ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<event_info_key>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  lea    D6(%rip),%rsi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strcmp@plt>
  test   %eax,%eax
  je     L9
L20:
  lea    D19(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L18
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L9:
  mov    %r12,%rdi
  call   <readsnumf@plt>
  xor    %esi,%esi
  mov    %eax,%edi
  xor    %eax,%eax
  call   <event_octave@plt>
  jmp    L20
  nopl   (%rax)
L18:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <event_specific@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type_guessed: string
name: extract_symbolic_data
arguments: tag: D19
type_guessed: string

tool: D6: octave

tool: D19: MIDI

assistant: ```c
void event_info_key(key, value) char *key;
char *value;
{
  int num;
  if (strcmp(key, "octave") == 0) {
    num = readsnumf(value);
    event_octave(num, 0);
  };
  if (strcmp(key, "MIDI") == 0)
    event_specific(key, value);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<event_info_key>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  lea    D8(%rip),%rsi
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strcmp@plt>
  test   %eax,%eax
  je     L7
L3:
  lea    D18(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L11
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L7:
  mov    %r12,%rdi
  call   <readsnumf@plt>
  xor    %esi,%esi
  mov    %eax,%edi
  xor    %eax,%eax
  call   <event_octave@plt>
  jmp    L3
  nopl   (%rax)
L11:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <event_specific@plt>
```

tool_calls: name: read_data
arguments: tag: D8
type_infered: string
name: read_data
arguments: tag: D18
type_infered: string

tool: D8: octave

tool: D18: MIDI

assistant: ```c
void event_info_key(key, value) char *key;
char *value;
{
  int num;
  if (strcmp(key, "octave") == 0) {
    num = readsnumf(value);
    event_octave(num, 0);
  };
  if (strcmp(key, "MIDI") == 0)
    event_specific(key, value);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x18(%rbp)
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movq   $0x0,-0x20(%rbp)
  jmp    L47
L89:
  lea    -0x28(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  cmp    %rax,%rdx
  jle    L79
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L87
L79:
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  cmp    %rax,%rdx
  jne    L99
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L39
L99:
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L87:
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  imul   %rax,%rdx
  mov    -0x10(%rbp),%rax
  imul   %rax,%rax
  mov    %rax,%rcx
  mov    %rdx,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L57
  mov    -0x30(%rbp),%rax
  cqto
  idivq  -0x10(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L57
  mov    -0x28(%rbp),%rax
  cqto
  idivq  -0x10(%rbp)
  mov    %rdx,%rax
  test   %rax,%rax
  jne    L57
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  imul   %rax,%rdx
  mov    -0x10(%rbp),%rax
  imul   %rax,%rax
  mov    %rax,%rcx
  mov    %rdx,%rax
  cqto
  idiv   %rcx
  mov    %rax,-0x18(%rbp)
  jmp    L4
L57:
  subq   $0x1,-0x10(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L87
L4:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movq   $0x0,-0x18(%rbp)
L39:
  addq   $0x1,-0x20(%rbp)
L47:
  mov    -0x38(%rbp),%rax
  cmp    %rax,-0x20(%rbp)
  jl     L89
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D17
type: string
name: disassemble_data
arguments: data_tag: D22
type: string
name: disassemble_data
arguments: data_tag: D30
type: string
name: disassemble_data
arguments: data_tag: D15
type: string

tool: D17: %ld

tool: D22: %ld %ld

tool: D30: 1

tool: D15: %ld


assistant: ```c
int main() {
  long int t, l, b, i, j, k = 0, min;
  scanf("%ld", &t);
  for (i = 0; i < t; i++) {
    scanf("%ld %ld", &l, &b);
    if (l > b)
      min = b;
    else if (l == b) {
      printf("1\n");
      continue;
    } else
      min = l;
    do {
      if ((l * b) % (min * min) == 0 && l % min == 0 && b % min == 0) {
        k = (l * b) / (min * min);
        break;
      } else
        min--;
    } while (k == 0);
    printf("%ld\n", k);
