ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<dsp_protocol_Q152linear>:
  endbr64
  mov    %esi,%ecx
  movzwl %di,%edi
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  shl    %cl,%eax
  cvtsi2sd %edi,%xmm0
  pxor   %xmm1,%xmm1
  mulsd  D27(%rip),%xmm0
  cvtsi2sd %eax,%xmm1
  mulsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  mulsd  D14(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  cvttss2si %xmm0,%eax
  movzwl %ax,%ecx
  cvtsi2ss %ecx,%xmm1
  subss  %xmm1,%xmm0
  comiss D5(%rip),%xmm0
  jbe    L5
  add    $0x1,%eax
L5:
  mov    %ax,(%rdx)
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D27
type_guessed: byte[8]
name: unpack_data
arguments: data_label: D14
type_guessed: f64
name: unpack_data
arguments: data_label: D5
type_guessed: f32

tool: {"D27": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x3f]}

tool: D14: 100

tool: D5: 0.5

assistant: ```c
void dsp_protocol_Q152linear(const unsigned short int scale, const unsigned short int power2, unsigned short int *output) {
  float result = scale * 1.0 / 0x8000 * (1 << power2) * 100.0;
  do {
  } while (0);
  *output = (short int)(result);
  if ((result - *output) > 0.5)
    (*output)++;
  do {
  } while (0);
  do {
  } while (0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<BNCkHz>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movq   $0x63,-0x8(%rbp)
  mov    $0xb,%edi
  call   <BNCIO@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long int BNCkHz() {
  long int stat = 99;
  BNCIO(11);
  return (stat);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<BNCkHz>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xb,%edi
  call   <BNCIO@plt>
  mov    $0x63,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
long int BNCkHz() {
  long int stat = 99;
  BNCIO(11);
  return (stat);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<BNCkHz>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xb,%edi
  call   <BNCIO@plt>
  mov    $0x63,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
long int BNCkHz() {
  long int stat = 99;
  BNCIO(11);
  return (stat);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<BNCkHz>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xb,%edi
  call   <BNCIO@plt>
  mov    $0x63,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
long int BNCkHz() {
  long int stat = 99;
  BNCIO(11);
  return (stat);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<conv_basic>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    %r8,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L14
L44:
  movl   $0x0,-0x4(%rbp)
  jmp    L28
L34:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm2
  mov    -0x8(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mulss  %xmm2,%xmm0
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L28:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L15
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L34
L15:
  addl   $0x1,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L44
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void conv_basic(const float *restrict a, int n, const float *restrict f, int m, float *restrict b) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m && j <= i; j++)
      b[i] += f[j] * a[i - j];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<conv_basic>:
  endbr64
  test   %esi,%esi
  jle    L38
  push   %rbx
  mov    %rdx,%r10
  mov    %esi,%ebx
  mov    $0x0,%r11d
  jmp    L71
L14:
  movss  %xmm1,(%r8,%r11,4)
L70:
  add    $0x1,%r11
  add    $0x4,%rdi
  cmp    %rbx,%r11
  je     L19
L71:
  mov    %r11d,%r9d
  test   %ecx,%ecx
  jle    L70
  test   %r11d,%r11d
  js     L70
