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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0x14(%rbp)
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jle    L58
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L58:
  movl   $0x2,-0x10(%rbp)
  jmp    L71
L65:
  movl   $0x2,-0xc(%rbp)
  jmp    L83
L29:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  je     L40
  addl   $0x1,-0xc(%rbp)
L83:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L29
  jmp    L4
L40:
  nop
L4:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L48
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L48:
  addl   $0x1,-0x14(%rbp)
L71:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L65
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L86
  call   <__stack_chk_fail@plt>
L86:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D10
data_type: string
name: unpack_data
arguments: data_tag: D44
data_type: string
name: unpack_data
arguments: data_tag: D38
data_type: string
name: unpack_data
arguments: data_tag: D13
data_type: string
name: unpack_data
arguments: data_tag: D37
data_type: string

tool: D10: Enter the number of prime numbers required

tool: D44: %d

tool: D38: First %d prime numbers are :


tool: D13: 2

tool: D37: %d


assistant: ```c
int main() {
  int class;
  int n, i = 3, count, c;
  printf("Enter the number of prime numbers required\n");
  scanf("%d", &n);
  if (n >= 1) {
    printf("First %d prime numbers are :\n", n);
    printf("2\n");
  }
  for (count = 2; count <= n;) {
    for (c = 2; c <= i - 1; c++) {
      if (i % c == 0)
        break;
    }
    if (c == i) {
      printf("%d\n", i);
      count++;
    }
    i++;
  }
  return 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   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D28(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D46(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jg     L93
L92:
  mov    $0x2,%ebp
  mov    $0x3,%ebx
  lea    D23(%rip),%r12
  cmpl   $0x1,0x4(%rsp)
  jg     L61
L11:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L93:
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <puts@plt>
  jmp    L92
L14:
  je     L37
L4:
  add    $0x1,%ebx
  cmp    %ebp,0x4(%rsp)
  jl     L11
L61:
  cmp    $0x2,%ebx
  jle    L14
  test   $0x1,%bl
  je     L4
  mov    $0x2,%ecx
L84:
  add    $0x1,%ecx
  cmp    %ebx,%ecx
  je     L37
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L84
  jmp    L4
L37:
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebp
  jmp    L4
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D28
type_guessed: string
name: retrieve_data
arguments: tag: D46
type_guessed: string
name: retrieve_data
arguments: tag: D23
type_guessed: string
name: retrieve_data
arguments: tag: D38
type_guessed: string
name: retrieve_data
arguments: tag: D13
type_guessed: string

tool: D28: Enter the number of prime numbers required

tool: D46: %d

tool: D23: %d


tool: D38: First %d prime numbers are :


tool: D13: 2

assistant: ```c
int main() {
  int class;
  int n, i = 3, count, c;
  printf("Enter the number of prime numbers required\n");
  scanf("%d", &n);
  if (n >= 1) {
    printf("First %d prime numbers are :\n", n);
    printf("2\n");
  }
  for (count = 2; count <= n;) {
    for (c = 2; c <= i - 1; c++) {
      if (i % c == 0)
        break;
    }
    if (c == i) {
      printf("%d\n", i);
      count++;
    }
    i++;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  lea    D19(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D10(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L47
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <puts@plt>
L47:
  cmpl   $0x1,0x4(%rsp)
  jle    L3
  mov    $0x2,%ebx
  mov    $0x3,%r12d
  lea    D41(%rip),%rbp
  nopl   0x0(%rax)
L35:
  mov    $0x2,%ecx
  nopl   (%rax)
L11:
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L13
  add    $0x1,%ecx
  cmp    %r12d,%ecx
  jne    L11
  mov    %r12d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
L13:
  add    $0x1,%r12d
  cmp    %ebx,0x4(%rsp)
  jge    L35
L3:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D19
type_guessed: string
name: read_data
arguments: tag: D10
type_guessed: string
name: read_data
arguments: tag: D16
type_guessed: string
name: read_data
arguments: tag: D15
type_guessed: string
name: read_data
arguments: tag: D41
type_guessed: string

tool: D19: Enter the number of prime numbers required

tool: D10: %d

tool: D16: First %d prime numbers are :