    k = 0;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  lea    D19(%rip),%rdi
  call   <scanf@plt>
  cmpq   $0x0,(%rsp)
  jle    L118
  mov    $0x0,%ebx
  lea    0x10(%rsp),%r12
  lea    D34(%rip),%rbp
  lea    D18(%rip),%r13
  lea    D9(%rip),%r14
  jmp    L45
L77:
  mov    %r9,%rcx
  jmp    L112
L103:
  mov    %r14,%rdi
  call   <puts@plt>
  jmp    L7
L75:
  sub    $0x1,%rcx
L40:
  mov    %rcx,%rdi
  imul   %rcx,%rdi
  mov    %rsi,%rax
  cqto
  idiv   %rdi
  test   %rdx,%rdx
  jne    L75
  mov    %r8,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L75
  mov    %r9,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L75
  mov    %rsi,%rax
  cqto
  idiv   %rdi
  mov    %rax,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L7:
  add    $0x1,%rbx
  cmp    %rbx,(%rsp)
  jle    L118
L45:
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%r8
  mov    0x10(%rsp),%r9
  cmp    %r9,%r8
  jg     L77
  mov    %r8,%rcx
  je     L103
L112:
  mov    %r8,%rsi
  imul   %r9,%rsi
  jmp    L40
L118:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L113
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L113:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D19
type_guessed: string
name: extract_symbolic_data
arguments: label: D34
type_guessed: string
name: extract_symbolic_data
arguments: label: D18
type_guessed: string
name: extract_symbolic_data
arguments: label: D9
type_guessed: string

tool: D19: %ld

tool: D34: %ld %ld

tool: D18: %ld


tool: D9: 1

assistant: ```c
int main() {
  long int t, l, b, i, j, k = 0, min;
  scanf("%ld", &t);
  for (i = 0; i < t; i++) {
    scanf("%ld %ld", &l, &b);
    if (l > b)
      min = b;
    else if (l == b) {
      printf("1\n");
      continue;
    } else
      min = l;
    do {
      if ((l * b) % (min * min) == 0 && l % min == 0 && b % min == 0) {
        k = (l * b) / (min * min);
        break;
      } else
        min--;
    } while (k == 0);
    printf("%ld\n", k);
    k = 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r15
  lea    D0(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  cmpq   $0x0,(%rsp)
  jle    L66
  xor    %ebx,%ebx
  lea    0x10(%rsp),%r13
  lea    0x8(%rsp),%r12
  lea    D25(%rip),%rbp
  lea    D39(%rip),%r14
  lea    D13(%rip),%r15
  nopl   (%rax)
L90:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%r8
  mov    0x10(%rsp),%r9
  cmp    %r9,%r8
  jg     L14
  mov    %r8,%rcx
  je     L47
L20:
  mov    %r8,%rdi
  imul   %r9,%rdi
  jmp    L59
  nopl   (%rax)
L100:
  sub    $0x1,%rcx
L59:
  mov    %rcx,%rsi
  mov    %rdi,%rax
  imul   %rcx,%rsi
  cqto
  idiv   %rsi
  mov    %rax,%rsi
  test   %rdx,%rdx
  jne    L100
  mov    %r8,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L100
  mov    %r9,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L100
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L11:
  add    $0x1,%rbx
  cmp    %rbx,(%rsp)
  jg     L90
L66:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L14:
  mov    %r9,%rcx
  jmp    L20
L47:
  mov    %r15,%rdi
  call   <puts@plt>
  jmp    L11
L58:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D0
type_infered: string
name: disassemble_data
arguments: label: D25
type_infered: string
name: disassemble_data
arguments: label: D39
type_infered: string
name: disassemble_data
arguments: label: D13
type_infered: string

tool: D0: %ld

tool: D25: %ld %ld

tool: D39: %ld


tool: D13: 1

assistant: ```c
int main() {
  long int t, l, b, i, j, k = 0, min;
  scanf("%ld", &t);
  for (i = 0; i < t; i++) {
    scanf("%ld %ld", &l, &b);
    if (l > b)
      min = b;
    else if (l == b) {
      printf("1\n");
      continue;
    } else
      min = l;
    do {
      if ((l * b) % (min * min) == 0 && l % min == 0 && b % min == 0) {
        k = (l * b) / (min * min);
        break;
      } else
        min--;
    } while (k == 0);
    printf("%ld\n", k);
    k = 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r15
  lea    D35(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  cmpq   $0x0,(%rsp)
  jle    L88
  xor    %ebx,%ebx
  lea    0x10(%rsp),%r13
  lea    0x8(%rsp),%r12
  lea    D3(%rip),%rbp
  lea    D38(%rip),%r14
  lea    D19(%rip),%r15
  nopl   (%rax)
L40:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%r8
