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

```asm
<CalcNN>:
  endbr64
  push   %r12
  mov    %edi,%r8d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L40
  mov    %rsp,%rsi
  mov    $0x1f4,%ecx
  xor    %eax,%eax
  mov    %rsi,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x6,(%rsp)
  cmp    $0x3,%r8d
  jle    L45
  mov    $0x6,%edx
  mov    $0x1,%edi
  mov    $0x4,%r9d
  mov    $0xcccccccd,%ecx
  lea    0x4(%rsp),%r10
  nopl   0x0(%rax)
L84:
  lea    -0x1(%rdi),%eax
  mov    %rsi,%r11
  lea    (%r10,%rax,4),%r12
  xor    %eax,%eax
  jmp    L42
  xchg   %ax,%ax
L102:
  mov    (%r11),%edx
L42:
  imul   %r9d,%edx
  add    $0x4,%r11
  add    %edx,%eax
  mov    %rax,%rdx
  imul   %rcx,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ebx
  add    %ebx,%ebx
  sub    %ebx,%edx
  mov    %edx,-0x4(%r11)
  cmp    %r12,%r11
  jne    L102
  test   %eax,%eax
  je     L70
  lea    0x1(%rdi),%r11d
  movslq %r11d,%r11
  nopl   0x0(%rax,%rax,1)
L128:
  mov    %eax,%edx
  mov    %r11d,%edi
  imul   %rcx,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%ebx
  add    %ebx,%ebx
  sub    %ebx,%eax
  mov    %eax,-0x4(%rsi,%r11,4)
  mov    %edx,%eax
  add    $0x1,%r11
  test   %edx,%edx
  jne    L128
L70:
  cmp    $0x3e7,%edi
  jg     L119
  add    $0x1,%r9d
  cmp    %r9d,%r8d
  jl     L108
  mov    (%rsp),%edx
  jmp    L84
L119:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L126
  add    $0xfb0,%rsp
  lea    D14(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10a0 <printf@plt>
L108:
  movslq %edi,%rax
  movb   $0x0,0x0(%rbp,%rax,1)
  lea    (%rsi,%rax,4),%rdx
  lea    -0x1(%rdi),%eax
  lea    0x1(%rbp,%rax,1),%rcx
  cs nopw 0x0(%rax,%rax,1)
L106:
  movzbl -0x4(%rdx),%eax
  add    $0x1,%rbp
  sub    $0x4,%rdx
  add    $0x30,%eax
  mov    %al,-0x1(%rbp)
  cmp    %rbp,%rcx
  jne    L106
L75:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L126
  add    $0xfb0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L40:
  call   <CalcMM@plt>
  mov    %eax,%edx
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L126
  add    $0xfb0,%rsp
  mov    %rbp,%rdi
  lea    D17(%rip),%rsi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10b0 <sprintf@plt>
L45:
  mov    $0x36,%eax
  mov    %ax,0x0(%rbp)
  jmp    L75
L126:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D14
type_guessed: string
name: retrieve_data
arguments: tag: D17
type_guessed: string

tool: D14: overflow!


