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
  mov    $0x0,%eax
  call   <showcur@plt>
  mov    $0x186a0000,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <showcur@plt>
  mov    $0x5,%edi
  call   <mymalloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D7
type: string

tool: D7:  %p %p


assistant: ```c
int main() {
  showcur();
  void *k = malloc(409600000);
  showcur();
  void *p = mymalloc(5);
  printf(" %p %p\n", k, p);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <showcur@plt>
  mov    $0x186a0000,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
  call   <showcur@plt>
  mov    $0x5,%edi
  call   <mymalloc@plt>
  mov    %rax,%rdx
  mov    %rbx,%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
type_infered: string

tool: D3:  %p %p


assistant: ```c
int main() {
  showcur();
  void *k = malloc(409600000);
  showcur();
  void *p = mymalloc(5);
  printf(" %p %p\n", k, p);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <showcur@plt>
  mov    $0x186a0000,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <showcur@plt>
  mov    $0x5,%edi
  call   <mymalloc@plt>
  mov    %rbp,%rsi
  lea    D3(%rip),%rdi
  mov    %rax,%rdx
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

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

tool: D3:  %p %p


assistant: ```c
int main() {
  showcur();
  void *k = malloc(409600000);
  showcur();
  void *p = mymalloc(5);
  printf(" %p %p\n", k, p);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <showcur@plt>
  mov    $0x186a0000,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  call   <showcur@plt>
  mov    $0x5,%edi
  call   <mymalloc@plt>
  mov    %rbp,%rsi
  lea    D9(%rip),%rdi
  mov    %rax,%rdx
  call   <printf@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D9
type_guessed: string

tool: D9:  %p %p


assistant: ```c
int main() {
  showcur();
  void *k = malloc(409600000);
  showcur();
  void *p = mymalloc(5);
  printf(" %p %p\n", k, p);
  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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0x61,-0xc(%rbp)
  mov    $0xffffffff,%edi
  call   <close@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <chroot@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rdi
  call   <pipe@plt>
  mov    -0x10(%rbp),%eax
  lea    -0xc(%rbp),%rcx
  mov    $0x2,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    -0x14(%rbp),%eax
  lea    -0xa(%rbp),%rcx
  mov    $0x2,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    $0x0,%edi
  call   <_exit@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: disassemble_data
arguments: tag: D7
datatype: string

tool: D7: .

assistant: ```c
int main(void) {
  int fd[2];
  char buf1[2] = "a";
  char buf2[2];
  close(-1);
  chroot(".");
  pipe(fd);
  write(fd[1], buf1, sizeof(buf1));
  read(fd[0], buf2, sizeof(buf2));
  _exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movw   $0x61,0x14(%rsp)
  mov    $0xffffffff,%edi
  call   <close@plt>
  lea    D1(%rip),%rdi
  call   <chroot@plt>
  lea    0xc(%rsp),%rdi
  call   <pipe@plt>
  lea    0x14(%rsp),%rsi
  mov    $0x2,%edx
  mov    0x10(%rsp),%edi
  call   <write@plt>
  lea    0x16(%rsp),%rsi
  mov    $0x2,%edx
  mov    0xc(%rsp),%edi
  call   <read@plt>
  mov    $0x0,%edi
  call   <_exit@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D1
datatype: string

tool: D1: .

assistant: ```c
int main(void) {
  int fd[2];
  char buf1[2] = "a";
  char buf2[2];
  close(-1);
  chroot(".");
  pipe(fd);
  write(fd[1], buf1, sizeof(buf1));
  read(fd[0], buf2, sizeof(buf2));
  _exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0xffffffff,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x61,%eax
  mov    %ax,0x14(%rsp)
  call   <close@plt>
  lea    D5(%rip),%rdi
  call   <chroot@plt>
  lea    0xc(%rsp),%rdi
  call   <pipe@plt>
  mov    0x10(%rsp),%edi
  lea    0x14(%rsp),%rsi
  mov    $0x2,%edx
  call   <write@plt>
  mov    0xc(%rsp),%edi
  lea    0x16(%rsp),%rsi
  mov    $0x2,%edx
  call   <read@plt>
  xor    %edi,%edi
  call   <_exit@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

