ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<flip_C1>:
  endbr64
  movabs $0x4000000000000000,%rdx
  mov    %rdi,%rax
  push   %r14
  movabs $0x20202020202000,%r14
  push   %r13
  shr    %rax
  movabs $0x2000000000000000,%r13
  push   %r12
  mov    %rdi,%r12
  and    %rsi,%rax
  shl    $0x7,%r12
  and    %rdx,%rax
  push   %rbp
  mov    %rsi,%rbp
  movabs $0x40000000000000,%rdx
  and    %rsi,%r12
  push   %rbx
  mov    %rdi,%rbx
  movabs $0x202020202020,%rdi
  and    %rdx,%r12
  movabs $0x1fdfdfdfdfdfc000,%rdx
  or     %rax,%r12
  mov    %rsi,%rax
  and    %rbx,%rdi
  and    %r14,%rax
  or     %rdx,%rax
  mov    %rdi,%rdx
  shl    $0x8,%rdx
  add    %rdx,%rax
  test   %r13,%rax
  jne    L48
L45:
  movabs $0x80402010000,%rdi
  mov    %rbp,%rax
  movabs $0x10080402000000,%r14
  movabs $0x1feff7fbfc000000,%rdx
  movabs $0x2000000000000000,%r13
  and    %rbx,%rdi
  and    %r14,%rax
  or     %rdx,%rax
  mov    %rdi,%rdx
  shl    $0x9,%rdx
  add    %rdx,%rax
  test   %r13,%rax
  jne    L25
L43:
  movabs $0xf00000000000000,%rax
  and    %rax,%rbx
  lea    0x0(%rbp,%rbx,2),%rax
  bt     $0x3d,%rax
  jb     L36
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L48:
  call   <last_bit@plt>
  mov    $0x2,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  sub    %rdx,%r13
  and    %r14,%r13
  or     %r13,%r12
  jmp    L45
  nopl   (%rax)
L36:
  mov    %rbx,%rdi
  call   <last_bit@plt>
  mov    $0x2,%edx
  pop    %rbx
  pop    %rbp
  mov    %eax,%ecx
  movabs $0x2000000000000000,%rax
  shl    %cl,%rdx
  sub    %rdx,%rax
  or     %rax,%r12
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L25:
  call   <last_bit@plt>
  mov    $0x2,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  sub    %rdx,%r13
  and    %r14,%r13
  or     %r13,%r12
  jmp    L43
```

assistant: ```c
unsigned long long flip_C1(const unsigned long long discs_player, const unsigned long long discs_opponent) {
  unsigned long long my_discs, opp_discs, r;
  unsigned long long flipped = (discs_opponent & 0X4000000000000000ULL) & (discs_player >> 1);
  flipped |= (discs_opponent & 0X0040000000000000ULL) & (discs_player << 7);
  my_discs = discs_player & 0X0000202020202020ULL;
  opp_discs = (discs_opponent & 0X0020202020202000ULL) | 0X1FDFDFDFDFDFC000ULL;
  r = ((my_discs << 8) + opp_discs) & 0X2000000000000000ULL;
  if (r != 0) {
    flipped |= (0X2000000000000000ULL - (0X2ULL << last_bit(my_discs))) & 0X0020202020202000ULL;
  }
  my_discs = discs_player & 0X0000080402010000ULL;
  opp_discs = (discs_opponent & 0X0010080402000000ULL) | 0X1FEFF7FBFC000000ULL;
  r = ((my_discs << 9) + opp_discs) & 0X2000000000000000ULL;
  if (r != 0) {
    flipped |= (0X2000000000000000ULL - (0X2ULL << last_bit(my_discs))) & 0X0010080402000000ULL;
  }
  my_discs = discs_player & 0X0F00000000000000ULL;
  r = ((my_discs << 1) + discs_opponent) & 0X2000000000000000ULL;
  if (r != 0) {
    flipped |= 0X2000000000000000ULL - (0X2ULL << last_bit(my_discs));
  }
  return flipped;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<flip_C1>:
  endbr64
  movabs $0x4000000000000000,%rdx
  mov    %rdi,%rax
  push   %r14
  movabs $0x20202020202000,%r14
  push   %r13
  shr    %rax
  movabs $0x2000000000000000,%r13
  push   %r12
  mov    %rdi,%r12
  and    %rsi,%rax
  shl    $0x7,%r12
  and    %rdx,%rax
  push   %rbp
  mov    %rsi,%rbp
  movabs $0x40000000000000,%rdx
  and    %rsi,%r12
  push   %rbx
  mov    %rdi,%rbx
  movabs $0x202020202020,%rdi
  and    %rdx,%r12
  movabs $0x1fdfdfdfdfdfc000,%rdx
  or     %rax,%r12
  mov    %rsi,%rax
  and    %rbx,%rdi
  and    %r14,%rax
  or     %rdx,%rax
  mov    %rdi,%rdx
  shl    $0x8,%rdx
  add    %rdx,%rax
  test   %r13,%rax
  jne    L32
L37:
  movabs $0x80402010000,%rdi
  mov    %rbp,%rax
  movabs $0x10080402000000,%r14
  movabs $0x1feff7fbfc000000,%rdx
  movabs $0x2000000000000000,%r13
  and    %rbx,%rdi
  and    %r14,%rax
  or     %rdx,%rax
  mov    %rdi,%rdx
  shl    $0x9,%rdx
  add    %rdx,%rax
  test   %r13,%rax
  jne    L18
L14:
  movabs $0xf00000000000000,%rax
  and    %rax,%rbx
  lea    0x0(%rbp,%rbx,2),%rax
  bt     $0x3d,%rax
  jb     L28
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   (%rax)
L32:
  call   <last_bit@plt>
  mov    $0x2,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  sub    %rdx,%r13
  and    %r14,%r13
  or     %r13,%r12
  jmp    L37
  nopl   (%rax)
L28:
  mov    %rbx,%rdi
  call   <last_bit@plt>
  mov    $0x2,%edx
  pop    %rbx
  pop    %rbp
  mov    %eax,%ecx
  movabs $0x2000000000000000,%rax
  shl    %cl,%rdx
  sub    %rdx,%rax
  or     %rax,%r12
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L18:
  call   <last_bit@plt>
  mov    $0x2,%edx
  mov    %eax,%ecx
  shl    %cl,%rdx
  sub    %rdx,%r13
  and    %r14,%r13
  or     %r13,%r12
  jmp    L14
```

