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

```asm
<main>:
  endbr64
  push   %rbp
  lea    D39(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xd(%rsp),%rbp
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  mov    $0x7,%edx
  lea    D8(%rip),%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L4
  lea    D0(%rip),%rdi
  call   <printf@plt>
L19:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L4:
  lea    D31(%rip),%rdi
  call   <printf@plt>
  jmp    L19
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D39
data_type: string
name: read_data
arguments: tag: D8
data_type: string
name: read_data
arguments: tag: D0
data_type: string
name: read_data
arguments: tag: D31
data_type: string

tool: D39: Enter password: 

tool: D8: letmein

tool: D0: Performing tasks that require password...


tool: D31: Incorrect Password!


assistant: ```c
int main(void) {
  char buffer[11];
  int correct_password = 0;
  printf("Enter password: ");
  gets(buffer);
  if (strncmp(buffer, "letmein", 7) == 0)
    correct_password = 1;
  else
    printf("Incorrect Password!\n");
  if (correct_password != 0)
    printf("Performing tasks that require password...\n");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D20(%rip),%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xd(%rsp),%rbp
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  mov    $0x7,%edx
  lea    D33(%rip),%rsi
  mov    %rbp,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L26
  lea    D14(%rip),%rdi
  call   <printf@plt>
L27:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L26:
  lea    D26(%rip),%rdi
  call   <printf@plt>
  jmp    L27
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D20
datatype: string
name: disassemble_data
arguments: data_label: D33
datatype: string
name: disassemble_data
arguments: data_label: D14
datatype: string
name: disassemble_data
arguments: data_label: D26
datatype: string

tool: D20: Enter password: 

tool: D33: letmein

tool: D14: Performing tasks that require password...


tool: D26: Incorrect Password!


assistant: ```c
int main(void) {
  char buffer[11];
  int correct_password = 0;
  printf("Enter password: ");
  gets(buffer);
  if (strncmp(buffer, "letmein", 7) == 0)
    correct_password = 1;
  else
    printf("Incorrect Password!\n");
  if (correct_password != 0)
    printf("Performing tasks that require password...\n");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<__strspn_c1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L16
L9:
  addq   $0x1,-0x8(%rbp)
L16:
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  je     L9
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long int __strspn_c1(const char *__s, signed int __accept) {
  unsigned long int __result = (unsigned long int)0;
  while ((signed int)__s[(signed long int)__result] == __accept) {
    __result = __result + 1ul;
    __result;
  }
  return __result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<__strspn_c1>:
  endbr64
  movsbl (%rdi),%eax
  cmp    %eax,%esi
  jne    L13
  mov    $0x0,%eax
L4:
  add    $0x1,%rax
  movsbl (%rdi,%rax,1),%edx
  cmp    %esi,%edx
  je     L4
  ret
L13:
  mov    $0x0,%eax
  ret
```

assistant: ```c
unsigned long int __strspn_c1(const char *__s, signed int __accept) {
  unsigned long int __result = (unsigned long int)0;
  while ((signed int)__s[(signed long int)__result] == __accept) {
    __result = __result + 1ul;
    __result;
  }
  return __result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<__strspn_c1>:
  endbr64
  movsbl (%rdi),%eax
  cmp    %eax,%esi
  mov    $0x0,%eax
  jne    L3
L14:
  add    $0x1,%rax
  movsbl (%rdi,%rax,1),%edx
  cmp    %esi,%edx
  je     L14
  ret
  nopl   (%rax)
L3:
  ret
```

assistant: ```c
unsigned long int __strspn_c1(const char *__s, signed int __accept) {
  unsigned long int __result = (unsigned long int)0;
  while ((signed int)__s[(signed long int)__result] == __accept) {
    __result = __result + 1ul;
    __result;
  }
  return __result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<__strspn_c1>:
  endbr64
  movsbl (%rdi),%eax
  cmp    %eax,%esi
  mov    $0x0,%eax
  jne    L9
L1:
  add    $0x1,%rax
  movsbl (%rdi,%rax,1),%edx
  cmp    %esi,%edx
  je     L1
  ret
  nopl   (%rax)
L9:
  ret
```

assistant: ```c
unsigned long int __strspn_c1(const char *__s, signed int __accept) {
  unsigned long int __result = (unsigned long int)0;
  while ((signed int)__s[(signed long int)__result] == __accept) {
    __result = __result + 1ul;
    __result;
  }
  return __result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sub_reg32_reg32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    $0x29,%edi
  call   <put8@plt>
  mov    -0x8(%rbp),%eax
  shl    $0x3,%eax
  or     -0x4(%rbp),%eax
  or     $0xc0,%al
  mov    %eax,%edi
  call   <put8@plt>
  nop
  leave
  ret
```

assistant: ```c
void sub_reg32_reg32(int reg1, int reg2) {
  put8(0x29);
  put8((reg2 << 3) | reg1 | 0xC0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sub_reg32_reg32>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  mov    $0x29,%edi
  call   <put8@plt>
  lea    0x0(,%rbx,8),%edi
  or     %ebp,%edi
  or     $0xc0,%dil
  call   <put8@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void sub_reg32_reg32(int reg1, int reg2) {
  put8(0x29);
  put8((reg2 << 3) | reg1 | 0xC0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sub_reg32_reg32>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  mov    $0x29,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <put8@plt>
  lea    0x0(,%rbx,8),%edi
  add    $0x8,%rsp
  or     %ebp,%edi
  pop    %rbx
  pop    %rbp
  or     $0xc0,%dil
  jmp    1050 <put8@plt>
```

assistant: ```c
void sub_reg32_reg32(int reg1, int reg2) {
  put8(0x29);
  put8((reg2 << 3) | reg1 | 0xC0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sub_reg32_reg32>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  mov    $0x29,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <put8@plt>
  lea    0x0(,%rbx,8),%edi
  add    $0x8,%rsp
  or     %ebp,%edi
  pop    %rbx
  pop    %rbp
  or     $0xc0,%dil
  jmp    1050 <put8@plt>
```

assistant: ```c
void sub_reg32_reg32(int reg1, int reg2) {
  put8(0x29);
  put8((reg2 << 3) | reg1 | 0xC0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<srandom_r>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L101
  mov    %rsi,%rbx
  mov    (%rsi),%eax
  cmp    $0x4,%eax
  ja     L70
  mov    0x8(%rsi),%rcx
  test   %edi,%edi
  mov    $0x1,%edx
  cmove  %edx,%edi
  mov    %edi,(%rcx)
  test   %eax,%eax
  je     L28
  mov    0x10(%rsi),%r8d
  movslq %r8d,%rsi
  cmp    $0x1,%r8d
  jle    L32
  mov    $0x1,%edx
L95:
  movslq %edi,%rax
  imul   $0xffffffff834e0b5f,%rax,%rax
  shr    $0x20,%rax
  add    %edi,%eax
  sar    $0x10,%eax
  mov    %edi,%r9d
  sar    $0x1f,%r9d
  sub    %r9d,%eax
  imul   $0x1f31d,%eax,%r9d
  sub    %r9d,%edi
  imul   $0x41a7,%edi,%edi
  imul   $0xfffff4ec,%eax,%eax
  add    %eax,%edi
  lea    0x7fffffff(%rdi),%eax
  cmovs  %eax,%edi
  mov    %edi,(%rcx,%rdx,4)
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L95
  mov    0x20(%rbx),%rax
  lea    (%rcx,%rax,4),%rax
  mov    %rax,0x18(%rbx)
  mov    %rcx,0x28(%rbx)
  lea    (%r8,%r8,4),%eax
  lea    -0x1(%rax,%rax,1),%ebp
L108:
  lea    0x4(%rsp),%r12
L7:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <random_r@plt>
  sub    $0x1,%ebp
  jns    L7
  mov    $0x0,%eax
L28:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L84
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L32:
  mov    0x20(%rbx),%rax
  lea    (%rcx,%rax,4),%rax
  mov    %rax,0x18(%rbx)
  mov    %rcx,0x28(%rbx)
  lea    (%r8,%r8,4),%eax
  lea    -0x1(%rax,%rax,1),%ebp
  test   %ebp,%ebp
  jns    L108
  mov    $0x0,%eax
  jmp    L28
L101:
  mov    $0xffffffff,%eax
  jmp    L28
L70:
  mov    $0xffffffff,%eax
  jmp    L28
L84:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int srandom_r(unsigned int seed, struct random_data *buf) {
  int type;
  int *state;
  long int i;
  int word;
  int *dst;
  int kc;
  if (buf == NULL)
    goto fail;
  type = buf->rand_type;
  if ((unsigned int)type >= 5)
