ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  movl   $0x0,-0x4(%rbp)
  jmp    L16
L39:
  cmpl   $0x5,-0x4(%rbp)
  je     L15
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L33
L15:
  nop
L33:
  addl   $0x1,-0x4(%rbp)
L16:
  cmpl   $0xa,-0x4(%rbp)
  jle    L39
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D3
type: string

tool: D3: %d


assistant: ```c
int main(void) {
  for (int i = 0; i <= 10; i++) {
    if (i == 5) {
      continue;
    }
    printf("%d\n", i);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebp
  mov    $0x0,%ebx
  lea    D8(%rip),%r12
  jmp    L1
L4:
  add    $0x1,%ebx
  add    $0x1,%ebp
L1:
  cmp    $0x5,%ebx
  je     L4
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  cmp    $0xa,%ebp
  jle    L4
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: tag: D8
data_type: string

tool: D8: %d


assistant: ```c
int main(void) {
  for (int i = 0; i <= 10; i++) {
    if (i == 5) {
      continue;
    }
    printf("%d\n", i);
  }
  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
  push   %rbp
  xor    %esi,%esi
  lea    D9(%rip),%rbp
  push   %rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L14:
  lea    0x1(%rsi),%ebx
  cmp    $0x5,%esi
  je     L17
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0xb,%ebx
  je     L8
L17:
  mov    %ebx,%esi
  jmp    L14
  nopl   0x0(%rax)
L8:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
data_type: string

tool: D9: %d


assistant: ```c
int main(void) {
  for (int i = 0; i <= 10; i++) {
    if (i == 5) {
      continue;
    }
    printf("%d\n", i);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  xor    %esi,%esi
  lea    D3(%rip),%rbp
  push   %rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L21:
  lea    0x1(%rsi),%ebx
  cmp    $0x5,%esi
  je     L24
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0xb,%ebx
  je     L5
L24:
  mov    %ebx,%esi
  jmp    L21
  nopl   0x0(%rax)
L5:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
datatype: string

