ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<model_from_file>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <filesize@plt>
  cmp    $0xffffffffffffffff,%rax
  je     L0
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rbx,%rdx
  mov    %rax,%rsi
  mov    %rbp,%rdi
  call   <read_file_to_mem@plt>
  mov    %rbp,%rdi
  call   <get_filename_ext@plt>
  mov    %rax,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <model_from_mem_buf@plt>
  mov    %rax,%rbx
  mov    %r12,%rdi
  call   <free@plt>
L9:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  mov    $0x0,%ebx
  jmp    L9
```

assistant: ```c
struct model *model_from_file(const char *fpath) {
  long filesz = filesize(fpath);
  if (filesz == -1)
    return 0;
  unsigned char *data_buf = malloc(filesz);
  read_file_to_mem(fpath, data_buf, filesz);
  const char *ext = get_filename_ext(fpath);
  struct model *m = model_from_mem_buf(data_buf, filesz, ext);
  free(data_buf);
  return m;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<model_from_file>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  call   <filesize@plt>
  cmp    $0xffffffffffffffff,%rax
  je     L5
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rbp,%rdx
  mov    %r12,%rdi
  mov    %rax,%rsi
  mov    %rax,%r13
  call   <read_file_to_mem@plt>
  mov    %r12,%rdi
  call   <get_filename_ext@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %rax,%rdx
  call   <model_from_mem_buf@plt>
  mov    %r13,%rdi
  mov    %rax,%r12
  call   <free@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L5:
  xor    %r12d,%r12d
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
struct model *model_from_file(const char *fpath) {
  long filesz = filesize(fpath);
  if (filesz == -1)
    return 0;
  unsigned char *data_buf = malloc(filesz);
  read_file_to_mem(fpath, data_buf, filesz);
  const char *ext = get_filename_ext(fpath);
  struct model *m = model_from_mem_buf(data_buf, filesz, ext);
  free(data_buf);
  return m;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<model_from_file>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  call   <filesize@plt>
  cmp    $0xffffffffffffffff,%rax
  je     L2
  mov    %rax,%rdi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rbp,%rdx
  mov    %r12,%rdi
  mov    %rax,%rsi
  mov    %rax,%r13
  call   <read_file_to_mem@plt>
  mov    %r12,%rdi
  call   <get_filename_ext@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  mov    %rax,%rdx
  call   <model_from_mem_buf@plt>
  mov    %r13,%rdi
  mov    %rax,%r12
  call   <free@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L2:
  xor    %r12d,%r12d
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
struct model *model_from_file(const char *fpath) {
  long filesz = filesize(fpath);
  if (filesz == -1)
    return 0;
  unsigned char *data_buf = malloc(filesz);
  read_file_to_mem(fpath, data_buf, filesz);
  const char *ext = get_filename_ext(fpath);
  struct model *m = model_from_mem_buf(data_buf, filesz, ext);
  free(data_buf);
  return m;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<register_device_store>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  mov    $0x0,%eax
  call   <register_input_dev@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
ssize_t register_device_store(struct kobject *kobj, struct kobj_attribute *attr, const char *buf, size_t count) {
  register_input_dev();
  return 0;
}
```
user: What is the c source code of the assembly code below?

