ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<strtol>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x18,%rsp
  jmp    L141
  nopl   0x0(%rax)
L50:
  add    $0x1,%r14
L141:
  movsbl (%r14),%edi
  call   <isspace@plt>
  mov    %rax,%r15
  test   %rax,%rax
  jne    L50
  movsbl (%r14),%ebx
  cmp    $0x2d,%bl
  je     L22
  movq   $0x1,0x8(%rsp)
  cmp    $0x2b,%bl
  je     L74
L5:
  test   $0xffffffef,%ebp
  jne    L103
  cmp    $0x30,%bl
  je     L118
  test   %ebp,%ebp
  jne    L103
  mov    $0xa,%ebp
L103:
  test   %bl,%bl
  je     L136
L174:
  movslq %ebp,%r12
  jmp    L203
  nopw   0x0(%rax,%rax,1)
L122:
  sub    $0x30,%ebx
  cmp    %ebx,%ebp
  jle    L206
L159:
  imul   %r12,%r15
  movslq %ebx,%rbx
  add    $0x1,%r14
  add    %rbx,%r15
  movsbl (%r14),%ebx
  test   %bl,%bl
  je     L206
L203:
  mov    %ebx,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L122
  mov    %ebx,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L206
  mov    %ebx,%edi
  call   <isupper@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0x20,%eax
  add    $0x37,%eax
  sub    %eax,%ebx
  cmp    %ebx,%ebp
  jg     L159
  xchg   %ax,%ax
L206:
  imul   0x8(%rsp),%r15
L136:
  test   %r13,%r13
  je     L143
  mov    %r14,0x0(%r13)
L143:
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L74:
  movsbl 0x1(%r14),%ebx
  add    $0x1,%r14
  jmp    L5
  cs nopw 0x0(%rax,%rax,1)
L22:
  movsbl 0x1(%r14),%ebx
  add    $0x1,%r14
  movq   $0xffffffffffffffff,0x8(%rsp)
  jmp    L5
  nopw   0x0(%rax,%rax,1)
L118:
  movzbl 0x1(%r14),%eax
  mov    %eax,%edx
  and    $0xffffffdf,%edx
  cmp    $0x58,%dl
  jne    L168
  movsbl 0x2(%r14),%ebx
  mov    $0x10,%ebp
  add    $0x2,%r14
  jmp    L103
L168:
  test   %ebp,%ebp
  jne    L174
  add    $0x1,%r14
  movsbl %al,%ebx
  mov    $0x8,%ebp
  jmp    L103
```

assistant: ```c
long int strtol(const char *nptr, char **endptr, int base) {
  int sign = 1;
  long int result = 0;
  int digit;
  while (isspace(*nptr))
    ++nptr;
  if (*nptr == '-') {
    sign = -1;
    ++nptr;
  } else if (*nptr == '+')
    ++nptr;
  if ((base == 0 || base == 16) && *nptr == '0' && (*(nptr + 1) == 'x' || *(nptr + 1) == 'X')) {
    nptr += 2;
    base = 16;
  } else if (base == 0 && *nptr == '0') {
    ++nptr;
    base = 8;
  } else if (base == 0)
    base = 10;
  while (*nptr) {
    digit = *nptr;
    if (isdigit(digit))
      digit -= '0';
    else if (isalpha(digit))
      digit -= isupper(digit) ? 'A' - 10 : 'a' - 10;
    else
      break;
    if (digit >= base)
      break;
    result *= base;
    result += digit;
    ++nptr;
  }
  if (endptr)
    *endptr = (char *)nptr;
  return sign * result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strtol>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x18,%rsp
  jmp    L97
  nopl   0x0(%rax)
L10:
  add    $0x1,%r14
L97:
  movsbl (%r14),%edi
  call   <isspace@plt>
  mov    %rax,%r15
  test   %rax,%rax
  jne    L10
  movsbl (%r14),%ebx
  cmp    $0x2d,%bl
  je     L15
  movq   $0x1,0x8(%rsp)
  cmp    $0x2b,%bl
  je     L178
L132:
  test   $0xffffffef,%ebp
  jne    L60
  cmp    $0x30,%bl
  je     L129
  test   %ebp,%ebp
  jne    L60
  mov    $0xa,%ebp
L60:
  test   %bl,%bl
  je     L206
L188:
  movslq %ebp,%r12
  jmp    L83
  nopw   0x0(%rax,%rax,1)
L34:
  sub    $0x30,%ebx
  cmp    %ebp,%ebx
  jge    L103
L110:
  imul   %r12,%r15
  movslq %ebx,%rbx
  add    $0x1,%r14
  add    %rbx,%r15
  movsbl (%r14),%ebx
  test   %bl,%bl
  je     L103
L83:
  mov    %ebx,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L34
  mov    %ebx,%edi
  call   <isalpha@plt>
  test   %rax,%rax
  je     L103
  mov    %ebx,%edi
  call   <isupper@plt>
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0x20,%eax
  add    $0x37,%eax
  sub    %eax,%ebx
  cmp    %ebp,%ebx
  jl     L110
  xchg   %ax,%ax
L103:
  imul   0x8(%rsp),%r15
L206:
  test   %r13,%r13
  je     L111
  mov    %r14,0x0(%r13)
L111:
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L178:
  movsbl 0x1(%r14),%ebx
  add    $0x1,%r14
  jmp    L132
  cs nopw 0x0(%rax,%rax,1)
L15:
  movsbl 0x1(%r14),%ebx
  add    $0x1,%r14
  movq   $0xffffffffffffffff,0x8(%rsp)
  jmp    L132
  nopw   0x0(%rax,%rax,1)
L129:
  movzbl 0x1(%r14),%eax
  mov    %eax,%edx
  and    $0xffffffdf,%edx
  cmp    $0x58,%dl
  jne    L196
  movsbl 0x2(%r14),%ebx
  mov    $0x10,%ebp
  add    $0x2,%r14
  jmp    L60
L196:
  test   %ebp,%ebp
  jne    L188
  add    $0x1,%r14
  movsbl %al,%ebx
  mov    $0x8,%ebp
  jmp    L60
```

