ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<test_udiv>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7,%esi
  mov    $0xfffffdea,%edi
  call   <udiv_i16@plt>
  cmp    $0x2446,%eax
  jne    L8
  mov    $0x40,%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  mov    $0x11f,%esi
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <fail_real@plt>
```

assistant: ```c
int test_udiv(void) {
  if (udiv_i16(-534, 7) != 9286)
    return fail_real(0, 287);
  return 0x40;
}
```
user: What is the c source code of the assembly code below?

```asm
<test_udiv>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7,%esi
  mov    $0xfffffdea,%edi
  call   <udiv_i16@plt>
  cmp    $0x2446,%eax
  jne    L9
  mov    $0x40,%eax
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  mov    $0x11f,%esi
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <fail_real@plt>
```

assistant: ```c
int test_udiv(void) {
  if (udiv_i16(-534, 7) != 9286)
    return fail_real(0, 287);
  return 0x40;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ulton>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  lea    D1(%rip),%rax
  mov    %rax,-0x8(%rbp)
  subq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
L0:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  div    %rcx
  subq   $0x1,-0x8(%rbp)
  lea    D6(%rip),%rax
  movzbl (%rdx,%rax,1),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  div    %rsi
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L0
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type_guessed: string

tool: D6: 0123456789ABCDEF

assistant: ```c
char *ulton(n, base)
unsigned long n;
int base;
{
  char *p;
  static char buf[20];
  p = &buf[sizeof(buf)];
  *--p = '\0';
  do {
    *--p = "0123456789ABCDEF"[n % base];
    n /= base;
  } while (n != 0);
  return p;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ulton>:
  endbr64
  movb   $0x0,D12(%rip)
  lea    D12(%rip),%rcx
  movslq %esi,%rsi
  lea    D7(%rip),%r8
L0:
  sub    $0x1,%rcx
  mov    %rdi,%rax
  mov    $0x0,%edx
  div    %rsi
  movzbl (%r8,%rdx,1),%edx
  mov    %dl,(%rcx)
  mov    %rdi,%rdx
  mov    %rax,%rdi
  cmp    %rdx,%rsi
  jbe    L0
  mov    %rcx,%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
datatype: string