tool: D15: 2

tool: D41: %d


assistant: ```c
int main() {
  int class;
  int n, i = 3, count, c;
  printf("Enter the number of prime numbers required\n");
  scanf("%d", &n);
  if (n >= 1) {
    printf("First %d prime numbers are :\n", n);
    printf("2\n");
  }
  for (count = 2; count <= n;) {
    for (c = 2; c <= i - 1; c++) {
      if (i % c == 0)
        break;
    }
    if (c == i) {
      printf("%d\n", i);
      count++;
    }
    i++;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  lea    D23(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jle    L24
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D30(%rip),%rdi
  call   <puts@plt>
  mov    0x4(%rsp),%esi
L24:
  mov    $0x2,%ebx
  mov    $0x3,%r12d
  lea    D38(%rip),%rbp
  cmp    $0x1,%esi
  jle    L31
  cs nopw 0x0(%rax,%rax,1)
L56:
  mov    $0x2,%ecx
  nopl   (%rax)
L51:
  mov    %r12d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L3
  add    $0x1,%ecx
  cmp    %r12d,%ecx
  jne    L51
  mov    %r12d,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  mov    0x4(%rsp),%esi
L3:
  add    $0x1,%r12d
  cmp    %esi,%ebx
  jle    L56
L31:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D23
type_infered: string
name: parse_data
arguments: tag: D13
type_infered: string
name: parse_data
arguments: tag: D42
type_infered: string
name: parse_data
arguments: tag: D30
type_infered: string
name: parse_data
arguments: tag: D38
type_infered: string

tool: D23: Enter the number of prime numbers required

tool: D13: %d

tool: D42: First %d prime numbers are :


tool: D30: 2

tool: D38: %d


assistant: ```c
int main() {
  int class;
  int n, i = 3, count, c;
  printf("Enter the number of prime numbers required\n");
  scanf("%d", &n);
  if (n >= 1) {
    printf("First %d prime numbers are :\n", n);
    printf("2\n");
  }
  for (count = 2; count <= n;) {
    for (c = 2; c <= i - 1; c++) {
      if (i % c == 0)
        break;
    }
    if (c == i) {
      printf("%d\n", i);
      count++;
    }
    i++;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  D81(%rip),%xmm0
  movsd  %xmm0,-0x70(%rbp)
  movsd  D23(%rip),%xmm0
  movsd  %xmm0,-0x68(%rbp)
  movsd  D85(%rip),%xmm0
  movsd  %xmm0,-0x60(%rbp)
  movsd  D42(%rip),%xmm0
  movsd  %xmm0,-0x58(%rbp)
  movsd  D23(%rip),%xmm0
  movsd  %xmm0,-0x50(%rbp)
  movsd  D108(%rip),%xmm0
  movsd  %xmm0,-0x48(%rbp)
  movsd  D10(%rip),%xmm0
  movsd  %xmm0,-0x40(%rbp)
  movsd  D50(%rip),%xmm0
  movsd  %xmm0,-0x38(%rbp)
  movsd  D10(%rip),%xmm0
  movsd  %xmm0,-0x30(%rbp)
  movsd  D85(%rip),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  movsd  D114(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  D39(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <solve@plt>
  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: data_label: D81
type_guessed: byte[8]
name: read_data
arguments: data_label: D23
type_guessed: f64
name: read_data
arguments: data_label: D85
type_guessed: f64
name: read_data
arguments: data_label: D42
type_guessed: f64
name: read_data
arguments: data_label: D108
type_guessed: byte[8]
name: read_data
arguments: data_label: D10
type_guessed: f64
name: read_data
arguments: data_label: D50
type_guessed: byte[8]
name: read_data
arguments: data_label: D114
type_guessed: byte[8]
name: read_data
arguments: data_label: D39
type_guessed: f64

