ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ieee_80_to_double>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movzbl (%rdi),%edx
  shl    $0x8,%edx
  movzbl 0x1(%rdi),%ebx
  or     %edx,%ebx
  mov    %ebx,%r12d
  sar    $0xf,%r12w
  and    $0x1,%r12d
  and    $0x7fff,%bx
  movzbl 0x2(%rdi),%edx
  shl    $0x8,%edx
  movzbl 0x3(%rdi),%ecx
  or     %edx,%ecx
  shl    $0x8,%ecx
  movzbl 0x4(%rdi),%edx
  or     %ecx,%edx
  shl    $0x8,%edx
  movzbl 0x5(%rdi),%ebp
  or     %edx,%ebp
  movzbl 0x6(%rdi),%edx
  shl    $0x8,%edx
  movzbl 0x7(%rdi),%ecx
  or     %edx,%ecx
  shl    $0x8,%ecx
  movzbl 0x8(%rdi),%edx
  or     %ecx,%edx
  shl    $0x8,%edx
  movzbl 0x9(%rdi),%edi
  or     %edx,%edi
  mov    %ebp,%eax
  or     %edi,%eax
  jne    L37
  test   %bx,%bx
  jne    L37
  pxor   %xmm0,%xmm0
  test   %r12b,%r12b
  jne    L37
L36:
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L37:
  call   <myUlongToDouble@plt>
  mulsd  D20(%rip),%xmm0
  movsd  %xmm0,0x8(%rsp)
  mov    %ebp,%edi
  call   <myUlongToDouble@plt>
  mulsd  D35(%rip),%xmm0
  movapd %xmm0,%xmm3
  addsd  0x8(%rsp),%xmm3
  movsd  %xmm3,0x8(%rsp)
  movswl %bx,%ebx
  pxor   %xmm1,%xmm1
  cvtsi2sd %ebx,%xmm1
  subsd  D8(%rip),%xmm1
  movsd  D37(%rip),%xmm0
  call   <pow@plt>
  mulsd  0x8(%rsp),%xmm0
  test   %r12b,%r12b
  je     L36
  xorpd  D10(%rip),%xmm0
  jmp    L36
```

tool_calls: name: extract_symbolic_data
arguments: label: D20
type_infered: f32[2]
name: extract_symbolic_data
arguments: label: D35
type_infered: f32[2]
name: extract_symbolic_data
arguments: label: D8
type_infered: f64
name: extract_symbolic_data
arguments: label: D37
type_infered: f64
name: extract_symbolic_data
arguments: label: D10
type_infered: f32[4]

tool: {"D20": [0.0,0.0078125]}

tool: {"D35": [0.0,0.125]}

tool: D8: 16383

tool: D37: 2

tool: {"D10": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double ieee_80_to_double(unsigned char *p) {
  unsigned char sign;
  short lexp = 0;
  unsigned int mant1 = 0;
  unsigned int mant0 = 0;
  double val;
  lexp = *p++;
  lexp <<= 8;
  lexp |= *p++;
  sign = (lexp & 0x8000) ? 1 : 0;
  lexp &= 0x7FFF;
  mant1 = *p++;
  mant1 <<= 8;
  mant1 |= *p++;
  mant1 <<= 8;
  mant1 |= *p++;
  mant1 <<= 8;
  mant1 |= *p++;
  mant0 = *p++;
  mant0 <<= 8;
  mant0 |= *p++;
  mant0 <<= 8;
  mant0 |= *p++;
  mant0 <<= 8;
  mant0 |= *p++;
  if (mant1 == 0 && mant0 == 0 && lexp == 0 && sign == 0)
    return 0.0;
  else {
    val = myUlongToDouble(mant0) * pow(2.0, -63.0);
    val += myUlongToDouble(mant1) * pow(2.0, -31.0);
    val *= pow(2.0, ((double)lexp) - 16383.0);
    return sign ? -val : val;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ieee_80_to_double>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movzwl (%rdi),%eax
  mov    0x2(%rdi),%ebp
  mov    0x6(%rdi),%edi
  rol    $0x8,%ax
  bswap  %ebp
  mov    %eax,%r12d
  and    $0x7fff,%ax
  bswap  %edi
  sar    $0xf,%r12w
  mov    %eax,%ebx
  mov    %ebp,%eax
  and    $0x1,%r12d
  or     %edi,%eax
  jne    L19
  test   %bx,%bx
  jne    L19
  pxor   %xmm0,%xmm0
  test   %r12b,%r12b
  jne    L19
L34:
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L19:
  call   <myUlongToDouble@plt>
  mulsd  D37(%rip),%xmm0
  mov    %ebp,%edi
  movsd  %xmm0,0x8(%rsp)
  call   <myUlongToDouble@plt>
  mulsd  D46(%rip),%xmm0
  movswl %bx,%eax
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm3
