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
<strlequ>:
  endbr64
  movzbl (%rdi),%edx
  xor    %eax,%eax
  xor    %r8d,%r8d
  test   %dl,%dl
  jne    L1
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L11:
  lea    0x1(%rax),%r8d
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L0
L1:
  mov    %eax,%r8d
  cmp    %dl,(%rsi,%rax,1)
  je     L11
L0:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int strlequ(char *s, char *t) {
  int length = 0;
  while (*s && (*s++ == *t++))
    length++;
  return length;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strlequ>:
  endbr64
  movzbl (%rdi),%edx
  xor    %eax,%eax
  xor    %r8d,%r8d
  test   %dl,%dl
  jne    L7
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L24:
  lea    0x1(%rax),%r8d
  add    $0x1,%rax
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L4
L7:
  mov    %eax,%r8d
  cmp    %dl,(%rsi,%rax,1)
  je     L24
L4:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int strlequ(char *s, char *t) {
  int length = 0;
  while (*s && (*s++ == *t++))
    length++;
  return length;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gmul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x18(%rbp)
  movb   $0x0,-0x3(%rbp)
  movb   $0x0,-0x2(%rbp)
  jmp    L28
L3:
  movzbl -0x18(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L26
  movzbl -0x14(%rbp),%eax
  xor    %al,-0x3(%rbp)
L26:
  movzbl -0x14(%rbp),%eax
  and    $0xffffff80,%eax
  mov    %al,-0x1(%rbp)
  shlb   -0x14(%rbp)
  cmpb   $0x80,-0x1(%rbp)
  jne    L23
  xorb   $0x1b,-0x14(%rbp)
L23:
  shrb   -0x18(%rbp)
  movzbl -0x2(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x2(%rbp)
L28:
  cmpb   $0x7,-0x2(%rbp)
  jbe    L3
  movzbl -0x3(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char gmul(unsigned char a, unsigned char b) {
  unsigned char p = 0;
  unsigned char counter;
  unsigned char hi_bit_set;
  for (counter = 0; counter < 8; counter++) {
    if ((b & 1) == 1)
      p ^= a;
    hi_bit_set = (a & 0x80);
    a <<= 1;
    if (hi_bit_set == 0x80)
      a ^= 0x1b;
    b >>= 1;
  }
  return p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<gmul>:
  endbr64
  mov    $0x8,%edx
  mov    $0x0,%eax
L9:
  mov    %eax,%ecx
  xor    %edi,%ecx
  test   $0x1,%sil
  cmovne %ecx,%eax
  lea    (%rdi,%rdi,1),%ecx
  mov    %ecx,%r8d
  xor    $0x1b,%r8d
  test   %dil,%dil
  mov    %ecx,%edi
  cmovs  %r8d,%edi
  shr    %sil
  sub    $0x1,%dl
  jne    L9
  ret
```

assistant: ```c
unsigned char gmul(unsigned char a, unsigned char b) {
  unsigned char p = 0;
  unsigned char counter;
  unsigned char hi_bit_set;
  for (counter = 0; counter < 8; counter++) {
    if ((b & 1) == 1)
      p ^= a;
    hi_bit_set = (a & 0x80);
    a <<= 1;
    if (hi_bit_set == 0x80)
      a ^= 0x1b;
    b >>= 1;
  }
  return p;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<gmul>:
  endbr64
  mov    $0x8,%edx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %eax,%ecx
  xor    %edi,%ecx
  test   $0x1,%sil
  cmovne %ecx,%eax
  lea    (%rdi,%rdi,1),%ecx
  mov    %ecx,%r8d
  xor    $0x1b,%r8d
  test   %dil,%dil
  mov    %ecx,%edi
  cmovs  %r8d,%edi
  shr    %sil
  sub    $0x1,%dl
  jne    L6
  ret
```

assistant: ```c
unsigned char gmul(unsigned char a, unsigned char b) {
  unsigned char p = 0;
  unsigned char counter;
  unsigned char hi_bit_set;
  for (counter = 0; counter < 8; counter++) {
    if ((b & 1) == 1)
      p ^= a;
    hi_bit_set = (a & 0x80);
    a <<= 1;
    if (hi_bit_set == 0x80)
      a ^= 0x1b;
    b >>= 1;
  }
  return p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<gmul>:
  endbr64
  lea    (%rdi,%rdi,1),%ecx
  mov    %esi,%eax
  mov    %esi,%edx
  and    $0x1,%eax
  mov    %ecx,%esi
  cmovne %edi,%eax
  xor    $0x1b,%esi
  test   %dil,%dil
  cmovs  %esi,%ecx
  mov    %eax,%esi
  xor    %ecx,%esi
  test   $0x2,%dl
  cmovne %esi,%eax
  lea    (%rcx,%rcx,1),%esi
  mov    %esi,%edi
  xor    $0x1b,%edi
  test   %cl,%cl
  mov    %eax,%ecx
  cmovs  %edi,%esi
  xor    %esi,%ecx
  test   $0x4,%dl
  cmovne %ecx,%eax
  lea    (%rsi,%rsi,1),%ecx
  mov    %ecx,%edi
  xor    $0x1b,%edi
  test   %sil,%sil
  mov    %eax,%esi
  cmovs  %edi,%ecx
  xor    %ecx,%esi
  test   $0x8,%dl
  cmovne %esi,%eax
  lea    (%rcx,%rcx,1),%esi
  mov    %esi,%edi
  xor    $0x1b,%edi
  test   %cl,%cl
  mov    %eax,%ecx
  cmovs  %edi,%esi
  xor    %esi,%ecx
  test   $0x10,%dl
  cmovne %ecx,%eax
  lea    (%rsi,%rsi,1),%ecx
  mov    %ecx,%edi
  xor    $0x1b,%edi
  test   %sil,%sil
  mov    %eax,%esi
  cmovs  %edi,%ecx
  xor    %ecx,%esi
  test   $0x20,%dl
  cmovne %esi,%eax
  lea    (%rcx,%rcx,1),%esi
  mov    %esi,%edi
  xor    $0x1b,%edi
  test   %cl,%cl
  mov    %eax,%ecx
  cmovs  %edi,%esi
  xor    %esi,%ecx
  test   $0x40,%dl
  cmovne %ecx,%eax
  lea    (%rsi,%rsi,1),%ecx
  mov    %ecx,%edi
  xor    $0x1b,%edi
  test   %sil,%sil
  cmovs  %edi,%ecx
  xor    %eax,%ecx
  test   %dl,%dl
  cmovs  %ecx,%eax
  ret
```

assistant: ```c
unsigned char gmul(unsigned char a, unsigned char b) {
  unsigned char p = 0;
  unsigned char counter;
  unsigned char hi_bit_set;
  for (counter = 0; counter < 8; counter++) {
    if ((b & 1) == 1)
      p ^= a;
    hi_bit_set = (a & 0x80);
    a <<= 1;
    if (hi_bit_set == 0x80)
      a ^= 0x1b;
    b >>= 1;
  }
  return p;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mine>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1ac(%rbp)
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
L23:
  mov    -0x1ac(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1ac(%rbp)
  lea    -0x1a0(%rbp),%rdx
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  cmp    $0x1,%eax
  je     L23
  mov    -0x1ac(%rbp),%eax
  sub    $0x4,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x1a8(%rbp)
  cmpq   $0x0,-0x1a8(%rbp)
  je     L36
  mov    -0x1ac(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1a8(%rbp),%rax
  mov    %edx,(%rax)
  movl   $0x1,-0x1b0(%rbp)
  jmp    L31
L5:
  mov    -0x1b0(%rbp),%eax
  lea    -0x1(%rax),%ecx
  mov    -0x1b0(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x1a8(%rbp),%rax
  add    %rax,%rdx
  movslq %ecx,%rax
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x1b0(%rbp)
L31:
  mov    -0x1b0(%rbp),%eax
  cmp    -0x1ac(%rbp),%eax
  jl     L5
L36:
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x1a8(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movl   $0x1,-0x1ac(%rbp)
  jmp    L43
L56:
  mov    -0x1ac(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1ac(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x1a8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L43:
  mov    -0x1ac(%rbp),%eax
  cmp    -0x1b0(%rbp),%eax
  jl     L56
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D24
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D20
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D52
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D3
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D1
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D41
type_infered: string

tool: D24: Enter some numbers separated by whitespaces:

tool: D20: %d

tool: D52: Your numbers, in a dynamic allocated int array would be..