assistant: ```c
unsigned long long flip_C1(const unsigned long long discs_player, const unsigned long long discs_opponent) {
  unsigned long long my_discs, opp_discs, r;
  unsigned long long flipped = (discs_opponent & 0X4000000000000000ULL) & (discs_player >> 1);
  flipped |= (discs_opponent & 0X0040000000000000ULL) & (discs_player << 7);
  my_discs = discs_player & 0X0000202020202020ULL;
  opp_discs = (discs_opponent & 0X0020202020202000ULL) | 0X1FDFDFDFDFDFC000ULL;
  r = ((my_discs << 8) + opp_discs) & 0X2000000000000000ULL;
  if (r != 0) {
    flipped |= (0X2000000000000000ULL - (0X2ULL << last_bit(my_discs))) & 0X0020202020202000ULL;
  }
  my_discs = discs_player & 0X0000080402010000ULL;
  opp_discs = (discs_opponent & 0X0010080402000000ULL) | 0X1FEFF7FBFC000000ULL;
  r = ((my_discs << 9) + opp_discs) & 0X2000000000000000ULL;
  if (r != 0) {
    flipped |= (0X2000000000000000ULL - (0X2ULL << last_bit(my_discs))) & 0X0010080402000000ULL;
  }
  my_discs = discs_player & 0X0F00000000000000ULL;
  r = ((my_discs << 1) + discs_opponent) & 0X2000000000000000ULL;
  if (r != 0) {
    flipped |= 0X2000000000000000ULL - (0X2ULL << last_bit(my_discs));
  }
  return flipped;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <test_length@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  test_length();
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <test_length@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  test_length();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_length@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  test_length();
  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
  xor    %eax,%eax
  call   <test_length@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, const char *argv[]) {
  test_length();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   $0x7
  push   $0x6
  mov    $0x5,%r9d
  mov    $0x4,%r8d
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <test@plt>
  add    $0x10,%rsp
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  test(0, 1, 2, 3, 4, 5, 6, 7);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  push   $0x7
  push   $0x6
  mov    $0x5,%r9d
  mov    $0x4,%r8d
  mov    $0x3,%ecx
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    $0x0,%edi
  call   <test@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int main(void) {
  test(0, 1, 2, 3, 4, 5, 6, 7);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%ecx
  mov    $0x2,%edx
  xor    %edi,%edi
  push   $0x7
  mov    $0x5,%r9d
  mov    $0x4,%r8d
  mov    $0x1,%esi
  push   $0x6
  call   <test@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int main(void) {
  test(0, 1, 2, 3, 4, 5, 6, 7);
  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
  sub    $0x8,%rsp
  mov    $0x3,%ecx
  mov    $0x2,%edx
  xor    %edi,%edi
  push   $0x7
  mov    $0x5,%r9d
  mov    $0x4,%r8d
  mov    $0x1,%esi
  push   $0x6
  call   <test@plt>
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
```

assistant: ```c
int main(void) {
  test(0, 1, 2, 3, 4, 5, 6, 7);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D0
data_type: string
name: disassemble_data
arguments: tag: D17
data_type: string

tool: D0: 
Usage: xtemp


tool: D17:        xtemp -h



assistant: ```c
void Usage() {
  printf("\nUsage: xtemp\n");
  printf("       xtemp -h\n\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D15
data_type: string
name: read_data
arguments: tag: D16
data_type: string

tool: D15: 
Usage: xtemp


tool: D16:        xtemp -h



assistant: ```c
void Usage() {
  printf("\nUsage: xtemp\n");
  printf("       xtemp -h\n\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<Usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    D1(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D7
type_infered: string
name: retrieve_data
arguments: data_label: D1
type_infered: string

tool: D7: 
Usage: xtemp


tool: D1:        xtemp -h



assistant: ```c
void Usage() {
  printf("\nUsage: xtemp\n");
  printf("       xtemp -h\n\n");
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: label: D8
data_type: string
name: parse_data
arguments: label: D0
data_type: string

tool: D8: 
Usage: xtemp


tool: D0:        xtemp -h



assistant: ```c
void Usage() {
  printf("\nUsage: xtemp\n");
  printf("       xtemp -h\n\n");
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<voltea_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L19