    goto fail;
  state = buf->state;
  if (seed == 0)
    seed = 1;
  state[0] = seed;
  if (type == 0)
    goto done;
  dst = state;
  word = seed;
  kc = buf->rand_deg;
  for (i = 1; i < kc; ++i) {
    long int hi = word / 127773;
    long int lo = word % 127773;
    word = 16807 * lo - 2836 * hi;
    if (word < 0)
      word += 2147483647;
    *++dst = word;
  }
  buf->fptr = &state[buf->rand_sep];
  buf->rptr = &state[0];
  kc *= 10;
  while (--kc >= 0) {
    int discard;
    (void)random_r(buf, &discard);
  }
done:
  return 0;
fail:
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<srandom_r>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L94
  mov    (%rsi),%edx
  mov    %rsi,%rbp
  cmp    $0x4,%edx
  ja     L94
  test   %edi,%edi
  mov    $0x1,%eax
  mov    0x8(%rsi),%r8
  cmovne %edi,%eax
  mov    %eax,(%r8)
  test   %edx,%edx
  je     L53
  mov    0x20(%rsi),%rcx
  movslq 0x10(%rsi),%rdx
  lea    (%r8,%rcx,4),%rdi
  lea    (%rdx,%rdx,4),%ecx
  lea    -0x1(%rcx,%rcx,1),%ebx
  cmp    $0x1,%edx
  jle    L85
  lea    -0x4(%r8,%rdx,4),%rsi
  mov    %r8,%rcx
L34:
  movslq %eax,%rdx
  mov    %eax,%r9d
  imul   $0xffffffff834e0b5f,%rdx,%rdx
  sar    $0x1f,%r9d
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x10,%edx
  sub    %r9d,%edx
  imul   $0x1f31d,%edx,%r9d
  imul   $0xfffff4ec,%edx,%edx
  sub    %r9d,%eax
  imul   $0x41a7,%eax,%eax
  add    %edx,%eax
  lea    0x7fffffff(%rax),%edx
  cmovs  %edx,%eax
  add    $0x4,%rcx
  mov    %eax,(%rcx)
  cmp    %rsi,%rcx
  jne    L34
  mov    %rdi,0x18(%rbp)
  mov    %r8,0x28(%rbp)
L37:
  lea    0x4(%rsp),%r12
  cs nopw 0x0(%rax,%rax,1)
L77:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <random_r@plt>
  sub    $0x1,%ebx
  jns    L77
L53:
  xor    %eax,%eax
L58:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L41
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L85:
  mov    %rdi,0x18(%rsi)
  mov    %r8,0x28(%rsi)
  test   %ebx,%ebx
  js     L53
  jmp    L37
  xchg   %ax,%ax
L94:
  mov    $0xffffffff,%eax
  jmp    L58
L41:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int srandom_r(unsigned int seed, struct random_data *buf) {
  int type;
  int *state;
  long int i;
  int word;
  int *dst;
  int kc;
  if (buf == NULL)
    goto fail;
  type = buf->rand_type;
  if ((unsigned int)type >= 5)
    goto fail;
  state = buf->state;
  if (seed == 0)
    seed = 1;
  state[0] = seed;
  if (type == 0)
    goto done;
  dst = state;
  word = seed;
  kc = buf->rand_deg;
  for (i = 1; i < kc; ++i) {
    long int hi = word / 127773;
    long int lo = word % 127773;
    word = 16807 * lo - 2836 * hi;
    if (word < 0)
      word += 2147483647;
    *++dst = word;
  }
  buf->fptr = &state[buf->rand_sep];
  buf->rptr = &state[0];
  kc *= 10;
  while (--kc >= 0) {
    int discard;
    (void)random_r(buf, &discard);
  }
done:
  return 0;
fail:
  return -1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<srandom_r>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rsi,%rsi
  je     L44
  mov    (%rsi),%eax
  mov    %rsi,%rbp
  cmp    $0x4,%eax
  ja     L44
  test   %edi,%edi
  mov    $0x1,%edx
  mov    0x8(%rsi),%r8
  cmovne %edi,%edx
  mov    %edx,(%r8)
  test   %eax,%eax
  je     L39
  mov    0x20(%rsi),%rcx
  movslq 0x10(%rsi),%rax
  lea    (%r8,%rcx,4),%rdi
  lea    (%rax,%rax,4),%ecx
  lea    -0x1(%rcx,%rcx,1),%ebx
  cmp    $0x1,%eax
  jle    L46
  lea    -0x4(%r8,%rax,4),%rsi
  mov    %r8,%rcx
L96:
  movslq %edx,%rax
  mov    %edx,%r9d
  imul   $0xffffffff834e0b5f,%rax,%rax
  sar    $0x1f,%r9d
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x10,%eax
  sub    %r9d,%eax
  imul   $0xfffff4ec,%eax,%r9d
  imul   $0x1f31d,%eax,%eax
  sub    %eax,%edx
  imul   $0x41a7,%edx,%edx
  lea    (%r9,%rdx,1),%eax
  test   %eax,%eax
  lea    0x7fffffff(%rax),%edx
  cmovns %eax,%edx
  add    $0x4,%rcx
  mov    %edx,(%rcx)
  cmp    %rcx,%rsi
  jne    L96
  mov    %rdi,0x18(%rbp)
  mov    %r8,0x28(%rbp)
L23:
  lea    0x4(%rsp),%r12
  nopl   0x0(%rax)
L17:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <random_r@plt>
  sub    $0x1,%ebx