assistant: ```c
long int strtol(const char *nptr, char **endptr, int base) {
  int sign = 1;
  long int result = 0;
  int digit;
  while (isspace(*nptr))
    ++nptr;
  if (*nptr == '-') {
    sign = -1;
    ++nptr;
  } else if (*nptr == '+')
    ++nptr;
  if ((base == 0 || base == 16) && *nptr == '0' && (*(nptr + 1) == 'x' || *(nptr + 1) == 'X')) {
    nptr += 2;
    base = 16;
  } else if (base == 0 && *nptr == '0') {
    ++nptr;
    base = 8;
  } else if (base == 0)
    base = 10;
  while (*nptr) {
    digit = *nptr;
    if (isdigit(digit))
      digit -= '0';
    else if (isalpha(digit))
      digit -= isupper(digit) ? 'A' - 10 : 'a' - 10;
    else
      break;
    if (digit >= base)
      break;
    result *= base;
    result += digit;
    ++nptr;
  }
  if (endptr)
    *endptr = (char *)nptr;
  return sign * result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<show_baud>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  add    $0xffffff80,%eax
  cmp    $0x6,%eax
  ja     L53
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D11(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D11(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D75(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L35
L53:
  lea    D94(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
L35:
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D1
data_type: string
name: unpack_data
arguments: label: D34
data_type: string
name: unpack_data
arguments: label: D18
data_type: string
name: unpack_data
arguments: label: D16
data_type: string
name: unpack_data
arguments: label: D75
data_type: string
name: unpack_data
arguments: label: D21
data_type: string
name: unpack_data
arguments: label: D12
data_type: string
name: unpack_data
arguments: label: D5
data_type: string
name: unpack_data
arguments: label: D94
data_type: string

tool: D1: The baud rate is: 

