ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D7(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L19
  lea    0x4(%rsp),%r12
  lea    D4(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L17:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <getCount@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jne    L17
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D7
type: string
name: parse_data
arguments: data_label: D4
type: string

tool: D7: %d

tool: D4: %d


assistant: ```c
int main() {
  int t, n;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    printf("%d\n", getCount(n));
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D11(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <scanf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  je     L19
  lea    0x4(%rsp),%r12
  lea    D15(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <getCount@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    (%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,(%rsp)
  test   %eax,%eax
  jne    L11
L19:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D11
type_infered: string
name: retrieve_data
arguments: label: D15
type_infered: string

tool: D11: %d

tool: D15: %d


assistant: ```c
int main() {
  int t, n;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    printf("%d\n", getCount(n));
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D79(%rip),%rsi
  lea    D4(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%r14
  lea    0x4(%rsp),%rcx
  mov    %rsp,%rdx
  lea    D13(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fscanf@plt>
  mov    (%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%r13
  cmpl   $0x0,(%rsp)
  jle    L38
  mov    %rax,%rbx
  mov    $0x0,%r12d
  mov    $0x0,%ebp
  lea    D28(%rip),%r15
L58:
  mov    %rbx,%rdx
  mov    %r15,%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <fscanf@plt>
  movslq (%rbx),%rax
  add    %rax,%r12
  add    $0x1,%ebp
  add    $0x4,%rbx
  cmp    %ebp,(%rsp)
  jg     L58
L53:
  mov    %r14,%rdi
  call   <fclose@plt>
  movslq 0x4(%rsp),%rcx
  mov    %r12,%rax
  cqto
  idiv   %rcx
  mov    %eax,%ecx
  mov    %eax,0x4(%rsp)
  mov    (%rsp),%edx
  test   %edx,%edx
  jle    L8
  mov    %r13,%rax
  mov    %edx,%edx
  lea    0x0(%r13,%rdx,4),%rdi
  mov    $0x0,%ebx
  movslq %ecx,%rcx
  mov    $0x0,%esi
L61:
  movslq (%rax),%rdx
  add    %rdx,%rbx
  cmp    %rbx,%rcx
  cmove  %rsi,%rbx
  add    $0x4,%rax
  cmp    %rdi,%rax
  jne    L61
  lea    D102(%rip),%rsi
  lea    D17(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rbx,%rbx
  lea    D91(%rip),%rdi
  lea    D68(%rip),%rax
  cmove  %rax,%rdi
L56:
  mov    %rbp,%rsi
  call   <fputs@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L38:
  mov    $0x0,%r12d
  jmp    L53
L8:
  lea    D102(%rip),%rsi
  lea    D17(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  lea    D68(%rip),%rdi
  jmp    L56
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D79
type: string
name: extract_symbolic_data
arguments: tag: D4
type: string
name: extract_symbolic_data
arguments: tag: D13
type: string
name: extract_symbolic_data
arguments: tag: D28
type: string
name: extract_symbolic_data
arguments: tag: D102
type: string
name: extract_symbolic_data
arguments: tag: D17
type: string
name: extract_symbolic_data
arguments: tag: D91
type: string
name: extract_symbolic_data
arguments: tag: D68
type: string

