ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<sig_int>:
  endbr64
  sub    $0x18,%rsp
  lea    D14(%rip),%rdi
  call   <puts@plt>
  xor    %esi,%esi
L19:
  xor    %ecx,%ecx
  mov    $0xfa0,%edx
  nopl   (%rax)
L3:
  mov    0xc(%rsp),%eax
  add    %ecx,%eax
  add    %esi,%ecx
  mov    %eax,0xc(%rsp)
  sub    $0x1,%edx
  jne    L3
  add    $0x1,%esi
  cmp    $0x7530,%esi
  jne    L19
  mov    0xc(%rsp),%esi
  lea    D19(%rip),%rdi
  xor    %eax,%eax
  add    $0x18,%rsp
  jmp    1070 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: label: D14
type_guessed: string
name: retrieve_data
arguments: label: D19
type_guessed: string

tool: D14: 
sig_int starting

tool: D19: sig_int finished:%d


assistant: ```c
void sig_int(int signo) {
  int i, j;
  volatile int k;
  printf("\nsig_int starting\n");
  for (i = 0; i < 30000; i++)
    for (j = 0; j < 4000; j++)
      k += i * j;
  printf("sig_int finished:%d\n", k);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<calculator>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  mov    %edi,%eax
  mov    %al,-0x24(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsbl -0x24(%rbp),%eax
  cmp    $0x2f,%eax
  je     L79
  cmp    $0x2f,%eax
  jg     L38
  cmp    $0x2d,%eax
  je     L0
  cmp    $0x2d,%eax
  jg     L38
  cmp    $0x2a,%eax
  je     L77
  cmp    $0x2b,%eax
  jne    L38
  movsd  -0x18(%rbp),%xmm0
  addsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L99
L0:
  movsd  -0x18(%rbp),%xmm0
  subsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L99
L77:
  movsd  -0x18(%rbp),%xmm0
  mulsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L99
L79:
  movsd  -0x18(%rbp),%xmm0
  divsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L99
L38:
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  nop
L99:
  movsd  -0x8(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int calculator(double operand1, double operand2, char operator1) {
  double result1 = 0.0;
  switch (operator1) {
  case '+':
    result1 = operand1 + operand2;
    break;
  case '-':
    result1 = operand1 - operand2;
    break;
  case '*':
    result1 = operand1 * operand2;
    break;
  case '/':
    result1 = operand1 / operand2;
    break;
  default:
    result1 = 0;
    break;
  }
  return result1;
}
```
user: What is the c source code of the assembly code below:

```asm
<calculator>:
  endbr64
  cmp    $0x2d,%dil
  je     L42
  jg     L3
  cmp    $0x2a,%dil
  je     L94
  cmp    $0x2b,%dil
  jne    L81
  addsd  %xmm1,%xmm0
  jmp    L38
L81:
  pxor   %xmm0,%xmm0
  jmp    L38
L3:
  cmp    $0x2f,%dil
  jne    L8
  divsd  %xmm1,%xmm0
  jmp    L38
L8:
  pxor   %xmm0,%xmm0
L38:
  cvttsd2si %xmm0,%eax
  ret
L42:
  subsd  %xmm1,%xmm0
  jmp    L38
L94:
  mulsd  %xmm1,%xmm0
  jmp    L38
```

assistant: ```c
int calculator(double operand1, double operand2, char operator1) {
  double result1 = 0.0;
  switch (operator1) {
  case '+':
    result1 = operand1 + operand2;
    break;
  case '-':
    result1 = operand1 - operand2;
    break;
  case '*':
    result1 = operand1 * operand2;
    break;
  case '/':
    result1 = operand1 / operand2;
    break;
  default:
    result1 = 0;
    break;
  }
  return result1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<calculator>:
  endbr64
  cmp    $0x2d,%dil
  je     L48
  jg     L44
  cmp    $0x2a,%dil
  je     L41
  cmp    $0x2b,%dil
  jne    L23
  addsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
  nopl   0x0(%rax)
L44:
  cmp    $0x2f,%dil
  jne    L23
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L23:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L41:
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
  nopl   0x0(%rax)
L48:
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
```

assistant: ```c
int calculator(double operand1, double operand2, char operator1) {
  double result1 = 0.0;
  switch (operator1) {
  case '+':
    result1 = operand1 + operand2;
    break;
  case '-':
    result1 = operand1 - operand2;
    break;
  case '*':
    result1 = operand1 * operand2;
    break;
  case '/':
    result1 = operand1 / operand2;
    break;
  default:
    result1 = 0;
    break;
  }
  return result1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<calculator>:
  endbr64
  cmp    $0x2d,%dil
  je     L21
  jg     L40
