ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RotatePointX>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movss  %xmm0,-0x14(%rbp)
  movss  %xmm1,-0x18(%rbp)
  movss  %xmm2,-0x1c(%rbp)
  movss  %xmm3,-0x20(%rbp)
  movss  %xmm4,-0x24(%rbp)
  movss  -0x14(%rbp),%xmm0
  subss  -0x1c(%rbp),%xmm0
  movss  %xmm0,-0x8(%rbp)
  movss  -0x18(%rbp),%xmm0
  subss  -0x20(%rbp),%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x24(%rbp),%eax
  movd   %eax,%xmm0
  call   <Cos@plt>
  movd   %xmm0,%eax
  movd   %eax,%xmm5
  mulss  -0x8(%rbp),%xmm5
  movss  %xmm5,-0x28(%rbp)
  mov    -0x24(%rbp),%eax
  movd   %eax,%xmm0
  call   <Sin@plt>
  movd   %xmm0,%eax
  movd   %eax,%xmm1
  mulss  -0x4(%rbp),%xmm1
  movss  -0x28(%rbp),%xmm0
  subss  %xmm1,%xmm0
  addss  -0x1c(%rbp),%xmm0
  leave
  ret
```

assistant: ```c
float RotatePointX(float x, float y, float centerX, float centerY, float degrees) {
  float dx = x - centerX;
  float dy = y - centerY;
  return (Cos(degrees) * dx) - (Sin(degrees) * dy) + centerX;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<RotatePointX>:
  endbr64
  sub    $0x18,%rsp
  movss  %xmm2,0x8(%rsp)
  subss  %xmm2,%xmm0
  movss  %xmm0,(%rsp)
  subss  %xmm3,%xmm1
  movss  %xmm1,0x4(%rsp)
  movss  %xmm4,0xc(%rsp)
  movaps %xmm4,%xmm0
  call   <Cos@plt>
  movaps %xmm0,%xmm5
  mulss  (%rsp),%xmm5
  movss  %xmm5,(%rsp)
  movss  0xc(%rsp),%xmm0
  call   <Sin@plt>
  movss  0x4(%rsp),%xmm1
  mulss  %xmm0,%xmm1
  movss  (%rsp),%xmm0
  subss  %xmm1,%xmm0
  addss  0x8(%rsp),%xmm0
  add    $0x18,%rsp
  ret
```

assistant: ```c
float RotatePointX(float x, float y, float centerX, float centerY, float degrees) {
  float dx = x - centerX;
  float dy = y - centerY;
  return (Cos(degrees) * dx) - (Sin(degrees) * dy) + centerX;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<RotatePointX>:
  endbr64
  subss  %xmm2,%xmm0
  movaps %xmm1,%xmm5
  sub    $0x18,%rsp
  subss  %xmm3,%xmm5
  movss  %xmm2,0xc(%rsp)
  movss  %xmm4,0x8(%rsp)
  movss  %xmm0,(%rsp)
  movaps %xmm4,%xmm0
  movss  %xmm5,0x4(%rsp)
  call   <Cos@plt>
  movss  (%rsp),%xmm1
  movss  0x8(%rsp),%xmm4
  mulss  %xmm0,%xmm1
  movaps %xmm4,%xmm0
  movss  %xmm1,(%rsp)
  call   <Sin@plt>
  movss  0x4(%rsp),%xmm3
  movss  (%rsp),%xmm1
  movss  0xc(%rsp),%xmm2
  add    $0x18,%rsp
  mulss  %xmm0,%xmm3
  subss  %xmm3,%xmm1
  movaps %xmm1,%xmm0
  addss  %xmm2,%xmm0
  ret
```

assistant: ```c
float RotatePointX(float x, float y, float centerX, float centerY, float degrees) {
  float dx = x - centerX;
  float dy = y - centerY;
  return (Cos(degrees) * dx) - (Sin(degrees) * dy) + centerX;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RotatePointX>:
  endbr64
  subss  %xmm2,%xmm0
  movaps %xmm1,%xmm5
  sub    $0x18,%rsp
  subss  %xmm3,%xmm5
  movss  %xmm2,0xc(%rsp)
  movss  %xmm4,0x8(%rsp)
  movss  %xmm0,(%rsp)
  movaps %xmm4,%xmm0
  movss  %xmm5,0x4(%rsp)