tool: D3: %d: 

tool: D1: %d, 

tool: D41: 


assistant: ```c
void mine(void) {
  int array[100];
  int y, x = 0;
  puts("Enter some numbers separated by whitespaces:");
  while (scanf("%d", &array[x++]) == 1)
    ;
  int *ptr = malloc(x - 1 * sizeof(int));
  if (ptr != NULL) {
    ptr[0] = x - 1;
    for (y = 1; y < x; y++)
      ptr[y] = array[y - 1];
  }
  puts("Your numbers, in a dynamic allocated int array would be..");
  printf("%d: ", ptr[0]);
  x = 1;
  while (x < y) {
    printf("%d, ", ptr[x++]);
  }
  puts("\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mine>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1a0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  lea    D52(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbp
  mov    $0x0,%ebx
  lea    D11(%rip),%r14
L56:
  mov    %ebx,%r12d
  add    $0x1,%ebx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  add    $0x4,%rbp
  cmp    $0x1,%eax
  je     L56
  lea    -0x3(%r12),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L27
  mov    %r12d,(%rax)
  cmp    $0x1,%ebx
  jle    L11
  mov    $0x0,%eax
L24:
  mov    (%rsp,%rax,4),%edx
  mov    %edx,0x4(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r12d
  jg     L24
  lea    0x1(%r12),%r13d
  test   %r12d,%r12d
  mov    $0x2,%eax
  cmovle %eax,%r13d
L27:
  lea    D53(%rip),%rdi
  call   <puts@plt>
  mov    0x0(%rbp),%esi
  lea    D69(%rip),%rdi
  call   <printf@plt>
  cmp    $0x1,%r13d
  jle    L6
  lea    0x4(%rbp),%rbx
  lea    -0x2(%r13),%eax
  lea    0x8(%rbp,%rax,4),%r12
  lea    D57(%rip),%rbp
L41:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L41
L6:
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  add    $0x1a0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L11:
  lea    D53(%rip),%rdi
  call   <puts@plt>
  mov    0x0(%rbp),%esi
  lea    D69(%rip),%rdi
  call   <printf@plt>
  jmp    L6
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D52
type_infered: string
name: retrieve_data
arguments: tag: D11
type_infered: string
name: retrieve_data
arguments: tag: D53
type_infered: string
name: retrieve_data
arguments: tag: D69
type_infered: string
name: retrieve_data
arguments: tag: D57
type_infered: string
name: retrieve_data
arguments: tag: D0
type_infered: string

tool: D52: Enter some numbers separated by whitespaces:

tool: D11: %d

tool: D53: Your numbers, in a dynamic allocated int array would be..

tool: D69: %d: 

tool: D57: %d, 

tool: D0: 


assistant: ```c
void mine(void) {
  int array[100];
  int y, x = 0;
  puts("Enter some numbers separated by whitespaces:");
  while (scanf("%d", &array[x++]) == 1)
    ;
  int *ptr = malloc(x - 1 * sizeof(int));
  if (ptr != NULL) {
    ptr[0] = x - 1;
    for (y = 1; y < x; y++)
      ptr[y] = array[y - 1];
  }
  puts("Your numbers, in a dynamic allocated int array would be..");
  printf("%d: ", ptr[0]);
  x = 1;
  while (x < y) {
    printf("%d, ", ptr[x++]);
  }
  puts("\n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<mine>:
  endbr64
  push   %r15
  lea    D7(%rip),%rdi
  lea    D39(%rip),%r15
  push   %r14
  push   %r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x1a8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x198(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <puts@plt>
  mov    %r13,%rbx
  nopl   0x0(%rax)
L40:
  mov    %rbx,%rsi
  mov    %r15,%rdi
  mov    %r12d,%ebp
  add    $0x4,%rbx
  call   <scanf@plt>
  lea    0x1(%r12),%r12d
  cmp    $0x1,%eax
  je     L40
  lea    -0x3(%rbp),%edi
  call   <malloc@plt>
  mov    %rax,%r15
  test   %rax,%rax
  je     L69
  mov    %ebp,(%rax)
  cmp    $0x1,%r12d
  je     L7
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L30:
  mov    0x0(%r13,%rax,4),%edx
  mov    %edx,0x4(%r15,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%ebp
  jg     L30
  test   %ebp,%ebp
  mov    $0x1,%eax
  cmovle %eax,%ebp
  lea    0x1(%rbp),%r14d
L69:
  lea    D47(%rip),%rdi
  lea    0x4(%r15),%rbx
  call   <puts@plt>
  mov    (%r15),%esi
  lea    D14(%rip),%rdi
  lea    D42(%rip),%rbp
  call   <printf@plt>
  lea    -0x2(%r14),%eax
  lea    0x8(%r15,%rax,4),%r12
  nopw   0x0(%rax,%rax,1)
L53:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L53
L64:
  lea    D30(%rip),%rdi
  call   <puts@plt>
  mov    0x198(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x1a8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L7:
  lea    D47(%rip),%rdi
  call   <puts@plt>
  mov    (%r15),%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  jmp    L64
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D7
type_infered: string
name: parse_data
arguments: data_tag: D39
type_infered: string
name: parse_data
arguments: data_tag: D47
type_infered: string
name: parse_data
arguments: data_tag: D14
type_infered: string
name: parse_data
arguments: data_tag: D42
type_infered: string
name: parse_data
arguments: data_tag: D30
type_infered: string

tool: D7: Enter some numbers separated by whitespaces:

tool: D39: %d

tool: D47: Your numbers, in a dynamic allocated int array would be..