tool: D17: %u

assistant: ```c
void CalcNN(int n, char *pOut) {
  if (n < 2) {
    sprintf(pOut, "%u", CalcMM(n));
    return;
  }
  unsigned int res[1000] = {6};
  int i, j, len, carry;
  for (i = 4, len = 1; i <= n; ++i) {
    for (j = 0, carry = 0; j < len; ++j) {
      res[j] = res[j] * i + carry;
      carry = res[j] / 10;
      res[j] %= 10;
    }
    for (; carry; carry /= 10)
      res[len++] = carry % 10;
    if (len >= 1000) {
      printf("overflow!\n");
      return;
    }
  }
  i = 0;
  pOut[len] = '\0';
  while (len-- > 0)
    pOut[i++] = res[len] + '0';
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<CalcNN>:
  endbr64
  push   %r12
  mov    %edi,%r8d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L47
  mov    %rsp,%r9
  mov    $0x1f4,%ecx
  xor    %eax,%eax
  mov    %r9,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x6,(%rsp)
  cmp    $0x3,%r8d
  jle    L16
  mov    $0x1,%edi
  mov    $0x4,%r10d
  lea    0x4(%rsp),%r11
  mov    $0xcccccccd,%esi
  nopw   0x0(%rax,%rax,1)
L95:
  lea    -0x1(%rdi),%eax
  mov    %r9,%rcx
  lea    (%r11,%rax,4),%r12
  xor    %eax,%eax
  nopl   0x0(%rax)
L82:
  mov    (%rcx),%edx
  add    $0x4,%rcx
  imul   %r10d,%edx
  add    %edx,%eax
  mov    %rax,%rdx
  imul   %rsi,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ebx
  add    %ebx,%ebx
  sub    %ebx,%edx
  mov    %edx,-0x4(%rcx)
  cmp    %rcx,%r12
  jne    L82
  test   %eax,%eax
  je     L63
  lea    0x1(%rdi),%ecx
  movslq %ecx,%rcx
L103:
  mov    %eax,%edx
  mov    %ecx,%edi
  imul   %rsi,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%ebx
  add    %ebx,%ebx
  sub    %ebx,%eax
  mov    %eax,-0x4(%r9,%rcx,4)
  mov    %edx,%eax
  add    $0x1,%rcx
  test   %edx,%edx
  jne    L103
L63:
  cmp    $0x3e7,%edi
  jg     L46
  add    $0x1,%r10d
  cmp    %r10d,%r8d
  jge    L95
  movslq %edi,%rax
  movb   $0x0,0x0(%rbp,%rax,1)
  lea    (%r9,%rax,4),%rdx
  lea    -0x1(%rdi),%eax
  lea    0x1(%rbp,%rax,1),%rcx
  nopw   0x0(%rax,%rax,1)
L75:
  movzbl -0x4(%rdx),%eax
  add    $0x1,%rbp
  sub    $0x4,%rdx
  add    $0x30,%eax
  mov    %al,-0x1(%rbp)
  cmp    %rbp,%rcx
  jne    L75
L70:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0xfb0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L46:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0xfb0,%rsp
  lea    D17(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10a0 <printf@plt>
L47:
  call   <CalcMM@plt>
  mov    %eax,%edx
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0xfb0,%rsp
  mov    %rbp,%rdi
  lea    D11(%rip),%rsi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10b0 <sprintf@plt>
L16:
  mov    $0x36,%eax
  mov    %ax,(%rsi)
  jmp    L70
L64:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D17
type: string
name: parse_data
arguments: label: D11
type: string

tool: D17: overflow!


tool: D11: %u

assistant: ```c
void CalcNN(int n, char *pOut) {
  if (n < 2) {
    sprintf(pOut, "%u", CalcMM(n));
    return;
  }
  unsigned int res[1000] = {6};
  int i, j, len, carry;
  for (i = 4, len = 1; i <= n; ++i) {
    for (j = 0, carry = 0; j < len; ++j) {
      res[j] = res[j] * i + carry;
      carry = res[j] / 10;
      res[j] %= 10;
    }
    for (; carry; carry /= 10)
      res[len++] = carry % 10;
    if (len >= 1000) {
      printf("overflow!\n");
      return;
    }
  }
  i = 0;
  pOut[len] = '\0';
  while (len-- > 0)
    pOut[i++] = res[len] + '0';
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<memmem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %rcx,-0x50(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x50(%rbp),%rax
  cmp    -0x40(%rbp),%rax
  ja     L76
  cmpq   $0x0,-0x50(%rbp)
  je     L76
  cmpq   $0x0,-0x40(%rbp)
  jne    L113
L76:
  mov    $0x0,%eax
