ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1a0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x32,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,-0x1ac(%rbp)
  jmp    L120
L72:
  lea    -0x1b0(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1b0(%rbp),%eax
  cmp    $0xffffffff,%eax
  je     L124
  movl   $0x0,-0x1a8(%rbp)
  jmp    L142
L3:
  addl   $0x1,-0x1a8(%rbp)
L142:
  mov    -0x1a8(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1b0(%rbp),%eax
  cmp    %eax,%edx
  jge    L107
  mov    -0x1a8(%rbp),%eax
  cmp    -0x1ac(%rbp),%eax
  jl     L3
L107:
  mov    -0x1a8(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1b0(%rbp),%eax
  cmp    %eax,%edx
  jne    L90
  mov    -0x1a8(%rbp),%eax
  cmp    -0x1ac(%rbp),%eax
  jge    L90
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L120
L90:
  mov    -0x1ac(%rbp),%eax
  mov    %eax,-0x1a4(%rbp)
  jmp    L15
L7:
  mov    -0x1a4(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  subl   $0x1,-0x1a4(%rbp)
L15:
  mov    -0x1a4(%rbp),%eax
  cmp    -0x1a8(%rbp),%eax
  jg     L7
  mov    -0x1b0(%rbp),%edx
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
  addl   $0x1,-0x1ac(%rbp)
  movl   $0x0,-0x1a4(%rbp)
  jmp    L88
L2:
  mov    -0x1a4(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x1a4(%rbp)
L88:
  mov    -0x1a4(%rbp),%eax
  cmp    -0x1ac(%rbp),%eax
  jl     L2
  mov    $0xa,%edi
  call   <putchar@plt>
L120:
  cmpl   $0x63,-0x1ac(%rbp)
  jle    L72
  jmp    L81
L124:
  nop
L81:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L84
  call   <__stack_chk_fail@plt>
L84:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D20
type: string
name: parse_data
arguments: tag: D19
type: string
name: parse_data
arguments: tag: D17
type: string

tool: D20: %d

tool: D19: duplicate entry

tool: D17: %d 

assistant: ```c
int main(void) {
  int data[100] = {
      0,
  };
  int size = 0;
  while (size < 100) {
    int input, i, j;
    {
      scanf("%d", &input);
      if (input == -1)
        break;
    }
    {
      for (i = 0; input > data[i] && i < size; ++i)
        ;
      if (input == data[i] && i < size) {
        printf("duplicate entry\n");
        continue;
      }
    }
    {
      for (j = size; j > i; --j)
        data[j] = data[j - 1];
      data[j] = input;
      ++size;
    }
    {
      for (j = 0; j < size; ++j)
        printf("%d ", data[j]);
      printf("\n");
    }
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  mov    $0x32,%ecx
  rep stos %rax,%es:(%rdi)
  mov    $0x0,%ebp
  lea    0xc(%rsp),%r14
  lea    D27(%rip),%r15
  lea    D3(%rip),%r12
  jmp    L71
L78:
  mov    $0x0,%edi
  jmp    L139
L46:
  cmp    %edi,%ebp
  jle    L72
  movslq %ebp,%rcx
  shl    $0x2,%rcx
  lea    0x10(%rsp,%rcx,1),%rax
  add    %r14,%rcx
  lea    -0x1(%rbp),%edx
  sub    %edi,%edx
  shl    $0x2,%rdx
  sub    %rdx,%rcx
L64:
  mov    -0x4(%rax),%edx
  mov    %edx,(%rax)
  sub    $0x4,%rax
  cmp    %rcx,%rax
  jne    L64
L27:
  movslq %edi,%rdi
  mov    %r8d,0x10(%rsp,%rdi,4)
  lea    0x1(%rbp),%r13d
  test   %r13d,%r13d
  jle    L121
  lea    0x10(%rsp),%rbx
  mov    %ebp,%ebp
  lea    0x14(%rsp,%rbp,4),%rbp
L145:
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L145
L121:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %r13d,%ebp
L144:
  cmp    $0x63,%ebp
  jg     L24
L71:
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%r8d
  cmp    $0xffffffff,%r8d
  je     L24
  mov    0x10(%rsp),%edx
  test   %ebp,%ebp
  jle    L78
  cmp    %edx,%r8d
  jle    L78
  mov    $0x1,%eax
L159:
  mov    %eax,%edi
  mov    0x10(%rsp,%rax,4),%edx
  cmp    %eax,%ebp
  setg   %sil
  cmp    %edx,%r8d
  setg   %cl
  add    $0x1,%rax
  test   %cl,%sil
  jne    L159
L139:
  cmp    %edi,%ebp
  jle    L46
  cmp    %edx,%r8d
  jne    L46
  lea    D17(%rip),%rdi
  call   <puts@plt>
  jmp    L144
L72:
  mov    %ebp,%edi
  jmp    L27
L24:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x1b8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D27
data_type: string
name: disassemble_data
arguments: tag: D3
data_type: string
name: disassemble_data
arguments: tag: D17
data_type: string