  cmp    $0x2a,%dil
  je     L4
  cmp    $0x2b,%dil
  jne    L17
  addsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
  nopl   0x0(%rax)
L40:
  cmp    $0x2f,%dil
  jne    L17
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
  nopl   0x0(%rax)
L21:
  subsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  ret
```

assistant: ```c
int calculator(double operand1, double operand2, char operator1) {
  double result1 = 0.0;
  switch (operator1) {
  case '+':
    result1 = operand1 + operand2;
    break;
  case '-':
    result1 = operand1 - operand2;
    break;
  case '*':
    result1 = operand1 * operand2;
    break;
  case '/':
    result1 = operand1 / operand2;
    break;
  default:
    result1 = 0;
    break;
  }
  return result1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x350,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x344(%rbp)
  lea    -0x348(%rbp),%rdx
  lea    -0x34c(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  movl   $0x0,-0x340(%rbp)
  jmp    L82
L72:
  lea    -0x1a0(%rbp),%rdx
  mov    -0x340(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x340(%rbp)
L82:
  mov    -0x34c(%rbp),%eax
  cmp    %eax,-0x340(%rbp)
  jl     L72
  mov    -0x348(%rbp),%edx
  mov    -0x34c(%rbp),%eax
  cmp    %eax,%edx
  jge    L120
  movl   $0x0,-0x33c(%rbp)
  jmp    L95
L17:
  mov    -0x33c(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x33c(%rbp),%eax
  cltq
  mov    %edx,-0x330(%rbp,%rax,4)
  addl   $0x1,-0x33c(%rbp)
L95:
  mov    -0x348(%rbp),%eax
  cmp    %eax,-0x33c(%rbp)
  jle    L17
  mov    -0x348(%rbp),%eax
  cltq
  mov    -0x330(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L7
L120:
  movl   $0x0,-0x338(%rbp)
  jmp    L79
L58:
  mov    -0x338(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x338(%rbp),%eax
  cltq
  mov    %edx,-0x330(%rbp,%rax,4)
  addl   $0x1,-0x338(%rbp)
L79:
  mov    -0x34c(%rbp),%eax
  cmp    %eax,-0x338(%rbp)
  jl     L58
  mov    -0x34c(%rbp),%eax
  mov    %eax,-0x338(%rbp)
  jmp    L104
L20:
  movl   $0x1,-0x334(%rbp)
  jmp    L98
L117:
  mov    -0x338(%rbp),%eax
  sub    -0x334(%rbp),%eax
  cltq
  mov    -0x330(%rbp,%rax,4),%eax
  add    %eax,-0x344(%rbp)
  addl   $0x1,-0x334(%rbp)
L98:
  mov    -0x34c(%rbp),%eax
  cmp    %eax,-0x334(%rbp)
  jle    L117
  mov    -0x338(%rbp),%eax
  cltq
  mov    -0x344(%rbp),%edx
  mov    %edx,-0x330(%rbp,%rax,4)
  addl   $0x1,-0x338(%rbp)
L104:
  mov    -0x348(%rbp),%eax
  cmp    %eax,-0x338(%rbp)
  jle    L20
  mov    -0x348(%rbp),%eax
  cltq
  mov    -0x330(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L7:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L62
  call   <__stack_chk_fail@plt>
L62:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D15
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D29
type_guessed: string

tool: D15: %d %d

tool: D29: %d

assistant: ```c
int main() {
  int sum = 0, d, N, a[100], b[100];
  scanf("%d %d", &d, &N);
  for (int i = 0; i < d; i++)
    scanf("%d", &b[i]);
  if (N < d) {
    for (int i = 0; i <= N; i++) {
      a[i] = b[i];
    }
    printf("%d", a[N]);
  } else {
    int i;
    for (i = 0; i < d; i++) {
      a[i] = b[i];
    }
    for (i = d; i <= N; i++) {
      for (int j = 1; j <= d; j++)
        sum = sum + a[i - j];
      a[i] = sum;
    }
    printf("%d", a[N]);
  }
  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
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x340,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D19(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%esi
  test   %esi,%esi
  jle    L61
  lea    0x1a0(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D3(%rip),%r12
L15:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0x8(%rsp),%esi
  add    $0x4,%rbp
  cmp    %ebx,%esi
  jg     L15
  mov    0xc(%rsp),%r9d
  cmp    %r9d,%esi
  jg     L40
  test   %esi,%esi
  jle    L116
  lea    -0x1(%rsi),%ecx
  mov    $0x0,%eax
L76:
  mov    0x1a0(%rsp,%rax,4),%edx
  mov    %edx,0x10(%rsp,%rax,4)
  mov    %rax,%rdx
  add    $0x1,%rax
  cmp    %rcx,%rdx
  jne    L76
L116:
  movslq %esi,%rax
  shl    $0x2,%rax
  lea    0x10(%rsp,%rax,1),%rdi
  lea    0xc(%rsp,%rax,1),%rcx
  lea    -0x1(%rsi),%eax
  shl    $0x2,%rax
  sub    %rax,%rcx
  mov    %esi,%r8d
  mov    $0x0,%edx
  jmp    L102
L11:
  add    -0x4(%rax),%edx
  sub    $0x4,%rax
  cmp    %rcx,%rax
  jne    L11
L121:
  mov    %edx,(%rdi)
  add    $0x1,%r8d
  add    $0x4,%rdi
  add    $0x4,%rcx
  cmp    %r8d,%r9d
  jl     L64
L102:
  mov    %rdi,%rax
  test   %esi,%esi
  jg     L11
  jmp    L121
L64:
  movslq %r9d,%r9
  mov    0x10(%rsp,%r9,4),%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L54
L61:
  mov    0xc(%rsp),%r9d
  cmp    %esi,%r9d
  jge    L116
L40:
  test   %r9d,%r9d
  js     L141
  lea    0x1(%r9),%ecx
  shl    $0x2,%rcx
  mov    $0x0,%eax
L127:
  mov    0x1a0(%rsp,%rax,1),%edx
  mov    %edx,0x10(%rsp,%rax,1)
  add    $0x4,%rax
  cmp    %rax,%rcx
  jne    L127
L141:
  movslq %r9d,%r9
  mov    0x10(%rsp,%r9,4),%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
L54:
  mov    0x338(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L96
  mov    $0x0,%eax
  add    $0x340,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L96:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D19
type: string
name: extract_symbolic_data
arguments: data_tag: D3
type: string

tool: D19: %d %d

tool: D3: %d

assistant: ```c
int main() {
  int sum = 0, d, N, a[100], b[100];
  scanf("%d %d", &d, &N);
  for (int i = 0; i < d; i++)
    scanf("%d", &b[i]);
  if (N < d) {
    for (int i = 0; i <= N; i++) {
      a[i] = b[i];
    }
    printf("%d", a[N]);
  } else {
    int i;
    for (i = 0; i < d; i++) {
      a[i] = b[i];
    }
    for (i = d; i <= N; i++) {
      for (int j = 1; j <= d; j++)
        sum = sum + a[i - j];
      a[i] = sum;
    }
    printf("%d", a[N]);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  lea    D19(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x348,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r13
  lea    0x8(%rsp),%rsi
  mov    %r13,%rdx
  call   <scanf@plt>
  mov    0x8(%rsp),%ebp
  test   %ebp,%ebp
  jle    L77
  lea    0x1a0(%rsp),%r14
  xor    %ebx,%ebx
  lea    D3(%rip),%r12
  mov    %r14,%r15
  nopl   0x0(%rax)
L6:
  mov    %r15,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <scanf@plt>
  mov    0x8(%rsp),%ebp
  add    $0x4,%r15
  cmp    %ebx,%ebp
  jg     L6
  movslq 0xc(%rsp),%r15
  cmp    %r15d,%ebp
  jle    L96
  test   %r15d,%r15d
  js     L3
  lea    0x10(%rsp),%rdi
  lea    0x4(,%r15,4),%rdx
  mov    %r14,%rsi
  call   <memcpy@plt>
L3:
  mov    0x10(%rsp,%r15,4),%esi
  mov    %r12,%rdi
  call   <printf@plt>
  mov    0x338(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x348,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L96:
  lea    0x10(%rsp),%rcx
  movslq %ebp,%rbx
  test   %ebp,%ebp
  jle    L91
  mov    %rcx,%rdi
  lea    0x0(,%rbx,4),%rdx
  mov    %r14,%rsi
  call   <memcpy@plt>
  mov    %rax,%rcx
L91:
  shl    $0x2,%rbx
  lea    -0x1(%rbp),%eax
  lea    0x1(%r15),%r8d
  mov    %ebp,%edi
  lea    (%rcx,%rbx,1),%rsi
  shl    $0x2,%rax
  lea    0x0(%r13,%rbx,1),%rcx
  xor    %edx,%edx
  sub    %rax,%rcx
  nopl   0x0(%rax)
L0:
  mov    %rsi,%rax
  test   %ebp,%ebp
  jle    L57
  nop
L85:
  add    -0x4(%rax),%edx
  sub    $0x4,%rax
  cmp    %rcx,%rax
  jne    L85
L57:
  add    $0x1,%edi
  mov    %edx,(%rsi)
  add    $0x4,%rcx
  add    $0x4,%rsi
  cmp    %r8d,%edi
  jne    L0
  jmp    L3
L77:
  movslq 0xc(%rsp),%r15
  lea    D3(%rip),%r12
  cmp    %ebp,%r15d
  jl     L3
  lea    0x10(%rsp),%rcx
  movslq %ebp,%rbx
  jmp    L91
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D19
datatype: string
name: retrieve_data
arguments: data_label: D3
datatype: string

tool: D19: %d %d

tool: D3: %d

assistant: ```c
int main() {
  int sum = 0, d, N, a[100], b[100];
  scanf("%d %d", &d, &N);
  for (int i = 0; i < d; i++)
    scanf("%d", &b[i]);
  if (N < d) {
    for (int i = 0; i <= N; i++) {
      a[i] = b[i];
    }
    printf("%d", a[N]);
  } else {
    int i;
    for (i = 0; i < d; i++) {
      a[i] = b[i];
    }
    for (i = d; i <= N; i++) {
      for (int j = 1; j <= d; j++)
        sum = sum + a[i - j];
      a[i] = sum;
    }
    printf("%d", a[N]);
  }
  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
<udie>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0xf4000(%rsp),%r11
L15:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L15
  sub    $0x320,%rsp
  mov    %rdi,-0xf4318(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L7