  movss  (%r8,%r11,4),%xmm1
  mov    %rdi,%rdx
  mov    $0x1,%eax
L50:
  movss  -0x4(%r10,%rax,4),%xmm0
  mulss  (%rdx),%xmm0
  addss  %xmm0,%xmm1
  mov    %eax,%esi
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %esi,%ecx
  jle    L14
  cmp    %r9d,%esi
  jle    L50
  jmp    L14
L19:
  pop    %rbx
  ret
L38:
  ret
```

assistant: ```c
void conv_basic(const float *restrict a, int n, const float *restrict f, int m, float *restrict b) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m && j <= i; j++)
      b[i] += f[j] * a[i - j];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<conv_basic>:
  endbr64
  test   %esi,%esi
  jle    L6
  movslq %esi,%r11
  xor    %r10d,%r10d
  xchg   %ax,%ax
L43:
  test   %ecx,%ecx
  jle    L32
  movss  (%r8,%r10,4),%xmm1
  mov    %rdi,%rsi
  mov    $0x1,%eax
  nopw   0x0(%rax,%rax,1)
L38:
  movss  -0x4(%rdx,%rax,4),%xmm0
  mulss  (%rsi),%xmm0
  mov    %eax,%r9d
  sub    $0x4,%rsi
  add    $0x1,%rax
  addss  %xmm0,%xmm1
  cmp    %r9d,%ecx
  jle    L27
  cmp    %r10d,%r9d
  jle    L38
L27:
  movss  %xmm1,(%r8,%r10,4)
L32:
  add    $0x1,%r10
  add    $0x4,%rdi
  cmp    %r10,%r11
  jne    L43
L6:
  ret
```

assistant: ```c
void conv_basic(const float *restrict a, int n, const float *restrict f, int m, float *restrict b) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m && j <= i; j++)
      b[i] += f[j] * a[i - j];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<conv_basic>:
  endbr64
  test   %esi,%esi
  jle    L2
  movslq %esi,%r11
  xor    %r10d,%r10d
  xchg   %ax,%ax
L26:
  test   %ecx,%ecx
  jle    L3
  movss  (%r8,%r10,4),%xmm1
  mov    %rdi,%rsi
  mov    $0x1,%eax
  nopw   0x0(%rax,%rax,1)
L17:
  movss  -0x4(%rdx,%rax,4),%xmm0
  mulss  (%rsi),%xmm0
  mov    %eax,%r9d
  sub    $0x4,%rsi
  add    $0x1,%rax
  addss  %xmm0,%xmm1
  cmp    %r9d,%ecx
  jle    L44
  cmp    %r10d,%r9d
  jle    L17
L44:
  movss  %xmm1,(%r8,%r10,4)
L3:
  add    $0x1,%r10
  add    $0x4,%rdi
  cmp    %r10,%r11
  jne    L26
L2:
  ret
```

assistant: ```c
void conv_basic(const float *restrict a, int n, const float *restrict f, int m, float *restrict b) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m && j <= i; j++)
      b[i] += f[j] * a[i - j];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<inet_pton>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  cmpl   $0x80,-0x4(%rbp)
  je     L26
  cmpl   $0x81,-0x4(%rbp)
  jne    L9
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <inet_pton4@plt>
  jmp    L17
L26:
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <inet_pton6@plt>
  jmp    L17
L9:
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    D0(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
L17:
  leave
  ret
```

assistant: ```c
int inet_pton(int af, const char *src, void *dst) {
  switch (af) {
  case AF_INET:
    return inet_pton4(src, dst);
  case AF_INET6:
    return inet_pton6(src, dst);
  default:
    errno = EAFNOSUPPORT;
    return -1;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<inet_pton>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%eax
  mov    %rsi,%rdi
  cmp    $0x80,%eax
  je     L18
  cmp    $0x81,%eax
  jne    L8
  mov    %rdx,%rsi
  call   <inet_pton4@plt>
L5:
  add    $0x8,%rsp
  ret
L18:
  mov    %rdx,%rsi
  call   <inet_pton6@plt>
  jmp    L5
L8:
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  jmp    L5
```

assistant: ```c
int inet_pton(int af, const char *src, void *dst) {
  switch (af) {
  case AF_INET:
    return inet_pton4(src, dst);
  case AF_INET6:
    return inet_pton6(src, dst);
  default:
    errno = EAFNOSUPPORT;
    return -1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<inet_pton>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%rdi