tool: D79: r

tool: D4: INP.TXT

tool: D13: %d %d

tool: D28: %d

tool: D102: w

tool: D17: OUT.TXT

tool: D91: NO


tool: D68: YES


assistant: ```c
int main() {
  FILE *f = fopen("INP.TXT", "r");
  int n, k, *a, i;
  long long s = 0;
  fscanf(f, "%d %d", &n, &k);
  a = malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    fscanf(f, "%d", a + i);
    s += a[i];
  }
  fclose(f);
  k = s / k;
  s = 0;
  for (i = 0; i < n; i++) {
    s += a[i];
    if (s == k)
      s = 0;
  }
  f = fopen("OUT.TXT", "w");
  fputs(s ? "NO\n" : "YES\n", f);
  fclose(f);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  lea    D40(%rip),%rsi
  lea    D19(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  lea    0x4(%rsp),%rcx
  mov    %rsp,%rdx
  lea    D30(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,%r15
  xor    %eax,%eax
  call   <fscanf@plt>
  mov    (%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    (%rsp),%eax
  test   %eax,%eax
  jle    L64
  mov    %rbx,%rbp
  xor    %r12d,%r12d
  lea    D93(%rip),%r14
  xor    %r13d,%r13d
  xchg   %ax,%ax
L72:
  mov    %rbp,%rdx
  mov    %r14,%rsi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <fscanf@plt>
  movslq 0x0(%rbp),%rax
  add    $0x1,%r13d
  add    $0x4,%rbp
  add    %rax,%r12
  cmp    %r13d,(%rsp)
  jg     L72
L75:
  mov    %r15,%rdi
  call   <fclose@plt>
  movslq 0x4(%rsp),%rcx
  mov    %r12,%rax
  cqto
  idiv   %rcx
  mov    (%rsp),%ecx
  mov    %eax,0x4(%rsp)
  test   %ecx,%ecx
  jle    L26
  mov    %rbx,%rdx
  lea    (%rbx,%rcx,4),%rdi
  cltq
  xor    %ebx,%ebx
  xor    %esi,%esi
  nopw   0x0(%rax,%rax,1)
L8:
  movslq (%rdx),%rcx
  add    %rcx,%rbx
  cmp    %rax,%rbx
  cmove  %rsi,%rbx
  add    $0x4,%rdx
  cmp    %rdi,%rdx
  jne    L8
  lea    D5(%rip),%rdi
  lea    D10(%rip),%rsi
  call   <fopen@plt>
  lea    D52(%rip),%rdi
  mov    %rax,%rbp
  test   %rbx,%rbx
  je     L17
L37:
  mov    %rbp,%rsi
  call   <fputs@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  lea    D10(%rip),%rsi
  lea    D5(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
L17:
  lea    D83(%rip),%rdi
  jmp    L37
L64:
  xor    %r12d,%r12d
  jmp    L75
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D40
data_type: string
name: retrieve_data
arguments: label: D19
data_type: string
name: retrieve_data
arguments: label: D30
data_type: string
name: retrieve_data
arguments: label: D93
data_type: string
name: retrieve_data
arguments: label: D5
data_type: string
name: retrieve_data
arguments: label: D10
data_type: string
name: retrieve_data
arguments: label: D52
data_type: string
name: retrieve_data
arguments: label: D83
data_type: string

tool: D40: r

tool: D19: INP.TXT

tool: D30: %d %d

tool: D93: %d

tool: D5: OUT.TXT

tool: D10: w

tool: D52: NO


tool: D83: YES


assistant: ```c
int main() {
  FILE *f = fopen("INP.TXT", "r");
  int n, k, *a, i;
  long long s = 0;
  fscanf(f, "%d %d", &n, &k);
  a = malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    fscanf(f, "%d", a + i);
    s += a[i];
  }
  fclose(f);
  k = s / k;
  s = 0;
  for (i = 0; i < n; i++) {
    s += a[i];
    if (s == k)
      s = 0;
  }
  f = fopen("OUT.TXT", "w");
  fputs(s ? "NO\n" : "YES\n", f);
  fclose(f);
  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   %r15
  lea    D21(%rip),%rsi
  lea    D35(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  lea    0x4(%rsp),%rcx
  mov    %rsp,%rdx
  lea    D95(%rip),%rsi
  mov    %rax,%rdi
  mov    %rax,%r15
  xor    %eax,%eax
  call   <fscanf@plt>
  mov    (%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    (%rsp),%eax
  test   %eax,%eax
  jle    L51
  mov    %rbx,%rbp
  xor    %r12d,%r12d
  lea    D65(%rip),%r14
  xor    %r13d,%r13d
  xchg   %ax,%ax
L71:
  mov    %rbp,%rdx
  mov    %r14,%rsi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <fscanf@plt>
  movslq 0x0(%rbp),%rax
  add    $0x1,%r13d
  add    $0x4,%rbp
  add    %rax,%r12
  cmp    %r13d,(%rsp)
  jg     L71
L48:
  mov    %r15,%rdi
  call   <fclose@plt>
  movslq 0x4(%rsp),%rcx
  mov    %r12,%rax
  cqto
  idiv   %rcx
  mov    (%rsp),%ecx
  mov    %eax,0x4(%rsp)
  test   %ecx,%ecx
  jle    L20
  cltq
  mov    %rbx,%rdx
  lea    (%rbx,%rcx,4),%rsi
L23:
  xor    %ebx,%ebx
L54:
  movslq (%rdx),%rcx
  add    $0x4,%rdx
  add    %rcx,%rbx
  cmp    %rax,%rbx
  je     L32
  cmp    %rdx,%rsi
  jne    L54
  lea    D24(%rip),%rdi
  lea    D96(%rip),%rsi
  call   <fopen@plt>
  lea    D5(%rip),%rdi
  mov    %rax,%rbp
  test   %rbx,%rbx
  je     L19
L56:
  mov    %rbp,%rsi
  call   <fputs@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L90
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L32:
  cmp    %rdx,%rsi
  jne    L23
L20:
  lea    D96(%rip),%rsi
  lea    D24(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
L19:
  lea    D83(%rip),%rdi
  jmp    L56
L51:
  xor    %r12d,%r12d
  jmp    L48
L90:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D21
datatype: string
name: disassemble_data
arguments: data_label: D35
datatype: string
name: disassemble_data
arguments: data_label: D95
datatype: string
name: disassemble_data
arguments: data_label: D65
datatype: string
name: disassemble_data
arguments: data_label: D24
datatype: string
name: disassemble_data
arguments: data_label: D96
datatype: string
name: disassemble_data
arguments: data_label: D5
datatype: string
name: disassemble_data
arguments: data_label: D83
datatype: string