  call   <Cos@plt>
  movss  (%rsp),%xmm1
  movss  0x8(%rsp),%xmm4
  mulss  %xmm0,%xmm1
  movaps %xmm4,%xmm0
  movss  %xmm1,(%rsp)
  call   <Sin@plt>
  movss  0x4(%rsp),%xmm3
  movss  (%rsp),%xmm1
  movss  0xc(%rsp),%xmm2
  add    $0x18,%rsp
  mulss  %xmm0,%xmm3
  subss  %xmm3,%xmm1
  movaps %xmm1,%xmm0
  addss  %xmm2,%xmm0
  ret
```

assistant: ```c
float RotatePointX(float x, float y, float centerX, float centerY, float degrees) {
  float dx = x - centerX;
  float dy = y - centerY;
  return (Cos(degrees) * dx) - (Sin(degrees) * dy) + centerX;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<base64_char2id>:
  endbr64
  mov    $0x0,%eax
  mov    $0x41,%edx
  lea    D6(%rip),%rsi
L13:
  mov    %eax,%ecx
  cmp    %dl,%dil
  je     L4
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L13
  mov    $0xffffffff,%ecx
L4:
  mov    %ecx,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
type_infered: string

tool: D6: ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/

assistant: ```c
int base64_char2id(char c) {
  static const char *table = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
  int i;
  for (i = 0; table[i] != '\0'; i++) {
    if (table[i] == c)
      return i;
  }
  return -1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<base64_char2id>:
  endbr64
  xor    %eax,%eax
  mov    $0x41,%edx
  lea    D1(%rip),%rcx
  jmp    L16
  nopl   0x0(%rax)
L1:
  add    $0x1,%rax
  movzbl (%rcx,%rax,1),%edx
  test   %dl,%dl
  je     L27
L16:
  mov    %eax,%r8d
  cmp    %dl,%dil
  jne    L1
  mov    %r8d,%eax
  ret
L27:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type_guessed: string

tool: D1: ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/

assistant: ```c
int base64_char2id(char c) {
  static const char *table = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
  int i;
  for (i = 0; table[i] != '\0'; i++) {
    if (table[i] == c)
      return i;
  }
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<base64_char2id>:
  endbr64
  xor    %eax,%eax
  mov    $0x41,%edx
  lea    D1(%rip),%rcx
  jmp    L24
  nopl   0x0(%rax)
L10:
  add    $0x1,%rax
  movzbl (%rcx,%rax,1),%edx
  test   %dl,%dl
  je     L14
L24:
  mov    %eax,%r8d
  cmp    %dl,%dil
  jne    L10
  mov    %r8d,%eax
  ret
L14:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
type_guessed: string

tool: D1: ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/

assistant: ```c
int base64_char2id(char c) {
  static const char *table = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
  int i;
  for (i = 0; table[i] != '\0'; i++) {
    if (table[i] == c)
      return i;
  }
  return -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
<ratt>:
  endbr64
  mov    %rsi,%rdx
  lea    0x7(%rdi),%r8d
  test   %edi,%edi
  cmovns %edi,%r8d
  sar    $0x3,%r8d
  mov    %edi,%eax
  sar    $0x1f,%eax
  shr    $0x1d,%eax
  lea    (%rdi,%rax,1),%esi