  jmp    L153
L113:
  cmpq   $0x1,-0x50(%rbp)
  je     L46
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L152
  movq   $0x2,-0x18(%rbp)
  movq   $0x1,-0x10(%rbp)
  jmp    L33
L152:
  movq   $0x1,-0x18(%rbp)
  movq   $0x2,-0x10(%rbp)
L33:
  movq   $0x0,-0x20(%rbp)
  jmp    L30
L143:
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L126
  mov    -0x18(%rbp),%rax
  add    %rax,-0x20(%rbp)
  jmp    L30
L126:
  mov    -0x50(%rbp),%rax
  lea    -0x2(%rax),%rdx
  mov    -0x20(%rbp),%rax
  lea    0x2(%rax),%rcx
  mov    -0x28(%rbp),%rax
  add    %rax,%rcx
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %eax,%eax
  jne    L112
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x28(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L112
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  jmp    L153
L112:
  mov    -0x10(%rbp),%rax
  add    %rax,-0x20(%rbp)
L30:
  mov    -0x40(%rbp),%rax
  sub    -0x50(%rbp),%rax
  cmp    %rax,-0x20(%rbp)
  jbe    L143
  jmp    L105
L46:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L140
  mov    -0x28(%rbp),%rax
  jmp    L153
L140:
  addq   $0x1,-0x28(%rbp)
  subq   $0x1,-0x40(%rbp)
  cmpq   $0x0,-0x40(%rbp)
  jne    L46
L105:
  mov    $0x0,%eax
L153:
  leave
  ret
```

assistant: ```c
void *memmem(const void *haystack, size_t n, const void *needle, size_t m) {
  const unsigned char *y = (const unsigned char *)haystack;
  const unsigned char *x = (const unsigned char *)needle;
  size_t j, k, l;
  if (m > n || !m || !n)
    return NULL;
  if (1 != m) {
    if (x[0] == x[1]) {
      k = 2;
      l = 1;
    } else {
      k = 1;
      l = 2;
    }
    j = 0;
    while (j <= n - m) {
      if (x[1] != y[j + 1]) {
        j += k;
      } else {
        if (!memcmp(x + 2, y + j + 2, m - 2) && x[0] == y[j])
          return (void *)&y[j];
        j += l;
      }
    }
  } else {
    do {
      if (*y == *x)
        return (void *)y;
      y++;
    } while (--n);
  }
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memmem>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdx,%r12
  cmp    %rsi,%rcx
  seta   %al
  test   %rcx,%rcx
  sete   %dl
  or     %dl,%al
  jne    L108
  mov    %rdi,%rbp
  mov    %rsi,%r13
  test   %rsi,%rsi
  je     L108
  cmp    $0x1,%rcx
  je     L62
  movzbl (%r12),%eax
  cmp    0x1(%r12),%al
  setne  %r15b
  movzbl %r15b,%r15d
  add    $0x1,%r15
  cmp    0x1(%r12),%al
  sete   %r14b
  movzbl %r14b,%r14d
  add    $0x1,%r14
  sub    %rcx,%r13
  mov    $0x0,%ebx
  lea    -0x2(%rcx),%rax
  mov    %rax,0x8(%rsp)
  jmp    L35
L28:
  lea    0x2(%rbp,%rbx,1),%rsi
  lea    0x2(%r12),%rdi
  mov    0x8(%rsp),%rdx
  call   <memcmp@plt>
  test   %eax,%eax
  jne    L113
  lea    0x0(%rbp,%rbx,1),%rax
  movzbl (%rax),%edi
  cmp    %dil,(%r12)
  je     L37
L113:
  add    %r15,%rbx
L68:
  cmp    %r13,%rbx
  ja     L117
L35:
  movzbl 0x1(%rbp,%rbx,1),%eax
  cmp    %al,0x1(%r12)
  je     L28
  add    %r14,%rbx
  jmp    L68
L117:
  mov    $0x0,%eax
L37:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L62:
  movzbl (%r12),%edx
  mov    %rdi,%rax
  add    %rsi,%rbp
L13:
  cmp    %dl,(%rax)
  je     L37
  add    $0x1,%rax
  cmp    %rbp,%rax
  jne    L13
  mov    $0x0,%eax
  jmp    L37
L108:
  mov    $0x0,%eax
  jmp    L37
```

assistant: ```c
void *memmem(const void *haystack, size_t n, const void *needle, size_t m) {
  const unsigned char *y = (const unsigned char *)haystack;
  const unsigned char *x = (const unsigned char *)needle;
  size_t j, k, l;
  if (m > n || !m || !n)
    return NULL;
  if (1 != m) {
    if (x[0] == x[1]) {
      k = 2;
      l = 1;
    } else {
      k = 1;
      l = 2;
    }
    j = 0;
    while (j <= n - m) {
      if (x[1] != y[j + 1]) {
        j += k;
      } else {