tool: D21: r

tool: D35: INP.TXT

tool: D95: %d %d

tool: D65: %d

tool: D24: OUT.TXT

tool: D96: w

tool: D5: NO


tool: D83: YES


assistant: ```c
int main() {
  FILE *f = fopen("INP.TXT", "r");
  int n, k, *a, i;
  long long s = 0;
  fscanf(f, "%d %d", &n, &k);
  a = malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    fscanf(f, "%d", a + i);
    s += a[i];
  }
  fclose(f);
  k = s / k;
  s = 0;
  for (i = 0; i < n; i++) {
    s += a[i];
    if (s == k)
      s = 0;
  }
  f = fopen("OUT.TXT", "w");
  fputs(s ? "NO\n" : "YES\n", f);
  fclose(f);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<dcd11>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x2a,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0xf,-0x14(%rbp)
  jbe    L13
  mov    $0x0,%eax
  jmp    L35
L13:
  mov    -0x14(%rbp),%eax
  shl    $0x1e,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  shl    $0x1c,%eax
  and    $0xc0000000,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L80
L30:
  movl   $0x0,-0x18(%rbp)
  jmp    L44
L56:
  mov    -0x1c(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x3b,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x3f,-0x14(%rbp)
  jbe    L75
  mov    $0x0,%eax
  jmp    L35
L75:
  mov    -0x1c(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,4),%esi
  mov    -0x18(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%ecx
  shl    %cl,%eax
  or     %eax,%esi
  mov    %esi,%edx
  mov    -0x1c(%rbp),%eax
  cltq
  mov    %edx,-0x10(%rbp,%rax,4)
  addl   $0x1,-0x18(%rbp)
L44:
  cmpl   $0x4,-0x18(%rbp)
  jle    L56
  addl   $0x1,-0x1c(%rbp)
L80:
  cmpl   $0x1,-0x1c(%rbp)
  jle    L30
  lea    -0x10(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    $0x1,%eax
L35:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L11
  call   <__stack_chk_fail@plt>
L11:
  leave
  ret
```

assistant: ```c
int dcd11(double *to, const char *s) {
  unsigned int qw[2], k = s[0] - 42u;
  if (k > 15u)
    return 0;
  qw[0] = (k & 3u) << 30;
  qw[1] = (k & 12u) << 28;
  int i, j;
  for (i = 0; i < 2; i++) {
    for (j = 0; j < 5; j++) {
      if ((k = s[5 * i + j + 1] - 59) > 63u)
        return 0;
      else
        qw[i] |= k << (6 * j);
    }
  }
  return memcpy(to, qw, 8), 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<dcd11>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movsbl (%rsi),%eax
  sub    $0x2a,%eax
  mov    $0x0,%edx
  cmp    $0xf,%eax
  ja     L20
  mov    %eax,%r8d
  shl    $0x1e,%r8d
  shl    $0x1c,%eax
  and    $0xc0000000,%eax
  mov    %eax,0x4(%rsp)
  lea    0x1(%rsi),%rdx
  mov    $0x0,%ecx
L60:
  movsbl (%rdx),%eax
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L22
  shl    %cl,%eax
  or     %eax,%r8d
  add    $0x1,%rdx
  add    $0x6,%ecx
  cmp    $0x1e,%ecx
  jne    L60
  mov    %r8d,(%rsp)
  add    $0x6,%rsi
  mov    $0x0,%ecx
L30:
  movsbl (%rsi),%eax
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L22
  shl    %cl,%eax
  or     %eax,0x4(%rsp)
  add    $0x1,%rsi
  add    $0x6,%ecx
  cmp    $0x1e,%ecx