tool: D3: %d


assistant: ```c
int main(void) {
  for (int i = 0; i <= 10; i++) {
    if (i == 5) {
      continue;
    }
    printf("%d\n", i);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ext4_resize_project>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %rdx,-0x58(%rbp)
  mov    %rcx,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    -0x34(%rbp),%rdx
  mov    -0x50(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <get_project@plt>
  test   %rax,%rax
  je     L1
  mov    $0xffffffff,%eax
  jmp    L38
L1:
  lea    -0x30(%rbp),%rax
  mov    $0x18,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    D18(%rip),%rax
  mov    (%rax),%rdx
  mov    -0x58(%rbp),%rax
  add    %rdx,%rax
  lea    -0x1(%rax),%rcx
  mov    D18(%rip),%rax
  mov    (%rax),%rbx
  mov    %rcx,%rax
  mov    $0x0,%edx
  div    %rbx
  mov    %rax,-0x30(%rbp)
  mov    -0x60(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    D48(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x34(%rbp),%ebx
  mov    D46(%rip),%rax
  mov    (%rax),%eax
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <QCMD@plt>
  mov    %eax,%edi
  lea    -0x30(%rbp),%rdx
  mov    -0x48(%rbp),%rax
  mov    %rdx,%rcx
  mov    %ebx,%edx
  mov    %rax,%rsi
  call   <quotactl@plt>
  test   %rax,%rax
  je     L10
  mov    $0xffffffff,%eax
  jmp    L38
L10:
  mov    D47(%rip),%rax
  mov    (%rax),%eax
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <QCMD@plt>
  mov    %eax,%edi
  mov    -0x48(%rbp),%rax
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %rax,%rsi
  call   <quotactl@plt>
  mov    $0x0,%eax
L38:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int ext4_resize_project(const char *device, const char *path, unsigned long long max_bytes, unsigned long long max_inodes) {
  struct if_dqblk quota;
  unsigned project;
  if (get_project(path, &project))
    return -1;
  memset(&quota, 0, sizeof(quota));
  quota.dqb_bhardlimit = (max_bytes + QIF_DQBLKSIZE - 1) / QIF_DQBLKSIZE;
  quota.dqb_ihardlimit = max_inodes;
  quota.dqb_valid = QIF_LIMITS;
  if (quotactl(QCMD(Q_SETQUOTA, 2), device, project, (caddr_t)&quota))
    return -1;
  quotactl(QCMD(Q_SYNC, 2), device, 0, NULL);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ext4_resize_project>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rdi
  mov    %rdx,%r12
  mov    %rcx,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <get_project@plt>
  test   %rax,%rax
  jne    L2
  lea    0x10(%rsp),%r13
  mov    $0x18,%edx
  mov    $0x0,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  mov    D28(%rip),%rax
  mov    (%rax),%rcx
  lea    -0x1(%rcx,%r12,1),%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rax,0x10(%rsp)
  mov    %rbp,0x18(%rsp)
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x20(%rsp)
  mov    0xc(%rsp),%ebp
  mov    $0x2,%esi
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  call   <QCMD@plt>
  mov    %eax,%edi
  mov    %r13,%rcx
  mov    %ebp,%edx
  mov    %rbx,%rsi
  call   <quotactl@plt>
  test   %rax,%rax
  jne    L25
  mov    $0x2,%esi
  mov    D31(%rip),%rax
  mov    (%rax),%edi
  call   <QCMD@plt>
  mov    %eax,%edi
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    %rbx,%rsi
  call   <quotactl@plt>
  mov    $0x0,%eax
L44:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  mov    $0xffffffff,%eax
  jmp    L44
L25:
  mov    $0xffffffff,%eax
  jmp    L44
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ext4_resize_project(const char *device, const char *path, unsigned long long max_bytes, unsigned long long max_inodes) {
  struct if_dqblk quota;
  unsigned project;
  if (get_project(path, &project))
    return -1;
  memset(&quota, 0, sizeof(quota));
  quota.dqb_bhardlimit = (max_bytes + QIF_DQBLKSIZE - 1) / QIF_DQBLKSIZE;
  quota.dqb_ihardlimit = max_inodes;
  quota.dqb_valid = QIF_LIMITS;
  if (quotactl(QCMD(Q_SETQUOTA, 2), device, project, (caddr_t)&quota))
    return -1;
  quotactl(QCMD(Q_SYNC, 2), device, 0, NULL);
  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
<ext4_resize_project>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <get_project@plt>
  test   %rax,%rax
  jne    L1
  lea    0x10(%rsp),%r13
  xor    %esi,%esi
  mov    $0x18,%edx
  mov    %r13,%rdi
  call   <memset@plt>
  mov    D38(%rip),%rax
  xor    %edx,%edx
  mov    $0x2,%esi
  mov    %rbx,0x18(%rsp)
  mov    (%rax),%rcx
  lea    -0x1(%rcx,%r12,1),%rax
  mov    0xc(%rsp),%r12d
  div    %rcx
  mov    %rax,0x10(%rsp)
  mov    D25(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x20(%rsp)
  mov    D10(%rip),%rax
  mov    (%rax),%edi
  call   <QCMD@plt>
  mov    %r13,%rcx
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %eax,%edi
  call   <quotactl@plt>
  test   %rax,%rax
  jne    L1
  mov    D31(%rip),%rax
  mov    $0x2,%esi
  mov    (%rax),%edi
  call   <QCMD@plt>
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %rbp,%rsi
  mov    %eax,%edi
  call   <quotactl@plt>
  xor    %eax,%eax
L12:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  mov    $0xffffffff,%eax
  jmp    L12
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ext4_resize_project(const char *device, const char *path, unsigned long long max_bytes, unsigned long long max_inodes) {
  struct if_dqblk quota;
  unsigned project;
  if (get_project(path, &project))
    return -1;
  memset(&quota, 0, sizeof(quota));
  quota.dqb_bhardlimit = (max_bytes + QIF_DQBLKSIZE - 1) / QIF_DQBLKSIZE;
  quota.dqb_ihardlimit = max_inodes;
  quota.dqb_valid = QIF_LIMITS;
  if (quotactl(QCMD(Q_SETQUOTA, 2), device, project, (caddr_t)&quota))
    return -1;
  quotactl(QCMD(Q_SYNC, 2), device, 0, NULL);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ext4_resize_project>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x48,%rsp
  mov    %rcx,0x8(%rsp)
  lea    0x1c(%rsp),%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <get_project@plt>
  test   %rax,%rax
  jne    L21
  lea    0x20(%rsp),%r12
  xor    %esi,%esi
  mov    $0x18,%edx
  mov    %r12,%rdi
  call   <memset@plt>
  mov    D3(%rip),%rax
  xor    %edx,%edx
  mov    0x1c(%rsp),%r13d
  mov    $0x2,%esi
  mov    (%rax),%rcx
  lea    -0x1(%rcx,%rbx,1),%rax
  div    %rcx
  movq   %rax,%xmm0
  mov    D2(%rip),%rax
  movhps 0x8(%rsp),%xmm0
  mov    (%rax),%eax
  movaps %xmm0,0x20(%rsp)
  mov    %eax,0x30(%rsp)
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <QCMD@plt>
  mov    %r12,%rcx
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %eax,%edi
  call   <quotactl@plt>
  test   %rax,%rax
  jne    L21
  mov    D21(%rip),%rax
  mov    $0x2,%esi
  mov    (%rax),%edi
  call   <QCMD@plt>
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    %rbp,%rsi
  mov    %eax,%edi
  call   <quotactl@plt>
  xor    %eax,%eax
L30:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  xchg   %ax,%ax
L21:
  mov    $0xffffffff,%eax
  jmp    L30
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ext4_resize_project(const char *device, const char *path, unsigned long long max_bytes, unsigned long long max_inodes) {
  struct if_dqblk quota;
  unsigned project;
  if (get_project(path, &project))
    return -1;
  memset(&quota, 0, sizeof(quota));
  quota.dqb_bhardlimit = (max_bytes + QIF_DQBLKSIZE - 1) / QIF_DQBLKSIZE;
  quota.dqb_ihardlimit = max_inodes;
  quota.dqb_valid = QIF_LIMITS;
  if (quotactl(QCMD(Q_SETQUOTA, 2), device, project, (caddr_t)&quota))
    return -1;
  quotactl(QCMD(Q_SYNC, 2), device, 0, NULL);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<printCharBitwise>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L3:
  movsbl -0x14(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  and    $0x80,%eax
  test   %eax,%eax
  setne  %al
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L4:
  cmpl   $0x7,-0x4(%rbp)
  jle    L3
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D6
data_type: string