  and    $0x7,%esi
  sub    %eax,%esi
  lea    0x1(%r8),%eax
  cmp    $0x7,%eax
  jg     L7
  lea    0x0(,%rax,8),%edi
  lea    (%rsi,%rdi,1),%ecx
  mov    $0x1,%eax
  shl    %cl,%rax
  bt     %rcx,%rdx
  jb     L143
  lea    0x8(%rdi,%rsi,1),%ecx
  lea    0x40(%rsi),%r10d
  mov    $0x1,%r11d
L74:
  cmp    %r10d,%ecx
  je     L143
  mov    %ecx,%edi
  mov    %r11,%r9
  shl    %cl,%r9
  or     %r9,%rax
  add    $0x8,%ecx
  bt     %rdi,%rdx
  jae    L74
L143:
  mov    %r8d,%edi
  sub    $0x1,%edi
  js     L138
  shl    $0x3,%edi
  lea    (%rsi,%rdi,1),%ecx
  mov    $0x1,%r9d
  shl    %cl,%r9
  or     %r9,%rax
  bt     %rcx,%rdx
  jb     L138
  lea    -0x8(%rdi,%rsi,1),%ecx
  lea    -0x8(%rsi),%r10d
  mov    $0x1,%r11d
L118:
  cmp    %r10d,%ecx
  je     L138
  mov    %ecx,%edi
  mov    %r11,%r9
  shl    %cl,%r9
  or     %r9,%rax
  sub    $0x8,%ecx
  bt     %rdi,%rdx
  jae    L118
L138:
  lea    0x1(%rsi),%ecx
  cmp    $0x7,%ecx
  jg     L40
  lea    0x0(,%r8,8),%r10d
  add    %r10d,%ecx
  mov    $0x1,%edi
  shl    %cl,%rdi
  or     %rdi,%rax
  bt     %rcx,%rdx
  jb     L40
  lea    0x2(%r10,%rsi,1),%ecx
  add    $0x8,%r10d
  mov    $0x1,%r11d
L14:
  cmp    %r10d,%ecx
  je     L40
  mov    %ecx,%edi
  mov    %r11,%r9
  shl    %cl,%r9
  or     %r9,%rax
  add    $0x1,%ecx
  bt     %rdi,%rdx
  jae    L14
L40:
  mov    %esi,%ecx
  sub    $0x1,%ecx
  js     L95
  shl    $0x3,%r8d
  add    %r8d,%ecx
  mov    $0x1,%edi
  shl    %cl,%rdi
  or     %rdi,%rax
  bt     %rcx,%rdx
  jb     L95
  lea    -0x2(%r8,%rsi,1),%ecx
  lea    -0x1(%r8),%esi
  mov    $0x1,%r9d
L28:
  cmp    %ecx,%esi
  je     L95
  mov    %ecx,%edi
  mov    %r9,%r8
  shl    %cl,%r8
  or     %r8,%rax
  sub    $0x1,%ecx
  bt     %rdi,%rdx
  jae    L28
L95:
  ret
L7:
  mov    $0x0,%eax
  jmp    L143
```

assistant: ```c
uint64_t ratt(int sq, uint64_t block) {
  uint64_t result = 0ULL;
  int rk = sq / 8, fl = sq % 8, r, f;
  for (r = rk + 1; r <= 7; r++) {
    result |= (1ULL << (fl + r * 8));
    if (block & (1ULL << (fl + r * 8)))
      break;
  }
  for (r = rk - 1; r >= 0; r--) {
    result |= (1ULL << (fl + r * 8));
    if (block & (1ULL << (fl + r * 8)))
      break;
  }
  for (f = fl + 1; f <= 7; f++) {
    result |= (1ULL << (f + rk * 8));
    if (block & (1ULL << (f + rk * 8)))
      break;
  }
  for (f = fl - 1; f >= 0; f--) {
    result |= (1ULL << (f + rk * 8));
    if (block & (1ULL << (f + rk * 8)))
      break;
  }
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ratt>:
  endbr64
  test   %edi,%edi
  lea    0x7(%rdi),%r9d
  mov    %edi,%eax
  mov    %rsi,%rdx
  cmovns %edi,%r9d
  sar    $0x1f,%eax
  shr    $0x1d,%eax
  add    %eax,%edi
  sar    $0x3,%r9d
  and    $0x7,%edi
  sub    %eax,%edi
  lea    0x1(%r9),%eax
  cmp    $0x7,%eax
  jg     L66
  lea    0x0(,%rax,8),%esi
  mov    $0x1,%eax
  lea    (%rdi,%rsi,1),%ecx
  shl    %cl,%rax
  bt     %rcx,%rdx