  jne    L30
  mov    %rsp,%rsi
  mov    $0x8,%edx
  call   <memcpy@plt>
  mov    $0x1,%edx
L20:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  mov    %edx,%eax
  add    $0x18,%rsp
  ret
L22:
  mov    $0x0,%edx
  jmp    L20
L65:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dcd11(double *to, const char *s) {
  unsigned int qw[2], k = s[0] - 42u;
  if (k > 15u)
    return 0;
  qw[0] = (k & 3u) << 30;
  qw[1] = (k & 12u) << 28;
  int i, j;
  for (i = 0; i < 2; i++) {
    for (j = 0; j < 5; j++) {
      if ((k = s[5 * i + j + 1] - 59) > 63u)
        return 0;
      else
        qw[i] |= k << (6 * j);
    }
  }
  return memcpy(to, qw, 8), 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<dcd11>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movsbl (%rsi),%eax
  sub    $0x2a,%eax
  cmp    $0xf,%eax
  ja     L66
  mov    %eax,%r8d
  shl    $0x1c,%eax
  lea    0x1(%rsi),%rdx
  xor    %ecx,%ecx
  and    $0xc0000000,%eax
  shl    $0x1e,%r8d
  mov    %eax,0x4(%rsp)
L53:
  movsbl (%rdx),%eax
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L66
  shl    %cl,%eax
  add    $0x6,%ecx
  add    $0x1,%rdx
  or     %eax,%r8d
  cmp    $0x1e,%ecx
  jne    L53
  mov    %r8d,(%rsp)
  add    $0x6,%rsi
  xor    %ecx,%ecx
L2:
  movsbl (%rsi),%eax
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L66
  shl    %cl,%eax
  add    $0x6,%ecx
  or     %eax,0x4(%rsp)
  add    $0x1,%rsi
  cmp    $0x1e,%ecx
  jne    L2
  mov    %rsp,%rsi
  mov    $0x8,%edx
  call   <memcpy@plt>
  mov    $0x1,%eax
  jmp    L62
  nopl   0x0(%rax,%rax,1)
L66:
  xor    %eax,%eax
L62:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x18,%rsp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dcd11(double *to, const char *s) {
  unsigned int qw[2], k = s[0] - 42u;
  if (k > 15u)
    return 0;
  qw[0] = (k & 3u) << 30;
  qw[1] = (k & 12u) << 28;
  int i, j;
  for (i = 0; i < 2; i++) {
    for (j = 0; j < 5; j++) {
      if ((k = s[5 * i + j + 1] - 59) > 63u)
        return 0;
      else
        qw[i] |= k << (6 * j);
    }
  }
  return memcpy(to, qw, 8), 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<dcd11>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rdx
  mov    %rdx,0x8(%rsp)
  xor    %edx,%edx
  movsbl (%rsi),%edx
  sub    $0x2a,%edx
  cmp    $0xf,%edx
  ja     L84
  mov    %rsi,%rax
  mov    %edx,%esi
  shl    $0x1c,%edx
  movsbl 0x1(%rax),%ecx
  and    $0xc0000000,%edx
  shl    $0x1e,%esi
  mov    %edx,0x4(%rsp)
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L84
  or     %ecx,%esi
  movsbl 0x2(%rax),%ecx
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L84
  shl    $0x6,%ecx
  or     %esi,%ecx
  movsbl 0x3(%rax),%esi
  sub    $0x3b,%esi
  cmp    $0x3f,%esi
  ja     L84
  shl    $0xc,%esi
  or     %ecx,%esi
  movsbl 0x4(%rax),%ecx
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L84
  shl    $0x12,%ecx
  or     %esi,%ecx
  movsbl 0x5(%rax),%esi
  sub    $0x3b,%esi
  cmp    $0x3f,%esi
  ja     L84
  shl    $0x18,%esi
  or     %ecx,%esi