tool: D6: %d

assistant: ```c
void printCharBitwise(char c) {
  int i;
  for (i = 0; i < 8; i++) {
    printf("%d", !!((c << i) & 0x80));
  }
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printCharBitwise>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    $0x0,%ebx
  lea    D3(%rip),%r12
L1:
  movsbl %bpl,%esi
  mov    %ebx,%ecx
  shl    %cl,%esi
  shr    $0x7,%esi
  and    $0x1,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x8,%ebx
  jne    L1
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D3
type_guessed: string

tool: D3: %d

assistant: ```c
void printCharBitwise(char c) {
  int i;
  for (i = 0; i < 8; i++) {
    printf("%d", !!((c << i) & 0x80));
  }
  printf("\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<printCharBitwise>:
  endbr64
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  movsbl %dil,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   (%rax)
L8:
  mov    %ebx,%ecx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  shl    %cl,%esi
  add    $0x1,%ebx
  shr    $0x7,%esi
  and    $0x1,%esi
  call   <printf@plt>
  cmp    $0x8,%ebx
  jne    L8
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
```

tool_calls: name: read_data
arguments: label: D7
type_infered: string

tool: D7: %d

assistant: ```c
void printCharBitwise(char c) {
  int i;
  for (i = 0; i < 8; i++) {
    printf("%d", !!((c << i) & 0x80));
  }
  printf("\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<printCharBitwise>:
  endbr64
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  movsbl %dil,%ebp
  push   %rbx
  xor    %ebx,%ebx
  nopl   (%rax)
L4:
  mov    %ebx,%ecx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  shl    %cl,%esi
  add    $0x1,%ebx
  shr    $0x7,%esi
  and    $0x1,%esi
  call   <printf@plt>
  cmp    $0x8,%ebx
  jne    L4
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
```