tool: D34: 300


tool: D18: 600


tool: D16: 1200


tool: D75: 1800


tool: D21: 2400


tool: D12: 4800


tool: D5: 9600


tool: D94: Fast


assistant: ```c
void show_baud(int speed) {
  printf("The baud rate is: ");
  switch (speed) {
  case B300:
    printf("300\n");
    break;
  case B600:
    printf("600\n");
    break;
  case B1200:
    printf("1200\n");
    break;
  case B1800:
    printf("1800\n");
    break;
  case B2400:
    printf("2400\n");
    break;
  case B4800:
    printf("4800\n");
    break;
  case B9600:
    printf("9600\n");
    break;
  default:
    printf("Fast\n");
    break;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<show_baud>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    -0x80(%rbx),%edi
  cmp    $0x6,%edi
  ja     L56
  mov    %edi,%edi
  lea    D91(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D42(%rip),%rdi
  call   <printf@plt>
L77:
  pop    %rbx
  ret
  lea    D28(%rip),%rdi
  call   <printf@plt>
  jmp    L77
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L77
  lea    D90(%rip),%rdi
  call   <printf@plt>
  jmp    L77
  lea    D43(%rip),%rdi
  call   <printf@plt>
  jmp    L77
  lea    D68(%rip),%rdi
  call   <printf@plt>
  jmp    L77
  lea    D31(%rip),%rdi
  call   <printf@plt>
  jmp    L77
L56:
  lea    D60(%rip),%rdi
  call   <printf@plt>
  jmp    L77
```

tool_calls: name: unpack_data
arguments: tag: D4
data_type: string
name: unpack_data
arguments: tag: D42
data_type: string
name: unpack_data
arguments: tag: D28
data_type: string
name: unpack_data
arguments: tag: D14
data_type: string
name: unpack_data
arguments: tag: D90
data_type: string
name: unpack_data
arguments: tag: D43
data_type: string
name: unpack_data
arguments: tag: D68
data_type: string
name: unpack_data
arguments: tag: D31
data_type: string
name: unpack_data
arguments: tag: D60
data_type: string

tool: D4: The baud rate is: 

tool: D42: 300


tool: D28: 600


tool: D14: 1200


tool: D90: 1800


tool: D43: 2400


tool: D68: 4800


tool: D31: 9600


tool: D60: Fast


assistant: ```c
void show_baud(int speed) {
  printf("The baud rate is: ");
  switch (speed) {
  case B300:
    printf("300\n");
    break;
  case B600:
    printf("600\n");
    break;
  case B1200:
    printf("1200\n");
    break;
  case B1800:
    printf("1800\n");
    break;
  case B2400:
    printf("2400\n");
    break;
  case B4800:
    printf("4800\n");
    break;
  case B9600:
    printf("9600\n");
    break;
  default:
    printf("Fast\n");
    break;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<show_baud>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    D6(%rip),%rdi
  call   <printf@plt>
  lea    -0x80(%rbx),%edi
  cmp    $0x6,%edi
  ja     L7
  lea    D30(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D82(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D14(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D95(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D59(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D37(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D29(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D9(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
L7:
  lea    D8(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: data_tag: D6
type_infered: string
name: read_data
arguments: data_tag: D82
type_infered: string
name: read_data
arguments: data_tag: D14
type_infered: string
name: read_data
arguments: data_tag: D95
type_infered: string
name: read_data
arguments: data_tag: D59
type_infered: string
name: read_data
arguments: data_tag: D37
type_infered: string
name: read_data
arguments: data_tag: D29
type_infered: string
name: read_data
arguments: data_tag: D9
type_infered: string
name: read_data
arguments: data_tag: D8
type_infered: string

tool: D6: The baud rate is: 

tool: D82: 1800


tool: D14: 1200


tool: D95: 9600


tool: D59: 600


tool: D37: 4800


tool: D29: 300


tool: D9: 2400


tool: D8: Fast


assistant: ```c
void show_baud(int speed) {
  printf("The baud rate is: ");
  switch (speed) {
  case B300:
    printf("300\n");
    break;
  case B600:
    printf("600\n");
    break;
  case B1200:
    printf("1200\n");
    break;
  case B1800:
    printf("1800\n");
    break;
  case B2400:
    printf("2400\n");
    break;
  case B4800:
    printf("4800\n");
    break;
  case B9600:
    printf("9600\n");
    break;
  default:
    printf("Fast\n");
    break;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<show_baud>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    D91(%rip),%rdi
  call   <printf@plt>
  lea    -0x80(%rbx),%edi
  cmp    $0x6,%edi
  ja     L7
  lea    D15(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D25(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D43(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D92(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D72(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D63(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D3(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
  nopl   (%rax)
  lea    D20(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
L7:
  lea    D16(%rip),%rdi
  pop    %rbx
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_label: D91
type: string
name: parse_data
arguments: data_label: D25
type: string
name: parse_data
arguments: data_label: D43
type: string
name: parse_data
arguments: data_label: D92
type: string
name: parse_data
arguments: data_label: D72
type: string
name: parse_data
arguments: data_label: D63
type: string
name: parse_data
arguments: data_label: D3
type: string
name: parse_data
arguments: data_label: D20
type: string
name: parse_data
arguments: data_label: D16
type: string

tool: D91: The baud rate is: 

tool: D25: 1800


tool: D43: 1200


tool: D92: 9600


tool: D72: 600


tool: D63: 4800


tool: D3: 300


tool: D20: 2400


tool: D16: Fast


assistant: ```c
void show_baud(int speed) {
  printf("The baud rate is: ");
  switch (speed) {
  case B300:
    printf("300\n");
    break;
  case B600:
    printf("600\n");
    break;
  case B1200:
    printf("1200\n");
    break;
  case B1800:
    printf("1800\n");
    break;
  case B2400:
    printf("2400\n");
    break;
  case B4800:
    printf("4800\n");
    break;
  case B9600:
    printf("9600\n");
    break;
  default:
    printf("Fast\n");
    break;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cdft>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  cmpl   $0x4,-0x4(%rbp)
  jle    L2
  cmpl   $0x0,-0x8(%rbp)
  js     L3
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <bitrv2@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <cftfsub@plt>
  jmp    L31
L3:
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <bitrv2conj@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <cftbsub@plt>
  jmp    L31
L2:
  cmpl   $0x4,-0x4(%rbp)
  jne    L31
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <cftfsub@plt>
L31:
  nop
  leave
  ret
```