        if (!memcmp(x + 2, y + j + 2, m - 2) && x[0] == y[j])
          return (void *)&y[j];
        j += l;
      }
    }
  } else {
    do {
      if (*y == *x)
        return (void *)y;
      y++;
    } while (--n);
  }
  return NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<memmem>:
  endbr64
  cmp    %rsi,%rcx
  ja     L52
  test   %rcx,%rcx
  je     L52
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  movzbl (%rdx),%edx
  cmp    $0x1,%rcx
  je     L76
  movzbl 0x1(%r12),%eax
  xor    %edi,%edi
  lea    -0x2(%rcx),%r14
  cmp    %dl,%al
  setne  %dil
  xor    %r13d,%r13d
  add    $0x1,%rdi
  cmp    %dl,%al
  sete   %r13b
  mov    %rdi,0x8(%rsp)
  sub    %rcx,%rbp
  xor    %ebx,%ebx
  add    $0x1,%r13
  jmp    L13
  nopl   0x0(%rax)
L85:
  add    %r13,%rbx
  cmp    %rbx,%rbp
  jb     L58
L45:
  movzbl 0x1(%r12),%eax
L13:
  cmp    %al,0x1(%r15,%rbx,1)
  jne    L85
  lea    0x2(%r15,%rbx,1),%rsi
  lea    0x2(%r12),%rdi
  mov    %r14,%rdx
  call   <memcmp@plt>
  test   %eax,%eax
  jne    L29
  lea    (%r15,%rbx,1),%rax
  movzbl (%rax),%edi
  cmp    %dil,(%r12)
  je     L3
L29:
  add    0x8(%rsp),%rbx
  cmp    %rbx,%rbp
  jae    L45
  nopw   0x0(%rax,%rax,1)
L58:
  xor    %eax,%eax
L3:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L76:
  add    %rdi,%rbp
  mov    %rdi,%rax
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L49:
  add    $0x1,%rax
  cmp    %rax,%rbp
  je     L58
L11:
  cmp    %dl,(%rax)
  jne    L49
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L52:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memmem(const void *haystack, size_t n, const void *needle, size_t m) {
  const unsigned char *y = (const unsigned char *)haystack;
  const unsigned char *x = (const unsigned char *)needle;
  size_t j, k, l;
  if (m > n || !m || !n)
    return NULL;
  if (1 != m) {
    if (x[0] == x[1]) {
      k = 2;
      l = 1;
    } else {
      k = 1;
      l = 2;
    }
    j = 0;
    while (j <= n - m) {
      if (x[1] != y[j + 1]) {
        j += k;
      } else {
        if (!memcmp(x + 2, y + j + 2, m - 2) && x[0] == y[j])
          return (void *)&y[j];
        j += l;
      }
    }
  } else {
    do {
      if (*y == *x)
        return (void *)y;
      y++;
    } while (--n);
  }
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memmem>:
  endbr64
  cmp    %rsi,%rcx
  ja     L113
  test   %rcx,%rcx
  je     L113
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x18,%rsp
  movzbl (%rdx),%edx
  cmp    $0x1,%rcx
  je     L70
  movzbl 0x1(%r12),%eax
  xor    %edi,%edi
  lea    -0x2(%rcx),%r14
  cmp    %dl,%al
  setne  %dil
  xor    %r13d,%r13d
  add    $0x1,%rdi
  cmp    %dl,%al
  sete   %r13b
  mov    %rdi,0x8(%rsp)
  sub    %rcx,%rbp
  xor    %ebx,%ebx
  add    $0x1,%r13
  jmp    L16
  nopl   0x0(%rax)
L4:
  add    %r13,%rbx
  cmp    %rbx,%rbp
  jb     L6
L111:
  movzbl 0x1(%r12),%eax
L16:
  cmp    %al,0x1(%r15,%rbx,1)
  jne    L4
  lea    0x2(%r15,%rbx,1),%rsi
  lea    0x2(%r12),%rdi
  mov    %r14,%rdx
  call   <memcmp@plt>
  test   %eax,%eax
  jne    L34
  lea    (%r15,%rbx,1),%rax
  movzbl (%rax),%edi
  cmp    %dil,(%r12)
  je     L8
L34:
  add    0x8(%rsp),%rbx
  cmp    %rbx,%rbp
  jae    L111
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
L8:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L70:
  add    %rdi,%rbp
  mov    %rdi,%rax
  jmp    L79
  nopl   0x0(%rax,%rax,1)
L56:
  add    $0x1,%rax
  cmp    %rax,%rbp
  je     L6
L79:
  cmp    %dl,(%rax)
  jne    L56
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L113:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memmem(const void *haystack, size_t n, const void *needle, size_t m) {
  const unsigned char *y = (const unsigned char *)haystack;