tool: D27: %d

tool: D3: %d 

tool: D17: duplicate entry

assistant: ```c
int main(void) {
  int data[100] = {
      0,
  };
  int size = 0;
  while (size < 100) {
    int input, i, j;
    {
      scanf("%d", &input);
      if (input == -1)
        break;
    }
    {
      for (i = 0; input > data[i] && i < size; ++i)
        ;
      if (input == data[i] && i < size) {
        printf("duplicate entry\n");
        continue;
      }
    }
    {
      for (j = size; j > i; --j)
        data[j] = data[j - 1];
      data[j] = input;
      ++size;
    }
    {
      for (j = 0; j < size; ++j)
        printf("%d ", data[j]);
      printf("\n");
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x32,%ecx
  push   %r14
  lea    D2(%rip),%r14
  push   %r13
  lea    D19(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x1d8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1c8(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rbp
  lea    0x2c(%rsp),%r15
  mov    %rbp,%rdi
  rep stos %rax,%es:(%rdi)
  lea    0x34(%rsp),%rax
  mov    %rax,0x18(%rsp)
  nopw   0x0(%rax,%rax,1)
L49:
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0x2c(%rsp),%r12d
  cmp    $0xffffffff,%r12d
  je     L115
  test   %ebx,%ebx
  mov    0x30(%rsp),%edx
  setg   %al
  mov    %eax,%esi
  cmp    %edx,%r12d
  jle    L134
  test   %al,%al
  je     L134
  mov    $0x1,%eax
  jmp    L140
  nopl   (%rax)
L129:
  test   %sil,%sil
  je     L4
L140:
  mov    0x0(%rbp,%rax,4),%edx
  cmp    %eax,%ebx
  mov    %eax,%ecx
  setg   %sil
  add    $0x1,%rax
  cmp    %edx,%r12d
  jg     L129
L4:
  cmp    %edx,%r12d
  jne    L35
  test   %sil,%sil
  jne    L77
L35:
  cmp    %ebx,%ecx
  jge    L43
  mov    %ebx,%eax
  movslq %ebx,%r8
  mov    %ecx,0x14(%rsp)
  sub    %ecx,%eax
  mov    %r8,0x8(%rsp)
  lea    -0x1(%rax),%edx
  mov    %rdx,%rax
  lea    0x4(,%rdx,4),%rdx
  neg    %rax
  shl    $0x2,%rax
  lea    -0x4(%rax,%r8,4),%rsi
  lea    (%rax,%r8,4),%rdi
  add    %rbp,%rsi
  add    %rbp,%rdi
  call   <memmove@plt>
  movslq 0x14(%rsp),%rcx
  mov    0x8(%rsp),%r8
L38:
  lea    0x1(%rbx),%eax
  mov    %r12d,0x30(%rsp,%rcx,4)
  mov    %rbp,%rbx
  mov    %eax,0x8(%rsp)
  mov    0x18(%rsp),%rax
  lea    (%rax,%r8,4),%r12
  nopl   0x0(%rax,%rax,1)
L75:
  mov    (%rbx),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L75
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%ebx
L116:
  cmp    $0x64,%ebx
  jne    L49
L115:
  mov    0x1c8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L131
  add    $0x1d8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L77:
  lea    D0(%rip),%rdi
  call   <puts@plt>
  jmp    L116
  cs nopw 0x0(%rax,%rax,1)
L134:
  xor    %ecx,%ecx
  jmp    L4
L43:
  movslq %ebx,%rcx
  movslq %ebx,%r8
  jmp    L38
L131:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D2
type: string
name: unpack_data
arguments: data_label: D19
type: string
name: unpack_data
arguments: data_label: D0
type: string