  mov    %esi,(%rsp)
  movsbl 0x6(%rax),%esi
  sub    $0x3b,%esi
  cmp    $0x3f,%esi
  ja     L84
  movsbl 0x7(%rax),%ecx
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L84
  shl    $0x6,%ecx
  or     %edx,%ecx
  movsbl 0x8(%rax),%edx
  or     %esi,%ecx
  sub    $0x3b,%edx
  cmp    $0x3f,%edx
  ja     L84
  shl    $0xc,%edx
  or     %ecx,%edx
  movsbl 0x9(%rax),%ecx
  sub    $0x3b,%ecx
  cmp    $0x3f,%ecx
  ja     L84
  movsbl 0xa(%rax),%eax
  shl    $0x12,%ecx
  or     %edx,%ecx
  sub    $0x3b,%eax
  cmp    $0x3f,%eax
  ja     L84
  shl    $0x18,%eax
  mov    %rsp,%rsi
  mov    $0x8,%edx
  or     %ecx,%eax
  mov    %eax,0x4(%rsp)
  call   <memcpy@plt>
  mov    $0x1,%eax
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L84:
  xor    %eax,%eax
L32:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L48
  add    $0x18,%rsp
  ret
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dcd11(double *to, const char *s) {
  unsigned int qw[2], k = s[0] - 42u;
  if (k > 15u)
    return 0;
  qw[0] = (k & 3u) << 30;
  qw[1] = (k & 12u) << 28;
  int i, j;
  for (i = 0; i < 2; i++) {
    for (j = 0; j < 5; j++) {
      if ((k = s[5 * i + j + 1] - 59) > 63u)
        return 0;
      else
        qw[i] |= k << (6 * j);
    }
  }
  return memcpy(to, qw, 8), 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<atoi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  nop
L20:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  cmpb   $0x20,-0x9(%rbp)
  je     L20
  cmpb   $0x9,-0x9(%rbp)
  je     L20
  cmpb   $0x2d,-0x9(%rbp)
  jne    L24
  movl   $0xffffffff,-0x4(%rbp)
  jmp    L42
L24:
  movl   $0x1,-0x4(%rbp)
  cmpb   $0x2b,-0x9(%rbp)
  je     L42
  subq   $0x1,-0x18(%rbp)
  jmp    L42
L71:
  mov    -0x8(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  movsbl -0x9(%rbp),%eax
  add    %edx,%eax
  sub    $0x30,%eax
  mov    %eax,-0x8(%rbp)
L42:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  cmpb   $0x2f,-0x9(%rbp)
  jle    L74
  cmpb   $0x39,-0x9(%rbp)
  jle    L71
L74:
  mov    -0x4(%rbp),%eax
  imul   -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int atoi(char *s) {
  int i = 0, n;
  char c;
  while ((c = *s++) == ' ' || c == '\t')
    ;
  if (c == '-')
    n = -1;
  else {
    n = 1;
    if (c != '+')
      s--;
  }
  while ((c = *s++) >= '0' && c <= '9')
    i = i * 10 + c - '0';
  return n * i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<atoi>:
  endbr64
L18:
  mov    %rdi,%rcx
  add    $0x1,%rdi
  movzbl -0x1(%rdi),%edx
  cmp    $0x9,%dl
  je     L18
  cmp    $0x20,%dl
  je     L18
  mov    $0xffffffff,%eax
  cmp    $0x2d,%dl
  je     L36
  cmp    $0x2b,%dl
  cmovne %rcx,%rdi
  mov    $0x1,%eax
L36:
  lea    0x1(%rdi),%rsi
  movzbl (%rdi),%edx
  lea    -0x30(%rdx),%edi
  mov    $0x0,%ecx
  cmp    $0x9,%dil
  ja     L1
L46:
  lea    (%rcx,%rcx,4),%ecx
  movsbl %dl,%edx
  lea    -0x30(%rdx,%rcx,2),%ecx
  add    $0x1,%rsi
  movzbl -0x1(%rsi),%edx
  lea    -0x30(%rdx),%edi
  cmp    $0x9,%dil
  jbe    L46
L1:
  imul   %ecx,%eax
  ret
```