  const unsigned char *x = (const unsigned char *)needle;
  size_t j, k, l;
  if (m > n || !m || !n)
    return NULL;
  if (1 != m) {
    if (x[0] == x[1]) {
      k = 2;
      l = 1;
    } else {
      k = 1;
      l = 2;
    }
    j = 0;
    while (j <= n - m) {
      if (x[1] != y[j + 1]) {
        j += k;
      } else {
        if (!memcmp(x + 2, y + j + 2, m - 2) && x[0] == y[j])
          return (void *)&y[j];
        j += l;
      }
    }
  } else {
    do {
      if (*y == *x)
        return (void *)y;
      y++;
    } while (--n);
  }
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<read_fifo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xa0,%rsp
  mov    %edi,-0x84(%rbp)
  mov    %rsi,-0x90(%rbp)
  mov    %rdx,-0x98(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x70(%rbp),%rcx
  mov    -0x84(%rbp),%eax
  mov    $0x64,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x74(%rbp)
  mov    -0x74(%rbp),%eax
  cltq
  movb   $0x0,-0x70(%rbp,%rax,1)
  cmpl   $0x0,-0x74(%rbp)
  jne    L3
  mov    -0x98(%rbp),%rdx
  mov    -0x84(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <FD_CLR@plt>
  mov    -0x84(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    $0x0,%eax
  jmp    L41
L3:
  lea    -0x70(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L11
  lea    -0x70(%rbp),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    -0x90(%rbp),%rdx
  mov    (%rdx),%edx
  add    %edx,%eax
  mov    %eax,%edx
  mov    -0x90(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L27
L11:
  mov    -0x84(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
L27:
  mov    -0x74(%rbp),%eax
L41:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
int read_fifo(int fd, int *sum, fd_set *read_set) {
  char buf[100];
  int bytes = read(fd, buf, 100);
  buf[bytes] = '\0';
  if (bytes == 0) {
    FD_CLR(fd, read_set);
    close(fd);
    return 0;
  }
  if (strlen(buf)) {
    *sum += atoi(buf);
  } else {
    close(fd);
  }
  return bytes;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<read_fifo>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0x64,%edx
  call   <read@plt>
  mov    %eax,%ebx
  cltq
  movb   $0x0,(%rsp,%rax,1)
  test   %ebx,%ebx
  je     L48
  cmpb   $0x0,(%rsp)
  je     L5
  mov    %rsp,%rdi
  call   <atoi@plt>
  add    %eax,(%r12)
L23:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  mov    %ebx,%eax
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L48:
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <FD_CLR@plt>
  mov    %ebp,%edi
  call   <close@plt>
  jmp    L23
L5:
  mov    %ebp,%edi
  call   <close@plt>
  jmp    L23
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_fifo(int fd, int *sum, fd_set *read_set) {
  char buf[100];
  int bytes = read(fd, buf, 100);
  buf[bytes] = '\0';
  if (bytes == 0) {
    FD_CLR(fd, read_set);
    close(fd);
    return 0;
  }
  if (strlen(buf)) {
    *sum += atoi(buf);
  } else {
    close(fd);
  }
  return bytes;
}
```
user: What is the c source code of the assembly code below:

```asm
<read_fifo>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  mov    $0x64,%edx
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rsi
  call   <read@plt>
  mov    %eax,%r12d
  cltq
  movb   $0x0,(%rsp,%rax,1)
  test   %r12d,%r12d
  je     L5
  cmpb   $0x0,(%rsp)
  jne    L1
  mov    %ebp,%edi
  call   <close@plt>
L12:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  add    $0x70,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %r14,%rdi
  call   <atoi@plt>
  add    %eax,(%rbx)
  jmp    L12
  nopl   0x0(%rax)
L5:
  mov    %ebp,%edi
  mov    %r13,%rsi
  call   <FD_CLR@plt>
  mov    %ebp,%edi
  call   <close@plt>
  jmp    L12
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_fifo(int fd, int *sum, fd_set *read_set) {