tool: D7: 0123456789ABCDEF

assistant: ```c
char *ulton(n, base)
unsigned long n;
int base;
{
  char *p;
  static char buf[20];
  p = &buf[sizeof(buf)];
  *--p = '\0';
  do {
    *--p = "0123456789ABCDEF"[n % base];
    n /= base;
  } while (n != 0);
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ulton>:
  endbr64
  movb   $0x0,D1(%rip)
  movslq %esi,%rsi
  lea    D1(%rip),%r8
  lea    D27(%rip),%rcx
  nopl   0x0(%rax)
L8:
  mov    %rdi,%rax
  xor    %edx,%edx
  sub    $0x1,%r8
  div    %rsi
  movzbl (%rcx,%rdx,1),%edx
  mov    %dl,(%r8)
  mov    %rdi,%rdx
  mov    %rax,%rdi
  cmp    %rdx,%rsi
  jbe    L8
  mov    %r8,%rax
  ret
```

tool_calls: name: parse_data
arguments: label: D27
type_infered: string

tool: D27: 0123456789ABCDEF

assistant: ```c
char *ulton(n, base)
unsigned long n;
int base;
{
  char *p;
  static char buf[20];
  p = &buf[sizeof(buf)];
  *--p = '\0';
  do {
    *--p = "0123456789ABCDEF"[n % base];
    n /= base;
  } while (n != 0);
  return p;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ulton>:
  endbr64
  movb   $0x0,D0(%rip)
  movslq %esi,%rsi
  lea    D0(%rip),%r8
  lea    D5(%rip),%rcx
  nopl   0x0(%rax)
L3:
  mov    %rdi,%rax
  xor    %edx,%edx
  sub    $0x1,%r8
  div    %rsi
  movzbl (%rcx,%rdx,1),%edx
  mov    %dl,(%r8)
  mov    %rdi,%rdx
  mov    %rax,%rdi
  cmp    %rdx,%rsi
  jbe    L3
  mov    %r8,%rax
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
data_type: string

tool: D5: 0123456789ABCDEF

assistant: ```c
char *ulton(n, base)
unsigned long n;
int base;
{
  char *p;
  static char buf[20];
  p = &buf[sizeof(buf)];
  *--p = '\0';
  do {
    *--p = "0123456789ABCDEF"[n % base];
    n /= base;
  } while (n != 0);
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x4,-0xc(%rbp)
  movl   $0xfffffffb,-0x8(%rbp)
  movl   $0x11,-0x4(%rbp)
  movl   $0xfffffffe,-0x24(%rbp)
  jmp    L118
L12:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  jmp    L104
L11:
  movl   $0xd,-0x1c(%rbp)
  jmp    L124
L26:
  mov    -0x24(%rbp),%eax
  lea    0x2(%rax),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%eax
  add    $0x5,%eax
  add    %edx,%eax
  lea    0x0(,%rax,4),%edx
  mov    -0x1c(%rbp),%eax
  sub    $0xd,%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  add    $0x9,%eax
  cmp    %eax,%ecx
  je     L83
  addl   $0x1,-0x18(%rbp)
L83:
  addl   $0x1,-0x1c(%rbp)
L124:
  mov    -0x1c(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L26
  addl   $0x1,-0x20(%rbp)
L104:
  cmpl   $0xfffffffe,-0x20(%rbp)
  jl     L11
  addl   $0x1,-0x24(%rbp)
L118:
  mov    -0x24(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L12
  movl   $0xfffffffe,-0x24(%rbp)
  jmp    L98
L33:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x20(%rbp)
  jmp    L63
L14:
  movl   $0xd,-0x1c(%rbp)
  jmp    L4
L138:
  mov    -0x24(%rbp),%eax
  lea    0x2(%rax),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%eax
  add    $0x5,%eax
  add    %edx,%eax
  lea    0x0(,%rax,4),%edx
  mov    -0x1c(%rbp),%eax
  sub    $0xd,%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x14(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  add    $0x9,%eax
  cmp    %eax,%ecx
  je     L123
  addl   $0x1,-0x10(%rbp)
L123:
  addl   $0x1,-0x1c(%rbp)
L4:
  mov    -0x1c(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L138
  addl   $0x1,-0x20(%rbp)
L63:
  cmpl   $0xfffffffe,-0x20(%rbp)
  jl     L14
  addl   $0x1,-0x24(%rbp)
L98:
  mov    -0x24(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L33
  mov    -0x18(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  je     L53
  mov    $0x0,%eax
  call   <abort@plt>
L53:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int i, j, k, l = 0, f = 0, x = 0;
  int m1 = 4, m2 = -5, m3 = 17;
#pragma acc parallel copy(l)
#pragma acc loop seq collapse(3) reduction(+ : l)
  for (i = -2; i < m1; i++)
    for (j = m2; j < -2; j++) {
      for (k = 13; k < m3; k++) {
        if ((i + 2) * 12 + (j + 5) * 4 + (k - 13) != 9 + f++)
          l++;
      }
    }
  for (i = -2; i < m1; i++)
    for (j = m2; j < -2; j++) {
      for (k = 13; k < m3; k++) {
        if ((i + 2) * 12 + (j + 5) * 4 + (k - 13) != 9 + f++)
          x++;
      }
    }
  if (l != x)
    abort();
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0xc,%ebx
  mov    $0x0,%r11d
  mov    $0x0,%edx
  jmp    L114
L112:
  add    $0xc,%r11d
  add    $0xc,%ebx
  cmp    $0x48,%r11d
  je     L13
L114:
  mov    %r11d,%r10d
  lea    0xd(%r11),%esi
  mov    %r11d,%edi
L3:
  mov    %edi,%r9d
  lea    0x9(%rdi),%eax
  lea    -0x9(%r10),%r8d
L12:
  cmp    %r9d,%r8d
  setne  %cl
  movzbl %cl,%ecx
  add    %ecx,%edx
  add    $0x1,%eax
  cmp    %esi,%eax
  jne    L12
  add    $0x4,%edi
  add    $0x4,%esi
  add    $0x4,%r10d
  cmp    %ebx,%edi
  jne    L3
  jmp    L112
L85:
  add    $0xc,%ebx
  cmp    $0x9c,%ebx
  je     L31
L51:
  lea    -0xc(%rbx),%r8d
  lea    0x1(%rbx),%edi
  lea    -0x54(%rbx),%r11d
L119:
  mov    %r8d,%r10d
  lea    0x9(%r8),%eax
  lea    -0x9(%r11),%r9d
L7:
  cmp    %r9d,%r10d
  setne  %sil
  movzbl %sil,%esi
  add    %esi,%ecx
  add    $0x1,%eax
  cmp    %eax,%edi
  jne    L7
  add    $0x4,%r8d
  add    $0x4,%edi
  add    $0x4,%r11d
  cmp    %ebx,%r8d
  jne    L119
  jmp    L85
L13:
  mov    $0x54,%ebx
  mov    $0x0,%ecx
  jmp    L51
L31:
  cmp    %ecx,%edx
  jne    L22
L113:
  mov    $0x0,%eax
  pop    %rbx
  ret
L22:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L113
```