tool: D2: %d

tool: D19: %d 

tool: D0: duplicate entry

assistant: ```c
int main(void) {
  int data[100] = {
      0,
  };
  int size = 0;
  while (size < 100) {
    int input, i, j;
    {
      scanf("%d", &input);
      if (input == -1)
        break;
    }
    {
      for (i = 0; input > data[i] && i < size; ++i)
        ;
      if (input == data[i] && i < size) {
        printf("duplicate entry\n");
        continue;
      }
    }
    {
      for (j = size; j > i; --j)
        data[j] = data[j - 1];
      data[j] = input;
      ++size;
    }
    {
      for (j = 0; j < size; ++j)
        printf("%d ", data[j]);
      printf("\n");
    }
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<tgamma>:
  endbr64
  push   %r15
  push   %r14
  push   %rbx
  sub    $0x10,%rsp
  subsd  D29(%rip),%xmm0
  movsd  %xmm0,0x8(%rsp)
  addsd  D10(%rip),%xmm0
  addsd  D3(%rip),%xmm0
  movq   %xmm0,%rbx
  mov    $0x1,%eax
  movsd  D66(%rip),%xmm4
  movsd  %xmm4,(%rsp)
  lea    D36(%rip),%rdx
L8:
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  addsd  0x8(%rsp),%xmm1
  movsd  (%rdx,%rax,8),%xmm0
  divsd  %xmm1,%xmm0
  addsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  add    $0x1,%rax
  cmp    $0xb,%rax
  jne    L8
  mov    $0x6,%edi
  call   <sqrt@plt>
  movq   %xmm0,%r14
  movsd  0x8(%rsp),%xmm1
  addsd  D3(%rip),%xmm1
  movq   %rbx,%xmm0
  call   <pow@plt>
  movq   %xmm0,%r15
  movq   %rbx,%xmm0
  xorpd  D43(%rip),%xmm0
  call   <exp@plt>
  movapd %xmm0,%xmm1
  movq   %r14,%xmm0
  movq   %r15,%xmm5
  mulsd  %xmm5,%xmm0
  mulsd  %xmm1,%xmm0
  mulsd  (%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_label: D29
type: f64
name: parse_data
arguments: data_label: D10
type: f32[2]
name: parse_data
arguments: data_label: D3
type: f64
name: parse_data
arguments: data_label: D66
type: f64
name: parse_data
arguments: data_label: D36
type: f64
name: parse_data
arguments: data_label: D43
type: f32[4]