  char buf[100];
  int bytes = read(fd, buf, 100);
  buf[bytes] = '\0';
  if (bytes == 0) {
    FD_CLR(fd, read_set);
    close(fd);
    return 0;
  }
  if (strlen(buf)) {
    *sum += atoi(buf);
  } else {
    close(fd);
  }
  return bytes;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_fifo>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdx,%r13
  mov    $0x64,%edx
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  mov    %r14,%rsi
  call   <read@plt>
  mov    %eax,%r12d
  cltq
  movb   $0x0,(%rsp,%rax,1)
  test   %r12d,%r12d
  je     L42
  cmpb   $0x0,(%rsp)
  jne    L17
  mov    %ebp,%edi
  call   <close@plt>
L12:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  add    $0x70,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %r14,%rdi
  call   <atoi@plt>
  add    %eax,(%rbx)
  jmp    L12
  nopl   0x0(%rax)
L42:
  mov    %ebp,%edi
  mov    %r13,%rsi
  call   <FD_CLR@plt>
  mov    %ebp,%edi
  call   <close@plt>
  jmp    L12
L47:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_fifo(int fd, int *sum, fd_set *read_set) {
  char buf[100];
  int bytes = read(fd, buf, 100);
  buf[bytes] = '\0';
  if (bytes == 0) {
    FD_CLR(fd, read_set);
    close(fd);
    return 0;
  }
  if (strlen(buf)) {
    *sum += atoi(buf);
  } else {
    close(fd);
  }
  return bytes;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<decToInt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L12
L24:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x2f,-0x4(%rbp)
  jbe    L41
  cmpl   $0x39,-0x4(%rbp)
  ja     L41
  mov    -0x4(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L55
L41:
  cmpl   $0x5f,-0x4(%rbp)
  jne    L18
  nop
L55:
  addq   $0x1,-0x18(%rbp)
L12:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L24
  jmp    L11
L18:
  nop
L11:
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int decToInt(const char *src) {
  int val = 0;
  int cval = 0;
  unsigned chr;
  for (; *src != '\0'; ++src) {
    chr = *src;
    if ('0' <= chr && chr <= '9') {
      cval = chr - '0';
    } else if (chr == '_') {
      continue;
    } else {
      break;
    }
    val = (val * 10) + cval;
  }
  return val;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<decToInt>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L13
  mov    $0x0,%ecx
  jmp    L18
L36:
  cmp    $0x5f,%eax
  je     L67
L42:
  mov    %ecx,%eax
  ret
L67:
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L42
L18:
  movsbl %al,%eax
  lea    -0x30(%rax),%edx
  cmp    $0x9,%edx
  ja     L36
  lea    (%rcx,%rcx,4),%eax
  lea    (%rdx,%rax,2),%ecx
  jmp    L67
L13:
  mov    $0x0,%ecx
  jmp    L42
```

assistant: ```c
int decToInt(const char *src) {
  int val = 0;
  int cval = 0;
  unsigned chr;
  for (; *src != '\0'; ++src) {
    chr = *src;
    if ('0' <= chr && chr <= '9') {
      cval = chr - '0';
    } else if (chr == '_') {
      continue;
    } else {
      break;
    }
    val = (val * 10) + cval;
  }
  return val;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<decToInt>:
  endbr64
  movsbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L13
L6:
  lea    -0x30(%rax),%edx
  cmp    $0x9,%edx
  ja     L37
  lea    (%r8,%r8,4),%eax
  lea    (%rdx,%rax,2),%r8d
L12:
  movsbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L6
L13:
  mov    %r8d,%eax
  ret
  xchg   %ax,%ax
L37:
  cmp    $0x5f,%eax
  je     L12
  mov    %r8d,%eax
  ret
```

assistant: ```c
int decToInt(const char *src) {
  int val = 0;
  int cval = 0;
  unsigned chr;
  for (; *src != '\0'; ++src) {
    chr = *src;
    if ('0' <= chr && chr <= '9') {
      cval = chr - '0';
    } else if (chr == '_') {
      continue;
    } else {
      break;
    }
    val = (val * 10) + cval;
  }
  return val;
}
```