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L7:
  mov    %fs:0x28,%rax
  mov    %rax,-0xb8(%rbp)
  xor    %eax,%eax
  mov    -0xf4318(%rbp),%rdx
  mov    -0xf430c(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xf430c(%rbp),%edx
  mov    -0xf4318(%rbp),%rcx
  lea    -0xf4300(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <vsprintf@plt>
  mov    -0xf430c(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  lea    -0xf4300(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0xf4308(%rbp)
  lea    -0xf4300(%rbp),%rdx
  mov    -0xf4308(%rbp),%rax
  add    %rdx,%rax
  movb   $0xa,(%rax)
  mov    -0xf4308(%rbp),%rax
  add    $0x1,%rax
  movb   $0x0,-0xf4300(%rbp,%rax,1)
  mov    D1(%rip),%rax
  mov    (%rax),%edx
  lea    -0xf4300(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <fputs@plt>
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <exit@plt>
  nop
  mov    -0xb8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
void udie(const char *fmt, ...) {
  char buf[(1000 * 1000)];
  va_list args;
  va_start(args, fmt);
  vsprintf(buf, fmt, args);
  va_end(args);
  size_t len = strlen(buf);
  buf[len] = '\n';
  buf[len + 1] = 0;
  fputs(buf, stderr);
  exit(EXIT_FAILURE);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<udie>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0xf4000(%rsp),%r11
L17:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L17
  sub    $0x250,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0xf4248(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rsi
  mov    $0x0,%r12d
  mov    %r12d,%edi
  call   <va_start@plt>
  mov    %rsp,%rbp
  mov    %r12d,%edx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <vsprintf@plt>
  mov    %r12d,%edi
  call   <va_end@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  movb   $0xa,(%rsp,%rax,1)
  movb   $0x0,0x1(%rsp,%rax,1)
  mov    D2(%rip),%rax
  mov    (%rax),%esi
  mov    %rbp,%rdi
  call   <fputs@plt>
  mov    D17(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  mov    0xf4248(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0xf4250,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void udie(const char *fmt, ...) {
  char buf[(1000 * 1000)];
  va_list args;
  va_start(args, fmt);
  vsprintf(buf, fmt, args);
  va_end(args);
  size_t len = strlen(buf);
  buf[len] = '\n';
  buf[len + 1] = 0;
  fputs(buf, stderr);
  exit(EXIT_FAILURE);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<udie>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  lea    -0xf4000(%rsp),%r11
L10:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L10
  sub    $0x250,%rsp
  xor    %r13d,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0xf4248(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rsi
  mov    %rsp,%r12
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <vsprintf@plt>
  mov    %r13d,%edi
  call   <va_end@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  movb   $0xa,(%rsp,%rax,1)
  movb   $0x0,0x1(%rsp,%rax,1)
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  call   <fputs@plt>
  mov    D13(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  mov    0xf4248(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0xf4250,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void udie(const char *fmt, ...) {
  char buf[(1000 * 1000)];
  va_list args;
  va_start(args, fmt);
  vsprintf(buf, fmt, args);
  va_end(args);
  size_t len = strlen(buf);
  buf[len] = '\n';
  buf[len + 1] = 0;
  fputs(buf, stderr);
  exit(EXIT_FAILURE);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<udie>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  lea    -0xf4000(%rsp),%r11
L3:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L3
  sub    $0x250,%rsp
  xor    %r13d,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0xf4248(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rsi
  mov    %rsp,%r12
  mov    %rdi,%rbp
  mov    %r13d,%edi
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <vsprintf@plt>
  mov    %r13d,%edi
  call   <va_end@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  movb   $0xa,(%rsp,%rax,1)
  movb   $0x0,0x1(%rsp,%rax,1)
  mov    D19(%rip),%rax
  mov    (%rax),%esi
  call   <fputs@plt>