assistant: ```c
void cdft(int n, int isgn, double *a) {
  void bitrv2(int n, double *a);
  void bitrv2conj(int n, double *a);
  void cftfsub(int n, double *a);
  void cftbsub(int n, double *a);
  if (n > 4) {
    if (isgn >= 0) {
      bitrv2(n, a);
      cftfsub(n, a);
    } else {
      bitrv2conj(n, a);
      cftbsub(n, a);
    }
  } else if (n == 4) {
    cftfsub(n, a);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cdft>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%rbp
  cmp    $0x4,%edi
  jle    L58
  mov    %edi,%ebx
  test   %esi,%esi
  js     L27
  mov    %rdx,%rsi
  call   <bitrv2@plt>
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <cftfsub@plt>
  jmp    L47
L27:
  mov    %rdx,%rsi
  call   <bitrv2conj@plt>
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <cftbsub@plt>
  jmp    L47
L58:
  je     L42
L47:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L42:
  mov    %rdx,%rsi
  mov    $0x4,%edi
  call   <cftfsub@plt>
  jmp    L47
```

assistant: ```c
void cdft(int n, int isgn, double *a) {
  void bitrv2(int n, double *a);
  void bitrv2conj(int n, double *a);
  void cftfsub(int n, double *a);
  void cftbsub(int n, double *a);
  if (n > 4) {
    if (isgn >= 0) {
      bitrv2(n, a);
      cftfsub(n, a);
    } else {
      bitrv2conj(n, a);
      cftbsub(n, a);
    }
  } else if (n == 4) {
    cftfsub(n, a);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cdft>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  sub    $0x8,%rsp
  cmp    $0x4,%edi
  jle    L10
  test   %esi,%esi
  mov    %edi,%ebp
  mov    %rdx,%rsi
  js     L6
  call   <bitrv2@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    10b0 <cftfsub@plt>
  nopw   0x0(%rax,%rax,1)
L10:
  je     L1
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  call   <bitrv2conj@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    10a0 <cftbsub@plt>
  xchg   %ax,%ax
L1:
  add    $0x8,%rsp
  mov    %rdx,%rsi
  mov    $0x4,%edi
  pop    %rbp
  pop    %r12
  jmp    10b0 <cftfsub@plt>
```