  mov    %rdx,%rsi
  cmp    $0x80,%eax
  je     L2
  cmp    $0x81,%eax
  jne    L10
  jmp    1070 <inet_pton4@plt>
  nop
L2:
  jmp    1060 <inet_pton6@plt>
  nopl   (%rax)
L10:
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    D14(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int inet_pton(int af, const char *src, void *dst) {
  switch (af) {
  case AF_INET:
    return inet_pton4(src, dst);
  case AF_INET6:
    return inet_pton6(src, dst);
  default:
    errno = EAFNOSUPPORT;
    return -1;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<inet_pton>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%rdi
  mov    %rdx,%rsi
  cmp    $0x80,%eax
  je     L4
  cmp    $0x81,%eax
  jne    L1
  jmp    1070 <inet_pton4@plt>
  nop
L4:
  jmp    1060 <inet_pton6@plt>
  nopl   (%rax)
L1:
  mov    D9(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int inet_pton(int af, const char *src, void *dst) {
  switch (af) {
  case AF_INET:
    return inet_pton4(src, dst);
  case AF_INET6:
    return inet_pton6(src, dst);
  default:
    errno = EAFNOSUPPORT;
    return -1;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<divideNeg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  imul   %eax,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  lea    0x3ffff(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x12,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type: string

tool: D2: divide negative value by power-of-2 = %d (0x%x)


assistant: ```c
void divideNeg(int x, int y) {
  int sum = x * x - y * y;
  int rem = sum / (1 << 18);
  printf("divide negative value by power-of-2 = %d (0x%x)\n", rem, rem);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<divideNeg>:
  endbr64
  sub    $0x8,%rsp
  imul   %edi,%edi
  imul   %esi,%esi
  sub    %esi,%edi
  lea    0x3ffff(%rdi),%esi
  cmovns %edi,%esi
  sar    $0x12,%esi
  mov    %esi,%edx
  lea    D5(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
type: string

tool: D5: divide negative value by power-of-2 = %d (0x%x)


assistant: ```c
void divideNeg(int x, int y) {
  int sum = x * x - y * y;
  int rem = sum / (1 << 18);
  printf("divide negative value by power-of-2 = %d (0x%x)\n", rem, rem);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<divideNeg>:
  endbr64
  imul   %esi,%esi
  imul   %edi,%edi
  sub    %esi,%edi
  lea    0x3ffff(%rdi),%esi
  cmovns %edi,%esi
  lea    D4(%rip),%rdi
  sar    $0x12,%esi
  mov    %esi,%edx
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D4
type_guessed: string

tool: D4: divide negative value by power-of-2 = %d (0x%x)


assistant: ```c
void divideNeg(int x, int y) {
  int sum = x * x - y * y;
  int rem = sum / (1 << 18);
  printf("divide negative value by power-of-2 = %d (0x%x)\n", rem, rem);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<divideNeg>:
  endbr64
  imul   %esi,%esi
  imul   %edi,%edi
  sub    %esi,%edi
  lea    0x3ffff(%rdi),%esi
  cmovns %edi,%esi
  lea    D4(%rip),%rdi
  sar    $0x12,%esi
  mov    %esi,%edx
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: label: D4
type_guessed: string

tool: D4: divide negative value by power-of-2 = %d (0x%x)


assistant: ```c
void divideNeg(int x, int y) {
  int sum = x * x - y * y;
  int rem = sum / (1 << 18);
  printf("divide negative value by power-of-2 = %d (0x%x)\n", rem, rem);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x5,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L5
  mov    $0xa,%eax
  jmp    L19
L5:
  mov    $0x14,%eax
L19:
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int x = 5;