  mov    0x10(%rsp),%r9
  cmp    %r9,%r8
  jg     L90
  mov    %r8,%rcx
  je     L74
L8:
  mov    %r8,%rdi
  imul   %r9,%rdi
  jmp    L33
  nopl   (%rax)
L94:
  sub    $0x1,%rcx
L33:
  mov    %rcx,%rsi
  mov    %rdi,%rax
  imul   %rcx,%rsi
  cqto
  idiv   %rsi
  mov    %rax,%rsi
  test   %rdx,%rdx
  jne    L94
  mov    %r8,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L94
  mov    %r9,%rax
  cqto
  idiv   %rcx
  test   %rdx,%rdx
  jne    L94
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L97:
  add    $0x1,%rbx
  cmp    %rbx,(%rsp)
  jg     L40
L88:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L90:
  mov    %r9,%rcx
  jmp    L8
L74:
  mov    %r15,%rdi
  call   <puts@plt>
  jmp    L97
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D35
type_infered: string
name: disassemble_data
arguments: label: D3
type_infered: string
name: disassemble_data
arguments: label: D38
type_infered: string
name: disassemble_data
arguments: label: D19
type_infered: string

tool: D35: %ld

tool: D3: %ld %ld

tool: D38: %ld


tool: D19: 1

assistant: ```c
int main() {
  long int t, l, b, i, j, k = 0, min;
  scanf("%ld", &t);
  for (i = 0; i < t; i++) {
    scanf("%ld %ld", &l, &b);
    if (l > b)
      min = b;
    else if (l == b) {
      printf("1\n");
      continue;
    } else
      min = l;
    do {
      if ((l * b) % (min * min) == 0 && l % min == 0 && b % min == 0) {
        k = (l * b) / (min * min);
        break;
      } else
        min--;
    } while (k == 0);
    printf("%ld\n", k);
    k = 0;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<newtonRecursive>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x8(%rbp)
  movsd  %xmm1,-0x10(%rbp)
  movsd  %xmm2,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  subsd  -0x10(%rbp),%xmm0
  movq   D7(%rip),%xmm1
  andpd  %xmm0,%xmm1
  movsd  -0x8(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jb     L0
  mov    -0x18(%rbp),%rax
  jmp    L5
L0:
  movsd  -0x18(%rbp),%xmm0
  mulsd  %xmm0,%xmm0
  movapd %xmm0,%xmm1
  addsd  -0x10(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  divsd  %xmm2,%xmm1
  movapd %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x8(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <newtonRecursive>
  movq   %xmm0,%rax
L5:
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D7
type_guessed: qword

tool: D7: 9223372036854775807

assistant: ```c
double newtonRecursive(double e, double x, double recursiveAnswer) {
  if (fabs(recursiveAnswer * recursiveAnswer - x) <= e) {
    return recursiveAnswer;
  } else {
    recursiveAnswer = (recursiveAnswer * recursiveAnswer + x) / (2 * recursiveAnswer);
    return 1.0 * (newtonRecursive(recursiveAnswer, x, e));
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<newtonRecursive>:
  endbr64
  movapd %xmm2,%xmm4
  mulsd  %xmm2,%xmm4
  movapd %xmm4,%xmm3
  subsd  %xmm1,%xmm3
  andpd  D2(%rip),%xmm3
  comisd %xmm3,%xmm0
  jb     L2
  movapd %xmm2,%xmm0
  ret
L2:
  sub    $0x8,%rsp
  addsd  %xmm2,%xmm2
  addsd  %xmm1,%xmm4
  divsd  %xmm2,%xmm4
  movapd %xmm0,%xmm2
  movapd %xmm4,%xmm0
  call   <newtonRecursive>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
datatype: qword[2]

tool: {"D2": [0x7fffffffffffffff,0x0000000000000000]}

assistant: ```c
double newtonRecursive(double e, double x, double recursiveAnswer) {
  if (fabs(recursiveAnswer * recursiveAnswer - x) <= e) {
    return recursiveAnswer;
  } else {
    recursiveAnswer = (recursiveAnswer * recursiveAnswer + x) / (2 * recursiveAnswer);
    return 1.0 * (newtonRecursive(recursiveAnswer, x, e));
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<newtonRecursive>:
  endbr64
  movapd %xmm2,%xmm3
  movq   D7(%rip),%xmm6
  mulsd  %xmm2,%xmm3
  movapd %xmm3,%xmm4
  subsd  %xmm1,%xmm4
  andpd  %xmm6,%xmm4
  comisd %xmm4,%xmm0
  jb     L26
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L27:
  movapd %xmm5,%xmm0
L26:
  addsd  %xmm1,%xmm3
  addsd  %xmm2,%xmm2
  movapd %xmm3,%xmm5
  movapd %xmm0,%xmm3
  divsd  %xmm2,%xmm5
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm3
  movapd %xmm3,%xmm4
  subsd  %xmm1,%xmm4
  andpd  %xmm6,%xmm4
  comisd %xmm4,%xmm5
  jb     L27
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  movapd %xmm2,%xmm0
  ret
```