  jb     L127
  lea    0x8(%rsi,%rdi,1),%ecx
  lea    0x40(%rdi),%r10d
  mov    $0x1,%r11d
  jmp    L1
  nopl   0x0(%rax)
L101:
  mov    %r11,%r8
  mov    %ecx,%esi
  shl    %cl,%r8
  add    $0x8,%ecx
  or     %r8,%rax
  bt     %rsi,%rdx
  jb     L127
L1:
  cmp    %r10d,%ecx
  jne    L101
L127:
  mov    %r9d,%esi
  sub    $0x1,%esi
  js     L75
  shl    $0x3,%esi
  mov    $0x1,%r8d
  lea    (%rdi,%rsi,1),%ecx
  shl    %cl,%r8
  or     %r8,%rax
  bt     %rcx,%rdx
  jb     L75
  lea    -0x8(%rsi,%rdi,1),%ecx
  lea    -0x8(%rdi),%r10d
  mov    $0x1,%r11d
  jmp    L28
  nopl   0x0(%rax)
L123:
  mov    %r11,%r8
  mov    %ecx,%esi
  shl    %cl,%r8
  sub    $0x8,%ecx
  or     %r8,%rax
  bt     %rsi,%rdx
  jb     L75
L28:
  cmp    %r10d,%ecx
  jne    L123
L75:
  lea    0x1(%rdi),%ecx
  lea    -0x1(%rdi),%esi
  shl    $0x3,%r9d
  cmp    $0x7,%edi
  je     L154
  add    %r9d,%ecx
  mov    $0x1,%r8d
  shl    %cl,%r8
  or     %r8,%rax
  bt     %rcx,%rdx
  jb     L73
  lea    0x2(%r9,%rdi,1),%ecx
  lea    0x8(%r9),%r10d
  mov    $0x1,%r11d
  jmp    L121
  xchg   %ax,%ax
L21:
  mov    %r11,%r8
  mov    %ecx,%edi
  shl    %cl,%r8
  add    $0x1,%ecx
  or     %r8,%rax
  bt     %rdi,%rdx
  jb     L73
L121:
  cmp    %r10d,%ecx
  jne    L21
L73:
  test   %esi,%esi
  js     L14
L154:
  lea    (%rsi,%r9,1),%ecx
  mov    $0x1,%edi
  shl    %cl,%rdi
  or     %rdi,%rax
  bt     %rcx,%rdx
  jb     L14
  mov    $0x1,%r8d
  jmp    L70
  nopw   0x0(%rax,%rax,1)
L161:
  lea    (%rsi,%r9,1),%ecx
  mov    %r8,%rdi
  shl    %cl,%rdi
  or     %rdi,%rax
  bt     %rcx,%rdx
  jb     L14
L70:
  sub    $0x1,%esi
  jae    L161
L14:
  ret
L66:
  xor    %eax,%eax
  jmp    L127
```

assistant: ```c
uint64_t ratt(int sq, uint64_t block) {
  uint64_t result = 0ULL;
  int rk = sq / 8, fl = sq % 8, r, f;
  for (r = rk + 1; r <= 7; r++) {
    result |= (1ULL << (fl + r * 8));
    if (block & (1ULL << (fl + r * 8)))
      break;
  }
  for (r = rk - 1; r >= 0; r--) {
    result |= (1ULL << (fl + r * 8));
    if (block & (1ULL << (fl + r * 8)))
      break;
  }
  for (f = fl + 1; f <= 7; f++) {
    result |= (1ULL << (f + rk * 8));
    if (block & (1ULL << (f + rk * 8)))
      break;
  }
  for (f = fl - 1; f >= 0; f--) {
    result |= (1ULL << (f + rk * 8));
    if (block & (1ULL << (f + rk * 8)))
      break;
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getRow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jg     L42
  mov    -0x8(%rbp),%rax
  movl   $0x1,(%rax)
  mov    -0x8(%rbp),%rax
  jmp    L17
L42:
  movl   $0x1,-0x14(%rbp)
  jmp    L38
L8:
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movl   $0x1,(%rax)
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  mov    -0xc(%rbp),%eax
  sub    $0x2,%eax
  mov    %eax,-0x10(%rbp)
  jmp    L0
L50:
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%ecx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x8(%rbp),%rax
