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

```asm
<isNum>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <integerAtmt@plt>
  test   %rax,%rax
  jne    L3
L19:
  xor    %eax,%eax
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L3:
  lea    D15(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L19
  mov    %rbp,%rdi
  lea    D17(%rip),%rsi
  call   <strcmp@plt>
  pop    %rbp
  neg    %rax
  sbb    %eax,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D15
type_guessed: string
name: disassemble_data
arguments: data_label: D17
type_guessed: string

tool: D15: Vrai

tool: D17: Faux

assistant: ```c
int isNum(char *str) {
  if ((integerAtmt(str) == 0) || (strcmp(str, "Vrai") == 0) || (strcmp(str, "Faux") == 0)) {
    return 0;
  }
  return -1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<isNum>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  call   <integerAtmt@plt>
  test   %rax,%rax
  jne    L18
L8:
  xor    %eax,%eax
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L18:
  lea    D6(%rip),%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L8
  mov    %rbp,%rdi
  lea    D1(%rip),%rsi
  call   <strcmp@plt>
  pop    %rbp
  neg    %rax
  sbb    %eax,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D6
type_guessed: string
name: disassemble_data
arguments: data_label: D1
type_guessed: string

tool: D6: Vrai

tool: D1: Faux

assistant: ```c
int isNum(char *str) {
  if ((integerAtmt(str) == 0) || (strcmp(str, "Vrai") == 0) || (strcmp(str, "Faux") == 0)) {
    return 0;
  }
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<sort1_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
  jmp    L24
L7:
  movl   $0x0,-0x14(%rbp)
  jmp    L33
L2:
  mov    -0x14(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L21
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x14(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x14(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
L21:
  addl   $0x1,-0x14(%rbp)
L33:
  mov    -0x2c(%rbp),%eax
  sub    -0x18(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x14(%rbp)
  jl     L2
  addl   $0x1,-0x18(%rbp)
L24:
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x18(%rbp)
  jl     L7
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

assistant: ```c
void sort1_1(char (*str)[8], int n) {
  char ct[8];
  int i, j;
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < n - i - 1; j++) {
      if (strcmp(str[j], str[j + 1]) > 0) {
        strcpy(ct, str[j]);
        strcpy(str[j], str[j + 1]);
        strcpy(str[j + 1], ct);
      }
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sort1_1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    -0x1(%rsi),%r14d
  test   %r14d,%r14d
  jle    L9
  lea    0x8(%rdi),%r15
  lea    0x10(%rsp),%r13
  jmp    L2
L45:
  cmp    %r12,%rbx
  je     L86
L81:
  mov    %rbx,%rbp
  add    $0x8,%rbx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L45
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  jmp    L45
L86:
  sub    $0x1,%r14d
  je     L9
L2:
  test   %r14d,%r14d
  jle    L86
  mov    0x8(%rsp),%rbx
  lea    -0x1(%r14),%eax
  lea    (%r15,%rax,8),%r12
  jmp    L81
L9:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L55:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sort1_1(char (*str)[8], int n) {
  char ct[8];
  int i, j;
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < n - i - 1; j++) {
      if (strcmp(str[j], str[j + 1]) > 0) {
        strcpy(ct, str[j]);
        strcpy(str[j], str[j + 1]);
        strcpy(str[j + 1], ct);
      }
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sort1_1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    -0x1(%rsi),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %r13d,%r13d
  jle    L29
  lea    0x8(%rdi),%r15
  lea    0x10(%rsp),%r12
  nopl   0x0(%rax)
L12:
  lea    -0x1(%r13),%eax
  mov    0x8(%rsp),%rbx