L5:
  mov    -0xc(%rbp),%eax
  cltq
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  sub    %rdx,%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x10(%rbp)
L19:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L5
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *voltea_string(char *s) {
  char *p;
  int c = strlen(s), i;
  p = malloc(strlen(s));
  i = 0;
  while (i < c) {
    *(p + i) = *(s + c - i - 1);
    i++;
  }
  return p;
}
```
user: What is the c source code of the assembly code below:

```asm
<voltea_string>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L2
  movslq %ebx,%rsi
  lea    -0x1(%rbp,%rsi,1),%rdx
  mov    %rax,%rcx
  lea    -0x2(%rbp,%rsi,1),%rdi
  lea    -0x1(%rbx),%esi
  sub    %rsi,%rdi
L4:
  movzbl (%rdx),%esi
  mov    %sil,(%rcx)
  sub    $0x1,%rdx
  add    $0x1,%rcx
  cmp    %rdi,%rdx
  jne    L4
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *voltea_string(char *s) {
  char *p;
  int c = strlen(s), i;
  p = malloc(strlen(s));
  i = 0;
  while (i < c) {
    *(p + i) = *(s + c - i - 1);
    i++;
  }
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<voltea_string>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L1
  movslq %ebx,%rsi
  mov    %rax,%rcx
  lea    -0x2(%rbp,%rsi,1),%rdi
  lea    -0x1(%rbp,%rsi,1),%rdx
  lea    -0x1(%rbx),%esi
  sub    %rsi,%rdi
  nopw   0x0(%rax,%rax,1)
L15:
  movzbl (%rdx),%esi
  sub    $0x1,%rdx
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  cmp    %rdi,%rdx
  jne    L15
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *voltea_string(char *s) {
  char *p;
  int c = strlen(s), i;
  p = malloc(strlen(s));
  i = 0;
  while (i < c) {
    *(p + i) = *(s + c - i - 1);
    i++;
  }
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<voltea_string>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L12
  movslq %ebx,%rsi
  mov    %rax,%rcx
  lea    -0x2(%rbp,%rsi,1),%rdi
  lea    -0x1(%rbp,%rsi,1),%rdx
  lea    -0x1(%rbx),%esi
  sub    %rsi,%rdi
  nopw   0x0(%rax,%rax,1)
L1:
  movzbl (%rdx),%esi
  sub    $0x1,%rdx
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  cmp    %rdi,%rdx
  jne    L1
L12:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *voltea_string(char *s) {
  char *p;
  int c = strlen(s), i;
  p = malloc(strlen(s));
  i = 0;
  while (i < c) {
    *(p + i) = *(s + c - i - 1);
    i++;
  }
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bits_disjoint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L27:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  and    %edx,%eax
  test   %eax,%eax
  je     L37
  mov    $0x1,%eax
  jmp    L5
L37:
  addl   $0x1,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L27
  mov    $0x0,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int bits_disjoint(const int *a, const int *b, const int bit_size) {
  for (int i = 0; i < bit_size; ++i)
    if (a[i] & b[i])
      return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<bits_disjoint>:
  endbr64
  test   %edx,%edx
  jle    L18
  mov    %edx,%edx
  mov    $0x0,%ecx
L26:
  mov    (%rdi,%rcx,4),%eax
  and    (%rsi,%rcx,4),%eax
  jne    L8
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L26
  ret
L18:
  mov    $0x0,%eax
  ret
L8:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int bits_disjoint(const int *a, const int *b, const int bit_size) {
  for (int i = 0; i < bit_size; ++i)
    if (a[i] & b[i])
      return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bits_disjoint>:
  endbr64
  test   %edx,%edx
  jle    L7
  movslq %edx,%rdx
  xor    %ecx,%ecx
  jmp    L15
  nop
L8:
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L18
L15:
  mov    (%rdi,%rcx,4),%eax
  and    (%rsi,%rcx,4),%eax
  je     L8
  mov    $0x1,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  ret
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
int bits_disjoint(const int *a, const int *b, const int bit_size) {
  for (int i = 0; i < bit_size; ++i)
    if (a[i] & b[i])
      return 1;
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bits_disjoint>:
  endbr64
  test   %edx,%edx
  jle    L6
  movslq %edx,%rdx
  xor    %ecx,%ecx
  jmp    L17
  nop
L13:
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L28
L17:
  mov    (%rdi,%rcx,4),%eax
  and    (%rsi,%rcx,4),%eax
  je     L13
  mov    $0x1,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L28:
  ret
L6:
  xor    %eax,%eax
  ret
```