tool: D29: 1

tool: {"D10": [0.0,2.53125]}

tool: D3: 0.5

tool: D66: 1.0000000000000002

tool: D36: 1.0000000000000002

tool: {"D43": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double tgamma(double z) {
  static unsigned int c = 9;
  static double lanczos_coefficients[] = {1.000000000000000174663, 5716.400188274341379136, -14815.30426768413909044, 14291.49277657478554025, -6348.160217641458813289, 1301.608286058321874105, -108.1767053514369634679, 2.605696505611755827729, -0.7423452510201416151527e-2, 0.5384136432509564062961e-7, -0.4023533141268236372067e-8};
  double return_val = 0.0;
  z -= 1.0;
  double temp = z + c + 0.5;
  double A = lanczos_coefficients[0];
  int i = 0;
  for (i = 1; i < c + 2; ++i) {
    A += lanczos_coefficients[i] / (z + i);
  }
  return_val = sqrt(2 * 3.14159265358979323846) * pow(temp, z + 0.5) * exp(-temp) * A;
  return return_val;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<tgamma>:
  endbr64
  push   %rbx
  movapd %xmm0,%xmm1
  mov    $0x1,%eax
  lea    D65(%rip),%rdx
  sub    $0x20,%rsp
  subsd  D45(%rip),%xmm1
  movsd  D72(%rip),%xmm0
  movsd  D70(%rip),%xmm2
  movsd  D28(%rip),%xmm3
  addsd  %xmm1,%xmm0
  addsd  D7(%rip),%xmm0
  movq   %xmm0,%rbx
  jmp    L16
  nopl   0x0(%rax)
L18:
  movsd  (%rdx,%rax,8),%xmm2
L16:
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  add    $0x1,%rax
  addsd  %xmm1,%xmm0
  divsd  %xmm0,%xmm2
  addsd  %xmm2,%xmm3
  cmp    $0xb,%rax
  jne    L18
  mov    $0x6,%edi
  movsd  %xmm3,0x18(%rsp)
  movsd  %xmm1,0x10(%rsp)
  call   <sqrt@plt>
  movsd  0x10(%rsp),%xmm1
  addsd  D7(%rip),%xmm1
  movsd  %xmm0,0x8(%rsp)
  movq   %rbx,%xmm0
  call   <pow@plt>
  movq   %rbx,%xmm4
  xorpd  D22(%rip),%xmm4
  movsd  %xmm0,0x10(%rsp)
  movapd %xmm4,%xmm0
  call   <exp@plt>
  movsd  0x8(%rsp),%xmm5
  mulsd  0x10(%rsp),%xmm5
  movapd %xmm0,%xmm1
  movsd  0x18(%rsp),%xmm3
  add    $0x20,%rsp
  pop    %rbx
  movapd %xmm5,%xmm0
  mulsd  %xmm1,%xmm0
  mulsd  %xmm3,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: label: D65
type_guessed: f64
name: disassemble_data
arguments: label: D45
type_guessed: f64
name: disassemble_data
arguments: label: D72
type_guessed: f32[2]
name: disassemble_data
arguments: label: D70
type_guessed: f64
name: disassemble_data
arguments: label: D28
type_guessed: f64
name: disassemble_data
arguments: label: D7
type_guessed: f64
name: disassemble_data
arguments: label: D22
type_guessed: f32[4]

tool: D65: 1.0000000000000002

tool: D45: 1

tool: {"D72": [0.0,2.53125]}

tool: D70: 5716.400188274341

tool: D28: 1.0000000000000002

tool: D7: 0.5

tool: {"D22": [0.0,-0.0,0.0,0.0]}

assistant: ```c
double tgamma(double z) {
  static unsigned int c = 9;
  static double lanczos_coefficients[] = {1.000000000000000174663, 5716.400188274341379136, -14815.30426768413909044, 14291.49277657478554025, -6348.160217641458813289, 1301.608286058321874105, -108.1767053514369634679, 2.605696505611755827729, -0.7423452510201416151527e-2, 0.5384136432509564062961e-7, -0.4023533141268236372067e-8};
  double return_val = 0.0;
  z -= 1.0;
  double temp = z + c + 0.5;
  double A = lanczos_coefficients[0];
  int i = 0;
  for (i = 1; i < c + 2; ++i) {
    A += lanczos_coefficients[i] / (z + i);
  }
  return_val = sqrt(2 * 3.14159265358979323846) * pow(temp, z + 0.5) * exp(-temp) * A;
  return return_val;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isNewWord>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  je     L19
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L19
  mov    $0x1,%eax
  jmp    L17
L19:
  mov    $0x0,%eax
L17:
  leave
  ret
```

assistant: ```c
int isNewWord(char *text, size_t pos) {
  if (isspace(text[pos]) && !isspace(text[pos + 1]))
    return 1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<isNewWord>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  movsbl (%rdi,%rsi,1),%edi
  call   <isspace@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L5
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  movsbl 0x1(%rbp,%rbx,1),%edi
  call   <isspace@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L3
```