  return x ? 10 : 20;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  mov    $0xa,%eax
  ret
```

assistant: ```c
int main() {
  int x = 5;
  return x ? 10 : 20;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  mov    $0xa,%eax
  ret
```

assistant: ```c
int main() {
  int x = 5;
  return x ? 10 : 20;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  mov    $0xa,%eax
  ret
```

assistant: ```c
int main() {
  int x = 5;
  return x ? 10 : 20;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverseWords>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  cmpb   $0x20,(%rdi)
  jne    L17
  mov    $0x1,%eax
L3:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x20,-0x1(%rbx,%rax,1)
  je     L3
L110:
  movslq %edx,%rax
  movzbl (%rbx,%rax,1),%eax
  test   %al,%al
  je     L138
  mov    $0x1,%edi
  mov    $0x0,%ecx
  jmp    L105
L17:
  mov    $0x0,%edx
  jmp    L110
L109:
  add    $0x1,%edx
  mov    %edi,%r8d
  mov    %al,(%rbx,%rcx,1)
L30:
  movslq %edx,%rax
  movzbl (%rbx,%rax,1),%eax
  add    $0x1,%rcx
  add    $0x1,%edi
  test   %al,%al
  je     L108
L105:
  mov    %ecx,%r8d
  cmp    $0x20,%al
  jne    L109
  movslq %edx,%rax
  movzbl (%rbx,%rax,1),%esi
  cmp    $0x20,%sil
  jne    L8
  add    $0x1,%edx
  movslq %edx,%rax
L74:
  mov    %eax,%edx
  add    $0x1,%rax
  movzbl -0x1(%rbx,%rax,1),%esi
  cmp    $0x20,%sil
  je     L74
L8:
  test   %sil,%sil
  je     L108
  mov    %edi,%r8d
  movb   $0x20,(%rbx,%rcx,1)
  jmp    L30
L138:
  mov    $0x0,%r8d
L108:
  movslq %r8d,%r8
  movb   $0x0,(%rbx,%r8,1)
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <reverse@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  je     L49
  mov    $0x0,%ebp
  mov    $0x0,%esi
  jmp    L60
L141:
  add    $0x1,%ebp
  movslq %ebp,%rax
  movzbl (%rbx,%rax,1),%eax
  test   %al,%al
  je     L102
L60:
  cmp    $0x20,%al
  jne    L141
  lea    -0x1(%rbp),%edx
  mov    %rbx,%rdi
  call   <reverse@plt>
  lea    0x1(%rbp),%esi
  mov    %esi,%ebp
  jmp    L141
L49:
  mov    $0x0,%ebp
  mov    $0x0,%esi
L102:
  lea    -0x1(%rbp),%edx
  mov    %rbx,%rdi
  call   <reverse@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void reverseWords(char *s) {
  int i = 0;
  int j = 0;
  while (s[i] == ' ' && s[i] != '\0') {
    i++;
  }
  while (s[i] != '\0') {
    if (s[i] == ' ') {
      while (s[i] == ' ' && s[i] != '\0')
        i++;
      if (s[i] == '\0') {
        break;
      } else {
        s[j++] = ' ';
      }
    } else {
      s[j++] = s[i++];
    }
  }
  s[j] = '\0';
  i = 0;
  j = strlen(s) - 1;
  reverse(s, i, j);
  j = 0;
  while (s[j] != '\0') {
    if (s[j] == ' ') {
      reverse(s, i, j - 1);
      i = j + 1;
      j++;
    }
    j++;
  }
  reverse(s, i, j - 1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<reverseWords>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  movzbl (%rdi),%edx
  cmp    $0x20,%dl
  jne    L163
  mov    $0x1,%eax
  nop
L16:
  mov    %eax,%ecx
  add    $0x1,%rax
  movzbl -0x1(%rbp,%rax,1),%edx
  cmp    $0x20,%dl
  je     L16
L10:
  test   %dl,%dl
  je     L53
  mov    %rbp,%rsi
  mov    $0x1,%r8d
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L56:
  add    $0x1,%ecx
L126:
  movslq %ecx,%rax
  mov    %dl,(%rsi)
  movslq %r8d,%r9
  add    $0x1,%rsi
  movzbl 0x0(%rbp,%rax,1),%edx
  add    $0x1,%r8d
  test   %dl,%dl
  je     L137
L1:
  mov    %rsi,%r9
  cmp    $0x20,%dl
  jne    L56
  movslq %ecx,%rax
  movzbl 0x0(%rbp,%rax,1),%edi
  cmp    $0x20,%dil
  jne    L160
  add    $0x1,%ecx
  movslq %ecx,%rax
  nopl   0x0(%rax)
L116:
  mov    %eax,%ecx
  add    $0x1,%rax
  movzbl -0x1(%rbp,%rax,1),%edi
  cmp    $0x20,%dil
  je     L116
L160:
  test   %dil,%dil
  jne    L126
  xchg   %ax,%ax