  cvtsi2sd %eax,%xmm1
  subsd  D45(%rip),%xmm1
  addsd  %xmm0,%xmm3
  movsd  D23(%rip),%xmm0
  movsd  %xmm3,0x8(%rsp)
  call   <pow@plt>
  mulsd  0x8(%rsp),%xmm0
  test   %r12b,%r12b
  je     L34
  xorpd  D34(%rip),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_label: D37
type_infered: f32[2]
name: read_data
arguments: data_label: D46
type_infered: f32[2]
name: read_data
arguments: data_label: D45
type_infered: f64
name: read_data
arguments: data_label: D23
type_infered: f64
name: read_data
arguments: data_label: D34
type_infered: f32[4]

tool: {"D37": [0.0,0.0078125]}

tool: {"D46": [0.0,0.125]}

tool: D45: 16383

tool: D23: 2

tool: {"D34": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double ieee_80_to_double(unsigned char *p) {
  unsigned char sign;
  short lexp = 0;
  unsigned int mant1 = 0;
  unsigned int mant0 = 0;
  double val;
  lexp = *p++;
  lexp <<= 8;
  lexp |= *p++;
  sign = (lexp & 0x8000) ? 1 : 0;
  lexp &= 0x7FFF;
  mant1 = *p++;
  mant1 <<= 8;
  mant1 |= *p++;
  mant1 <<= 8;
  mant1 |= *p++;
  mant1 <<= 8;
  mant1 |= *p++;
  mant0 = *p++;
  mant0 <<= 8;
  mant0 |= *p++;
  mant0 <<= 8;
  mant0 |= *p++;
  mant0 <<= 8;
  mant0 |= *p++;
  if (mant1 == 0 && mant0 == 0 && lexp == 0 && sign == 0)
    return 0.0;
  else {
    val = myUlongToDouble(mant0) * pow(2.0, -63.0);
    val += myUlongToDouble(mant1) * pow(2.0, -31.0);
    val *= pow(2.0, ((double)lexp) - 16383.0);
    return sign ? -val : val;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ieee_80_to_double>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  movzwl (%rdi),%eax
  mov    0x2(%rdi),%ebp
  mov    0x6(%rdi),%edi
  rol    $0x8,%ax
  bswap  %ebp
  mov    %eax,%r12d
  and    $0x7fff,%ax
  bswap  %edi
  sar    $0xf,%r12w
  mov    %eax,%ebx
  mov    %ebp,%eax
  and    $0x1,%r12d
  or     %edi,%eax
  jne    L38
  test   %bx,%bx
  jne    L38
  pxor   %xmm0,%xmm0
  test   %r12b,%r12b
  jne    L38
L2:
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L38:
  call   <myUlongToDouble@plt>
  mulsd  D26(%rip),%xmm0
  mov    %ebp,%edi
  movsd  %xmm0,0x8(%rsp)
  call   <myUlongToDouble@plt>
  mulsd  D20(%rip),%xmm0
  movswl %bx,%eax
  pxor   %xmm1,%xmm1
  movsd  0x8(%rsp),%xmm3
  cvtsi2sd %eax,%xmm1
  subsd  D44(%rip),%xmm1
  addsd  %xmm0,%xmm3
  movsd  D23(%rip),%xmm0
  movsd  %xmm3,0x8(%rsp)
  call   <pow@plt>
  mulsd  0x8(%rsp),%xmm0
  test   %r12b,%r12b
  je     L2
  xorpd  D41(%rip),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D26
datatype: f32[2]
name: retrieve_data
arguments: tag: D20
datatype: f32[2]
name: retrieve_data
arguments: tag: D44
datatype: f64
name: retrieve_data
arguments: tag: D23
datatype: f64
name: retrieve_data
arguments: tag: D41
datatype: f32[4]

tool: {"D26": [0.0,0.0078125]}

tool: {"D20": [0.0,0.125]}

tool: D44: 16383

tool: D23: 2

tool: {"D41": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double ieee_80_to_double(unsigned char *p) {
  unsigned char sign;
  short lexp = 0;
  unsigned int mant1 = 0;
  unsigned int mant0 = 0;
  double val;
  lexp = *p++;
  lexp <<= 8;
  lexp |= *p++;
  sign = (lexp & 0x8000) ? 1 : 0;
  lexp &= 0x7FFF;
  mant1 = *p++;
  mant1 <<= 8;
  mant1 |= *p++;
  mant1 <<= 8;
  mant1 |= *p++;
  mant1 <<= 8;
  mant1 |= *p++;
  mant0 = *p++;
  mant0 <<= 8;
  mant0 |= *p++;
  mant0 <<= 8;
  mant0 |= *p++;
  mant0 <<= 8;
  mant0 |= *p++;
  if (mant1 == 0 && mant0 == 0 && lexp == 0 && sign == 0)