  jns    L17
L39:
  xor    %eax,%eax
L75:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L54
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L46:
  mov    %rdi,0x18(%rsi)
  mov    %r8,0x28(%rsi)
  test   %ebx,%ebx
  js     L39
  jmp    L23
  xchg   %ax,%ax
L44:
  mov    $0xffffffff,%eax
  jmp    L75
L54:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int srandom_r(unsigned int seed, struct random_data *buf) {
  int type;
  int *state;
  long int i;
  int word;
  int *dst;
  int kc;
  if (buf == NULL)
    goto fail;
  type = buf->rand_type;
  if ((unsigned int)type >= 5)
    goto fail;
  state = buf->state;
  if (seed == 0)
    seed = 1;
  state[0] = seed;
  if (type == 0)
    goto done;
  dst = state;
  word = seed;
  kc = buf->rand_deg;
  for (i = 1; i < kc; ++i) {
    long int hi = word / 127773;
    long int lo = word % 127773;
    word = 16807 * lo - 2836 * hi;
    if (word < 0)
      word += 2147483647;
    *++dst = word;
  }
  buf->fptr = &state[buf->rand_sep];
  buf->rptr = &state[0];
  kc *= 10;
  while (--kc >= 0) {
    int discard;
    (void)random_r(buf, &discard);
  }
done:
  return 0;
fail:
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x400,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x3f8(%rbp)
  jmp    L46
L0:
  mov    -0x3f8(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x3f0(%rbp),%rcx
  mov    -0x3f8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %rax,%rdi
  call   <gets@plt>
  addl   $0x1,-0x3f8(%rbp)
L46:
  cmpl   $0x9,-0x3f8(%rbp)
  jle    L0
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <insertion_sort@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x3f4(%rbp)
  jmp    L3
L39:
  lea    -0x3f0(%rbp),%rcx
  mov    -0x3f4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addl   $0x1,-0x3f4(%rbp)
L3:
  cmpl   $0x9,-0x3f4(%rbp)
  jle    L39
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D18
type: string
name: read_data
arguments: tag: D10
type: string

tool: D18: Ingresa el %d nombre: 

tool: D10: 
Nombres ordenados:

assistant: ```c
int main() {
  char nombres[10][100];
  for (int i = 0; i < 10; i++) {
    printf("Ingresa el %d nombre: ", i + 1);
    gets(nombres[i]);
  }
  insertion_sort(nombres);
  printf("\nNombres ordenados:\n");
  for (int i = 0; i < 10; i++)
    printf("%s\n", nombres[i]);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%r12
  mov    $0x0,%ebp
  lea    D18(%rip),%r13
L19:
  add    $0x1,%ebp
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  call   <gets@plt>
  add    $0x64,%r12
  cmp    $0xa,%ebp
  jne    L19
  mov    %rsp,%rdi
  call   <insertion_sort@plt>
  lea    D7(%rip),%rdi
  call   <puts@plt>
  lea    0x3e8(%rbx),%rbp
L9:
  mov    %rbx,%rdi
  call   <puts@plt>
  add    $0x64,%rbx
  cmp    %rbp,%rbx
  jne    L9
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    $0x0,%eax
  add    $0x3f8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D18
type_guessed: string
name: unpack_data
arguments: data_tag: D7
type_guessed: string

tool: D18: Ingresa el %d nombre: 

tool: D7: 
Nombres ordenados:

assistant: ```c
int main() {
  char nombres[10][100];
  for (int i = 0; i < 10; i++) {
    printf("Ingresa el %d nombre: ", i + 1);
    gets(nombres[i]);
  }
  insertion_sort(nombres);
  printf("\nNombres ordenados:\n");
  for (int i = 0; i < 10; i++)
    printf("%s\n", nombres[i]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D10(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  mov    %r13,%rbx
  mov    %r13,%r12
  nopl   0x0(%rax,%rax,1)
L21:
  add    $0x1,%ebp
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %ebp,%esi
  call   <printf@plt>
  mov    %r12,%rdi
  add    $0x64,%r12
  call   <gets@plt>
  cmp    $0xa,%ebp
  jne    L21
  mov    %r13,%rdi
  add    $0x3e8,%r13
  call   <insertion_sort@plt>
  lea    D6(%rip),%rdi
  call   <puts@plt>
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %rbx,%rdi
  add    $0x64,%rbx
  call   <puts@plt>
  cmp    %r13,%rbx
  jne    L8
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x3f0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D10