assistant: ```c
int main(void) {
  int i, j, k, l = 0, f = 0, x = 0;
  int m1 = 4, m2 = -5, m3 = 17;
#pragma acc parallel copy(l)
#pragma acc loop seq collapse(3) reduction(+ : l)
  for (i = -2; i < m1; i++)
    for (j = m2; j < -2; j++) {
      for (k = 13; k < m3; k++) {
        if ((i + 2) * 12 + (j + 5) * 4 + (k - 13) != 9 + f++)
          l++;
      }
    }
  for (i = -2; i < m1; i++)
    for (j = m2; j < -2; j++) {
      for (k = 13; k < m3; k++) {
        if ((i + 2) * 12 + (j + 5) * 4 + (k - 13) != 9 + f++)
          x++;
      }
    }
  if (l != x)
    abort();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  mov    $0xc,%r11d
  xor    %r10d,%r10d
  xor    %esi,%esi
  nop
L59:
  mov    %r10d,%r9d
  lea    0xd(%r10),%edi
  mov    %r10d,%edx
L62:
  lea    0x9(%rdx),%eax
  lea    -0x9(%r9),%r8d
L17:
  xor    %ecx,%ecx
  cmp    %edx,%r8d
  setne  %cl
  add    $0x1,%eax
  add    %ecx,%esi
  cmp    %edi,%eax
  jne    L17
  add    $0x4,%edx
  lea    0x4(%rax),%edi
  add    $0x4,%r9d
  cmp    %r11d,%edx
  jne    L62
  add    $0xc,%r10d
  lea    0xc(%rdx),%r11d
  cmp    $0x48,%r10d
  jne    L59
  mov    $0x54,%r11d
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L14:
  lea    -0xc(%r11),%edx
  lea    0x1(%r11),%r8d
  lea    -0x54(%r11),%r10d
L66:
  lea    0x9(%rdx),%eax
  lea    -0x9(%r10),%r9d
L52:
  xor    %edi,%edi
  cmp    %edx,%r9d
  setne  %dil
  add    $0x1,%eax
  add    %edi,%ecx
  cmp    %eax,%r8d
  jne    L52
  add    $0x4,%edx
  add    $0x4,%r8d
  add    $0x4,%r10d
  cmp    %r11d,%edx
  jne    L66
  lea    0xc(%rdx),%r11d
  cmp    $0x90,%edx
  jne    L14
  cmp    %ecx,%esi
  jne    L56
  xor    %eax,%eax
  ret
L56:
  push   %rax
  xor    %eax,%eax
  call   <abort@plt>
  xor    %eax,%eax
  pop    %rdx
  ret
```

assistant: ```c
int main(void) {
  int i, j, k, l = 0, f = 0, x = 0;
  int m1 = 4, m2 = -5, m3 = 17;
#pragma acc parallel copy(l)
#pragma acc loop seq collapse(3) reduction(+ : l)
  for (i = -2; i < m1; i++)
    for (j = m2; j < -2; j++) {
      for (k = 13; k < m3; k++) {
        if ((i + 2) * 12 + (j + 5) * 4 + (k - 13) != 9 + f++)
          l++;
      }
    }
  for (i = -2; i < m1; i++)
    for (j = m2; j < -2; j++) {
      for (k = 13; k < m3; k++) {
        if ((i + 2) * 12 + (j + 5) * 4 + (k - 13) != 9 + f++)
          x++;
      }
    }
  if (l != x)
    abort();
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

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

assistant: ```c
int main(void) {
  int i, j, k, l = 0, f = 0, x = 0;
  int m1 = 4, m2 = -5, m3 = 17;
#pragma acc parallel copy(l)
#pragma acc loop seq collapse(3) reduction(+ : l)
  for (i = -2; i < m1; i++)
    for (j = m2; j < -2; j++) {
      for (k = 13; k < m3; k++) {
        if ((i + 2) * 12 + (j + 5) * 4 + (k - 13) != 9 + f++)
          l++;
      }
    }
  for (i = -2; i < m1; i++)
    for (j = m2; j < -2; j++) {
      for (k = 13; k < m3; k++) {
        if ((i + 2) * 12 + (j + 5) * 4 + (k - 13) != 9 + f++)
          x++;
      }
    }
  if (l != x)
    abort();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<iTestStrBit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  lea    0x7(%rax),%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  sar    $0x3,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
  mov    -0x1c(%rbp),%eax
  cltd
  shr    $0x1d,%edx
  add    %edx,%eax
  and    $0x7,%eax
  sub    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  movzbl -0x1(%rbp),%edx
  mov    $0x7,%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L8
  mov    $0x1,%eax
  jmp    L15
L8:
  mov    $0x0,%eax
L15:
  pop    %rbp
  ret
```