  add    %rsi,%rax
  add    %ecx,%edx
  mov    %edx,(%rax)
  subl   $0x1,-0x10(%rbp)
L0:
  cmpl   $0x0,-0x10(%rbp)
  jg     L50
  addl   $0x1,-0x14(%rbp)
L38:
  mov    -0x14(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jle    L8
  mov    -0x8(%rbp),%rax
L17:
  leave
  ret
```

assistant: ```c
int *getRow(int rowIndex, int *returnSize) {
  int *row = malloc((rowIndex + 1) * sizeof(int));
  *returnSize = rowIndex + 1;
  int num = rowIndex + 1;
  if (rowIndex < 1) {
    row[0] = 1;
    return row;
  }
  int i, j;
  for (i = 1; i <= rowIndex; i++) {
    num = i + 1;
    row[0] = 1;
    row[num - 1] = 1;
    for (j = num - 2; j >= 1; j--) {
      row[j] = row[j - 1] + row[j];
    }
  }
  return row;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getRow>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %rsi,%rbp
  lea    0x1(%rdi),%r12d
  lea    0x0(,%r12,4),%edi
  call   <malloc@plt>
  mov    %r12d,0x0(%rbp)
  test   %ebx,%ebx
  jle    L46
  movl   $0x1,(%rax)
  movl   $0x1,0x4(%rax)
  lea    0x8(%rax),%rdi
  mov    $0x1,%esi
  jmp    L59
L46:
  movl   $0x1,(%rax)
  jmp    L11
L55:
  movl   $0x1,(%rax)
  movl   $0x1,(%rdi)
  test   %esi,%esi
  jle    L47
  lea    -0x4(%rdi),%rdx
L41:
  mov    -0x4(%rdx),%ecx
  add    %ecx,(%rdx)
  sub    $0x4,%rdx
  cmp    %rax,%rdx
  jne    L41
L47:
  add    $0x4,%rdi
  add    $0x1,%esi
L59:
  cmp    %ebx,%esi
  jne    L55
L11:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *getRow(int rowIndex, int *returnSize) {
  int *row = malloc((rowIndex + 1) * sizeof(int));
  *returnSize = rowIndex + 1;
  int num = rowIndex + 1;
  if (rowIndex < 1) {
    row[0] = 1;
    return row;
  }
  int i, j;
  for (i = 1; i <= rowIndex; i++) {
    num = i + 1;
    row[0] = 1;
    row[num - 1] = 1;
    for (j = num - 2; j >= 1; j--) {
      row[j] = row[j - 1] + row[j];
    }
  }
  return row;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getRow>:
  endbr64
  push   %r12
  lea    0x1(%rdi),%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%r12,4),%edi
  call   <malloc@plt>
  mov    %r12d,0x0(%rbp)
  test   %ebx,%ebx
  jle    L28
  movabs $0x100000001,%rdi
  lea    0x8(%rax),%rsi
  mov    %rdi,(%rax)
  mov    $0x1,%edi
  cmp    %ebx,%edi
  je     L9
L6:
  movl   $0x1,(%rax)
  lea    -0x4(%rsi),%rdx
  movl   $0x1,(%rsi)
L39:
  mov    -0x4(%rdx),%ecx
  add    %ecx,(%rdx)
  sub    $0x4,%rdx
  cmp    %rdx,%rax
  jne    L39
  add    $0x1,%edi
  add    $0x4,%rsi
  cmp    %ebx,%edi
  jne    L6
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L28:
  movl   $0x1,(%rax)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *getRow(int rowIndex, int *returnSize) {
  int *row = malloc((rowIndex + 1) * sizeof(int));
  *returnSize = rowIndex + 1;
  int num = rowIndex + 1;
  if (rowIndex < 1) {
    row[0] = 1;
    return row;
  }
  int i, j;
  for (i = 1; i <= rowIndex; i++) {
    num = i + 1;