tool_calls: name: read_data
arguments: label: D7
type: qword

tool: D7: 9223372036854775807

assistant: ```c
double newtonRecursive(double e, double x, double recursiveAnswer) {
  if (fabs(recursiveAnswer * recursiveAnswer - x) <= e) {
    return recursiveAnswer;
  } else {
    recursiveAnswer = (recursiveAnswer * recursiveAnswer + x) / (2 * recursiveAnswer);
    return 1.0 * (newtonRecursive(recursiveAnswer, x, e));
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<newtonRecursive>:
  endbr64
  movapd %xmm2,%xmm3
  movq   D4(%rip),%xmm6
  mulsd  %xmm2,%xmm3
  movapd %xmm3,%xmm4
  subsd  %xmm1,%xmm4
  andpd  %xmm6,%xmm4
  comisd %xmm4,%xmm0
  jb     L13
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L29:
  movapd %xmm5,%xmm0
L13:
  addsd  %xmm1,%xmm3
  addsd  %xmm2,%xmm2
  movapd %xmm3,%xmm5
  movapd %xmm0,%xmm3
  divsd  %xmm2,%xmm5
  movapd %xmm0,%xmm2
  mulsd  %xmm0,%xmm3
  movapd %xmm3,%xmm4
  subsd  %xmm1,%xmm4
  andpd  %xmm6,%xmm4
  comisd %xmm4,%xmm5
  jb     L29
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  movapd %xmm2,%xmm0
  ret
```

tool_calls: name: parse_data
arguments: label: D4
datatype: qword

tool: D4: 9223372036854775807

assistant: ```c
double newtonRecursive(double e, double x, double recursiveAnswer) {
  if (fabs(recursiveAnswer * recursiveAnswer - x) <= e) {
    return recursiveAnswer;
  } else {
    recursiveAnswer = (recursiveAnswer * recursiveAnswer + x) / (2 * recursiveAnswer);
    return 1.0 * (newtonRecursive(recursiveAnswer, x, e));
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<captura>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <fflush@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
type_infered: string

tool: D1: He conseguido capturar SIGALRM


assistant: ```c
void captura(int sennal) {
  printf("He conseguido capturar SIGALRM\n");
  fflush(NULL);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<captura>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <fflush@plt>
  add    $0x8,%rsp
  ret
```

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

tool: D7: He conseguido capturar SIGALRM


assistant: ```c
void captura(int sennal) {
  printf("He conseguido capturar SIGALRM\n");
  fflush(NULL);
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<captura>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <fflush@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D1
type_infered: string

tool: D1: He conseguido capturar SIGALRM


assistant: ```c
void captura(int sennal) {
  printf("He conseguido capturar SIGALRM\n");
  fflush(NULL);
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<captura>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <fflush@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
data_type: string