tool: D14: %d: 

tool: D42: %d, 

tool: D30: 


assistant: ```c
void mine(void) {
  int array[100];
  int y, x = 0;
  puts("Enter some numbers separated by whitespaces:");
  while (scanf("%d", &array[x++]) == 1)
    ;
  int *ptr = malloc(x - 1 * sizeof(int));
  if (ptr != NULL) {
    ptr[0] = x - 1;
    for (y = 1; y < x; y++)
      ptr[y] = array[y - 1];
  }
  puts("Your numbers, in a dynamic allocated int array would be..");
  printf("%d: ", ptr[0]);
  x = 1;
  while (x < y) {
    printf("%d, ", ptr[x++]);
  }
  puts("\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <some_library_function@plt>
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D1
data_type: string

tool: D1: some_library_function returned %d


assistant: ```c
int main(int argc, char *arg[]) {
  printf("some_library_function returned %d\n", some_library_function());
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <some_library_function@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D9
type: string

tool: D9: some_library_function returned %d


assistant: ```c
int main(int argc, char *arg[]) {
  printf("some_library_function returned %d\n", some_library_function());
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <some_library_function@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
datatype: string

tool: D6: some_library_function returned %d


assistant: ```c
int main(int argc, char *arg[]) {
  printf("some_library_function returned %d\n", some_library_function());
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <some_library_function@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
type_infered: string

tool: D5: some_library_function returned %d


assistant: ```c
int main(int argc, char *arg[]) {
  printf("some_library_function returned %d\n", some_library_function());
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<blas_sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L0:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L0
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double blas_sum(const double *X, const int N) {
  double sum = 0;
  for (int i = 0; i < N; ++i) {
    sum += X[i];
  }
  return sum;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<blas_sum>:
  endbr64
  test   %esi,%esi
  jle    L11
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x8(%rdi,%rdx,8),%rdx
  pxor   %xmm0,%xmm0
L6:
  addsd  (%rax),%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L6
  ret
L11:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double blas_sum(const double *X, const int N) {
  double sum = 0;
  for (int i = 0; i < N; ++i) {
    sum += X[i];
  }
  return sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<blas_sum>:
  endbr64
  test   %esi,%esi
  jle    L1
  lea    -0x1(%rsi),%eax
  pxor   %xmm0,%xmm0
  lea    0x8(%rdi,%rax,8),%rax
  nopl   0x0(%rax)
L14:
  addsd  (%rdi),%xmm0
  add    $0x8,%rdi
  cmp    %rax,%rdi
  jne    L14
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double blas_sum(const double *X, const int N) {
  double sum = 0;
  for (int i = 0; i < N; ++i) {
    sum += X[i];
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<blas_sum>:
  endbr64
  test   %esi,%esi
  jle    L47
  cmp    $0x1,%esi
  je     L45
  mov    %esi,%edx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    %edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nopl   0x0(%rax)
L3:
  movsd  (%rax),%xmm1
  add    $0x10,%rax
  addsd  %xmm1,%xmm0
  movsd  -0x8(%rax),%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L3
  mov    %esi,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%esi
  je     L10
L6:
  cltq
  addsd  (%rdi,%rax,8),%xmm0
  ret
  nopl   0x0(%rax)
L10:
  ret
  nopl   0x0(%rax)
L47:
  pxor   %xmm0,%xmm0
  ret
L45:
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  jmp    L6
```

assistant: ```c
double blas_sum(const double *X, const int N) {
  double sum = 0;
  for (int i = 0; i < N; ++i) {
    sum += X[i];
  }
  return sum;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<autohelperowl_attackpat220>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  mov    %edx,%ecx
  mov    %eax,%edx
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <offset@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <countstones@plt>
  cmp    $0x3,%eax
  jg     L29
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <countlib@plt>
  cmp    $0x1,%eax