    row[0] = 1;
    row[num - 1] = 1;
    for (j = num - 2; j >= 1; j--) {
      row[j] = row[j - 1] + row[j];
    }
  }
  return row;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getRow>:
  endbr64
  push   %r12
  lea    0x1(%rdi),%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%r12,4),%edi
  call   <malloc@plt>
  mov    %r12d,0x0(%rbp)
  mov    %rax,%r8
  test   %ebx,%ebx
  jle    L51
  mov    $0x14,%r9d
  lea    -0xc(%r8),%rcx
  mov    D3(%rip),%rax
  mov    $0x1,%r10d
  sub    %r8,%r9
  mov    %rax,(%r8)
  cmp    $0x1,%ebx
  je     L72
  nopl   0x0(%rax,%rax,1)
L83:
  lea    -0x1(%r10),%edx
  movl   $0x1,(%r8)
  mov    %r10d,%eax
  lea    0x1(%r10),%edi
  movl   $0x1,0x14(%rcx)
  cmp    $0x2,%edx
  jbe    L42
  mov    %r10d,%esi
  lea    0x4(%rcx),%rdx
  xor    %eax,%eax
  shr    $0x2,%esi
  neg    %rsi
  shl    $0x4,%rsi
  nopl   (%rax)
L77:
  movdqu (%rcx,%rax,1),%xmm2
  movdqu (%rdx,%rax,1),%xmm3
  pshufd $0x1b,%xmm2,%xmm0
  pshufd $0x1b,%xmm3,%xmm1
  paddd  %xmm1,%xmm0
  pshufd $0x1b,%xmm0,%xmm0
  movups %xmm0,(%rdx,%rax,1)
  sub    $0x10,%rax
  cmp    %rax,%rsi
  jne    L77
  mov    %r10d,%esi
  mov    %r10d,%eax
  and    $0xfffffffc,%esi
  and    $0x3,%eax
  and    $0x3,%r10d
  je     L55
L3:
  mov    %edi,%r11d
  sub    %esi,%r11d
  lea    -0x1(%r11),%r10d
  cmp    $0x2,%r11d
  je     L11
  add    %r9,%rcx
  neg    %rsi
  lea    (%rcx,%rsi,4),%rsi
  lea    -0x8(%r8,%rsi,1),%rcx
  movq   -0xc(%r8,%rsi,1),%xmm0
  movq   (%rcx),%xmm1
  pshufd $0xe1,%xmm0,%xmm0
  pshufd $0xe1,%xmm1,%xmm1
  paddd  %xmm1,%xmm0
  pshufd $0xe1,%xmm0,%xmm0
  movq   %xmm0,(%rcx)
  mov    %r10d,%ecx
  and    $0xfffffffe,%ecx
  sub    %ecx,%eax
  cmp    %ecx,%r10d
  je     L55
L11:
  cltq
  shl    $0x2,%rax
  mov    -0x4(%r8,%rax,1),%ecx
  add    %ecx,(%r8,%rax,1)
L55:
  mov    %edi,%r10d
  mov    %rdx,%rcx
  cmp    %ebx,%edi
  jne    L83
L72:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
L42:
  xor    %esi,%esi
  lea    0x4(%rcx),%rdx
  jmp    L3
L51:
  movl   $0x1,(%rax)
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
data_type: byte[8]

tool: {"D3": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

assistant: ```c
int *getRow(int rowIndex, int *returnSize) {
  int *row = malloc((rowIndex + 1) * sizeof(int));
  *returnSize = rowIndex + 1;
  int num = rowIndex + 1;
  if (rowIndex < 1) {
    row[0] = 1;
    return row;
  }
  int i, j;
  for (i = 1; i <= rowIndex; i++) {
    num = i + 1;
    row[0] = 1;
    row[num - 1] = 1;
    for (j = num - 2; j >= 1; j--) {
      row[j] = row[j - 1] + row[j];
    }
  }
  return row;
}
```
user: Translate the below assembly code into C programming language:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rsi
  mov    -0x34(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rcx
  mov    $0x0,%ebx
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%ebx
  mov    $0x0,%edx