tool: {"D81": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

tool: D23: 0.1

tool: D85: 0.2

tool: D42: 7.85

tool: {"D108": [0x00,0x00,0x00,0x00,0x00,0x00,0x1c,0x40]}

tool: D10: 0.3

tool: {"D50": [0xcd,0xcc,0xcc,0xcc,0xcc,0x4c,0x33,0xc0]}

tool: {"D114": [0x00,0x00,0x00,0x00,0x00,0x00,0x24,0x40]}

tool: D39: 71.4

assistant: ```c
int main(int argc, char const *argv[]) {
  double coefficient[3][4] = {{3, 0.1, 0.2, 7.85}, {0.1, 7, 0.3, -19.3}, {0.3, 0.2, 10, 71.4}};
  solve(coefficient);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    D22(%rip),%rax
  mov    %rax,(%rsp)
  movsd  D29(%rip),%xmm1
  movsd  %xmm1,0x8(%rsp)
  movsd  D9(%rip),%xmm0
  movsd  %xmm0,0x10(%rsp)
  mov    D44(%rip),%rax
  mov    %rax,0x18(%rsp)
  movsd  %xmm1,0x20(%rsp)
  mov    D50(%rip),%rax
  mov    %rax,0x28(%rsp)
  movsd  D64(%rip),%xmm1
  movsd  %xmm1,0x30(%rsp)
  mov    D38(%rip),%rax
  mov    %rax,0x38(%rsp)
  movsd  %xmm1,0x40(%rsp)
  movsd  %xmm0,0x48(%rsp)
  mov    D30(%rip),%rax
  mov    %rax,0x50(%rsp)
  mov    D70(%rip),%rax
  mov    %rax,0x58(%rsp)
  mov    %rsp,%rdi
  call   <solve@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x78,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D22
type_guessed: byte[8]
name: read_data
arguments: label: D29
type_guessed: f64
name: read_data
arguments: label: D9
type_guessed: f64
name: read_data
arguments: label: D44
type_guessed: f64
name: read_data
arguments: label: D50
type_guessed: byte[8]
name: read_data
arguments: label: D64
type_guessed: f64
name: read_data
arguments: label: D38
type_guessed: byte[8]
name: read_data
arguments: label: D30
type_guessed: byte[8]
name: read_data
arguments: label: D70
type_guessed: f64

tool: {"D22": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

tool: D29: 0.1

tool: D9: 0.2

tool: D44: 7.85

tool: {"D50": [0x00,0x00,0x00,0x00,0x00,0x00,0x1c,0x40]}

tool: D64: 0.3

tool: {"D38": [0xcd,0xcc,0xcc,0xcc,0xcc,0x4c,0x33,0xc0]}

tool: {"D30": [0x00,0x00,0x00,0x00,0x00,0x00,0x24,0x40]}

tool: D70: 71.4

assistant: ```c
int main(int argc, char const *argv[]) {
  double coefficient[3][4] = {{3, 0.1, 0.2, 7.85}, {0.1, 7, 0.3, -19.3}, {0.3, 0.2, 10, 71.4}};
  solve(coefficient);
  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    $0x78,%rsp
  movsd  D46(%rip),%xmm1
  movsd  D50(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    D45(%rip),%rax
  mov    %rsp,%rdi
  movsd  %xmm1,0x8(%rsp)
  mov    %rax,(%rsp)
  mov    D86(%rip),%rax
  movsd  %xmm1,0x20(%rsp)
  movsd  D52(%rip),%xmm1
  mov    %rax,0x18(%rsp)
  mov    D65(%rip),%rax
  movsd  %xmm0,0x10(%rsp)
  mov    %rax,0x28(%rsp)
  mov    D9(%rip),%rax
  movsd  %xmm1,0x30(%rsp)
  mov    %rax,0x38(%rsp)
  mov    D41(%rip),%rax
  movsd  %xmm1,0x40(%rsp)
  mov    %rax,0x50(%rsp)
  mov    D60(%rip),%rax
  movsd  %xmm0,0x48(%rsp)
  mov    %rax,0x58(%rsp)
  call   <solve@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x78,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D46
type: f64
name: unpack_data
arguments: tag: D50
type: f64
name: unpack_data
arguments: tag: D45
type: byte[8]
name: unpack_data
arguments: tag: D86
type: f64
name: unpack_data
arguments: tag: D52
type: f64
name: unpack_data
arguments: tag: D65
type: byte[8]
name: unpack_data
arguments: tag: D9
type: byte[8]
name: unpack_data
arguments: tag: D41
type: byte[8]
name: unpack_data
arguments: tag: D60
type: f64

tool: D46: 0.1

tool: D50: 0.2

tool: {"D45": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40]}

tool: D86: 7.85

tool: D52: 0.3

tool: {"D65": [0x00,0x00,0x00,0x00,0x00,0x00,0x1c,0x40]}

tool: {"D9": [0xcd,0xcc,0xcc,0xcc,0xcc,0x4c,0x33,0xc0]}

tool: {"D41": [0x00,0x00,0x00,0x00,0x00,0x00,0x24,0x40]}

tool: D60: 71.4

assistant: ```c
int main(int argc, char const *argv[]) {
  double coefficient[3][4] = {{3, 0.1, 0.2, 7.85}, {0.1, 7, 0.3, -19.3}, {0.3, 0.2, 10, 71.4}};
  solve(coefficient);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  movapd D37(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movaps %xmm0,(%rsp)
  movapd D35(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  movapd D0(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movapd D32(%rip),%xmm0
  movaps %xmm0,0x30(%rsp)
  movapd D29(%rip),%xmm0
  movaps %xmm0,0x40(%rsp)
  movapd D53(%rip),%xmm0
  movaps %xmm0,0x50(%rsp)
  call   <solve@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x78,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D37
type: byte[16]
name: read_data
arguments: data_tag: D35
type: f64[2]
name: read_data
arguments: data_tag: D0
type: f64[2]
name: read_data
arguments: data_tag: D32
type: f64[2]
name: read_data
arguments: data_tag: D29
type: f64[2]
name: read_data
arguments: data_tag: D53
type: byte[16]