    return 0.0;
  else {
    val = myUlongToDouble(mant0) * pow(2.0, -63.0);
    val += myUlongToDouble(mant1) * pow(2.0, -31.0);
    val *= pow(2.0, ((double)lexp) - 16383.0);
    return sign ? -val : val;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<f0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  sub    $0xa,%eax
  cmp    $0xfffffff5,%eax
  jbe    L5
  addl   $0x1,-0x4(%rbp)
L5:
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int f0(unsigned x) {
  switch (x) {
  default:
    x += 1;
  case 10 ... 0xFFFFFFFF:
    return 0;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<f0>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int f0(unsigned x) {
  switch (x) {
  default:
    x += 1;
  case 10 ... 0xFFFFFFFF:
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<f0>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int f0(unsigned x) {
  switch (x) {
  default:
    x += 1;
  case 10 ... 0xFFFFFFFF:
    return 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
<f0>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int f0(unsigned x) {
  switch (x) {
  default:
    x += 1;
  case 10 ... 0xFFFFFFFF:
    return 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <ubaci@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D29
datatype: string
name: extract_symbolic_data
arguments: data_tag: D27
datatype: string
name: extract_symbolic_data
arguments: data_tag: D14
datatype: string

tool: D29: Unesite n i a: 

tool: D27: %lf %d

tool: D14: %f

assistant: ```c
int main() {
  int n;
  int a;
  printf("Unesite n i a: ");
  scanf("%lf %d", &n, &a);
  printf("%f", ubaci(n, a));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D27(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  mov    (%rsp),%edi
  call   <ubaci@plt>
  lea    D16(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D27
type: string
name: retrieve_data
arguments: tag: D22
type: string
name: retrieve_data
arguments: tag: D16
type: string

tool: D27: Unesite n i a: 

tool: D22: %lf %d

tool: D16: %f

assistant: ```c
int main() {
  int n;
  int a;
  printf("Unesite n i a: ");
  scanf("%lf %d", &n, &a);
  printf("%f", ubaci(n, a));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D12(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%edi
  mov    0x4(%rsp),%esi
  call   <ubaci@plt>
  lea    D28(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D12
data_type: string
name: parse_data
arguments: data_tag: D15
data_type: string
name: parse_data
arguments: data_tag: D28
data_type: string

tool: D12: Unesite n i a: 

tool: D15: %lf %d

tool: D28: %f

assistant: ```c
int main() {
  int n;
  int a;
  printf("Unesite n i a: ");
  scanf("%lf %d", &n, &a);
  printf("%f", ubaci(n, a));
  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
  sub    $0x18,%rsp
  lea    D29(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%edi
  mov    0x4(%rsp),%esi
  call   <ubaci@plt>
  lea    D13(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D29
type_infered: string
name: read_data
arguments: data_tag: D18
type_infered: string
name: read_data
arguments: data_tag: D13
type_infered: string

tool: D29: Unesite n i a: 

tool: D18: %lf %d

tool: D13: %f

assistant: ```c
int main() {
  int n;
  int a;
  printf("Unesite n i a: ");
  scanf("%lf %d", &n, &a);
  printf("%f", ubaci(n, a));