assistant: ```c
int iTestStrBit(unsigned char *psStr, int iPosition) {
  unsigned char ucByte;
  ucByte = psStr[iPosition / 8];
  iPosition %= 8;
  ucByte >>= (7 - iPosition);
  if (ucByte & 0x01)
    return (1);
  return (0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<iTestStrBit>:
  endbr64
  lea    0x7(%rsi),%eax
  test   %esi,%esi
  cmovns %esi,%eax
  sar    $0x3,%eax
  cltq
  movzbl (%rdi,%rax,1),%eax
  mov    %esi,%edx
  sar    $0x1f,%edx
  shr    $0x1d,%edx
  add    %edx,%esi
  and    $0x7,%esi
  sub    %esi,%edx
  lea    0x7(%rdx),%ecx
  sar    %cl,%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
int iTestStrBit(unsigned char *psStr, int iPosition) {
  unsigned char ucByte;
  ucByte = psStr[iPosition / 8];
  iPosition %= 8;
  ucByte >>= (7 - iPosition);
  if (ucByte & 0x01)
    return (1);
  return (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<iTestStrBit>:
  endbr64
  test   %esi,%esi
  lea    0x7(%rsi),%eax
  mov    %esi,%edx
  cmovns %esi,%eax
  sar    $0x1f,%edx
  shr    $0x1d,%edx
  add    %edx,%esi
  sar    $0x3,%eax
  and    $0x7,%esi
  cltq
  movzbl (%rdi,%rax,1),%eax
  sub    %esi,%edx
  lea    0x7(%rdx),%ecx
  sar    %cl,%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
int iTestStrBit(unsigned char *psStr, int iPosition) {
  unsigned char ucByte;
  ucByte = psStr[iPosition / 8];
  iPosition %= 8;
  ucByte >>= (7 - iPosition);
  if (ucByte & 0x01)
    return (1);
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<iTestStrBit>:
  endbr64
  test   %esi,%esi
  lea    0x7(%rsi),%eax
  mov    %esi,%edx
  cmovns %esi,%eax
  sar    $0x1f,%edx
  shr    $0x1d,%edx
  add    %edx,%esi
  sar    $0x3,%eax
  and    $0x7,%esi
  cltq
  movzbl (%rdi,%rax,1),%eax
  sub    %esi,%edx
  lea    0x7(%rdx),%ecx
  sar    %cl,%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
int iTestStrBit(unsigned char *psStr, int iPosition) {
  unsigned char ucByte;
  ucByte = psStr[iPosition / 8];
  iPosition %= 8;
  ucByte >>= (7 - iPosition);
  if (ucByte & 0x01)
    return (1);
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%edx
  mov    $0x0,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x7d0,%ecx
  mov    $0xc8,%edx
  mov    $0x64,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
type_infered: string
name: extract_symbolic_data
arguments: label: D3
type_infered: string