assistant: ```c
void cdft(int n, int isgn, double *a) {
  void bitrv2(int n, double *a);
  void bitrv2conj(int n, double *a);
  void cftfsub(int n, double *a);
  void cftbsub(int n, double *a);
  if (n > 4) {
    if (isgn >= 0) {
      bitrv2(n, a);
      cftfsub(n, a);
    } else {
      bitrv2conj(n, a);
      cftbsub(n, a);
    }
  } else if (n == 4) {
    cftfsub(n, a);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<cdft>:
  endbr64
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  sub    $0x8,%rsp
  cmp    $0x4,%edi
  jle    L29
  test   %esi,%esi
  mov    %edi,%ebp
  mov    %rdx,%rsi
  js     L20
  call   <bitrv2@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    10b0 <cftfsub@plt>
  nopw   0x0(%rax,%rax,1)
L29:
  je     L18
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  call   <bitrv2conj@plt>
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  jmp    10a0 <cftbsub@plt>
  xchg   %ax,%ax
L18:
  add    $0x8,%rsp
  mov    %rdx,%rsi
  mov    $0x4,%edi
  pop    %rbp
  pop    %r12
  jmp    10b0 <cftfsub@plt>
```

assistant: ```c
void cdft(int n, int isgn, double *a) {
  void bitrv2(int n, double *a);
  void bitrv2conj(int n, double *a);
  void cftfsub(int n, double *a);
  void cftbsub(int n, double *a);
  if (n > 4) {
    if (isgn >= 0) {
      bitrv2(n, a);
      cftfsub(n, a);
    } else {
      bitrv2conj(n, a);
      cftbsub(n, a);
    }
  } else if (n == 4) {
    cftfsub(n, a);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<string_copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  nop
L4:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L4
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void string_copy(char *f_dst_str, char *f_src_str) {
  while ((*f_dst_str++ = *f_src_str++))
    ;
  *f_dst_str = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<string_copy>:
  endbr64
L9:
  add    $0x1,%rsi
  add    $0x1,%rdi
  movzbl -0x1(%rsi),%eax
  mov    %al,-0x1(%rdi)
  test   %al,%al
  jne    L9
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void string_copy(char *f_dst_str, char *f_src_str) {
  while ((*f_dst_str++ = *f_src_str++))
    ;
  *f_dst_str = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<string_copy>:
  endbr64
  nopl   0x0(%rax)
L8:
  movzbl (%rsi),%eax
  add    $0x1,%rsi
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  test   %al,%al
  jne    L8
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void string_copy(char *f_dst_str, char *f_src_str) {
  while ((*f_dst_str++ = *f_src_str++))
    ;
  *f_dst_str = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<string_copy>:
  endbr64
  nopl   0x0(%rax)
L8:
  movzbl (%rsi),%eax
  add    $0x1,%rsi
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  test   %al,%al
  jne    L8
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void string_copy(char *f_dst_str, char *f_src_str) {
  while ((*f_dst_str++ = *f_src_str++))
    ;
  *f_dst_str = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverseBits_fun_short>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %ax,-0x14(%rbp)
  movzwl -0x14(%rbp),%eax
  mov    %ax,-0x6(%rbp)
  movl   $0xf,-0x4(%rbp)
  movzwl -0x14(%rbp),%eax
  shr    %ax
  mov    %ax,-0x14(%rbp)
  jmp    L7
L1:
  movswl -0x6(%rbp),%eax
  add    %eax,%eax
  mov    %eax,%edx
  movzwl -0x14(%rbp),%eax
  and    $0x1,%eax
  or     %edx,%eax
  mov    %ax,-0x6(%rbp)
  subl   $0x1,-0x4(%rbp)
  sarw   -0x14(%rbp)
L7:
  cmpw   $0x0,-0x14(%rbp)
  jne    L1
  movswl -0x6(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
short reverseBits_fun_short(short x) {
  short r = x;
  int i = 15;
  for (x = x >> 1 & 0x7fff; x; x >>= 1) {
    r = (r << 1) | (x & 1);
    --i;
  }
  return r << i;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<reverseBits_fun_short>:
  endbr64
  mov    %edi,%eax
  shr    %di
  mov    %edi,%edx
  je     L27
  mov    $0xf,%ecx
L28:
  add    %eax,%eax
  mov    %edx,%esi
  and    $0x1,%esi
  or     %esi,%eax
  sub    $0x1,%ecx
  sar    %dx
  jne    L28
L3:
  cwtl
  shl    %cl,%eax
  ret
L27:
  mov    $0xf,%ecx
  jmp    L3
```

assistant: ```c
short reverseBits_fun_short(short x) {
  short r = x;
  int i = 15;
  for (x = x >> 1 & 0x7fff; x; x >>= 1) {
    r = (r << 1) | (x & 1);
    --i;
  }
  return r << i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<reverseBits_fun_short>:
  endbr64
  mov    %edi,%eax
  shr    %di
  mov    $0xf,%ecx
  je     L10
L16:
  mov    %edi,%edx
  add    %eax,%eax
  sub    $0x1,%ecx
  and    $0x1,%edx
  or     %edx,%eax
  sar    %di
  jne    L16
L10:
  cwtl
  shl    %cl,%eax
  ret
```

assistant: ```c
short reverseBits_fun_short(short x) {
  short r = x;
  int i = 15;
  for (x = x >> 1 & 0x7fff; x; x >>= 1) {
    r = (r << 1) | (x & 1);
    --i;
  }
  return r << i;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverseBits_fun_short>:
  endbr64
  mov    %edi,%eax
  shr    %di
  mov    $0xf,%ecx
  je     L8
L18:
  mov    %edi,%edx
  add    %eax,%eax
  sub    $0x1,%ecx
  and    $0x1,%edx
  or     %edx,%eax
  sar    %di
  jne    L18
L8:
  cwtl
  shl    %cl,%eax
  ret
```

assistant: ```c
short reverseBits_fun_short(short x) {
  short r = x;
  int i = 15;
  for (x = x >> 1 & 0x7fff; x; x >>= 1) {
    r = (r << 1) | (x & 1);
    --i;
  }
  return r << i;
}
```