assistant: ```c
int isNewWord(char *text, size_t pos) {
  if (isspace(text[pos]) && !isspace(text[pos + 1]))
    return 1;
  else
    return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<isNewWord>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi,%rsi,1),%edi
  call   <isspace@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  je     L7
  movsbl 0x1(%rbp,%rbx,1),%edi
  call   <isspace@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L7:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int isNewWord(char *text, size_t pos) {
  if (isspace(text[pos]) && !isspace(text[pos + 1]))
    return 1;
  else
    return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<isNewWord>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi,%rsi,1),%edi
  call   <isspace@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  je     L8
  movsbl 0x1(%rbp,%rbx,1),%edi
  call   <isspace@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
L8:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int isNewWord(char *text, size_t pos) {
  if (isspace(text[pos]) && !isspace(text[pos + 1]))
    return 1;
  else
    return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<read_UInt32Number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %edx,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %edx,%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int read_UInt32Number(char *p) {
  unsigned int rv;
  rv = 16777216 * (unsigned int)((unsigned char *)p)[0] + 65536 * (unsigned int)((unsigned char *)p)[1] + 256 * (unsigned int)((unsigned char *)p)[2] + (unsigned int)((unsigned char *)p)[3];
  return rv;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<read_UInt32Number>:
  endbr64
  movzbl (%rdi),%eax
  shl    $0x8,%eax
  movzbl 0x1(%rdi),%edx
  add    %edx,%eax
  shl    $0x8,%eax
  movzbl 0x2(%rdi),%edx
  add    %edx,%eax
  shl    $0x8,%eax
  movzbl 0x3(%rdi),%edx
  add    %edx,%eax
  ret
```

assistant: ```c
unsigned int read_UInt32Number(char *p) {
  unsigned int rv;
  rv = 16777216 * (unsigned int)((unsigned char *)p)[0] + 65536 * (unsigned int)((unsigned char *)p)[1] + 256 * (unsigned int)((unsigned char *)p)[2] + (unsigned int)((unsigned char *)p)[3];
  return rv;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_UInt32Number>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%edx
  shl    $0x8,%eax
  add    %edx,%eax
  movzbl 0x2(%rdi),%edx
  shl    $0x8,%eax
  add    %edx,%eax
  movzbl 0x3(%rdi),%edx
  shl    $0x8,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
unsigned int read_UInt32Number(char *p) {
  unsigned int rv;
  rv = 16777216 * (unsigned int)((unsigned char *)p)[0] + 65536 * (unsigned int)((unsigned char *)p)[1] + 256 * (unsigned int)((unsigned char *)p)[2] + (unsigned int)((unsigned char *)p)[3];
  return rv;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<read_UInt32Number>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%edx
  shl    $0x8,%eax
  add    %edx,%eax
  movzbl 0x2(%rdi),%edx
  shl    $0x8,%eax
  add    %edx,%eax
  movzbl 0x3(%rdi),%edx
  shl    $0x8,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
unsigned int read_UInt32Number(char *p) {
  unsigned int rv;
  rv = 16777216 * (unsigned int)((unsigned char *)p)[0] + 65536 * (unsigned int)((unsigned char *)p)[1] + 256 * (unsigned int)((unsigned char *)p)[2] + (unsigned int)((unsigned char *)p)[3];
  return rv;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nested_parallel_0>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  nop
  pop    %rbp
  ret
```

assistant: ```c
void nested_parallel_0(void) {
#pragma omp parallel
  {
#pragma omp parallel
    {}
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<nested_parallel_0>:
  endbr64
  ret
```

assistant: ```c
void nested_parallel_0(void) {
#pragma omp parallel
  {
#pragma omp parallel
    {}
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<nested_parallel_0>:
  endbr64
  ret
```