assistant: ```c
int atoi(char *s) {
  int i = 0, n;
  char c;
  while ((c = *s++) == ' ' || c == '\t')
    ;
  if (c == '-')
    n = -1;
  else {
    n = 1;
    if (c != '+')
      s--;
  }
  while ((c = *s++) >= '0' && c <= '9')
    i = i * 10 + c - '0';
  return n * i;
}
```
user: What is the c source code of the assembly code below?

```asm
<atoi>:
  endbr64
  nopl   0x0(%rax)
L41:
  mov    %rdi,%rax
  movsbl (%rdi),%edx
  add    $0x1,%rdi
  cmp    $0x9,%dl
  je     L41
  cmp    $0x20,%dl
  je     L41
  cmp    $0x2d,%dl
  je     L57
  mov    $0x1,%esi
  cmp    $0x2b,%dl
  jne    L65
  movsbl 0x1(%rax),%edx
  lea    0x2(%rax),%rdi
L65:
  lea    -0x30(%rdx),%ecx
  xor    %eax,%eax
  cmp    $0x9,%cl
  ja     L30
  nopl   (%rax)
L35:
  lea    (%rax,%rax,4),%eax
  add    $0x1,%rdi
  lea    -0x30(%rdx,%rax,2),%eax
  movsbl -0x1(%rdi),%edx
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  jbe    L35
  imul   %esi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L57:
  movsbl 0x1(%rax),%edx
  lea    0x2(%rax),%rdi
  mov    $0xffffffff,%esi
  jmp    L65
L30:
  ret
```

assistant: ```c
int atoi(char *s) {
  int i = 0, n;
  char c;
  while ((c = *s++) == ' ' || c == '\t')
    ;
  if (c == '-')
    n = -1;
  else {
    n = 1;
    if (c != '+')
      s--;
  }
  while ((c = *s++) >= '0' && c <= '9')
    i = i * 10 + c - '0';
  return n * i;
}
```
user: What is the c source code of the assembly code below:

```asm
<atoi>:
  endbr64
  nopl   0x0(%rax)
L49:
  mov    %rdi,%rax
  movsbl (%rdi),%edx
  add    $0x1,%rdi
  cmp    $0x9,%dl
  je     L49
  cmp    $0x20,%dl
  je     L49
  cmp    $0x2d,%dl
  je     L65
  mov    $0x1,%esi
  cmp    $0x2b,%dl
  jne    L8
  movsbl 0x1(%rax),%edx
  lea    0x2(%rax),%rdi
L8:
  lea    -0x30(%rdx),%ecx
  xor    %eax,%eax
  cmp    $0x9,%cl
  ja     L61
  nopl   (%rax)
L21:
  lea    (%rax,%rax,4),%eax
  add    $0x1,%rdi
  lea    -0x30(%rdx,%rax,2),%eax
  movsbl -0x1(%rdi),%edx
  lea    -0x30(%rdx),%ecx
  cmp    $0x9,%cl
  jbe    L21
  imul   %esi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L65:
  movsbl 0x1(%rax),%edx
  lea    0x2(%rax),%rdi
  mov    $0xffffffff,%esi
  jmp    L8
L61:
  ret
```

assistant: ```c
int atoi(char *s) {
  int i = 0, n;
  char c;
  while ((c = *s++) == ' ' || c == '\t')
    ;
  if (c == '-')
    n = -1;
  else {
    n = 1;
    if (c != '+')
      s--;
  }
  while ((c = *s++) >= '0' && c <= '9')
    i = i * 10 + c - '0';
  return n * i;
}
```