ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<snprintf>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  mov    %rdx,%rsi
  push   %r12
  mov    %rdi,%r12
  mov    %r13d,%edi
  push   %rbp
  mov    %rdx,%rbp
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <vsprintf@plt>
  mov    %r13d,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int snprintf(char *str, size_t size, const char *format, ...) {
  int retval;
  va_list ap;
  va_start(ap, format);
  retval = vsprintf(str, format, ap);
  va_end(ap);
  return retval;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x3,(%rsp)
  movl   $0x3,0x4(%rsp)
  movl   $0x4,0x8(%rsp)
  movl   $0x2,0xc(%rsp)
  movl   $0x4,0x10(%rsp)
  movl   $0x4,0x14(%rsp)
  movl   $0x2,0x18(%rsp)
  movl   $0x4,0x1c(%rsp)
  movl   $0x4,0x20(%rsp)
  mov    %rsp,%rbx
  lea    0x24(%rsp),%rbp
  mov    %rbx,%rax
  jmp    L107
L100:
  add    $0x1,%edx
  jmp    L30
L110:
  test   %ecx,%ecx
  jle    L4
  cmp    %r12d,(%rax)
  je     L34
  test   %edx,%edx
  je     L176
  jmp    L53
L34:
  add    $0x1,%ecx
  jmp    L30
L4:
  test   %edx,%edx
  jne    L209
L176:
  mov    (%rax),%esi
  mov    $0x0,%edx
L30:
  add    $0x4,%rax
  cmp    %rbp,%rax
  je     L58
L107:
  test   %edx,%edx
  jle    L110
  cmp    %esi,(%rax)
  je     L100
  test   %ecx,%ecx
  jle    L209
  cmp    %r12d,(%rax)
  je     L34
L53:
  sub    $0x1,%edx
  sub    $0x1,%ecx
  jmp    L30
L209:
  test   %ecx,%ecx
  jne    L53
  mov    (%rax),%r12d
  jmp    L30
L58:
  mov    %rbx,%rax
  mov    $0x0,%edx
L144:
  cmp    %esi,(%rax)
  sete   %cl
  movzbl %cl,%ecx
  add    %ecx,%edx
  add    $0x4,%rax
  cmp    %rbp,%rax
  jne    L144
  cmp    $0x3,%edx
  jg     L206
L18:
  mov    $0x0,%eax
L124:
  cmp    %r12d,(%rbx)
  sete   %dl
  movzbl %dl,%edx
  add    %edx,%eax
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L124
  cmp    $0x3,%eax
  jg     L102
L203:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L147
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L206:
  lea    D12(%rip),%rdi
  call   <printf@plt>
  jmp    L18
L102:
  mov    %r12d,%esi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  jmp    L203
L147:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D12
type_infered: string