tool_calls: name: read_data
arguments: data_label: D7
type: string

tool: D7: %d

assistant: ```c
void printCharBitwise(char c) {
  int i;
  for (i = 0; i < 8; i++) {
    printf("%d", !!((c << i) & 0x80));
  }
  printf("\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ftello>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ftell@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
off_t ftello(FILE *stream) {
  long pos;
  pos = ftell(stream);
  return (off_t)pos;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<ftello>:
  endbr64
  sub    $0x8,%rsp
  call   <ftell@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
off_t ftello(FILE *stream) {
  long pos;
  pos = ftell(stream);
  return (off_t)pos;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ftello>:
  endbr64
  sub    $0x8,%rsp
  call   <ftell@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
off_t ftello(FILE *stream) {
  long pos;
  pos = ftell(stream);
  return (off_t)pos;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ftello>:
  endbr64
  sub    $0x8,%rsp
  call   <ftell@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
off_t ftello(FILE *stream) {
  long pos;
  pos = ftell(stream);
  return (off_t)pos;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cmd_negateBit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,%eax
  mov    %al,-0x8(%rbp)
  movzbl -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  test   %eax,%eax
  sete   %al
  movzbl %al,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movzbl -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x1,%eax
  test   %eax,%eax
  sete   %al
  movzbl %al,%edx
  movzbl -0x8(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <setBit@plt>
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
type_guessed: string

tool: D4: Neuer Wert des Bits %d: %d


assistant: ```c
unsigned char cmd_negateBit(int workonbit, unsigned char workingbyte) {
  printf("Neuer Wert des Bits %d: %d\n", workonbit, !((workingbyte >> workonbit) & 1));
  return setBit(workonbit, workingbyte, !((workingbyte >> workonbit) & 1));
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<cmd_negateBit>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  movzbl %sil,%ebp
  mov    %ebp,%eax
  mov    %edi,%ecx
  sar    %cl,%eax
  test   $0x1,%al
  sete   %r12b
  movzbl %r12b,%r12d
  mov    %r12d,%edx
  mov    %edi,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %ebx,%edi
  call   <setBit@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D3
data_type: string

tool: D3: Neuer Wert des Bits %d: %d


assistant: ```c
unsigned char cmd_negateBit(int workonbit, unsigned char workingbyte) {
  printf("Neuer Wert des Bits %d: %d\n", workonbit, !((workingbyte >> workonbit) & 1));
  return setBit(workonbit, workingbyte, !((workingbyte >> workonbit) & 1));
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<cmd_negateBit>:
  endbr64
  push   %r13
  movzbl %sil,%r13d
  mov    %edi,%ecx
  mov    %edi,%esi
  push   %r12
  mov    %r13d,%r12d
  sar    %cl,%r12d
  push   %rbp
  mov    %edi,%ebp
  lea    D2(%rip),%rdi
  not    %r12d
  and    $0x1,%r12d
  mov    %r12d,%edx
  call   <printf@plt>
  mov    %r12d,%edx
  mov    %r13d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <setBit@plt>
```

tool_calls: name: disassemble_data
arguments: label: D2
datatype: string

tool: D2: Neuer Wert des Bits %d: %d


assistant: ```c
unsigned char cmd_negateBit(int workonbit, unsigned char workingbyte) {
  printf("Neuer Wert des Bits %d: %d\n", workonbit, !((workingbyte >> workonbit) & 1));
  return setBit(workonbit, workingbyte, !((workingbyte >> workonbit) & 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
<cmd_negateBit>:
  endbr64
  push   %r13
  movzbl %sil,%r13d
  mov    %edi,%ecx
  mov    %edi,%esi
  push   %r12
  mov    %r13d,%r12d
  sar    %cl,%r12d
  push   %rbp
  mov    %edi,%ebp
  lea    D0(%rip),%rdi
  not    %r12d
  and    $0x1,%r12d
  mov    %r12d,%edx
  call   <printf@plt>
  mov    %r12d,%edx
  mov    %r13d,%esi
  mov    %ebp,%edi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <setBit@plt>
```