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

tool: D5: .

assistant: ```c
int main(void) {
  int fd[2];
  char buf1[2] = "a";
  char buf2[2];
  close(-1);
  chroot(".");
  pipe(fd);
  write(fd[1], buf1, sizeof(buf1));
  read(fd[0], buf2, sizeof(buf2));
  _exit(0);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    $0xffffffff,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    $0x61,%eax
  mov    %ax,0x14(%rsp)
  call   <close@plt>
  lea    D5(%rip),%rdi
  call   <chroot@plt>
  lea    0xc(%rsp),%rdi
  call   <pipe@plt>
  mov    0x10(%rsp),%edi
  lea    0x14(%rsp),%rsi
  mov    $0x2,%edx
  call   <write@plt>
  mov    0xc(%rsp),%edi
  lea    0x16(%rsp),%rsi
  mov    $0x2,%edx
  call   <read@plt>
  xor    %edi,%edi
  call   <_exit@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
data_type: string

tool: D5: .

assistant: ```c
int main(void) {
  int fd[2];
  char buf1[2] = "a";
  char buf2[2];
  close(-1);
  chroot(".");
  pipe(fd);
  write(fd[1], buf1, sizeof(buf1));
  read(fd[0], buf2, sizeof(buf2));
  _exit(0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<decode_bit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <decode_range@plt>
  test   %rax,%rax
  je     L4
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    -0x4(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L54
L4:
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  mov    %eax,%edx
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  imul   %edx,%eax
  mov    %eax,-0x4(%rbp)
L54:
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  shr    $0x3,%al
  sub    %eax,%edx
  mov    -0x38(%rbp),%rax
  mov    %dl,(%rax)
  cmpq   $0x0,-0x28(%rbp)
  je     L11
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
L11:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jbe    L14
  mov    -0x18(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x38(%rbp),%rax
  movzbl (%rax),%eax
  lea    0x1f(%rax),%edx
  mov    -0x38(%rbp),%rax
  mov    %dl,(%rax)
  cmpq   $0x0,-0x28(%rbp)
  je     L31
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
L31:
  mov    $0x1,%eax
  jmp    L33
L14:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  sub    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
L33:
  leave
  ret
```

assistant: ```c
int decode_bit(unsigned int *range, unsigned int *code, int *index, unsigned char **src, unsigned char *c) {
  unsigned int val = *range;
  if (decode_range(range, code, src))
    val *= (*c);
  else
    val = (val >> 8) * (*c);
  *c -= ((*c) >> 3);
  if (index)
    (*index) <<= 1;
  if (*code < val) {
    *range = val;
    *c += 31;
    if (index)
      (*index)++;
    return 1;
  } else {
    *code -= val;
    *range -= val;
    return 0;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<decode_bit>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r13
  mov    %rdx,%r12
  mov    %rcx,%rdx
  mov    %r8,%rbx
  mov    (%rdi),%r14d
  call   <decode_range@plt>
  test   %rax,%rax
  je     L30
  movzbl (%rbx),%eax
  imul   %r14d,%eax
L4:
  movzbl (%rbx),%edx
  mov    %edx,%ecx
  shr    $0x3,%cl
  sub    %ecx,%edx
  mov    %dl,(%rbx)
  test   %r12,%r12
  je     L24
  shll   (%r12)
  mov    0x0(%r13),%edx
  cmp    %edx,%eax
  ja     L34
L41:
  sub    %eax,%edx
  mov    %edx,0x0(%r13)
  sub    %eax,0x0(%rbp)
  mov    $0x0,%eax
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L30:
  shr    $0x8,%r14d
  movzbl (%rbx),%eax
  imul   %r14d,%eax
  jmp    L4
L34:
  mov    %eax,0x0(%rbp)
  addb   $0x1f,(%rbx)
  addl   $0x1,(%r12)
  mov    $0x1,%eax
  jmp    L9
L24:
  mov    0x0(%r13),%edx
  cmp    %edx,%eax
  jbe    L41
  mov    %eax,0x0(%rbp)
  addb   $0x1f,(%rbx)
  mov    $0x1,%eax
  jmp    L9
```