  jne    L29
  mov    -0x24(%rbp),%esi
  mov    -0x20(%rbp),%eax
  mov    $0x0,%ecx
  mov    $0x14,%edx
  mov    %eax,%edi
  call   <accurate_approxlib@plt>
  cmp    $0x1,%eax
  jle    L29
  mov    $0x1,%eax
  jmp    L4
L29:
  mov    $0x0,%eax
L4:
  leave
  ret
```

assistant: ```c
int autohelperowl_attackpat220(struct pattern *patt, int transformation, int move, int color, int action) {
  int A;
  (void)patt;
  (void)color;
  (void)action;
  A = offset(1, 0, move, transformation);
  return countstones(A) <= 3 && countlib(A) == 1 && accurate_approxlib(move, color, 20, NULL) > 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<autohelperowl_attackpat220>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edx,%ebp
  mov    %ecx,%r12d
  mov    %esi,%ecx
  mov    $0x0,%esi
  mov    $0x1,%edi
  call   <offset@plt>
  mov    %eax,%ebx
  mov    %eax,%edi
  call   <countstones@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    $0x3,%edx
  jle    L29
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  mov    %ebx,%edi
  call   <countlib@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    $0x1,%edx
  jne    L8
  mov    $0x0,%ecx
  mov    $0x14,%edx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <accurate_approxlib@plt>
  cmp    $0x1,%eax
  setg   %al
  movzbl %al,%eax
  jmp    L8
```

assistant: ```c
int autohelperowl_attackpat220(struct pattern *patt, int transformation, int move, int color, int action) {
  int A;
  (void)patt;
  (void)color;
  (void)action;
  A = offset(1, 0, move, transformation);
  return countstones(A) <= 3 && countlib(A) == 1 && accurate_approxlib(move, color, 20, NULL) > 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<autohelperowl_attackpat220>:
  endbr64
  push   %r13
  mov    $0x1,%edi
  mov    %ecx,%r13d
  mov    %esi,%ecx
  push   %r12
  xor    %esi,%esi
  mov    %edx,%r12d
  push   %rbp
  call   <offset@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <countstones@plt>
  cmp    $0x3,%eax
  jle    L10
L16:
  xor    %eax,%eax
L27:
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %ebp,%edi
  call   <countlib@plt>
  cmp    $0x1,%eax
  jne    L16
  xor    %ecx,%ecx
  mov    $0x14,%edx
  mov    %r13d,%esi
  mov    %r12d,%edi
  call   <accurate_approxlib@plt>
  cmp    $0x1,%eax
  setg   %al
  movzbl %al,%eax
  jmp    L27
```

assistant: ```c
int autohelperowl_attackpat220(struct pattern *patt, int transformation, int move, int color, int action) {
  int A;
  (void)patt;
  (void)color;
  (void)action;
  A = offset(1, 0, move, transformation);
  return countstones(A) <= 3 && countlib(A) == 1 && accurate_approxlib(move, color, 20, NULL) > 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<autohelperowl_attackpat220>:
  endbr64
  push   %r13
  mov    $0x1,%edi
  mov    %ecx,%r13d
  mov    %esi,%ecx
  push   %r12
  xor    %esi,%esi
  mov    %edx,%r12d
  push   %rbp
  call   <offset@plt>
  mov    %eax,%edi
  mov    %eax,%ebp
  call   <countstones@plt>
  cmp    $0x3,%eax
  jle    L25
L19:
  xor    %eax,%eax
L8:
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  mov    %ebp,%edi
  call   <countlib@plt>
  cmp    $0x1,%eax
  jne    L19
  xor    %ecx,%ecx
  mov    $0x14,%edx
  mov    %r13d,%esi
  mov    %r12d,%edi
  call   <accurate_approxlib@plt>
  cmp    $0x1,%eax
  setg   %al
  movzbl %al,%eax
  jmp    L8
```

assistant: ```c
int autohelperowl_attackpat220(struct pattern *patt, int transformation, int move, int color, int action) {
  int A;
  (void)patt;
  (void)color;
  (void)action;
  A = offset(1, 0, move, transformation);
  return countstones(A) <= 3 && countlib(A) == 1 && accurate_approxlib(move, color, 20, NULL) > 1;
}
```
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    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x30(%rbp)
  movl   $0x9,-0x2c(%rbp)
  movl   $0x2,-0x28(%rbp)
  movl   $0x8,-0x24(%rbp)
  movl   $0x3,-0x20(%rbp)
  movl   $0x7,-0x1c(%rbp)
  movl   $0x4,-0x18(%rbp)
  movl   $0x6,-0x14(%rbp)
  movl   $0x5,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0xa,-0x34(%rbp)
  lea    -0x30(%rbp),%rax
  mov    $0x9,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <quick_sort@plt>
  mov    -0x34(%rbp),%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
int main() {
  int a[] = {1, 9, 2, 8, 3, 7, 4, 6, 5, 0};
  int n = sizeof(a) / sizeof(int);
  quick_sort(a, 0, 9);
  print(a, n);
  return 0;
}
```