  mov    %rax,%r13
  lea    (%r15,%rax,8),%rbp
L4:
  mov    %rbx,%r14
  add    $0x8,%rbx
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L42
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
L42:
  cmp    %rbp,%rbx
  jne    L4
  test   %r13d,%r13d
  jne    L12
L29:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sort1_1(char (*str)[8], int n) {
  char ct[8];
  int i, j;
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < n - i - 1; j++) {
      if (strcmp(str[j], str[j + 1]) > 0) {
        strcpy(ct, str[j]);
        strcpy(str[j], str[j + 1]);
        strcpy(str[j + 1], ct);
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<sort1_1>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    -0x1(%rsi),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %r13d,%r13d
  jle    L3
  lea    0x8(%rdi),%r15
  lea    0x10(%rsp),%r12
  nopl   0x0(%rax)
L33:
  lea    -0x1(%r13),%eax
  mov    0x8(%rsp),%rbx
  mov    %rax,%r13
  lea    (%r15,%rax,8),%rbp
L30:
  mov    %rbx,%r14
  add    $0x8,%rbx
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jle    L34
  mov    %r14,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <strcpy@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
L34:
  cmp    %rbp,%rbx
  jne    L30
  test   %r13d,%r13d
  jne    L33
L3:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L40:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sort1_1(char (*str)[8], int n) {
  char ct[8];
  int i, j;
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < n - i - 1; j++) {
      if (strcmp(str[j], str[j + 1]) > 0) {
        strcpy(ct, str[j]);
        strcpy(str[j], str[j + 1]);
        strcpy(str[j + 1], ct);
      }
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<insertion_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L20
L19:
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  cltq
  imul   -0x38(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L32
L9:
  mov    -0x10(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    -0x40(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  test   %eax,%eax
  js     L10
  mov    -0x38(%rbp),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memswp@plt>
  mov    -0x38(%rbp),%rax
  neg    %rax
  add    %rax,-0x10(%rbp)
  subl   $0x1,-0x14(%rbp)
L32:
  cmpl   $0x0,-0x14(%rbp)
  jns    L9
  jmp    L45
L10:
  nop
L45:
  addl   $0x1,-0x18(%rbp)
L20:
  mov    -0x18(%rbp),%eax
  cltq
  cmp    %rax,-0x30(%rbp)
  ja     L19
  nop
  nop
  leave
  ret
```

assistant: ```c
void insertion_sort(void *base, size_t nel, size_t width, int (*compar)(const void *, const void *)) {
  char *p1;
  char *p2;
  int i;
  for (i = 1; i < nel; i++) {
    int j;
    p1 = (char *)base + width * (i - 1);
    for (j = i - 1; j >= 0; j--) {
      p2 = p1 + width;
      if (compar(p1, p2) < 0)
        break;
      memswp(p1, p2, width);
      p1 = p1 - width;
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<insertion_sort>:
  endbr64
  cmp    $0x1,%rsi
  jbe    L21
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdx,%rbp
  mov    %rcx,%r14
  mov    %rdi,(%rsp)
  lea    -0x1(%rsi),%rax
  mov    %rax,0x8(%rsp)
  mov    $0x0,%r15d
  jmp    L38
L9:
  add    $0x1,%r15
  add    %rbp,(%rsp)
  cmp    0x8(%rsp),%r15
  je     L19
L38:
  mov    %r15d,%r12d
  mov    (%rsp),%rbx
  test   %r15d,%r15d
  js     L9
L28:
  lea    (%rbx,%rbp,1),%r13
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   *%r14
  test   %eax,%eax
  js     L9
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <memswp@plt>
  sub    %rbp,%rbx
  sub    $0x1,%r12d
  cmp    $0xffffffff,%r12d
  jne    L28
  jmp    L9
L19:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  ret
```

assistant: ```c
void insertion_sort(void *base, size_t nel, size_t width, int (*compar)(const void *, const void *)) {
  char *p1;
  char *p2;
  int i;
  for (i = 1; i < nel; i++) {