L146:
  movb   $0x0,(%r9)
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  lea    -0x1(%rax),%edx
  call   <reverse@plt>
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L37
  xor    %edx,%edx
  xor    %esi,%esi
  jmp    L70
  nopw   0x0(%rax,%rax,1)
L118:
  lea    0x1(%r12),%edx
  movslq %edx,%rax
  movzbl 0x0(%rbp,%rax,1),%eax
  test   %al,%al
  je     L119
L70:
  mov    %edx,%r12d
  cmp    $0x20,%al
  jne    L118
  add    $0x1,%r12d
  sub    $0x1,%edx
  mov    %rbp,%rdi
  call   <reverse@plt>
  mov    %r12d,%esi
  jmp    L118
L37:
  mov    $0xffffffff,%r12d
  xor    %esi,%esi
L119:
  add    $0x8,%rsp
  mov    %r12d,%edx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <reverse@plt>
  cs nopw 0x0(%rax,%rax,1)
L137:
  add    %rbp,%r9
  jmp    L146
L163:
  xor    %ecx,%ecx
  jmp    L10
L53:
  mov    %rbp,%r9
  jmp    L146
```

assistant: ```c
void reverseWords(char *s) {
  int i = 0;
  int j = 0;
  while (s[i] == ' ' && s[i] != '\0') {
    i++;
  }
  while (s[i] != '\0') {
    if (s[i] == ' ') {
      while (s[i] == ' ' && s[i] != '\0')
        i++;
      if (s[i] == '\0') {
        break;
      } else {
        s[j++] = ' ';
      }
    } else {
      s[j++] = s[i++];
    }
  }
  s[j] = '\0';
  i = 0;
  j = strlen(s) - 1;
  reverse(s, i, j);
  j = 0;
  while (s[j] != '\0') {
    if (s[j] == ' ') {
      reverse(s, i, j - 1);
      i = j + 1;
      j++;
    }
    j++;
  }
  reverse(s, i, j - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<reverseWords>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movzbl (%rdi),%ecx
  cmp    $0x20,%cl
  jne    L166
  mov    $0x1,%eax
  nopl   0x0(%rax)
L117:
  mov    %eax,%edx
  add    $0x1,%rax
  movzbl -0x1(%rbp,%rax,1),%ecx
  cmp    $0x20,%cl
  je     L117
L168:
  test   %cl,%cl
  je     L67
  mov    %rbp,%rsi
  mov    $0x1,%r8d
  jmp    L134
  nopl   0x0(%rax,%rax,1)
L128:
  add    $0x1,%edx
  movslq %r8d,%r10
  movslq %edx,%rax
  add    %rbp,%rax
L156:
  mov    %cl,(%rsi)
  movzbl (%rax),%ecx
  add    $0x1,%r8d
  add    $0x1,%rsi
  test   %cl,%cl
  je     L159
L134:
  mov    %rsi,%r10
  cmp    $0x20,%cl
  jne    L128
  movslq %edx,%rax
  lea    0x1(%rbp,%rax,1),%rdi
  nopl   (%rax)
L4:
  mov    %rdi,%rax
  movzbl (%rdi),%r9d
  lea    0x1(%rdi),%rdi
  add    $0x1,%edx
  cmp    $0x20,%r9b
  je     L4
  test   %r9b,%r9b
  je     L150
  movslq %r8d,%r10
  jmp    L156
  cs nopw 0x0(%rax,%rax,1)
L159:
  add    %rbp,%r10
L150:
  movb   $0x0,(%r10)
  mov    %rbp,%rdi
  call   <strlen@plt>
  xor    %esi,%esi
  mov    %rbp,%rdi
  lea    -0x1(%rax),%edx
  call   <reverse@plt>
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L107
  xor    %r12d,%r12d
  xor    %esi,%esi
  jmp    L158
  nopl   0x0(%rax,%rax,1)
L55:
  lea    0x1(%r12),%edx
  movslq %edx,%rax
  movzbl 0x0(%rbp,%rax,1),%eax
  test   %al,%al
  je     L147
L125:
  mov    %edx,%r12d
L158:
  cmp    $0x20,%al
  jne    L55
  lea    -0x1(%r12),%edx
  mov    %rbp,%rdi
  lea    0x1(%r12),%ebx
  call   <reverse@plt>
  lea    0x2(%r12),%edx
  movslq %edx,%rax
  movzbl 0x0(%rbp,%rax,1),%eax
  test   %al,%al
  je     L163
  mov    %ebx,%esi
  jmp    L125
L163:
  mov    %ebx,%r12d
  mov    %ebx,%esi
L147:
  pop    %rbx
  mov    %r12d,%edx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <reverse@plt>
L166:
  xor    %edx,%edx
  jmp    L168
L107:
  mov    $0xffffffff,%r12d
  pop    %rbx
  mov    %rbp,%rdi
  xor    %esi,%esi
  mov    %r12d,%edx
  pop    %rbp
  pop    %r12
  jmp    1060 <reverse@plt>
L67:
  mov    %rbp,%r10
  jmp    L150
```

assistant: ```c
void reverseWords(char *s) {
  int i = 0;
  int j = 0;
  while (s[i] == ' ' && s[i] != '\0') {
    i++;
  }
  while (s[i] != '\0') {