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

tool: D0: Neuer Wert des Bits %d: %d


assistant: ```c
unsigned char cmd_negateBit(int workonbit, unsigned char workingbyte) {
  printf("Neuer Wert des Bits %d: %d\n", workonbit, !((workingbyte >> workonbit) & 1));
  return setBit(workonbit, workingbyte, !((workingbyte >> workonbit) & 1));
}
```
user: Translate the below assembly code into C programming language:

```asm
<func>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x3,-0x8(%rbp)
  movl   $0x5,-0x4(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int func(void) {
  int n = 3;
  int m = 5;
  n < m;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<func>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int func(void) {
  int n = 3;
  int m = 5;
  n < m;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int func(void) {
  int n = 3;
  int m = 5;
  n < m;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

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

assistant: ```c
int func(void) {
  int n = 3;
  int m = 5;
  n < m;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<mergeSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L26
L10:
  movl   $0x0,-0x14(%rbp)
  jmp    L42
L15:
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0x14(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <doMerge@plt>
  mov    -0x18(%rbp),%eax
  add    %eax,%eax
  add    %eax,-0x14(%rbp)
L42:
  mov    -0x18(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x2c(%rbp)
  jge    L15
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L48
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  sub    $0x1,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%esi
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <doMerge@plt>
L48:
  shll   -0x18(%rbp)
L26:
  mov    -0x18(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L10
  nop
  nop
  leave
  ret
```

assistant: ```c
void mergeSort(int *data, int size) {
  for (int gap = 1; gap < size; gap *= 2) {
    int left = 0;
    for (; left + 2 * gap - 1 < size; left += 2 * gap) {
      int midd = left + gap - 1;
      int right = left + 2 * gap - 1;
      doMerge(data, left, midd, right);
    }
    if (left + gap - 1 < size) {
      int midd = left + gap - 1;
      int right = size - 1;
      doMerge(data, left, midd, right);
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<mergeSort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r14
  mov    %esi,%r12d
  mov    $0x1,%r15d
  cmp    $0x1,%esi
  jg     L12
L26:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L45:
  cmp    %r15d,%r12d
  jle    L26
L12:
  mov    %r15d,%ebp
  add    %r15d,%r15d
  mov    %r15d,0xc(%rsp)
  mov    %r15d,%edx
  lea    -0x1(%rbp),%r13d
  sub    %r15d,%r13d
  cmp    %r15d,%r12d
  jl     L15
L55:
  mov    %edx,%ebx
  sub    0xc(%rsp),%ebx
  mov    %edx,0x8(%rsp)
  lea    -0x1(%rdx),%ecx
  add    %r13d,%edx
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <doMerge@plt>
  lea    (%rbx,%rbp,4),%edx
  cmp    %edx,%r12d
  jge    L55
L64:
  add    0x8(%rsp),%ebp
  cmp    %r12d,%ebp
  jg     L45
  lea    -0x1(%r12),%ecx
  lea    -0x1(%rbp),%edx
  mov    0x8(%rsp),%esi
  mov    %r14,%rdi
  call   <doMerge@plt>
  jmp    L45
L15:
  movl   $0x0,0x8(%rsp)
  jmp    L64
```

assistant: ```c
void mergeSort(int *data, int size) {
  for (int gap = 1; gap < size; gap *= 2) {
    int left = 0;
    for (; left + 2 * gap - 1 < size; left += 2 * gap) {
      int midd = left + gap - 1;
      int right = left + 2 * gap - 1;
      doMerge(data, left, midd, right);
    }
    if (left + gap - 1 < size) {
      int midd = left + gap - 1;
      int right = size - 1;
      doMerge(data, left, midd, right);
    }
  }
}
```