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<elf_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L6
L25:
  mov    -0x10(%rbp),%rax
  shl    $0x4,%rax
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %rcx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  and    $0xf0000000,%eax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L22
  mov    -0x8(%rbp),%rax
  shr    $0x18,%rax
  xor    %rax,-0x10(%rbp)
L22:
  mov    -0x8(%rbp),%rax
  not    %rax
  and    %rax,-0x10(%rbp)
L6:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L25
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long elf_hash(const unsigned char *name) {
  unsigned long h = 0, g;
  while (*name) {
    h = (h << 4) + *name++;
    if (g = h & 0xf0000000)
      h ^= g >> 24;
    h &= ~g;
  }
  return h;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<elf_hash>:
  endbr64
  movzbl (%rdi),%ecx
  add    $0x1,%rdi
  movzbl %cl,%edx
  mov    $0x0,%eax
  test   %cl,%cl
  jne    L4
  ret
L14:
  shl    $0x4,%rax
  add    $0x1,%rdi
  movzbl %dl,%edx
  add    %rax,%rdx
  mov    %rdx,%rax
  and    $0xf0000000,%eax
  mov    %rax,%rcx
  shr    $0x18,%rcx
  xor    %rdx,%rcx
  test   %rax,%rax
  cmovne %rcx,%rdx
L4:
  not    %rax
  and    %rdx,%rax
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L14
  ret
```

assistant: ```c
unsigned long elf_hash(const unsigned char *name) {
  unsigned long h = 0, g;
  while (*name) {
    h = (h << 4) + *name++;
    if (g = h & 0xf0000000)
      h ^= g >> 24;
    h &= ~g;
  }
  return h;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<elf_hash>:
  endbr64
  movzbl (%rdi),%edx
  xor    %eax,%eax
  add    $0x1,%rdi
  test   %dl,%dl
  jne    L28
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L3:
  shl    $0x4,%rax
  add    $0x1,%rdi
  add    %rax,%rdx
  mov    %rdx,%rax
  and    $0xf0000000,%eax
  mov    %rax,%rcx
  shr    $0x18,%rcx
  xor    %rdx,%rcx
  test   %rax,%rax
  cmovne %rcx,%rdx
L28:
  not    %rax
  and    %rdx,%rax
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L3
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  ret
```

assistant: ```c
unsigned long elf_hash(const unsigned char *name) {
  unsigned long h = 0, g;
  while (*name) {
    h = (h << 4) + *name++;
    if (g = h & 0xf0000000)
      h ^= g >> 24;
    h &= ~g;
  }
  return h;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<elf_hash>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L42
  movzbl 0x1(%rdi),%edx
  lea    0x1(%rdi),%rcx
  test   %dl,%dl
  je     L13
  nopw   0x0(%rax,%rax,1)
L23:
  shl    $0x4,%rax
  add    $0x1,%rcx
  add    %rdx,%rax
  mov    %rax,%rdx
  and    $0xf0000000,%edx
  je     L31
  mov    %rdx,%rsi
  not    %rdx
  shr    $0x18,%rsi
  xor    %rsi,%rax
  and    %rdx,%rax
  movzbl (%rcx),%edx
  test   %dl,%dl
  jne    L23
  ret
  xchg   %ax,%ax
L31:
  movzbl (%rcx),%edx
  test   %dl,%dl
  jne    L23
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  ret
  nopl   0x0(%rax)
L42:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned long elf_hash(const unsigned char *name) {
  unsigned long h = 0, g;
  while (*name) {
    h = (h << 4) + *name++;
    if (g = h & 0xf0000000)
      h ^= g >> 24;
    h &= ~g;
  }
  return h;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x5,-0x4(%rbp)
  jmp    L6
L7:
  mov    $0x0,%eax
  call   <load_and_test@plt>
  test   %rax,%rax
  je     L6