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

assistant: ```c
ssize_t register_device_store(struct kobject *kobj, struct kobj_attribute *attr, const char *buf, size_t count) {
  register_input_dev();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
ssize_t register_device_store(struct kobject *kobj, struct kobj_attribute *attr, const char *buf, size_t count) {
  register_input_dev();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

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

assistant: ```c
ssize_t register_device_store(struct kobject *kobj, struct kobj_attribute *attr, const char *buf, size_t count) {
  register_input_dev();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_input_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    D30(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <fflush@plt>
  mov    D48(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  mov    D10(%rip),%rax
  mov    (%rax),%edx
  mov    D48(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x8(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L5
L7:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L24
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L24:
  addl   $0x1,-0xc(%rbp)
L5:
  mov    D48(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L7
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *get_input_string() {
  fflush(stdout);
  char *input = malloc(sizeof(char) * BUFSIZ);
  assert(input);
  fgets(input, BUFSIZ, stdin);
  int i;
  for (i = 0; i < BUFSIZ; i++) {
    if (input[i] == '\n') {
      input[i] = '\0';
    }
  }
  return input;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_input_string>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D30(%rip),%rax
  mov    (%rax),%edi
  call   <fflush@plt>
  mov    D6(%rip),%rbp
  mov    0x0(%rbp),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <assert@plt>
  mov    D12(%rip),%rax
  mov    (%rax),%edx
  mov    0x0(%rbp),%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  cmpl   $0x0,0x0(%rbp)
  jle    L1
  mov    $0x0,%eax
  mov    D6(%rip),%rdx
  jmp    L49
L24:
  add    $0x1,%rax
  cmp    %eax,(%rdx)
  jle    L1
L49:
  cmpb   $0xa,(%rbx,%rax,1)
  jne    L24
  movb   $0x0,(%rbx,%rax,1)
  jmp    L24
L1:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *get_input_string() {
  fflush(stdout);
  char *input = malloc(sizeof(char) * BUFSIZ);
  assert(input);
  fgets(input, BUFSIZ, stdin);
  int i;
  for (i = 0; i < BUFSIZ; i++) {
    if (input[i] == '\n') {
      input[i] = '\0';
    }
  }
  return input;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_input_string>:
  endbr64
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  mov    D1(%rip),%rax
  mov    (%rax),%edi
  call   <fflush@plt>
  mov    D27(%rip),%rbx
  mov    (%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <assert@plt>
  mov    D9(%rip),%rax
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    (%rax),%edx
  call   <fgets@plt>
  mov    (%rbx),%eax
  test   %eax,%eax
  jle    L15
  xor    %eax,%eax
  nopl   (%rax)
L4:
  cmpb   $0xa,(%r12,%rax,1)
  jne    L29
  movb   $0x0,(%r12,%rax,1)
L29:
  add    $0x1,%rax
  cmp    %eax,(%rbx)
  jg     L4
L15:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *get_input_string() {
  fflush(stdout);
  char *input = malloc(sizeof(char) * BUFSIZ);
  assert(input);
  fgets(input, BUFSIZ, stdin);
  int i;
  for (i = 0; i < BUFSIZ; i++) {
    if (input[i] == '\n') {
      input[i] = '\0';
    }
  }
  return input;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<get_input_string>:
  endbr64
  push   %r12
  push   %rbx
  sub    $0x8,%rsp
  mov    D1(%rip),%rax
  mov    (%rax),%edi
  call   <fflush@plt>
  mov    D13(%rip),%rbx
  mov    (%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <assert@plt>
  mov    D14(%rip),%rax
  mov    (%rbx),%esi
  mov    %r12,%rdi
  mov    (%rax),%edx
  call   <fgets@plt>
  mov    (%rbx),%edx
  xor    %eax,%eax
  test   %edx,%edx
  jle    L22
  nopl   (%rax)
L17:
  cmpb   $0xa,(%r12,%rax,1)
  je     L19
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L17
L22:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  movb   $0x0,(%r12,%rax,1)
  mov    (%rbx),%edx
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L17
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *get_input_string() {
  fflush(stdout);
  char *input = malloc(sizeof(char) * BUFSIZ);
  assert(input);
  fgets(input, BUFSIZ, stdin);
  int i;
  for (i = 0; i < BUFSIZ; i++) {
    if (input[i] == '\n') {
      input[i] = '\0';
    }
  }
  return input;
}
```
user: What is the c source code of the assembly code below?