tool: D12: read = %d, orange = %d 


tool: D3: low = %d, medium = %d, high = %d 


assistant: ```c
int main(void) {
  enum spectrum { read, orange, yellow, green, blue, voiolet };
  enum spectrum color;
  printf("read = %d, orange = %d \n", read, orange);
  enum levels { low = 100, medium = 200, high = 2000 };
  printf("low = %d, medium = %d, high = %d \n", low, medium, high);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  mov    $0x0,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x7d0,%ecx
  mov    $0xc8,%edx
  mov    $0x64,%esi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D1
type_guessed: string
name: unpack_data
arguments: data_tag: D18
type_guessed: string

tool: D1: read = %d, orange = %d 


tool: D18: low = %d, medium = %d, high = %d 


assistant: ```c
int main(void) {
  enum spectrum { read, orange, yellow, green, blue, voiolet };
  enum spectrum color;
  printf("read = %d, orange = %d \n", read, orange);
  enum levels { low = 100, medium = 200, high = 2000 };
  printf("low = %d, medium = %d, high = %d \n", low, medium, high);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  xor    %esi,%esi
  xor    %eax,%eax
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    $0x7d0,%ecx
  xor    %eax,%eax
  mov    $0xc8,%edx
  mov    $0x64,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
data_type: string
name: retrieve_data
arguments: data_tag: D7
data_type: string

tool: D18: read = %d, orange = %d 


tool: D7: low = %d, medium = %d, high = %d 


assistant: ```c
int main(void) {
  enum spectrum { read, orange, yellow, green, blue, voiolet };
  enum spectrum color;
  printf("read = %d, orange = %d \n", read, orange);
  enum levels { low = 100, medium = 200, high = 2000 };
  printf("low = %d, medium = %d, high = %d \n", low, medium, high);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  xor    %esi,%esi
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    $0x7d0,%ecx
  xor    %eax,%eax
  mov    $0xc8,%edx
  mov    $0x64,%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D14
data_type: string
name: retrieve_data
arguments: label: D18
data_type: string