  div    %rbx
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L11:
  cmp    %rdx,%rsp
  je     L20
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L11
L20:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L38
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L38:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    -0x38(%rbp),%edx
  movslq %edx,%rdx
  movl   $0x1,(%rax,%rdx,4)
  mov    %rsi,%rsp
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void foo(int n, int i) {
  int A[n];
#pragma omp parallel firstprivate(A)
  { A[i] = 1; }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<foo>:
  endbr64
  ret
```

assistant: ```c
void foo(int n, int i) {
  int A[n];
#pragma omp parallel firstprivate(A)
  { A[i] = 1; }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<foo>:
  endbr64
  ret
```

assistant: ```c
void foo(int n, int i) {
  int A[n];
#pragma omp parallel firstprivate(A)
  { A[i] = 1; }
}
```
user: What is the c source code of the assembly code below?

```asm
<foo>:
  endbr64
  ret
```

assistant: ```c
void foo(int n, int i) {
  int A[n];
#pragma omp parallel firstprivate(A)
  { A[i] = 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
<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    L6
L15:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x14(%rbp)
L6:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jb     L15
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *memcpy(void *to, const void *from, unsigned int n) {
  unsigned int i;
  unsigned char *xto = (unsigned char *)to;
  unsigned char *xfrom = (unsigned char *)from;
  for (i = 0; i < n; i++) {
    *(xto++) = *(xfrom++);
  }
  return xto;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<memcpy>:
  endbr64
  test   %edx,%edx
  je     L2
  mov    %edx,%edx
  mov    $0x0,%ecx
L8:
  movzbl (%rsi,%rcx,1),%r8d
  mov    %r8b,(%rdi,%rcx,1)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L8
  lea    (%rdi,%rdx,1),%rax
  ret
L2:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memcpy(void *to, const void *from, unsigned int n) {
  unsigned int i;
  unsigned char *xto = (unsigned char *)to;
  unsigned char *xfrom = (unsigned char *)from;
  for (i = 0; i < n; i++) {
    *(xto++) = *(xfrom++);
  }
  return xto;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<memcpy>:
  endbr64
  test   %edx,%edx
  je     L14
  mov    %edx,%edx
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L6:
  movzbl (%rsi,%rcx,1),%r8d
  mov    %r8b,(%rdi,%rcx,1)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L6
  lea    (%rdi,%rcx,1),%rax
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memcpy(void *to, const void *from, unsigned int n) {
  unsigned int i;
  unsigned char *xto = (unsigned char *)to;
  unsigned char *xfrom = (unsigned char *)from;
  for (i = 0; i < n; i++) {
    *(xto++) = *(xfrom++);
  }
  return xto;
}
```