assistant: ```c
int bits_disjoint(const int *a, const int *b, const int bit_size) {
  for (int i = 0; i < bit_size; ++i)
    if (a[i] & b[i])
      return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x120,%rsp
  mov    %edi,-0x114(%rbp)
  mov    %rsi,-0x120(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x110(%rbp),%rax
  mov    $0x100,%esi
  mov    %rax,%rdi
  call   <getcwd@plt>
  lea    -0x110(%rbp),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

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

tool: D1: %s


assistant: ```c
int main(int argc, char **argv) {
  char dirname[256];
  getcwd(dirname, 256);
  printf("%s\n", dirname);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    $0x100,%esi
  mov    %rbx,%rdi
  call   <getcwd@plt>
  mov    %rbx,%rsi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x110,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D7
datatype: string

tool: D7: %s


assistant: ```c
int main(int argc, char **argv) {
  char dirname[256];
  getcwd(dirname, 256);
  printf("%s\n", dirname);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x100,%esi
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <getcwd@plt>
  mov    %rbp,%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x110,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D0
data_type: string

tool: D0: %s


assistant: ```c
int main(int argc, char **argv) {
  char dirname[256];
  getcwd(dirname, 256);
  printf("%s\n", dirname);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x100,%esi
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <getcwd@plt>
  mov    %rbp,%rsi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x110,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D9
type: string