assistant: ```c
int decode_bit(unsigned int *range, unsigned int *code, int *index, unsigned char **src, unsigned char *c) {
  unsigned int val = *range;
  if (decode_range(range, code, src))
    val *= (*c);
  else
    val = (val >> 8) * (*c);
  *c -= ((*c) >> 3);
  if (index)
    (*index) <<= 1;
  if (*code < val) {
    *range = val;
    *c += 31;
    if (index)
      (*index)++;
    return 1;
  } else {
    *code -= val;
    *range -= val;
    return 0;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<decode_bit>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    (%rdi),%r13d
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %rdx,%rbp
  mov    %rcx,%rdx
  push   %rbx
  mov    %rdi,%rbx
  call   <decode_range@plt>
  test   %rax,%rax
  jne    L35
  shr    $0x8,%r13d
L35:
  movzbl (%r12),%edx
  mov    %edx,%eax
  imul   %r13d,%edx
  mov    %eax,%ecx
  shr    $0x3,%cl
  sub    %ecx,%eax
  mov    %al,(%r12)
  test   %rbp,%rbp
  je     L22
  shll   0x0(%rbp)
  mov    (%r14),%eax
  cmp    %eax,%edx
  ja     L14
L34:
  sub    %edx,%eax
  mov    %eax,(%r14)
  xor    %eax,%eax
  sub    %edx,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L14:
  mov    %edx,(%rbx)
  mov    $0x1,%eax
  pop    %rbx
  addb   $0x1f,(%r12)
  addl   $0x1,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L22:
  mov    (%r14),%eax
  cmp    %eax,%edx
  jbe    L34
  mov    %edx,(%rbx)
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  addb   $0x1f,(%r12)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int decode_bit(unsigned int *range, unsigned int *code, int *index, unsigned char **src, unsigned char *c) {
  unsigned int val = *range;
  if (decode_range(range, code, src))
    val *= (*c);
  else
    val = (val >> 8) * (*c);
  *c -= ((*c) >> 3);
  if (index)
    (*index) <<= 1;
  if (*code < val) {
    *range = val;
    *c += 31;
    if (index)
      (*index)++;
    return 1;
  } else {
    *code -= val;
    *range -= val;
    return 0;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<decode_bit>:
  endbr64
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    (%rdi),%r13d
  push   %r12
  mov    %r8,%r12
  push   %rbp
  mov    %rdx,%rbp
  mov    %rcx,%rdx
  push   %rbx
  mov    %rdi,%rbx
  call   <decode_range@plt>
  test   %rax,%rax
  jne    L26
  shr    $0x8,%r13d
L26:
  movzbl (%r12),%edx
  mov    %edx,%eax
  imul   %r13d,%edx
  mov    %eax,%ecx
  shr    $0x3,%cl
  sub    %ecx,%eax
  mov    %al,(%r12)
  test   %rbp,%rbp
  je     L8
  shll   0x0(%rbp)
  mov    (%r14),%eax
  cmp    %eax,%edx
  ja     L4
L27:
  sub    %edx,%eax
  mov    %eax,(%r14)
  xor    %eax,%eax
  sub    %edx,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  mov    %edx,(%rbx)
  mov    $0x1,%eax
  pop    %rbx
  addb   $0x1f,(%r12)
  addl   $0x1,0x0(%rbp)
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L8:
  mov    (%r14),%eax
  cmp    %eax,%edx
  jbe    L27
  mov    %edx,(%rbx)
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  addb   $0x1f,(%r12)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int decode_bit(unsigned int *range, unsigned int *code, int *index, unsigned char **src, unsigned char *c) {
  unsigned int val = *range;
  if (decode_range(range, code, src))
    val *= (*c);
  else
    val = (val >> 8) * (*c);
  *c -= ((*c) >> 3);
  if (index)
    (*index) <<= 1;
  if (*code < val) {
    *range = val;
    *c += 31;
    if (index)
      (*index)++;
    return 1;
  } else {
    *code -= val;
    *range -= val;
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strstrc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,%eax
  mov    %al,-0x34(%rbp)
  movsbl -0x34(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L10
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  jmp    L51
L10:
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L11
L38:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L27
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x30(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x28(%rbp),%rax
  add    $0x1,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L27
  mov    -0x28(%rbp),%rax
  jmp    L51
L27:
  addq   $0x1,-0x28(%rbp)
  subq   $0x1,-0x18(%rbp)
L11:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jae    L38
  mov    $0x0,%eax
L51:
  leave
  ret
```

assistant: ```c
char *strstrc(const char *s, const char *p, const char t) {
  char *endptr;
  size_t slen, plen;
  endptr = strchr(s, t);
  if (!endptr)
    return strstr(s, p);
  plen = strlen(p);
  slen = endptr - s;
  while (slen >= plen) {
    if (*s == *p && strncmp(s + 1, p + 1, plen - 1) == 0)
      return (char *)s;
    s++;
    slen--;
  }
  return NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strstrc>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  movsbl %dl,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L42
  mov    %rax,%r12
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%r14
  sub    %r13,%r12
  cmp    %r12,%rax
  ja     L11
  mov    %r13,%rbx
  lea    -0x1(%rax),%r15
  jmp    L17
L42:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strstr@plt>
  mov    %rax,%rbx
  jmp    L23
L36:
  add    $0x1,%rbx
  lea    0x0(%r13,%r12,1),%rax
  sub    %rbx,%rax
  cmp    %r14,%rax
  jb     L51
L17:
  movzbl 0x0(%rbp),%eax
  cmp    %al,(%rbx)
  jne    L36
  lea    0x1(%rbp),%rsi
  lea    0x1(%rbx),%rdi
  mov    %r15,%rdx
  call   <strncmp@plt>
  test   %rax,%rax
  jne    L36
  jmp    L23
L51:
  mov    $0x0,%ebx
L23:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L11:
  mov    $0x0,%ebx
  jmp    L23
```