tool: D12: Majority element can be  %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int a[] = {3, 3, 4, 2, 4, 4, 2, 4, 4};
  int size = sizeof(a) / sizeof(a[0]);
  int c1, c2;
  int A, B;
  for (int i = 0; i < size; i++) {
    if (c1 > 0 && A == a[i])
      c1++;
    else if (c2 > 0 && B == a[i])
      c2++;
    else if (c1 == 0) {
      c1 = 0;
      A = a[i];
    } else if (c2 == 0) {
      c2 = 0;
      B = a[i];
    } else {
      c1--;
      c2--;
    }
  }
  int count1 = 0, count2 = 0;
  for (int i = 0; i < size; i++) {
    if (a[i] == A) {
      count1++;
    }
  }
  if (count1 >= ((size / 3) + 1))
    printf("Majority element can be  %d\n", A);
  for (int i = 0; i < size; i++) {
    if (a[i] == B) {
      count2++;
    }
  }
  if (count2 >= ((size / 3) + 1))
    printf("Majority element can be  %d\n", B);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x300000003,%rax
  lea    0x24(%rsp),%rcx
  movl   $0x4,0x20(%rsp)
  mov    %rax,(%rsp)
  movabs $0x200000004,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x400000004,%rax
  mov    %rax,0x10(%rsp)
  sub    $0x2,%rax
  mov    %rax,0x18(%rsp)
  mov    %rsp,%rax
  jmp    L106
  nopl   0x0(%rax)
L157:
  mov    (%rax),%r8d
  cmp    %esi,%r8d
  je     L212
  test   %edi,%edi
  jle    L248
  cmp    %r12d,%r8d
  je     L63
L94:
  sub    $0x1,%edx
  sub    $0x1,%edi
L132:
  add    $0x4,%rax
  cmp    %rax,%rcx
  je     L104
L106:
  test   %edx,%edx
  jg     L157
  test   %edi,%edi
  jle    L133
  mov    (%rax),%r8d
  cmp    %r12d,%r8d
  je     L63
  test   %edx,%edx
  jne    L94
  mov    %r8d,%esi
  xor    %edx,%edx
L72:
  add    $0x4,%rax
  cmp    %rax,%rcx
  jne    L106
L104:
  lea    0x4(%rsp),%rbp
  mov    $0x3,%edx
  xor    %eax,%eax
  mov    %rbp,%rbx
  jmp    L92
  nop
L131:
  mov    (%rbx),%edx
  add    $0x4,%rbx
L92:
  cmp    %edx,%esi
  sete   %dl
  movzbl %dl,%edx
  add    %edx,%eax
  cmp    %rbx,%rcx
  jne    L131
  cmp    $0x3,%eax
  jg     L14
L23:
  mov    $0x3,%edx
  xor    %eax,%eax
  jmp    L35
  nopl   0x0(%rax,%rax,1)
L153:
  mov    0x0(%rbp),%edx
  add    $0x4,%rbp
L35:
  cmp    %edx,%r12d
  sete   %dl
  movzbl %dl,%edx
  add    %edx,%eax
  cmp    %rbx,%rbp
  jne    L153
  cmp    $0x3,%eax
  jg     L148
L236:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L197
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L212:
  add    $0x1,%edx
  jmp    L132
  nopl   0x0(%rax,%rax,1)
L133:
  test   %edx,%edx
  jne    L248
  mov    (%rax),%esi
  xor    %edx,%edx
  jmp    L72
  nopl   (%rax)
L248:
  test   %edi,%edi
  jne    L94
  mov    (%rax),%r12d
  jmp    L132
L63:
  add    $0x1,%edi
  jmp    L132
L14:
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L23
L148:
  mov    %r12d,%esi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  jmp    L236
L197:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D10
type: string