tool: D9: %s


assistant: ```c
int main(int argc, char **argv) {
  char dirname[256];
  getcwd(dirname, 256);
  printf("%s\n", dirname);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<startsWith_nocase>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L87
  cmpq   $0x0,-0x30(%rbp)
  jne    L33
L87:
  mov    $0x0,%eax
  jmp    L38
L33:
  mov    -0x28(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jne    L102
  mov    $0x1,%eax
  jmp    L38
L102:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L110
  mov    $0x0,%eax
  jmp    L38
L110:
  movl   $0x0,-0x14(%rbp)
  jmp    L76
L127:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x16(%rbp)
  cmpb   $0x40,-0x16(%rbp)
  jle    L69
  cmpb   $0x5a,-0x16(%rbp)
  jg     L69
  movzbl -0x16(%rbp),%eax
  add    $0x20,%eax
  mov    %al,-0x16(%rbp)
L69:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x15(%rbp)
  cmpb   $0x40,-0x15(%rbp)
  jle    L60
  cmpb   $0x5a,-0x15(%rbp)
  jg     L60
  movzbl -0x15(%rbp),%eax
  add    $0x20,%eax
  mov    %al,-0x15(%rbp)
L60:
  movzbl -0x16(%rbp),%eax
  cmp    -0x15(%rbp),%al
  je     L135
  mov    $0x0,%eax
  jmp    L38
L135:
  addl   $0x1,-0x14(%rbp)
L76:
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x8(%rbp)
  ja     L127
  mov    $0x1,%eax
L38:
  leave
  ret
```

assistant: ```c
int startsWith_nocase(const char *str, const char *with) {
  if (str == NULL || with == NULL)
    return 0;
  if (str == with)
    return 1;
  size_t l1 = strlen(str);
  size_t l2 = strlen(with);
  if (l1 < l2)
    return 0;
  for (int i = 0; i < l2; i++) {
    char s1 = str[i];
    if (s1 >= 'A' && s1 <= 'Z')
      s1 += ' ';
    char s2 = with[i];
    if (s2 >= 'A' && s2 <= 'Z')
      s2 += ' ';
    if (s1 != s2) {
      return 0;
    }
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<startsWith_nocase>:
  endbr64
  test   %rdi,%rdi
  je     L91
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  test   %rsi,%rsi
  je     L50
  cmp    %rsi,%rdi
  je     L17
  call   <strlen@plt>
  mov    %rax,%r12
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    $0x0,%edx
  cmp    %rax,%r12
  jb     L25
  test   %rax,%rax
  je     L108
  mov    $0x0,%esi
L106:
  movzbl (%rbx,%rsi,1),%ecx
  lea    -0x41(%rcx),%edi
  lea    0x20(%rcx),%edx
  cmp    $0x1a,%dil
  cmovb  %edx,%ecx
  movzbl 0x0(%rbp,%rsi,1),%edx
  lea    -0x41(%rdx),%r8d
  lea    0x20(%rdx),%edi
  cmp    $0x1a,%r8b
  cmovb  %edi,%edx
  cmp    %dl,%cl
  jne    L40
  add    $0x1,%rsi
  cmp    %rsi,%rax
  jne    L106
  mov    $0x1,%edx
  jmp    L25
L50:
  mov    $0x0,%edx
L25:
  mov    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  mov    $0x1,%edx
  jmp    L25
L108:
  mov    $0x1,%edx
  jmp    L25
L40:
  mov    $0x0,%edx
  jmp    L25
L91:
  mov    $0x0,%edx
  mov    %edx,%eax
  ret
```

assistant: ```c
int startsWith_nocase(const char *str, const char *with) {
  if (str == NULL || with == NULL)
    return 0;
  if (str == with)
    return 1;
  size_t l1 = strlen(str);
  size_t l2 = strlen(with);
  if (l1 < l2)
    return 0;
  for (int i = 0; i < l2; i++) {
    char s1 = str[i];
    if (s1 >= 'A' && s1 <= 'Z')
      s1 += ' ';
    char s2 = with[i];
    if (s2 >= 'A' && s2 <= 'Z')
      s2 += ' ';
    if (s1 != s2) {
      return 0;
    }
  }
  return 1;
}
```