```asm
<safe_bcopy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jle    L20
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  je     L20
  mov    -0x20(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jb     L25
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x20(%rbp)
  jb     L68
L25:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  jmp    L74
L68:
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%r12
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  cmp    $0x3f,%rax
  jg     L36
L5:
  sub    $0x1,%r12
  sub    $0x1,%rbx
  movzbl (%r12),%eax
  mov    %al,(%rbx)
  cmp    -0x18(%rbp),%r12
  jne    L5
  jmp    L74
L36:
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  neg    %rax
  add    %rax,%rbx
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  neg    %rax
  add    %rax,%r12
  cmp    -0x20(%rbp),%rbx
  jb     L92
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <memmove@plt>
  jmp    L36
L92:
  nop
  mov    %rbx,%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memmove@plt>
  jmp    L74
L20:
  nop
L74:
  add    $0x20,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
void safe_bcopy(from, to, size) char *from, *to;
int size;
{
  if (size <= 0 || from == to)
    return;
  if (to < from || from + size <= to)
    bcopy(from, to, size);
  else {
    register char *endf = from + size;
    register char *endt = to + size;
    if (to - from < 64) {
      do
        *--endt = *--endf;
      while (endf != from);
    } else {
      for (;;) {
        endt -= (to - from);
        endf -= (to - from);
        if (endt < to)
          break;
        bcopy(endf, endt, to - from);
      }
      bcopy(from, to, endt - from);
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<safe_bcopy>:
  endbr64
  test   %edx,%edx
  jle    L11
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  cmp    %rsi,%rdi
  je     L0
  ja     L17
  movslq %edx,%r12
  lea    (%rdi,%r12,1),%r13
  cmp    %rsi,%r13
  ja     L94
L17:
  movslq %edx,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memmove@plt>
L0:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L94:
  add    %rsi,%r12
  mov    %rsi,%r15
  sub    %rdi,%r15
  cmp    $0x3f,%r15
  jg     L9
  sub    $0x1,%rbx
  sub    %r13,%rbx
  mov    $0xffffffffffffffff,%rax
L68:
  movzbl 0x0(%r13,%rax,1),%edx
  mov    %dl,(%r12,%rax,1)
  sub    $0x1,%rax
  cmp    %rax,%rbx
  jne    L68
  jmp    L0
L9:
  mov    %r15,%r14
  neg    %r14
  sub    %r15,%r12
  sub    %r15,%r13
  cmp    %r12,%rsi
  ja     L37
L29:
  mov    %r15,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <memmove@plt>
  add    %r14,%r12
  add    %r14,%r13
  cmp    %r12,%rbp
  jbe    L29
L37:
  mov    %r12,%rdx
  sub    %rbx,%rdx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <memmove@plt>
  jmp    L0
L11:
  ret
```

assistant: ```c
void safe_bcopy(from, to, size) char *from, *to;
int size;
{
  if (size <= 0 || from == to)
    return;
  if (to < from || from + size <= to)
    bcopy(from, to, size);
  else {
    register char *endf = from + size;
    register char *endt = to + size;
    if (to - from < 64) {
      do
        *--endt = *--endf;
      while (endf != from);
    } else {
      for (;;) {
        endt -= (to - from);
        endf -= (to - from);
        if (endt < to)
          break;
        bcopy(endf, endt, to - from);
      }
      bcopy(from, to, endt - from);
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<safe_bcopy>:
  endbr64
  test   %edx,%edx
  jle    L9
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  cmp    %rsi,%rdi
  je     L59
  movslq %edx,%rdx
  ja     L38
  lea    (%rdi,%rdx,1),%rbx
  cmp    %rsi,%rbx
  jbe    L38
  mov    %rsi,%r14
  lea    (%rsi,%rdx,1),%rcx