tool: D6: He conseguido capturar SIGALRM


assistant: ```c
void captura(int sennal) {
  printf("He conseguido capturar SIGALRM\n");
  fflush(NULL);
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<closeUninheritableFds>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L2:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  je     L49
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  je     L49
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L49
  mov    D15(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L49
  mov    D24(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L49
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
L49:
  addl   $0x1,-0x4(%rbp)
L18:
  cmpl   $0x3f,-0x4(%rbp)
  jle    L2
  nop
  nop
  leave
  ret
```

assistant: ```c
void closeUninheritableFds(int const stdinFd, int const stdoutFd) {
  int fd;
  for (fd = 0; fd < 64; ++fd) {
    if (fd == stdinFd) {
    } else if (fd == stdoutFd) {
    } else if (fd == STDIN_FILENO) {
    } else if (fd == STDOUT_FILENO) {
    } else if (fd == STDERR_FILENO) {
    } else {
      close(fd);
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<closeUninheritableFds>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r12d
  mov    %esi,%ebp
  mov    $0x0,%ebx
  jmp    L9
L73:
  mov    %ebx,%edi
  call   <close@plt>
L5:
  add    $0x1,%ebx
  cmp    $0x40,%ebx
  je     L69
L9:
  cmp    %ebx,%ebp
  je     L5
  cmp    %ebx,%r12d
  je     L5
  mov    D7(%rip),%rax
  cmp    %ebx,(%rax)
  je     L5
  mov    D8(%rip),%rax
  cmp    %ebx,(%rax)
  je     L5
  mov    D17(%rip),%rax
  cmp    %ebx,(%rax)
  je     L5
  jmp    L73
L69:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void closeUninheritableFds(int const stdinFd, int const stdoutFd) {
  int fd;
  for (fd = 0; fd < 64; ++fd) {
    if (fd == stdinFd) {
    } else if (fd == stdoutFd) {
    } else if (fd == STDIN_FILENO) {
    } else if (fd == STDOUT_FILENO) {
    } else if (fd == STDERR_FILENO) {
    } else {
      close(fd);
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<closeUninheritableFds>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nop
L59:
  cmp    %ebx,%ebp
  je     L33
  cmp    %ebx,%r12d
  je     L33
  mov    D9(%rip),%rax
  cmp    %ebx,(%rax)
  je     L33
  mov    D2(%rip),%rax
  cmp    %ebx,(%rax)
  je     L33
  mov    D10(%rip),%rax
  cmp    %ebx,(%rax)
  je     L33
  mov    %ebx,%edi
  call   <close@plt>
  nopl   0x0(%rax)
L33:
  add    $0x1,%ebx
  cmp    $0x40,%ebx
  jne    L59
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void closeUninheritableFds(int const stdinFd, int const stdoutFd) {
  int fd;
  for (fd = 0; fd < 64; ++fd) {
    if (fd == stdinFd) {
    } else if (fd == stdoutFd) {
    } else if (fd == STDIN_FILENO) {
    } else if (fd == STDOUT_FILENO) {
    } else if (fd == STDERR_FILENO) {
    } else {
      close(fd);
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<closeUninheritableFds>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nop
L38:
  cmp    %ebx,%ebp
  je     L35
  cmp    %ebx,%r12d
  je     L35
  mov    D11(%rip),%rax
  cmp    %ebx,(%rax)
  je     L35
  mov    D8(%rip),%rax
  cmp    %ebx,(%rax)
  je     L35
  mov    D16(%rip),%rax
  cmp    %ebx,(%rax)
  je     L35
  mov    %ebx,%edi
  call   <close@plt>
  nopl   0x0(%rax)
L35:
  add    $0x1,%ebx
  cmp    $0x40,%ebx
  jne    L38
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void closeUninheritableFds(int const stdinFd, int const stdoutFd) {
  int fd;
  for (fd = 0; fd < 64; ++fd) {
    if (fd == stdinFd) {
    } else if (fd == stdoutFd) {
    } else if (fd == STDIN_FILENO) {
    } else if (fd == STDOUT_FILENO) {
    } else if (fd == STDERR_FILENO) {
    } else {
      close(fd);
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<read>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    D15(%rip),%rax
  mov    (%rax),%edx
  mov    D14(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
ssize_t read(int fd, void *dest, size_t count) {
  (void)fd;
  (void)dest;
  (void)count;
  errno = ENODEV;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<read>:
  endbr64
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D13(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
ssize_t read(int fd, void *dest, size_t count) {
  (void)fd;
  (void)dest;
  (void)count;
  errno = ENODEV;
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read>:
  endbr64
  mov    D0(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
ssize_t read(int fd, void *dest, size_t count) {
  (void)fd;
  (void)dest;
  (void)count;
  errno = ENODEV;
  return -1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<read>:
  endbr64
  mov    D12(%rip),%rax
  mov    (%rax),%edx
  mov    D2(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
ssize_t read(int fd, void *dest, size_t count) {
  (void)fd;
  (void)dest;
  (void)count;
  errno = ENODEV;
  return -1;
}
```