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  jmp    L15
L6:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jg     L7
  mov    D7(%rip),%rax
  mov    (%rax),%eax
L15:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int count = 5;
  while (count-- > 0)
    if (load_and_test())
      return EXIT_FAILURE;
  return EXIT_SUCCESS;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x5,%ebx
L23:
  mov    $0x0,%eax
  call   <load_and_test@plt>
  test   %rax,%rax
  jne    L3
  sub    $0x1,%ebx
  jne    L23
  mov    D5(%rip),%rax
  mov    (%rax),%eax
L15:
  pop    %rbx
  ret
L3:
  mov    D16(%rip),%rax
  mov    (%rax),%eax
  jmp    L15
```

assistant: ```c
int main(int argc, char **argv) {
  int count = 5;
  while (count-- > 0)
    if (load_and_test())
      return EXIT_FAILURE;
  return EXIT_SUCCESS;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x5,%ebx
L15:
  xor    %eax,%eax
  call   <load_and_test@plt>
  test   %rax,%rax
  jne    L4
  sub    $0x1,%ebx
  jne    L15
  mov    D10(%rip),%rax
  pop    %rbx
  mov    (%rax),%eax
  ret
L4:
  mov    D14(%rip),%rax
  pop    %rbx
  mov    (%rax),%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int count = 5;
  while (count-- > 0)
    if (load_and_test())
      return EXIT_FAILURE;
  return EXIT_SUCCESS;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x5,%ebx
L13:
  xor    %eax,%eax
  call   <load_and_test@plt>
  test   %rax,%rax
  jne    L15
  sub    $0x1,%ebx
  jne    L13
  mov    D3(%rip),%rax
  pop    %rbx
  mov    (%rax),%eax
  ret
L15:
  mov    D1(%rip),%rax
  pop    %rbx
  mov    (%rax),%eax
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int count = 5;
  while (count-- > 0)
    if (load_and_test())
      return EXIT_FAILURE;
  return EXIT_SUCCESS;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<memcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L16
L19:
  mov    -0x14(%rbp),%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%ecx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x14(%rbp)
L16:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jb     L19
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *memcpy(void *out, void *in, unsigned int bytes) {
  unsigned int i;
  char *b1 = (char *)out;
  char *b2 = (char *)in;
  for (i = 0; i < bytes; i++) {
    b1[i] = b2[i];
  }
  return b1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<memcpy>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  je     L10
  mov    %edx,%edx
  mov    $0x0,%ecx
L11:
  movzbl (%rsi,%rcx,1),%r8d
  mov    %r8b,(%rax,%rcx,1)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L11
L10:
  ret
```

assistant: ```c
void *memcpy(void *out, void *in, unsigned int bytes) {
  unsigned int i;
  char *b1 = (char *)out;
  char *b2 = (char *)in;
  for (i = 0; i < bytes; i++) {
    b1[i] = b2[i];
  }
  return b1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memcpy>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  je     L11
  mov    %edx,%edx
  xor    %ecx,%ecx
  nop
L3:
  movzbl (%rsi,%rcx,1),%r8d
  mov    %r8b,(%rax,%rcx,1)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L3
L11:
  ret
```

assistant: ```c
void *memcpy(void *out, void *in, unsigned int bytes) {
  unsigned int i;
  char *b1 = (char *)out;
  char *b2 = (char *)in;
  for (i = 0; i < bytes; i++) {
    b1[i] = b2[i];
  }
  return b1;
}
```
user: What is the c source code of the assembly code below?