  sub    %rdi,%r14
  cmp    $0x3f,%r14
  jg     L54
  sub    $0x1,%rbp
  mov    $0xffffffffffffffff,%rax
  sub    %rbx,%rbp
  nop
L56:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  sub    $0x1,%rax
  cmp    %rax,%rbp
  jne    L56
L59:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L54:
  mov    %r14,%r13
  sub    %r14,%rcx
  sub    %r14,%rbx
  neg    %r13
  cmp    %rcx,%rsi
  ja     L8
  nopl   0x0(%rax)
L65:
  mov    %rbx,%rsi
  mov    %rcx,%rdi
  mov    %r14,%rdx
  add    %r13,%rbx
  call   <memmove@plt>
  mov    %rax,%rcx
  add    %r13,%rcx
  cmp    %rcx,%r12
  jbe    L65
L8:
  sub    %rbp,%rcx
  mov    %rcx,%rdx
L38:
  pop    %rbx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <memmove@plt>
  nopl   (%rax)
L9:
  ret
```

assistant: ```c
void safe_bcopy(from, to, size) char *from, *to;
int size;
{
  if (size <= 0 || from == to)
    return;
  if (to < from || from + size <= to)
    bcopy(from, to, size);
  else {
    register char *endf = from + size;
    register char *endt = to + size;
    if (to - from < 64) {
      do
        *--endt = *--endf;
      while (endf != from);
    } else {
      for (;;) {
        endt -= (to - from);
        endf -= (to - from);
        if (endt < to)
          break;
        bcopy(endf, endt, to - from);
      }
      bcopy(from, to, endt - from);
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<safe_bcopy>:
  endbr64
  test   %edx,%edx
  jle    L41
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  cmp    %rsi,%rdi
  je     L1
  movslq %edx,%rdx
  ja     L13
  lea    (%rdi,%rdx,1),%rbx
  cmp    %rsi,%rbx
  jbe    L13
  mov    %rsi,%r14
  lea    (%rsi,%rdx,1),%rcx
  sub    %rdi,%r14
  cmp    $0x3f,%r14
  jg     L2
  sub    $0x1,%rbp
  mov    $0xffffffffffffffff,%rax
  sub    %rbx,%rbp
  nop
L42:
  movzbl (%rbx,%rax,1),%edx
  mov    %dl,(%rcx,%rax,1)
  sub    $0x1,%rax
  cmp    %rax,%rbp
  jne    L42
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L2:
  mov    %r14,%r13
  sub    %r14,%rcx
  sub    %r14,%rbx
  neg    %r13
  cmp    %rcx,%rsi
  ja     L26
  nopl   0x0(%rax)
L3:
  mov    %rbx,%rsi
  mov    %rcx,%rdi
  mov    %r14,%rdx
  add    %r13,%rbx
  call   <memmove@plt>
  mov    %rax,%rcx
  add    %r13,%rcx
  cmp    %rcx,%r12
  jbe    L3
L26:
  sub    %rbp,%rcx
  mov    %rcx,%rdx
L13:
  pop    %rbx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <memmove@plt>
  nopl   (%rax)
L41:
  ret
```

assistant: ```c
void safe_bcopy(from, to, size) char *from, *to;
int size;
{
  if (size <= 0 || from == to)
    return;
  if (to < from || from + size <= to)
    bcopy(from, to, size);
  else {
    register char *endf = from + size;
    register char *endt = to + size;
    if (to - from < 64) {
      do
        *--endt = *--endf;
      while (endf != from);
    } else {
      for (;;) {
        endt -= (to - from);
        endf -= (to - from);
        if (endt < to)
          break;
        bcopy(endf, endt, to - from);
      }
      bcopy(from, to, endt - from);
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x75667265646e6f77,%rax
  mov    %rax,-0x3a(%rbp)
  movw   $0x6c,-0x32(%rbp)
  movabs $0x72616365636172,%rax
  mov    %rax,-0x42(%rbp)
  movw   $0x31,-0x44(%rbp)
  movabs $0x6c656d6572747865,%rax
  movabs $0x747320676e6f6c79,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x74697720676e6972,%rax
  movabs $0x7365636170732068,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movb   $0x0,-0x10(%rbp)
  lea    -0x3a(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x3a(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <rev_string@plt>
  lea    -0x42(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x42(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <rev_string@plt>