    if (s[i] == ' ') {
      while (s[i] == ' ' && s[i] != '\0')
        i++;
      if (s[i] == '\0') {
        break;
      } else {
        s[j++] = ' ';
      }
    } else {
      s[j++] = s[i++];
    }
  }
  s[j] = '\0';
  i = 0;
  j = strlen(s) - 1;
  reverse(s, i, j);
  j = 0;
  while (s[j] != '\0') {
    if (s[j] == ' ') {
      reverse(s, i, j - 1);
      i = j + 1;
      j++;
    }
    j++;
  }
  reverse(s, i, j - 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<test2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <dummy@plt>
  mov    0x8(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void __attribute__((no)) __attribute__((noclone)) * test2(void) {
  void *temp;
  dummy();
  temp = __builtin_return_address(0);
  return temp;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test2>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <dummy@plt>
  mov    0x8(%rsp),%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
void __attribute__((no)) __attribute__((noclone)) * test2(void) {
  void *temp;
  dummy();
  temp = __builtin_return_address(0);
  return temp;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test2>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <dummy@plt>
  mov    0x8(%rsp),%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
void __attribute__((no)) __attribute__((noclone)) * test2(void) {
  void *temp;
  dummy();
  temp = __builtin_return_address(0);
  return temp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test2>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <dummy@plt>
  mov    0x8(%rsp),%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
void __attribute__((no)) __attribute__((noclone)) * test2(void) {
  void *temp;
  dummy();
  temp = __builtin_return_address(0);
  return temp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getpwnam>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x0,%eax
  call   <endpwent@plt>
  nop
L1:
  mov    $0x0,%eax
  call   <getpwent@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L12
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L1
L12:
  mov    $0x0,%eax
  call   <endpwent@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
struct passwd *getpwnam(const char *name) {
  struct passwd *pw;
  endpwent();
  while ((pw = getpwent()) != NULL && strcmp(pw->pw_name, name) != 0) {
  }
  endpwent();
  return pw;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getpwnam>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x0,%eax
  call   <endpwent@plt>
L17:
  mov    $0x0,%eax
  call   <getpwent@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L2
  mov    %rbp,%rsi
  mov    (%rax),%edi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L17
L2:
  mov    $0x0,%eax
  call   <endpwent@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
struct passwd *getpwnam(const char *name) {
  struct passwd *pw;
  endpwent();
  while ((pw = getpwent()) != NULL && strcmp(pw->pw_name, name) != 0) {
  }
  endpwent();
  return pw;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getpwnam>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <endpwent@plt>
  jmp    L23
  nopw   0x0(%rax,%rax,1)
L15:
  mov    (%rax),%edi
  mov    %rbx,%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  je     L13
L23:
  xor    %eax,%eax
  call   <getpwent@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L15
L13:
  xor    %eax,%eax
  call   <endpwent@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
struct passwd *getpwnam(const char *name) {
  struct passwd *pw;
  endpwent();
  while ((pw = getpwent()) != NULL && strcmp(pw->pw_name, name) != 0) {
  }
  endpwent();
  return pw;
}
```