```asm
<memcpy>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rcx
  test   %edx,%edx
  je     L23
  lea    0x1(%rsi),%rdi
  mov    %r8,%rsi
  lea    -0x1(%rdx),%eax
  sub    %rdi,%rsi
  cmp    $0xe,%rsi
  jbe    L100
  cmp    $0x6,%eax
  jbe    L100
  mov    %edx,%esi
  cmp    $0xe,%eax
  jbe    L111
  shr    $0x4,%esi
  xor    %eax,%eax
  shl    $0x4,%rsi
  cs nopw 0x0(%rax,%rax,1)
L43:
  movdqu (%rcx,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rsi,%rax
  jne    L43
  mov    %edx,%eax
  and    $0xfffffff0,%eax
  test   $0xf,%dl
  je     L23
  mov    %edx,%esi
  sub    %eax,%esi
  lea    -0x1(%rsi),%edi
  cmp    $0x6,%edi
  jbe    L24
L52:
  mov    %eax,%edi
  mov    (%rcx,%rdi,1),%r9
  mov    %r9,(%r8,%rdi,1)
  mov    %esi,%edi
  and    $0xfffffff8,%edi
  add    %edi,%eax
  cmp    %edi,%esi
  je     L23
L24:
  mov    %eax,%esi
  movzbl (%rcx,%rsi,1),%edi
  mov    %dil,(%r8,%rsi,1)
  lea    0x1(%rax),%esi
  cmp    %esi,%edx
  jbe    L23
  movzbl (%rcx,%rsi,1),%edi
  mov    %dil,(%r8,%rsi,1)
  lea    0x2(%rax),%esi
  cmp    %esi,%edx
  jbe    L23
  movzbl (%rcx,%rsi,1),%edi
  mov    %dil,(%r8,%rsi,1)
  lea    0x3(%rax),%esi
  cmp    %esi,%edx
  jbe    L23
  movzbl (%rcx,%rsi,1),%edi
  mov    %dil,(%r8,%rsi,1)
  lea    0x4(%rax),%esi
  cmp    %esi,%edx
  jbe    L23
  movzbl (%rcx,%rsi,1),%edi
  mov    %dil,(%r8,%rsi,1)
  lea    0x5(%rax),%esi
  cmp    %esi,%edx
  jbe    L23
  movzbl (%rcx,%rsi,1),%edi
  add    $0x6,%eax
  mov    %dil,(%r8,%rsi,1)
  cmp    %eax,%edx
  jbe    L23
  movzbl (%rcx,%rax,1),%edx
  mov    %dl,(%r8,%rax,1)
L23:
  mov    %r8,%rax
  ret
  nopw   0x0(%rax,%rax,1)
L100:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L58:
  movzbl (%rcx,%rax,1),%esi
  mov    %sil,(%r8,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L58
  mov    %r8,%rax
  ret
L111:
  xor    %eax,%eax
  jmp    L52
```

assistant: ```c
void *memcpy(void *out, void *in, unsigned int bytes) {
  unsigned int i;
  char *b1 = (char *)out;
  char *b2 = (char *)in;
  for (i = 0; i < bytes; i++) {
    b1[i] = b2[i];
  }
  return b1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<pevent_read_number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x8,-0x24(%rbp)
  je     L50
  cmpl   $0x8,-0x24(%rbp)
  jg     L89
  cmpl   $0x4,-0x24(%rbp)
  je     L49
  cmpl   $0x4,-0x24(%rbp)
  jg     L89
  cmpl   $0x1,-0x24(%rbp)
  je     L20
  cmpl   $0x2,-0x24(%rbp)
  je     L53
  jmp    L89
L20:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  jmp    L109
L53:
  mov    -0x20(%rbp),%rax
  movzwl (%rax),%eax
  movzwl %ax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <__data2host2@plt>
  jmp    L109
L49:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <__data2host4@plt>
  jmp    L109
L50:
  mov    -0x20(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    $0x8,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <__data2host8@plt>
  cltq
  jmp    L109
L89:
  mov    $0x0,%eax
L109:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L51
  call   <__stack_chk_fail@plt>
L51:
  leave
  ret
```

assistant: ```c
unsigned long long pevent_read_number(struct pevent *pevent, const void *ptr, int size) {
  switch (size) {
  case 1:
    return *(unsigned char *)ptr;
  case 2:
    return __data2host2(pevent, *(unsigned short *)(ptr));
  case 4:
    return __data2host4(pevent, *(unsigned int *)(ptr));
  case 8:
    return ({
      unsigned long long __val;
      memcpy(&__val, (ptr), sizeof(unsigned long long));
      __data2host8(pevent, __val);
    });
  default:
    return 0;
  }
}
```