    int j;
    p1 = (char *)base + width * (i - 1);
    for (j = i - 1; j >= 0; j--) {
      p2 = p1 + width;
      if (compar(p1, p2) < 0)
        break;
      memswp(p1, p2, width);
      p1 = p1 - width;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<insertion_sort>:
  endbr64
  cmp    $0x1,%rsi
  jbe    L2
  push   %r15
  lea    -0x1(%rsi),%rax
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  mov    %rax,0x8(%rsp)
  xchg   %ax,%ax
L27:
  mov    (%rsp),%r15
  mov    %r13d,%ebp
  nopw   0x0(%rax,%rax,1)
L13:
  lea    (%r15,%rbx,1),%r14
  mov    %r15,%rdi
  mov    %r14,%rsi
  call   *%r12
  test   %eax,%eax
  js     L32
  mov    %r15,%rdi
  mov    %rbx,%rdx
  mov    %r14,%rsi
  sub    %rbx,%r15
  call   <memswp@plt>
  sub    $0x1,%ebp
  jae    L13
L32:
  add    %rbx,(%rsp)
  add    $0x1,%r13
  cmp    0x8(%rsp),%r13
  jne    L27
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  ret
```

assistant: ```c
void insertion_sort(void *base, size_t nel, size_t width, int (*compar)(const void *, const void *)) {
  char *p1;
  char *p2;
  int i;
  for (i = 1; i < nel; i++) {
    int j;
    p1 = (char *)base + width * (i - 1);
    for (j = i - 1; j >= 0; j--) {
      p2 = p1 + width;
      if (compar(p1, p2) < 0)
        break;
      memswp(p1, p2, width);
      p1 = p1 - width;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<insertion_sort>:
  endbr64
  cmp    $0x1,%rsi
  jbe    L10
  push   %r15
  lea    -0x1(%rsi),%rax
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rcx,%r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  mov    %rax,0x8(%rsp)
  xchg   %ax,%ax
L36:
  mov    (%rsp),%r15
  mov    %r13d,%ebp
  nopw   0x0(%rax,%rax,1)
L38:
  lea    (%r15,%rbx,1),%r14
  mov    %r15,%rdi
  mov    %r14,%rsi
  call   *%r12
  test   %eax,%eax
  js     L6
  mov    %r15,%rdi
  mov    %rbx,%rdx
  mov    %r14,%rsi
  sub    %rbx,%r15
  call   <memswp@plt>
  sub    $0x1,%ebp
  jae    L38
L6:
  add    %rbx,(%rsp)
  add    $0x1,%r13
  cmp    0x8(%rsp),%r13
  jne    L36
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L10:
  ret
```

assistant: ```c
void insertion_sort(void *base, size_t nel, size_t width, int (*compar)(const void *, const void *)) {
  char *p1;
  char *p2;
  int i;
  for (i = 1; i < nel; i++) {
    int j;
    p1 = (char *)base + width * (i - 1);
    for (j = i - 1; j >= 0; j--) {
      p2 = p1 + width;
      if (compar(p1, p2) < 0)
        break;
      memswp(p1, p2, width);
      p1 = p1 - width;
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<node_point>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <Tuple2_0@plt>
  mov    %rax,%rdx
  mov    -0x50(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <distant2_0@plt>
  mov    %eax,-0x3c(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,-0x38(%rbp)
  mov    -0x48(%rbp),%edx
  mov    -0x44(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <Tuple2_0@plt>
  mov    %rax,%rbx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <Tuple2_0@plt>
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <distant2_0@plt>
  mov    %eax,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x30(%rbp)
  mov    -0x38(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  setl   %al
  movzbl %al,%eax
  mov    %eax,-0x2c(%rbp)
  cmpl   $0x1,-0x2c(%rbp)
  jne    L26
  mov    -0x48(%rbp),%edx
  mov    -0x44(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <Tuple2_0@plt>
  mov    %rax,-0x28(%rbp)
  jmp    L6
L26:
  cmpl   $0x0,-0x2c(%rbp)
  jne    L6
  mov    -0x50(%rbp),%rax
  mov    %rax,-0x28(%rbp)
L6:
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x58(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    $0x1,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int node_point(int inX, int inY, struct Tuple2_Int_Int *point_at_last, struct Tuple2_Int_Int **output) {
  int _tmp001;
  struct Tuple2_Int_Int *_tmp000;