tool: D10: Majority element can be  %d


assistant: ```c
int main(int argc, char const *argv[]) {
  int a[] = {3, 3, 4, 2, 4, 4, 2, 4, 4};
  int size = sizeof(a) / sizeof(a[0]);
  int c1, c2;
  int A, B;
  for (int i = 0; i < size; i++) {
    if (c1 > 0 && A == a[i])
      c1++;
    else if (c2 > 0 && B == a[i])
      c2++;
    else if (c1 == 0) {
      c1 = 0;
      A = a[i];
    } else if (c2 == 0) {
      c2 = 0;
      B = a[i];
    } else {
      c1--;
      c2--;
    }
  }
  int count1 = 0, count2 = 0;
  for (int i = 0; i < size; i++) {
    if (a[i] == A) {
      count1++;
    }
  }
  if (count1 >= ((size / 3) + 1))
    printf("Majority element can be  %d\n", A);
  for (int i = 0; i < size; i++) {
    if (a[i] == B) {
      count2++;
    }
  }
  if (count2 >= ((size / 3) + 1))
    printf("Majority element can be  %d\n", B);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<interpo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %ecx,-0x30(%rbp)
  mov    -0x30(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  je     L28
  mov    -0x30(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  cmp    $0x1,%eax
  je     L28
  mov    -0x30(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  shr    %eax
  mov    %eax,-0x14(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x14(%rbp),%eax
  lea    (%rdx,%rax,1),%esi
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    $0x1,%ecx
  mov    %eax,%edi
  call   <Glcd_v_Line@plt>
  mov    -0x8(%rbp),%edx
  mov    -0x14(%rbp),%eax
  lea    (%rdx,%rax,1),%edi
  mov    -0xc(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    $0x1,%ecx
  mov    %eax,%esi
  call   <Glcd_v_Line@plt>
  jmp    L11
L28:
  mov    -0x30(%rbp),%ecx
  mov    -0x28(%rbp),%eax
  mov    $0x1,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <Glcd_Dot@plt>
  nop
L11:
  nop
  leave
  ret
```

assistant: ```c
void interpo(unsigned int past_i, unsigned int pres_i, unsigned int past_y, unsigned int pres_y) {
  unsigned int med, r_past_y, r_pres_y, r_past_i, r_pres_i;
  if ((pres_y - past_y) != 0 && (pres_y - past_y) != 1) {
    med = (pres_y - past_y) / 2;
    if (med < 0) {
      med = abs(med);
      r_past_i = pres_i;
      r_pres_i = past_i;
      r_past_y = pres_y;
      r_pres_y = past_y;
    } else {
      r_past_i = past_i;
      r_pres_i = pres_i;
      r_past_y = past_y;
      r_pres_y = pres_y;
    }
    Glcd_v_Line(r_past_y, r_past_y + med, r_past_i, 1);
    Glcd_v_Line(r_past_y + med, r_pres_y, r_pres_i, 1);
  } else {
    Glcd_Dot(pres_i, pres_y, 1);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<interpo>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebp
  mov    %ecx,%ebx
  cmp    %edx,%ecx
  je     L1
  mov    %edi,%r8d
  mov    %edx,%edi
  mov    %ecx,%eax
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L9
L1:
  mov    $0x1,%edx
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <Glcd_Dot@plt>
L16:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  shr    %eax
  lea    (%rax,%rdx,1),%r12d
  mov    $0x1,%ecx
  mov    %r8d,%edx
  mov    %r12d,%esi
  call   <Glcd_v_Line@plt>
  mov    $0x1,%ecx
  mov    %ebp,%edx
  mov    %ebx,%esi
  mov    %r12d,%edi
  call   <Glcd_v_Line@plt>
  jmp    L16
```