tool: D14: read = %d, orange = %d 


tool: D18: low = %d, medium = %d, high = %d 


assistant: ```c
int main(void) {
  enum spectrum { read, orange, yellow, green, blue, voiolet };
  enum spectrum color;
  printf("read = %d, orange = %d \n", read, orange);
  enum levels { low = 100, medium = 200, high = 2000 };
  printf("low = %d, medium = %d, high = %d \n", low, medium, high);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ecc200placementbit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    %edx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  mov    %r8d,-0x18(%rbp)
  mov    %r9d,-0x1c(%rbp)
  mov    0x10(%rbp),%eax
  mov    %al,-0x20(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L5
  mov    -0xc(%rbp),%eax
  add    %eax,-0x14(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x4,%eax
  cltd
  shr    $0x1d,%edx
  add    %edx,%eax
  and    $0x7,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    $0x4,%eax
  sub    %edx,%eax
  add    %eax,-0x18(%rbp)
L5:
  cmpl   $0x0,-0x18(%rbp)
  jns    L2
  mov    -0x10(%rbp),%eax
  add    %eax,-0x18(%rbp)
  mov    -0x10(%rbp),%eax
  add    $0x4,%eax
  cltd
  shr    $0x1d,%edx
  add    %edx,%eax
  and    $0x7,%eax
  sub    %edx,%eax
  mov    %eax,%edx
  mov    $0x4,%eax
  sub    %edx,%eax
  add    %eax,-0x14(%rbp)
L2:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L14
  mov    -0xc(%rbp),%eax
  sub    %eax,-0x14(%rbp)
L14:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  setl   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x18(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  setl   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x14(%rbp),%eax
  imul   -0x10(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x1c(%rbp),%eax
  lea    0x0(,%rax,8),%esi
  movsbl -0x20(%rbp),%edx
  mov    -0x14(%rbp),%eax
  imul   -0x10(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%eax
  add    %ecx,%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  add    %esi,%edx
  mov    %edx,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void ecc200placementbit(int *array, const int NR, const int NC, int r, int c, const int p, const char b) {
  if (r < 0) {
    r += NR;
    c += 4 - ((NR + 4) % 8);
  }
  if (c < 0) {
    c += NC;
    r += 4 - ((NC + 4) % 8);
  }
  if (r >= NR) {
    r -= NR;
  }
  assert(r < NR);
  assert(c < NC);
  assert(0 == array[r * NC + c]);
  array[r * NC + c] = (p << 3) + b;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ecc200placementbit>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%eax
  mov    %edx,%r15d
  mov    %ecx,%ebx
  mov    %r8d,%r14d
  mov    %r9d,%r12d
  mov    0x40(%rsp),%ebp
  test   %ecx,%ecx
  js     L1
L4:
  test   %r14d,%r14d
  js     L13
L25:
  mov    %ebx,%edx
  sub    %eax,%edx
  cmp    %eax,%ebx
  cmovge %edx,%ebx
  cmp    %eax,%ebx
  setl   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmp    %r15d,%r14d
  setl   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  imul   %r15d,%ebx
  add    %r14d,%ebx
  movslq %ebx,%rbx
  lea    0x0(%r13,%rbx,4),%rbx
  cmpl   $0x0,(%rbx)
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  movsbl %bpl,%ebp
  lea    0x0(%rbp,%r12,8),%eax
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L1:
  add    %esi,%ebx
  lea    0x4(%rsi),%ecx
  mov    %ecx,%edx
  sar    $0x1f,%edx
  shr    $0x1d,%edx
  add    %edx,%ecx
  and    $0x7,%ecx
  sub    %ecx,%edx
  lea    0x4(%r8,%rdx,1),%r14d
  jmp    L4
L13:
  add    %r15d,%r14d
  lea    0x4(%r15),%ecx
  mov    %ecx,%edx
  sar    $0x1f,%edx
  shr    $0x1d,%edx
  add    %edx,%ecx
  and    $0x7,%ecx
  sub    %ecx,%edx
  lea    0x4(%rbx,%rdx,1),%ebx
  jmp    L25
```

assistant: ```c
void ecc200placementbit(int *array, const int NR, const int NC, int r, int c, const int p, const char b) {
  if (r < 0) {
    r += NR;
    c += 4 - ((NR + 4) % 8);
  }
  if (c < 0) {
    c += NC;
    r += 4 - ((NC + 4) % 8);
  }
  if (r >= NR) {
    r -= NR;
  }
  assert(r < NR);
  assert(c < NC);
  assert(0 == array[r * NC + c]);
  array[r * NC + c] = (p << 3) + b;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ecc200placementbit>:
  endbr64
  push   %r15
  mov    %esi,%eax
  mov    %r9d,%r15d
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %r8d,%ebp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x8,%rsp
  mov    0x40(%rsp),%r12d
  test   %ecx,%ecx
  jns    L2
  lea    0x4(%rsi),%ecx
  add    %esi,%ebx
  mov    %ecx,%edx
  sar    $0x1f,%edx
  shr    $0x1d,%edx
  add    %edx,%ecx
  and    $0x7,%ecx
  sub    %ecx,%edx
  lea    0x4(%r8,%rdx,1),%ebp
L2:
  test   %ebp,%ebp
  jns    L17
  lea    0x4(%r14),%ecx
  add    %r14d,%ebp
  mov    %ecx,%edx
  sar    $0x1f,%edx
  shr    $0x1d,%edx
  add    %edx,%ecx
  and    $0x7,%ecx
  sub    %ecx,%edx
  lea    0x4(%rbx,%rdx,1),%ebx
L17:
  mov    %ebx,%edx
  movsbl %r12b,%r12d
  sub    %eax,%edx
  cmp    %eax,%ebx
  cmovge %edx,%ebx
  xor    %edi,%edi
  cmp    %eax,%ebx
  setl   %dil
  call   <assert@plt>
  xor    %edi,%edi
  cmp    %r14d,%ebp
  setl   %dil
  imul   %r14d,%ebx
  call   <assert@plt>
  xor    %edi,%edi
  add    %ebp,%ebx
  movslq %ebx,%rbx
  lea    0x0(%r13,%rbx,4),%rbx
  mov    (%rbx),%eax
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  lea    (%r12,%r15,8),%eax
  mov    %eax,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void ecc200placementbit(int *array, const int NR, const int NC, int r, int c, const int p, const char b) {
  if (r < 0) {
    r += NR;
    c += 4 - ((NR + 4) % 8);
  }
  if (c < 0) {
    c += NC;
    r += 4 - ((NC + 4) % 8);
  }
  if (r >= NR) {
    r -= NR;
  }
  assert(r < NR);
  assert(c < NC);
  assert(0 == array[r * NC + c]);
  array[r * NC + c] = (p << 3) + b;
}
```