assistant: ```c
char *strstrc(const char *s, const char *p, const char t) {
  char *endptr;
  size_t slen, plen;
  endptr = strchr(s, t);
  if (!endptr)
    return strstr(s, p);
  plen = strlen(p);
  slen = endptr - s;
  while (slen >= plen) {
    if (*s == *p && strncmp(s + 1, p + 1, plen - 1) == 0)
      return (char *)s;
    s++;
    slen--;
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strstrc>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  movsbl %dl,%esi
  push   %rbx
  sub    $0x18,%rsp
  call   <strchr@plt>
  test   %rax,%rax
  je     L33
  mov    %rax,%r12
  mov    %rbp,%rdi
  mov    %r14,%rbx
  call   <strlen@plt>
  sub    %r14,%r12
  mov    %rax,%r15
  lea    -0x1(%rax),%rcx
  cmp    %r12,%rax
  jbe    L48
  jmp    L43
  nopl   0x0(%rax)
L54:
  lea    (%r14,%r12,1),%rdx
  sub    %rbx,%rdx
  cmp    %r15,%rdx
  jb     L43
L48:
  movzbl (%rbx),%edx
  mov    %rbx,%r13
  add    $0x1,%rbx
  cmp    0x0(%rbp),%dl
  jne    L54
  mov    %rcx,%rdx
  lea    0x1(%rbp),%rsi
  mov    %rbx,%rdi
  mov    %rcx,0x8(%rsp)
  call   <strncmp@plt>
  mov    0x8(%rsp),%rcx
  test   %rax,%rax
  jne    L54
L20:
  add    $0x18,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L43:
  xor    %r13d,%r13d
  jmp    L20
  nopl   (%rax)
L33:
  add    $0x18,%rsp
  mov    %rbp,%rsi
  mov    %r14,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10b0 <strstr@plt>
```

assistant: ```c
char *strstrc(const char *s, const char *p, const char t) {
  char *endptr;
  size_t slen, plen;
  endptr = strchr(s, t);
  if (!endptr)
    return strstr(s, p);
  plen = strlen(p);
  slen = endptr - s;
  while (slen >= plen) {
    if (*s == *p && strncmp(s + 1, p + 1, plen - 1) == 0)
      return (char *)s;
    s++;
    slen--;
  }
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strstrc>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  movsbl %dl,%esi
  push   %rbx
  sub    $0x18,%rsp
  call   <strchr@plt>
  test   %rax,%rax
  je     L30
  mov    %rax,%r12
  mov    %rbp,%rdi
  mov    %r14,%rbx
  call   <strlen@plt>
  sub    %r14,%r12
  mov    %rax,%r15
  lea    -0x1(%rax),%rcx
  cmp    %r12,%rax
  jbe    L32
  jmp    L46
  nopl   0x0(%rax)
L29:
  lea    (%r14,%r12,1),%rdx
  sub    %rbx,%rdx
  cmp    %r15,%rdx
  jb     L46
L32:
  movzbl (%rbx),%edx
  mov    %rbx,%r13
  add    $0x1,%rbx
  cmp    0x0(%rbp),%dl
  jne    L29
  mov    %rcx,%rdx
  lea    0x1(%rbp),%rsi
  mov    %rbx,%rdi
  mov    %rcx,0x8(%rsp)
  call   <strncmp@plt>
  mov    0x8(%rsp),%rcx
  test   %rax,%rax
  jne    L29
L58:
  add    $0x18,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L46:
  xor    %r13d,%r13d
  jmp    L58
  nopl   (%rax)
L30:
  add    $0x18,%rsp
  mov    %rbp,%rsi
  mov    %r14,%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    10b0 <strstr@plt>
```