tool: {"D37": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40,0x9a,0x99,0x99,0x99,0x99,0x99,0xb9,0x3f]}

tool: {"D35": [0.2,7.85]}

tool: {"D0": [0.1,7.0]}

tool: {"D32": [0.3,-19.3]}

tool: {"D29": [0.3,0.2]}

tool: {"D53": [0x00,0x00,0x00,0x00,0x00,0x00,0x24,0x40,0x9a,0x99,0x99,0x99,0x99,0xd9,0x51,0x40]}

assistant: ```c
int main(int argc, char const *argv[]) {
  double coefficient[3][4] = {{3, 0.1, 0.2, 7.85}, {0.1, 7, 0.3, -19.3}, {0.3, 0.2, 10, 71.4}};
  solve(coefficient);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<escape_iteration_count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %rsi,-0x60(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x100,-0x28(%rbp)
  jmp    L15
L9:
  mov    -0x50(%rbp),%rax
  imul   %rax,%rax
  sar    $0x10,%rax
  mov    %rax,-0x40(%rbp)
  mov    -0x48(%rbp),%rax
  imul   %rax,%rax
  sar    $0x10,%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x40(%rbp),%rax
  sub    -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x50(%rbp),%rax
  imul   -0x48(%rbp),%rax
  sar    $0xf,%rax
  mov    %rax,%rdx
  mov    -0x60(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  addq   $0x1,-0x30(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x50(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x48(%rbp)
L15:
  mov    -0x40(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x4,%edi
  call   <int_to_fix@plt>
  cmp    %rax,%rbx
  jge    L37
  mov    -0x30(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L9
L37:
  mov    -0x30(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  je     L33
  mov    -0x30(%rbp),%rax
  jmp    L36
L33:
  mov    $0xffffffffffffffff,%rax
L36:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
long escape_iteration_count(long cx, long cy) {
  long zx = 0, zy = 0, zx2 = 0, zy2 = 0;
  unsigned long i = 0;
  unsigned long const iteration_limit = 256;
  while (((zx2 + zy2) < int_to_fix(4)) && (i < iteration_limit)) {
    long tx, ty;
    zx2 = (zx * zx) >> 16;
    zy2 = (zy * zy) >> 16;
    tx = cx + (zx2 - zy2);
    ty = cy + ((zx * zy) >> 15);
    i++;
    zx = tx;
    zy = ty;
  }
  return (i == iteration_limit) ? -1 : i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<escape_iteration_count>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r13
  mov    $0x0,%r12d
  mov    $0x0,%edx
  mov    $0x0,%eax
  mov    $0x0,%ebp
  mov    $0x0,%ebx
L10:
  lea    (%rax,%rdx,1),%r15
  mov    $0x4,%edi
  call   <int_to_fix@plt>
  cmp    $0xff,%r12
  ja     L26
  cmp    %rax,%r15
  jge    L26
  mov    %rbx,%rax
  imul   %rbx,%rax
  sar    $0x10,%rax
  mov    %rbp,%rdx
  imul   %rbp,%rdx
  sar    $0x10,%rdx
  mov    %rax,%rcx
  sub    %rdx,%rcx
  imul   %rbp,%rbx
  sar    $0xf,%rbx
  lea    (%rbx,%r13,1),%rbp
  add    $0x1,%r12
  lea    (%rcx,%r14,1),%rbx
  jmp    L10
L26:
  cmp    $0x100,%r12
  mov    $0xffffffffffffffff,%rax
  cmovne %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
long escape_iteration_count(long cx, long cy) {
  long zx = 0, zy = 0, zx2 = 0, zy2 = 0;
  unsigned long i = 0;
  unsigned long const iteration_limit = 256;
  while (((zx2 + zy2) < int_to_fix(4)) && (i < iteration_limit)) {
    long tx, ty;
    zx2 = (zx * zx) >> 16;
    zy2 = (zy * zy) >> 16;
    tx = cx + (zx2 - zy2);
    ty = cy + ((zx * zy) >> 15);
    i++;
    zx = tx;
    zy = ty;
  }
  return (i == iteration_limit) ? -1 : i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<escape_iteration_count>:
  endbr64
  push   %r15
  xor    %edx,%edx
  xor    %r15d,%r15d
  xor    %eax,%eax
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L24
  cs nopw 0x0(%rax,%rax,1)
L3:
  cmp    $0x100,%r15
  je     L5
  mov    %rbx,%rax
  mov    %r14,%rdx
  add    $0x1,%r15
  imul   %rbx,%rax
  imul   %r14,%rdx