assistant: ```c
void interpo(unsigned int past_i, unsigned int pres_i, unsigned int past_y, unsigned int pres_y) {
  unsigned int med, r_past_y, r_pres_y, r_past_i, r_pres_i;
  if ((pres_y - past_y) != 0 && (pres_y - past_y) != 1) {
    med = (pres_y - past_y) / 2;
    if (med < 0) {
      med = abs(med);
      r_past_i = pres_i;
      r_pres_i = past_i;
      r_past_y = pres_y;
      r_pres_y = past_y;
    } else {
      r_past_i = past_i;
      r_pres_i = pres_i;
      r_past_y = past_y;
      r_pres_y = pres_y;
    }
    Glcd_v_Line(r_past_y, r_past_y + med, r_past_i, 1);
    Glcd_v_Line(r_past_y + med, r_pres_y, r_pres_i, 1);
  } else {
    Glcd_Dot(pres_i, pres_y, 1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<interpo>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %ecx,%ebp
  cmp    %edx,%ecx
  je     L13
  mov    %ecx,%eax
  mov    %edi,%r8d
  mov    %edx,%edi
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L0
L13:
  mov    %ebp,%esi
  mov    %r12d,%edi
  pop    %rbp
  mov    $0x1,%edx
  pop    %r12
  pop    %r13
  jmp    1060 <Glcd_Dot@plt>
  nopl   0x0(%rax)
L0:
  shr    %eax
  mov    $0x1,%ecx
  lea    (%rax,%rdx,1),%r13d
  mov    %r8d,%edx
  mov    %r13d,%esi
  call   <Glcd_v_Line@plt>
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13d,%edi
  pop    %rbp
  mov    $0x1,%ecx
  pop    %r12
  pop    %r13
  jmp    1070 <Glcd_v_Line@plt>
```

assistant: ```c
void interpo(unsigned int past_i, unsigned int pres_i, unsigned int past_y, unsigned int pres_y) {
  unsigned int med, r_past_y, r_pres_y, r_past_i, r_pres_i;
  if ((pres_y - past_y) != 0 && (pres_y - past_y) != 1) {
    med = (pres_y - past_y) / 2;
    if (med < 0) {
      med = abs(med);
      r_past_i = pres_i;
      r_pres_i = past_i;
      r_past_y = pres_y;
      r_pres_y = past_y;
    } else {
      r_past_i = past_i;
      r_pres_i = pres_i;
      r_past_y = past_y;
      r_pres_y = pres_y;
    }
    Glcd_v_Line(r_past_y, r_past_y + med, r_past_i, 1);
    Glcd_v_Line(r_past_y + med, r_pres_y, r_pres_i, 1);
  } else {
    Glcd_Dot(pres_i, pres_y, 1);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<interpo>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %ecx,%ebp
  cmp    %edx,%ecx
  je     L17
  mov    %ecx,%eax
  mov    %edi,%r8d
  mov    %edx,%edi
  sub    %edx,%eax
  cmp    $0x1,%eax
  jne    L14
L17:
  mov    %ebp,%esi
  mov    %r12d,%edi
  pop    %rbp
  mov    $0x1,%edx
  pop    %r12
  pop    %r13
  jmp    1060 <Glcd_Dot@plt>
  nopl   0x0(%rax)
L14:
  shr    %eax
  mov    $0x1,%ecx
  lea    (%rax,%rdx,1),%r13d
  mov    %r8d,%edx
  mov    %r13d,%esi
  call   <Glcd_v_Line@plt>
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %r13d,%edi
  pop    %rbp
  mov    $0x1,%ecx
  pop    %r12
  pop    %r13
  jmp    1070 <Glcd_v_Line@plt>
```

assistant: ```c
void interpo(unsigned int past_i, unsigned int pres_i, unsigned int past_y, unsigned int pres_y) {
  unsigned int med, r_past_y, r_pres_y, r_past_i, r_pres_i;
  if ((pres_y - past_y) != 0 && (pres_y - past_y) != 1) {
    med = (pres_y - past_y) / 2;
    if (med < 0) {
      med = abs(med);
      r_past_i = pres_i;
      r_pres_i = past_i;
      r_past_y = pres_y;
      r_pres_y = past_y;
    } else {
      r_past_i = past_i;
      r_pres_i = pres_i;
      r_past_y = past_y;
      r_pres_y = pres_y;
    }
    Glcd_v_Line(r_past_y, r_past_y + med, r_past_i, 1);
    Glcd_v_Line(r_past_y + med, r_pres_y, r_pres_i, 1);
  } else {
    Glcd_Dot(pres_i, pres_y, 1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<spin_unlock>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <pthread_mutex_unlock@plt>
  test   %rax,%rax
  je     L1
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L1:
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D8
data_type: string