  _tmp001 = distant2_0(Tuple2_0(0, 0), point_at_last);
  if (1) {
    int _tmp003;
    int pvar0_d1 = _tmp001;
    struct Tuple2_Int_Int *_tmp002;
    _tmp003 = distant2_0(Tuple2_0(0, 0), Tuple2_0(inX, inY));
    if (1) {
      int _tmp005;
      int pvar1_d2 = _tmp003;
      struct Tuple2_Int_Int *_tmp004;
      _tmp005 = (pvar0_d1 < pvar1_d2);
      if (_tmp005 == 1) {
        _tmp004 = Tuple2_0(inX, inY);
      } else if (_tmp005 == 0) {
        _tmp004 = point_at_last;
      }
      _tmp002 = _tmp004;
    }
    _tmp000 = _tmp002;
  }
  *output = _tmp000;
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<node_point>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%r12d
  mov    %rdx,%rbx
  mov    %rcx,%r13
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <Tuple2_0@plt>
  mov    %rax,%rdi
  mov    %rbx,%rsi
  call   <distant2_0@plt>
  mov    %eax,%r14d
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <Tuple2_0@plt>
  mov    %rax,%r15
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <Tuple2_0@plt>
  mov    %rax,%rdi
  mov    %r15,%rsi
  call   <distant2_0@plt>
  cmp    %eax,%r14d
  jl     L19
L0:
  mov    %rbx,0x0(%r13)
  mov    $0x1,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L19:
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <Tuple2_0@plt>
  mov    %rax,%rbx
  jmp    L0
```

assistant: ```c
int node_point(int inX, int inY, struct Tuple2_Int_Int *point_at_last, struct Tuple2_Int_Int **output) {
  int _tmp001;
  struct Tuple2_Int_Int *_tmp000;
  _tmp001 = distant2_0(Tuple2_0(0, 0), point_at_last);
  if (1) {
    int _tmp003;
    int pvar0_d1 = _tmp001;
    struct Tuple2_Int_Int *_tmp002;
    _tmp003 = distant2_0(Tuple2_0(0, 0), Tuple2_0(inX, inY));
    if (1) {
      int _tmp005;
      int pvar1_d2 = _tmp003;
      struct Tuple2_Int_Int *_tmp004;
      _tmp005 = (pvar0_d1 < pvar1_d2);
      if (_tmp005 == 1) {
        _tmp004 = Tuple2_0(inX, inY);
      } else if (_tmp005 == 0) {
        _tmp004 = point_at_last;
      }
      _tmp002 = _tmp004;
    }
    _tmp000 = _tmp002;
  }
  *output = _tmp000;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<node_point>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %esi,%r12d
  xor    %esi,%esi
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <Tuple2_0@plt>
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <distant2_0@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  mov    %eax,%r14d
  call   <Tuple2_0@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  mov    %rax,%r15
  call   <Tuple2_0@plt>
  mov    %r15,%rsi
  mov    %rax,%rdi
  call   <distant2_0@plt>
  cmp    %eax,%r14d
  jl     L1
L2:
  mov    %rbx,0x0(%r13)
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <Tuple2_0@plt>
  mov    %rax,%rbx
  jmp    L2
```

assistant: ```c
int node_point(int inX, int inY, struct Tuple2_Int_Int *point_at_last, struct Tuple2_Int_Int **output) {
  int _tmp001;
  struct Tuple2_Int_Int *_tmp000;
  _tmp001 = distant2_0(Tuple2_0(0, 0), point_at_last);
  if (1) {
    int _tmp003;
    int pvar0_d1 = _tmp001;
    struct Tuple2_Int_Int *_tmp002;
    _tmp003 = distant2_0(Tuple2_0(0, 0), Tuple2_0(inX, inY));
    if (1) {
      int _tmp005;
      int pvar1_d2 = _tmp003;
      struct Tuple2_Int_Int *_tmp004;
      _tmp005 = (pvar0_d1 < pvar1_d2);
      if (_tmp005 == 1) {
        _tmp004 = Tuple2_0(inX, inY);
      } else if (_tmp005 == 0) {
        _tmp004 = point_at_last;
      }
      _tmp002 = _tmp004;
    }
    _tmp000 = _tmp002;
  }
  *output = _tmp000;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<node_point>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %esi,%r12d
  xor    %esi,%esi
  push   %rbp
  mov    %edi,%ebp
  xor    %edi,%edi
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x8,%rsp
  call   <Tuple2_0@plt>
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <distant2_0@plt>
  mov    %r12d,%esi