  imul   %r14,%rbx
  sar    $0x10,%rax
  sar    $0x10,%rdx
  mov    %rax,%rcx
  sar    $0xf,%rbx
  sub    %rdx,%rcx
  lea    (%rbx,%rbp,1),%r14
  lea    (%rcx,%r12,1),%rbx
L24:
  mov    $0x4,%edi
  lea    (%rax,%rdx,1),%r13
  call   <int_to_fix@plt>
  cmp    %rax,%r13
  jl     L3
  cmp    $0x100,%r15
  mov    $0xffffffffffffffff,%rax
  cmovne %r15,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  add    $0x8,%rsp
  mov    $0xffffffffffffffff,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
long escape_iteration_count(long cx, long cy) {
  long zx = 0, zy = 0, zx2 = 0, zy2 = 0;
  unsigned long i = 0;
  unsigned long const iteration_limit = 256;
  while (((zx2 + zy2) < int_to_fix(4)) && (i < iteration_limit)) {
    long tx, ty;
    zx2 = (zx * zx) >> 16;
    zy2 = (zy * zy) >> 16;
    tx = cx + (zx2 - zy2);
    ty = cy + ((zx * zy) >> 15);
    i++;
    zx = tx;
    zy = ty;
  }
  return (i == iteration_limit) ? -1 : i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<escape_iteration_count>:
  endbr64
  push   %r15
  xor    %edx,%edx
  xor    %r15d,%r15d
  xor    %eax,%eax
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L0:
  cmp    $0x100,%r15
  je     L7
  mov    %rbx,%rax
  mov    %r14,%rdx
  add    $0x1,%r15
  imul   %rbx,%rax
  imul   %r14,%rdx
  imul   %r14,%rbx
  sar    $0x10,%rax
  sar    $0x10,%rdx
  mov    %rax,%rcx
  sar    $0xf,%rbx
  sub    %rdx,%rcx
  lea    (%rbx,%rbp,1),%r14
  lea    (%rcx,%r12,1),%rbx
L16:
  mov    $0x4,%edi
  lea    (%rax,%rdx,1),%r13
  call   <int_to_fix@plt>
  cmp    %rax,%r13
  jl     L0
  cmp    $0x100,%r15
  mov    $0xffffffffffffffff,%rax
  cmovne %r15,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  add    $0x8,%rsp
  mov    $0xffffffffffffffff,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
long escape_iteration_count(long cx, long cy) {
  long zx = 0, zy = 0, zx2 = 0, zy2 = 0;
  unsigned long i = 0;
  unsigned long const iteration_limit = 256;
  while (((zx2 + zy2) < int_to_fix(4)) && (i < iteration_limit)) {
    long tx, ty;
    zx2 = (zx * zx) >> 16;
    zy2 = (zy * zy) >> 16;
    tx = cx + (zx2 - zy2);
    ty = cy + ((zx * zy) >> 15);
    i++;
    zx = tx;
    zy = ty;
  }
  return (i == iteration_limit) ? -1 : i;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<set_fpu_except_flags>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <feclearexcept@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <feraiseexcept@plt>
  nop
  leave
  ret
```

assistant: ```c
void set_fpu_except_flags(int set, int clear) {
  int exc_set = 0, exc_clr = 0;
  feclearexcept(exc_clr);
  feraiseexcept(exc_set);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<set_fpu_except_flags>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <feclearexcept@plt>
  mov    $0x0,%edi
  call   <feraiseexcept@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void set_fpu_except_flags(int set, int clear) {
  int exc_set = 0, exc_clr = 0;
  feclearexcept(exc_clr);
  feraiseexcept(exc_set);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<set_fpu_except_flags>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <feclearexcept@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1060 <feraiseexcept@plt>
```

assistant: ```c
void set_fpu_except_flags(int set, int clear) {
  int exc_set = 0, exc_clr = 0;
  feclearexcept(exc_clr);
  feraiseexcept(exc_set);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<set_fpu_except_flags>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <feclearexcept@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1060 <feraiseexcept@plt>
```

assistant: ```c
void set_fpu_except_flags(int set, int clear) {
  int exc_set = 0, exc_clr = 0;
  feclearexcept(exc_clr);
  feraiseexcept(exc_set);
}
```