tool: D8: pthread_mutex_unlock

assistant: ```c
void spin_unlock(pthread_mutex_t *lp) {
  if (pthread_mutex_unlock(lp) != 0) {
    perror("pthread_mutex_unlock");
    exit(-1);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<spin_unlock>:
  endbr64
  sub    $0x8,%rsp
  call   <pthread_mutex_unlock@plt>
  test   %rax,%rax
  jne    L7
L15:
  add    $0x8,%rsp
  ret
L7:
  lea    D3(%rip),%rdi
  call   <perror@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L15
```

tool_calls: name: extract_symbolic_data
arguments: label: D3
type: string

tool: D3: pthread_mutex_unlock

assistant: ```c
void spin_unlock(pthread_mutex_t *lp) {
  if (pthread_mutex_unlock(lp) != 0) {
    perror("pthread_mutex_unlock");
    exit(-1);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<spin_unlock>:
  endbr64
  sub    $0x8,%rsp
  call   <pthread_mutex_unlock@plt>
  test   %rax,%rax
  jne    L5
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  lea    D4(%rip),%rdi
  call   <perror@plt>
  mov    $0xffffffff,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
type: string

tool: D4: pthread_mutex_unlock

assistant: ```c
void spin_unlock(pthread_mutex_t *lp) {
  if (pthread_mutex_unlock(lp) != 0) {
    perror("pthread_mutex_unlock");
    exit(-1);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<spin_unlock>:
  endbr64
  sub    $0x8,%rsp
  call   <pthread_mutex_unlock@plt>
  test   %rax,%rax
  jne    L3
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  lea    D5(%rip),%rdi
  call   <perror@plt>
  mov    $0xffffffff,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: parse_data
arguments: label: D5
type_guessed: string

tool: D5: pthread_mutex_unlock

assistant: ```c
void spin_unlock(pthread_mutex_t *lp) {
  if (pthread_mutex_unlock(lp) != 0) {
    perror("pthread_mutex_unlock");
    exit(-1);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<mul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L9:
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L12
  mov    -0x18(%rbp),%eax
  xor    %eax,-0x4(%rbp)
L12:
  sarl   -0x14(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <mul2@plt>
  mov    %eax,-0x18(%rbp)
L2:
  cmpl   $0x0,-0x14(%rbp)
  jne    L9
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
uint8_t mul(uint8_t x, uint8_t y) {
  uint8_t z;
  z = 0;
  while (x != 0) {
    if (x & 1)
      z ^= y;
    x >>= 1;
    y = mul2(y);
  }
  return z;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mul>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  test   %edi,%edi
  je     L21
  mov    %esi,%edi
  mov    $0x0,%ebp
L2:
  mov    %ebp,%eax
  xor    %edi,%eax
  test   $0x1,%bl
  cmovne %eax,%ebp
  sar    %ebx
  call   <mul2@plt>
  mov    %eax,%edi
  test   %ebx,%ebx
  jne    L2
L0:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L21:
  mov    %edi,%ebp
  jmp    L0
```

assistant: ```c
uint8_t mul(uint8_t x, uint8_t y) {
  uint8_t z;
  z = 0;
  while (x != 0) {
    if (x & 1)
      z ^= y;
    x >>= 1;
    y = mul2(y);
  }
  return z;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mul>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %edi,%edi
  je     L18
  mov    %esi,%edi
  xor    %r12d,%r12d
  cs nopw 0x0(%rax,%rax,1)
L4:
  mov    %r12d,%eax
  xor    %edi,%eax
  test   $0x1,%bl
  cmovne %eax,%r12d
  sar    %ebx
  call   <mul2@plt>
  mov    %eax,%edi
  test   %ebx,%ebx
  jne    L4
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L18:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
uint8_t mul(uint8_t x, uint8_t y) {
  uint8_t z;
  z = 0;
  while (x != 0) {
    if (x & 1)
      z ^= y;
    x >>= 1;
    y = mul2(y);
  }
  return z;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mul>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %edi,%edi
  je     L0
  mov    %esi,%edi
  xor    %r12d,%r12d
  cs nopw 0x0(%rax,%rax,1)
L10:
  mov    %r12d,%eax
  xor    %edi,%eax
  test   $0x1,%bl
  cmovne %eax,%r12d
  sar    %ebx
  call   <mul2@plt>
  mov    %eax,%edi
  test   %ebx,%ebx
  jne    L10
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L0:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
uint8_t mul(uint8_t x, uint8_t y) {
  uint8_t z;
  z = 0;
  while (x != 0) {
    if (x & 1)
      z ^= y;
    x >>= 1;
    y = mul2(y);
  }
  return z;
}
```
user: Translate the below assembly code into C programming language:

```asm
<assert_true>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L8
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L8:
  mov    $0x1,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type_infered: string

tool: D0: %s


assistant: ```c
int assert_true(int given, char *msg) {
  if (!given) {
    printf("%s\n", msg);
    exit(1);
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<assert_true>:
  endbr64
  test   %edi,%edi
  je     L5
  mov    $0x1,%eax
  ret
L5:
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D8
type_infered: string

tool: D8: %s


assistant: ```c
int assert_true(int given, char *msg) {
  if (!given) {
    printf("%s\n", msg);
    exit(1);
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<assert_true>:
  endbr64
  test   %edi,%edi
  je     L8
  mov    $0x1,%eax
  ret
  xchg   %ax,%ax
L8:
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
type: string

tool: D3: %s


assistant: ```c
int assert_true(int given, char *msg) {
  if (!given) {
    printf("%s\n", msg);
    exit(1);
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<assert_true>:
  endbr64
  test   %edi,%edi
  je     L3
  mov    $0x1,%eax
  ret
  xchg   %ax,%ax
L3:
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D3
type_guessed: string

tool: D3: %s


assistant: ```c
int assert_true(int given, char *msg) {
  if (!given) {
    printf("%s\n", msg);
    exit(1);
  }
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x2,-0x18(%rbp)
  movl   $0x3,-0x14(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%ecx
  lea    -0x10(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <AsignarMem2D@plt>
  mov    -0x10(%rbp),%rax
  test   %rax,%rax
  jne    L1
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L15
L1:
  mov    -0x10(%rbp),%rax
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%ecx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <Visualizar@plt>
  mov    -0x10(%rbp),%rax
  mov    -0x18(%rbp),%edx
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <LiberarMem2D@plt>
  mov    $0x0,%eax
L15:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
datatype: string

tool: D1: Insuficiente memoria


assistant: ```c
main() {
  int filas = 2, cols = 3;
  double **m = NULL;
  AsignarMem2D(&m, filas, cols);
  if (m == NULL) {
    printf("Insuficiente memoria\n");
    return -1;
  }
  Visualizar(m, filas, cols);
  LiberarMem2D(m, filas);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  mov    %rsp,%rdi
  mov    $0x3,%edx
  mov    $0x2,%esi
  call   <AsignarMem2D@plt>
  mov    (%rsp),%rdi
  test   %rdi,%rdi
  je     L11
  mov    $0x3,%edx
  mov    $0x2,%esi
  call   <Visualizar@plt>
  mov    $0x2,%esi
  mov    (%rsp),%rdi
  call   <LiberarMem2D@plt>
  mov    $0x0,%eax
L10:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L20
  add    $0x18,%rsp
  ret
L11:
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L10
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D5
datatype: string

tool: D5: Insuficiente memoria


assistant: ```c
main() {
  int filas = 2, cols = 3;
  double **m = NULL;
  AsignarMem2D(&m, filas, cols);
  if (m == NULL) {
    printf("Insuficiente memoria\n");
    return -1;
  }
  Visualizar(m, filas, cols);
  LiberarMem2D(m, filas);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x3,%edx
  mov    $0x2,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movq   $0x0,(%rsp)
  call   <AsignarMem2D@plt>
  mov    (%rsp),%rdi
  test   %rdi,%rdi
  je     L20
  mov    $0x3,%edx
  mov    $0x2,%esi
  call   <Visualizar@plt>
  mov    (%rsp),%rdi
  mov    $0x2,%esi
  call   <LiberarMem2D@plt>
  xor    %eax,%eax
L3:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x18,%rsp
  ret
L20:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L3
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D9
type_infered: string

tool: D9: Insuficiente memoria


assistant: ```c
main() {
  int filas = 2, cols = 3;
  double **m = NULL;
  AsignarMem2D(&m, filas, cols);
  if (m == NULL) {
    printf("Insuficiente memoria\n");
    return -1;
  }
  Visualizar(m, filas, cols);
  LiberarMem2D(m, filas);
  return 0;
}
```