x07);
  else
    sprintf(str,
            "MOV   C, "
            "0x%02X"
            "                   ",
            bit);
}
```
user: What is the c source code of the assembly code below:

```asm
<sprint_mov_bit>:
  endbr64
  mov    %esi,%edx
  test   $0x80,%sil
  je     L0
  mov    %esi,%ecx
  and    $0xf8,%edx
  lea    D13(%rip),%rsi
  xor    %eax,%eax
  and    $0x7,%ecx
  jmp    1050 <sprintf@plt>
  nopl   (%rax)
L0:
  lea    D1(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <sprintf@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
type: string
name: retrieve_data
arguments: data_tag: D1
type: string

tool: D13: MOV   C, 0x%02X.%X                 

tool: D1: MOV   C, 0x%02X                   

assistant: ```c
void sprint_mov_bit(char *str, uint8_t bit) {
  if (bit & 0x80)
    sprintf(str,
            "MOV   C, "
            "0x%02X"
            ".%X                 ",
            bit & 0xF8, bit & 0x07);
  else
    sprintf(str,
            "MOV   C, "
            "0x%02X"
            "                   ",
            bit);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sprint_mov_bit>:
  endbr64
  mov    %esi,%edx
  test   $0x80,%sil
  je     L4
  mov    %esi,%ecx
  and    $0xf8,%edx
  lea    D13(%rip),%rsi
  xor    %eax,%eax
  and    $0x7,%ecx
  jmp    1050 <sprintf@plt>
  nopl   (%rax)
L4:
  lea    D6(%rip),%rsi
  xor    %eax,%eax
  jmp    1050 <sprintf@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D13
type_infered: string
name: disassemble_data
arguments: data_tag: D6
type_infered: string

tool: D13: MOV   C, 0x%02X.%X                 

tool: D6: MOV   C, 0x%02X                   

assistant: ```c
void sprint_mov_bit(char *str, uint8_t bit) {
  if (bit & 0x80)
    sprintf(str,
            "MOV   C, "
            "0x%02X"
            ".%X                 ",
            bit & 0xF8, bit & 0x07);
  else
    sprintf(str,
            "MOV   C, "
            "0x%02X"
            "                   ",
            bit);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L8
L58:
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x14(%rbp)
L8:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L58
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L34
L21:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jle    L69
  cmpl   $0x0,-0x14(%rbp)
  je     L36
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L36:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L69:
  addl   $0x1,-0x14(%rbp)
L34:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x14(%rbp)
  jl     L21
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L43
  call   <__stack_chk_fail@plt>
L43:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D9
type_guessed: string
name: parse_data
arguments: label: D44
type_guessed: string
name: parse_data
arguments: label: D49
type_guessed: string
name: parse_data
arguments: label: D68
type_guessed: string
name: parse_data
arguments: label: D20
type_guessed: string

tool: D9: Introduzca numero de elementos: 

tool: D44: %d

tool: D49: introduzca el elemento con indice %d: 

tool: D68: Elementos positivos: 

tool: D20: , 

assistant: ```c
int main() {
  int n, i;
  int *arr;
  printf("Introduzca numero de elementos: ");
  scanf("%d", &n);
  arr = (int *)malloc(sizeof(int) * n);
  for (i = 0; i < n; i++) {
    printf("introduzca el elemento con indice %d: ", i);
    scanf("%d", &(arr[i]));
  }
  printf("\n");
  printf("Elementos positivos: ");
  for (i = 0; i < n; i++) {
    if (*(arr + i) > 0) {
      if (i != 0)
        printf(", ");
      printf("%d", *(arr + i));
    }
  }
  free(arr);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D43(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D52(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  cmpl   $0x0,0x4(%rsp)
  jle    L44
  mov    %rax,%rbp
  mov    $0x0,%ebx
  lea    D23(%rip),%r14
  lea    D52(%rip),%r12
L11:
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0x4(%rsp)
  jg     L11
L44:
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    D31(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,0x4(%rsp)
  jle    L69
  mov    %r13,%rbx
  mov    $0x0,%ebp
  lea    D52(%rip),%r14
  lea    D61(%rip),%r15
  jmp    L20
L22:
  mov    (%r12),%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L31:
  add    $0x1,%ebp
  add    $0x4,%rbx
  cmp    %ebp,0x4(%rsp)
  jle    L69
L20:
  mov    %rbx,%r12
  cmpl   $0x0,(%rbx)
  jle    L31
  test   %ebp,%ebp
  je     L22
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L22
L69:
  mov    %r13,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L36:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D43
type_guessed: string
name: extract_symbolic_data
arguments: label: D52
type_guessed: string
name: extract_symbolic_data
arguments: label: D23
type_guessed: string
name: extract_symbolic_data
arguments: label: D31
type_guessed: string
name: extract_symbolic_data
arguments: label: D61
type_guessed: string

tool: D43: Introduzca numero de elementos: 

tool: D52: %d

tool: D23: introduzca el elemento con indice %d: 

tool: D31: Elementos positivos: 

tool: D61: , 

assistant: ```c
int main() {
  int n, i;
  int *arr;
  printf("Introduzca numero de elementos: ");
  scanf("%d", &n);
  arr = (int *)malloc(sizeof(int) * n);
  for (i = 0; i < n; i++) {
    printf("introduzca el elemento con indice %d: ", i);
    scanf("%d", &(arr[i]));
  }
  printf("\n");
  printf("Elementos positivos: ");
  for (i = 0; i < n; i++) {
    if (*(arr + i) > 0) {
      if (i != 0)
        printf(", ");
      printf("%d", *(arr + i));
    }
  }
  free(arr);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D31(%rip),%rdi
  lea    D12(%rip),%r14
  push   %r13
  push   %r12
  lea    D29(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    0x4(%rsp),%edx
  mov    %rax,%r13
  mov    %rax,%rbp
  test   %edx,%edx
  jle    L3
  nopw   0x0(%rax,%rax,1)
L68:
  mov    %ebx,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L68
L3:
  mov    $0xa,%edi
  mov    %r13,%rbp
  xor    %ebx,%ebx
  call   <putchar@plt>
  lea    D3(%rip),%rdi
  xor    %eax,%eax
  lea    D49(%rip),%r14
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jg     L55
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L44:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L1:
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0x4(%rsp)
  jle    L13
L55:
  mov    0x0(%rbp),%esi
  test   %esi,%esi
  jle    L1
  test   %ebx,%ebx
  je     L44
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x0(%rbp),%esi
  jmp    L44
  nopl   0x0(%rax)
L13:
  mov    %r13,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D31
type: string
name: unpack_data
arguments: label: D12
type: string
name: unpack_data
arguments: label: D29
type: string
name: unpack_data
arguments: label: D3
type: string
name: unpack_data
arguments: label: D49
type: string

tool: D31: Introduzca numero de elementos: 

tool: D12: introduzca el elemento con indice %d: 

tool: D29: %d

tool: D3: Elementos positivos: 

tool: D49: , 

assistant: ```c
int main() {
  int n, i;
  int *arr;
  printf("Introduzca numero de elementos: ");
  scanf("%d", &n);
  arr = (int *)malloc(sizeof(int) * n);
  for (i = 0; i < n; i++) {
    printf("introduzca el elemento con indice %d: ", i);
    scanf("%d", &(arr[i]));
  }
  printf("\n");
  printf("Elementos positivos: ");
  for (i = 0; i < n; i++) {
    if (*(arr + i) > 0) {
      if (i != 0)
        printf(", ");
      printf("%d", *(arr + i));
    }
  }
  free(arr);
  return 0;
}
```