assistant: ```c
char *strstrc(const char *s, const char *p, const char t) {
  char *endptr;
  size_t slen, plen;
  endptr = strchr(s, t);
  if (!endptr)
    return strstr(s, p);
  plen = strlen(p);
  slen = endptr - s;
  while (slen >= plen) {
    if (*s == *p && strncmp(s + 1, p + 1, plen - 1) == 0)
      return (char *)s;
    s++;
    slen--;
  }
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_most_significant_byte>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x18,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    -0x14(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int get_most_significant_byte(int num) {
  int shift_val = (sizeof(int) - 1) << 3;
  int num_right = num >> shift_val;
  return num_right & 0xFF;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<get_most_significant_byte>:
  endbr64
  mov    %edi,%eax
  shr    $0x18,%eax
  ret
```

assistant: ```c
int get_most_significant_byte(int num) {
  int shift_val = (sizeof(int) - 1) << 3;
  int num_right = num >> shift_val;
  return num_right & 0xFF;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_most_significant_byte>:
  endbr64
  mov    %edi,%eax
  shr    $0x18,%eax
  ret
```

assistant: ```c
int get_most_significant_byte(int num) {
  int shift_val = (sizeof(int) - 1) << 3;
  int num_right = num >> shift_val;
  return num_right & 0xFF;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<get_most_significant_byte>:
  endbr64
  mov    %edi,%eax
  shr    $0x18,%eax
  ret
```

assistant: ```c
int get_most_significant_byte(int num) {
  int shift_val = (sizeof(int) - 1) << 3;
  int num_right = num >> shift_val;
  return num_right & 0xFF;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<example__hash32_n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <hash32_n@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x3,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <hash32_n@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D2
type_infered: string
name: extract_symbolic_data
arguments: label: D16
type_infered: string

tool: 

tool: D16: abc

assistant: ```c
void example__hash32_n(void) {
  assert(hash32_n("", 0) == (2860486313));
  assert(hash32_n("abc", 3) == 2407100781);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<example__hash32_n>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  lea    D8(%rip),%rdi
  call   <hash32_n@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x3,%esi
  lea    D13(%rip),%rdi
  call   <hash32_n@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D8
type: string
name: unpack_data
arguments: tag: D13
type: string

tool: 

tool: D13: abc

assistant: ```c
void example__hash32_n(void) {
  assert(hash32_n("", 0) == (2860486313));
  assert(hash32_n("abc", 3) == 2407100781);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<example__hash32_n>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  lea    D0(%rip),%rdi
  call   <hash32_n@plt>
  xor    %edi,%edi
  call   <assert@plt>
  lea    D5(%rip),%rdi
  mov    $0x3,%esi
  call   <hash32_n@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1060 <assert@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D0
type_infered: string
name: retrieve_data
arguments: data_label: D5
type_infered: string

tool: 

tool: D5: abc

assistant: ```c
void example__hash32_n(void) {
  assert(hash32_n("", 0) == (2860486313));
  assert(hash32_n("abc", 3) == 2407100781);
}
```