  mov    %ebp,%edi
  mov    %eax,%r14d
  call   <Tuple2_0@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  mov    %rax,%r15
  call   <Tuple2_0@plt>
  mov    %r15,%rsi
  mov    %rax,%rdi
  call   <distant2_0@plt>
  cmp    %eax,%r14d
  jl     L1
L0:
  mov    %rbx,0x0(%r13)
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <Tuple2_0@plt>
  mov    %rax,%rbx
  jmp    L0
```

assistant: ```c
int node_point(int inX, int inY, struct Tuple2_Int_Int *point_at_last, struct Tuple2_Int_Int **output) {
  int _tmp001;
  struct Tuple2_Int_Int *_tmp000;
  _tmp001 = distant2_0(Tuple2_0(0, 0), point_at_last);
  if (1) {
    int _tmp003;
    int pvar0_d1 = _tmp001;
    struct Tuple2_Int_Int *_tmp002;
    _tmp003 = distant2_0(Tuple2_0(0, 0), Tuple2_0(inX, inY));
    if (1) {
      int _tmp005;
      int pvar1_d2 = _tmp003;
      struct Tuple2_Int_Int *_tmp004;
      _tmp005 = (pvar0_d1 < pvar1_d2);
      if (_tmp005 == 1) {
        _tmp004 = Tuple2_0(inX, inY);
      } else if (_tmp005 == 0) {
        _tmp004 = point_at_last;
      }
      _tmp002 = _tmp004;
    }
    _tmp000 = _tmp002;
  }
  *output = _tmp000;
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readsnumf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L20
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <skipspace@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <readnumf@plt>
  neg    %eax
  jmp    L0
L20:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <readnumf@plt>
L0:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L13
  call   <__stack_chk_fail@plt>
L13:
  leave
  ret
```

assistant: ```c
int readsnumf(s)
char *s;
{
  char *p;
  p = s;
  if (*p == '-') {
    p = p + 1;
    skipspace(&p);
    return (-readnumf(p));
  } else {
    return (readnumf(p));
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<readsnumf>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rdi,(%rsp)
  cmpb   $0x2d,(%rdi)
  je     L9
  mov    $0x0,%eax
  call   <readnumf@plt>
L0:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L28
  add    $0x18,%rsp
  ret
L9:
  add    $0x1,%rdi
  mov    %rdi,(%rsp)
  mov    %rsp,%rdi
  call   <skipspace@plt>
  mov    (%rsp),%rdi
  mov    $0x0,%eax
  call   <readnumf@plt>
  neg    %eax
  jmp    L0
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readsnumf(s)
char *s;
{
  char *p;
  p = s;
  if (*p == '-') {
    p = p + 1;
    skipspace(&p);
    return (-readnumf(p));
  } else {
    return (readnumf(p));
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readsnumf>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmpb   $0x2d,(%rdi)
  mov    %rdi,(%rsp)
  je     L16
  xor    %eax,%eax
  call   <readnumf@plt>
L19:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L24
  add    $0x18,%rsp
  ret
  nopl   (%rax)
L16:
  add    $0x1,%rdi
  mov    %rdi,(%rsp)
  mov    %rsp,%rdi
  call   <skipspace@plt>
  mov    (%rsp),%rdi
  xor    %eax,%eax
  call   <readnumf@plt>
  neg    %eax
  jmp    L19
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readsnumf(s)
char *s;
{
  char *p;
  p = s;
  if (*p == '-') {
    p = p + 1;
    skipspace(&p);
    return (-readnumf(p));
  } else {
    return (readnumf(p));
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<readsnumf>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  cmpb   $0x2d,(%rdi)
  mov    %rdi,(%rsp)
  je     L12
  xor    %eax,%eax
  call   <readnumf@plt>
L26:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x18,%rsp
  ret
  nopl   (%rax)
L12:
  add    $0x1,%rdi
  mov    %rdi,(%rsp)
  mov    %rsp,%rdi
  call   <skipspace@plt>
  mov    (%rsp),%rdi
  xor    %eax,%eax
  call   <readnumf@plt>
  neg    %eax
  jmp    L26
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int readsnumf(s)
char *s;
{
  char *p;
  p = s;
  if (*p == '-') {
    p = p + 1;
    skipspace(&p);
    return (-readnumf(p));
  } else {
    return (readnumf(p));
  }
}
```