  mov    D12(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  mov    0xf4248(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0xf4250,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void udie(const char *fmt, ...) {
  char buf[(1000 * 1000)];
  va_list args;
  va_start(args, fmt);
  vsprintf(buf, fmt, args);
  va_end(args);
  size_t len = strlen(buf);
  buf[len] = '\n';
  buf[len + 1] = 0;
  fputs(buf, stderr);
  exit(EXIT_FAILURE);
}
```
user: Translate the below assembly code into C programming language:

```asm
<help>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
type_guessed: string
name: retrieve_data
arguments: label: D14
type_guessed: string
name: retrieve_data
arguments: label: D6
type_guessed: string

tool: D4: usage: %s opts* infile outfile


tool: D14: 	-d Decode

tool: D6: 	-1 .. -9 Compression Level

assistant: ```c
int help(char *prg) {
  printf("usage: %s opts* infile outfile\n", prg);
  printf("\t-d Decode\n");
  printf("\t-1 .. -9 Compression Level\n");
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<help>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D5(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <puts@plt>
  lea    D20(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
data_type: string
name: disassemble_data
arguments: data_tag: D11
data_type: string
name: disassemble_data
arguments: data_tag: D20
data_type: string

tool: D5: usage: %s opts* infile outfile


tool: D11: 	-d Decode

tool: D20: 	-1 .. -9 Compression Level

assistant: ```c
int help(char *prg) {
  printf("usage: %s opts* infile outfile\n", prg);
  printf("\t-d Decode\n");
  printf("\t-1 .. -9 Compression Level\n");
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<help>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D14(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D18(%rip),%rdi
  call   <puts@plt>
  lea    D15(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D14
datatype: string
name: retrieve_data
arguments: data_tag: D18
datatype: string
name: retrieve_data
arguments: data_tag: D15
datatype: string

tool: D14: usage: %s opts* infile outfile


tool: D18: 	-d Decode

tool: D15: 	-1 .. -9 Compression Level

assistant: ```c
int help(char *prg) {
  printf("usage: %s opts* infile outfile\n", prg);
  printf("\t-d Decode\n");
  printf("\t-1 .. -9 Compression Level\n");
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<help>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D22(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D3(%rip),%rdi
  call   <puts@plt>
  lea    D12(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D22
type_guessed: string
name: read_data
arguments: tag: D3
type_guessed: string
name: read_data
arguments: tag: D12
type_guessed: string

tool: D22: usage: %s opts* infile outfile


tool: D3: 	-d Decode

tool: D12: 	-1 .. -9 Compression Level

assistant: ```c
int help(char *prg) {
  printf("usage: %s opts* infile outfile\n", prg);
  printf("\t-d Decode\n");
  printf("\t-1 .. -9 Compression Level\n");
  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
<min5>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jge    L16
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L16:
  mov    -0x28(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jge    L33
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L33:
  mov    -0x30(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jge    L36
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L36:
  mov    -0x38(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jge    L24
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x8(%rbp)
L24:
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long min5(long a, long b, long c, long d, long e) {
  long out = a;
  if (b < out) {
    out = b;
  }
  if (c < out) {
    out = c;
  }
  if (d < out) {
    out = d;
  }
  if (e < out) {
    out = e;
  }
  return out;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<min5>:
  endbr64
  cmp    %r8,%rcx
  cmovg  %r8,%rcx
  cmp    %rdx,%rcx
  cmovg  %rdx,%rcx
  cmp    %rsi,%rcx
  cmovg  %rsi,%rcx
  cmp    %rdi,%rcx
  mov    %rdi,%rax
  cmovle %rcx,%rax
  ret
```

assistant: ```c
long min5(long a, long b, long c, long d, long e) {
  long out = a;
  if (b < out) {
    out = b;
  }
  if (c < out) {
    out = c;
  }
  if (d < out) {
    out = d;
  }
  if (e < out) {
    out = e;
  }
  return out;
}
```