type_infered: string
name: unpack_data
arguments: tag: D6
type_infered: string

tool: D10: Ingresa el %d nombre: 

tool: D6: 
Nombres ordenados:

assistant: ```c
int main() {
  char nombres[10][100];
  for (int i = 0; i < 10; i++) {
    printf("Ingresa el %d nombre: ", i + 1);
    gets(nombres[i]);
  }
  insertion_sort(nombres);
  printf("\nNombres ordenados:\n");
  for (int i = 0; i < 10; i++)
    printf("%s\n", nombres[i]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  lea    D18(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  mov    %r13,%rbx
  mov    %r13,%r12
  nopl   0x0(%rax,%rax,1)
L19:
  add    $0x1,%ebp
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %ebp,%esi
  call   <printf@plt>
  mov    %r12,%rdi
  add    $0x64,%r12
  call   <gets@plt>
  cmp    $0xa,%ebp
  jne    L19
  mov    %r13,%rdi
  add    $0x3e8,%r13
  call   <insertion_sort@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
  nopl   0x0(%rax,%rax,1)
L18:
  mov    %rbx,%rdi
  add    $0x64,%rbx
  call   <puts@plt>
  cmp    %rbx,%r13
  jne    L18
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x3f0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D18
type: string
name: retrieve_data
arguments: tag: D1
type: string

tool: D18: Ingresa el %d nombre: 

tool: D1: 
Nombres ordenados:

assistant: ```c
int main() {
  char nombres[10][100];
  for (int i = 0; i < 10; i++) {
    printf("Ingresa el %d nombre: ", i + 1);
    gets(nombres[i]);
  }
  insertion_sort(nombres);
  printf("\nNombres ordenados:\n");
  for (int i = 0; i < 10; i++)
    printf("%s\n", nombres[i]);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<_float32_unpack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  and    $0x1fffff,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  and    $0x80000000,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  sar    $0x15,%rax
  and    $0x3ff,%eax
  mov    %rax,-0x8(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L7
  movsd  -0x10(%rbp),%xmm0
  movq   D6(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
L7:
  mov    -0x8(%rbp),%rax
  lea    -0x314(%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rdi
  movq   %rax,%xmm0
  call   <ldexp@plt>
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
type_infered: byte[8]

tool: {"D6": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
float _float32_unpack(long val) {
  double mant = val & 0x1fffff;
  int sign = val & 0x80000000;
  long exp = (val & 0x7fe00000L) >> 21;
  if (sign)
    mant = -mant;
  return (ldexp(mant, exp - (21 - 1) - 768));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_float32_unpack>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rdx
  mov    %rdi,%rax
  and    $0x1fffff,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  sar    $0x15,%rdi
  mov    %rdi,%rax
  and    $0x3ff,%eax
  test   %edx,%edx
  js     L19
L9:
  lea    -0x314(%rax),%rdi
  call   <ldexp@plt>
  add    $0x8,%rsp
  ret
L19:
  xorpd  D3(%rip),%xmm0
  jmp    L9
```

tool_calls: name: parse_data
arguments: data_label: D3
type: byte[16]

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
float _float32_unpack(long val) {
  double mant = val & 0x1fffff;
  int sign = val & 0x80000000;
  long exp = (val & 0x7fe00000L) >> 21;
  if (sign)
    mant = -mant;
  return (ldexp(mant, exp - (21 - 1) - 768));
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_float32_unpack>:
  endbr64
  mov    %rdi,%rax
  mov    %rdi,%rdx
  pxor   %xmm0,%xmm0
  sar    $0x15,%rdi
  and    $0x1fffff,%eax
  cvtsi2sd %rax,%xmm0
  mov    %rdi,%rax
  and    $0x3ff,%eax
  test   %edx,%edx
  jns    L0
  xorpd  D0(%rip),%xmm0
L0:
  lea    -0x314(%rax),%rdi
  jmp    1050 <ldexp@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D0
type_guessed: byte[16]

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
float _float32_unpack(long val) {
  double mant = val & 0x1fffff;
  int sign = val & 0x80000000;
  long exp = (val & 0x7fe00000L) >> 21;
  if (sign)
    mant = -mant;
  return (ldexp(mant, exp - (21 - 1) - 768));
}
```