assistant: ```c
void nested_parallel_0(void) {
#pragma omp parallel
  {
#pragma omp parallel
    {}
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<nested_parallel_0>:
  endbr64
  ret
```

assistant: ```c
void nested_parallel_0(void) {
#pragma omp parallel
  {
#pragma omp parallel
    {}
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<good>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    %r9d,-0x2c(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L61
L39:
  movl   $0x0,-0x8(%rbp)
  jmp    L56
L0:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L12
  mov    -0x1c(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  add    %ecx,%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  cmp    $0x1,%eax
  je     L92
L12:
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  je     L30
  mov    -0x1c(%rbp),%edx
  mov    -0xc(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x4,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  add    %ecx,%eax
  cltq
  mov    (%rdx,%rax,4),%eax
  test   %eax,%eax
  je     L92
L30:
  movl   $0x0,-0x4(%rbp)
  mov    $0x0,%eax
  jmp    L72
L92:
  addl   $0x1,-0x8(%rbp)
L56:
  mov    -0x8(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jle    L0
  addl   $0x1,-0xc(%rbp)
L61:
  mov    -0xc(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jle    L39
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L72
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  jmp    L72
L72:
  pop    %rbp
  ret
```

assistant: ```c
int good(int a[20][20], int i, int j, int m, int n, int d) {
  int p, q, check = -1;
  for (p = 0; p <= d; p++)
    for (q = 0; q <= d; q++)
      if (!(((p == q) && (a[i + p][j + q] == 1)) || ((p != q) && (a[i + p][j + q] == 0)))) {
        check = 0;
        return 0;
        break;
      }
  if (check == -1)
    return d + 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<good>:
  endbr64
  mov    %rdi,%r10
  mov    %edx,%r11d
  test   %r9d,%r9d
  js     L31
  mov    $0x0,%r8d
L67:
  movslq %esi,%rax
  lea    (%rax,%rax,4),%rdi
  shl    $0x4,%rdi
  add    %r10,%rdi
  mov    %r11d,%edx
  mov    $0x0,%eax
  jmp    L66
L41:
  movslq %edx,%rcx
  cmpl   $0x0,(%rdi,%rcx,4)
  jne    L50
L42:
  add    $0x1,%eax
  add    $0x1,%edx
  cmp    %eax,%r9d
  jl     L51
L66:
  cmp    %eax,%r8d
  jne    L41
  movslq %edx,%rcx
  cmpl   $0x1,(%rdi,%rcx,4)
  je     L42
  mov    $0x0,%eax
  ret
L51:
  add    $0x1,%r8d
  add    $0x1,%esi
  cmp    %r8d,%r9d
  jge    L67
L31:
  lea    0x1(%r9),%eax
  ret
L50:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int good(int a[20][20], int i, int j, int m, int n, int d) {
  int p, q, check = -1;
  for (p = 0; p <= d; p++)
    for (q = 0; q <= d; q++)
      if (!(((p == q) && (a[i + p][j + q] == 1)) || ((p != q) && (a[i + p][j + q] == 0)))) {
        check = 0;
        return 0;
        break;
      }
  if (check == -1)
    return d + 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<good>:
  endbr64
  mov    %rdi,%r8
  mov    %esi,%r10d
  test   %r9d,%r9d
  js     L36
  movslq %edx,%rdx
  xor    %esi,%esi
  lea    0x0(,%rdx,4),%rdi
L61:
  lea    (%rsi,%r10,1),%eax
  cltq
  lea    (%rax,%rax,4),%rcx
  xor    %eax,%eax
  shl    $0x4,%rcx
  add    %rdi,%rcx
  add    %r8,%rcx
  jmp    L30
  nopl   0x0(%rax)
L65:
  test   %edx,%edx
  jne    L0
L23:
  add    $0x1,%rax
  cmp    %eax,%r9d
  jl     L26
L30:
  mov    (%rcx,%rax,4),%edx
  cmp    %eax,%esi
  jne    L65
  cmp    $0x1,%edx
  je     L23
L0:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L26:
  add    $0x1,%esi
  cmp    %esi,%r9d
  jge    L61
L36:
  lea    0x1(%r9),%eax
  ret
```

assistant: ```c
int good(int a[20][20], int i, int j, int m, int n, int d) {
  int p, q, check = -1;
  for (p = 0; p <= d; p++)
    for (q = 0; q <= d; q++)
      if (!(((p == q) && (a[i + p][j + q] == 1)) || ((p != q) && (a[i + p][j + q] == 0)))) {
        check = 0;
        return 0;
        break;
      }
  if (check == -1)
    return d + 1;
}
```