  lea    -0x44(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x44(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <rev_string@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  lea    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <rev_string@plt>
  lea    -0x3a(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x42(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x44(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D15
datatype: string

tool: D15: %s


assistant: ```c
int main(int argc, char **args) {
  char test1[] = "wonderful";
  char test2[] = "racecar";
  char test3[] = "1";
  char test4[] = "extremelylong string with spaces";
  rev_string(test1, strlen(test1));
  rev_string(test2, strlen(test2));
  rev_string(test3, strlen(test3));
  rev_string(test4, strlen(test4));
  printf("%s\n", test1);
  printf("%s\n", test2);
  printf("%s\n", test3);
  printf("%s\n", test4);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movabs $0x75667265646e6f77,%rax
  mov    %rax,0x16(%rsp)
  movw   $0x6c,0x1e(%rsp)
  movabs $0x72616365636172,%rax
  mov    %rax,0xe(%rsp)
  movw   $0x31,0xc(%rsp)
  movabs $0x6c656d6572747865,%rax
  movabs $0x747320676e6f6c79,%rdx
  mov    %rax,0x20(%rsp)
  mov    %rdx,0x28(%rsp)
  movabs $0x74697720676e6972,%rax
  movabs $0x7365636170732068,%rdx
  mov    %rax,0x30(%rsp)
  mov    %rdx,0x38(%rsp)
  movb   $0x0,0x40(%rsp)
  lea    0x16(%rsp),%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <rev_string@plt>
  lea    0xe(%rsp),%r13
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %r13,%rdi
  call   <rev_string@plt>
  lea    0xc(%rsp),%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %r12,%rdi
  call   <rev_string@plt>
  lea    0x20(%rsp),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%esi
  mov    %rbp,%rdi
  call   <rev_string@plt>
  mov    %rbx,%rsi
  lea    D3(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D3
datatype: string

tool: D3: %s


assistant: ```c
int main(int argc, char **args) {
  char test1[] = "wonderful";
  char test2[] = "racecar";
  char test3[] = "1";
  char test4[] = "extremelylong string with spaces";
  rev_string(test1, strlen(test1));
  rev_string(test2, strlen(test2));
  rev_string(test3, strlen(test3));
  rev_string(test4, strlen(test4));
  printf("%s\n", test1);
  printf("%s\n", test2);
  printf("%s\n", test3);
  printf("%s\n", test4);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x31,%edx
  push   %r13
  push   %r12
  push   %rbp
  sub    $0x58,%rsp
  movdqa D12(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  movabs $0x75667265646e6f77,%rax
  lea    0x16(%rsp),%rbp
  mov    %dx,0xc(%rsp)
  lea    0xe(%rsp),%r14
  movaps %xmm0,0x20(%rsp)
  movdqa D7(%rip),%xmm0
  mov    %rbp,%rdi
  lea    0xc(%rsp),%r13
  movb   $0x0,0x40(%rsp)
  lea    0x20(%rsp),%r12
  movaps %xmm0,0x30(%rsp)
  mov    %rax,0x16(%rsp)
  mov    $0x6c,%eax
  mov    %ax,0x1e(%rsp)
  movabs $0x72616365636172,%rax
  mov    %rax,0xe(%rsp)
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <rev_string@plt>
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %r14,%rdi
  mov    %eax,%esi
  call   <rev_string@plt>
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %eax,%esi
  call   <rev_string@plt>
  mov    %r12,%rdi
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <rev_string@plt>
  mov    %rbp,%rsi
  lea    D28(%rip),%rbp
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D12
type_infered: byte[16]
name: unpack_data
arguments: data_label: D7
type_infered: byte[16]
name: unpack_data
arguments: data_label: D28
type_infered: string

tool: {"D12": [0x65,0x78,0x74,0x72,0x65,0x6d,0x65,0x6c,0x79,0x6c,0x6f,0x6e,0x67,0x20,0x73,0x74]}

tool: {"D7": [0x72,0x69,0x6e,0x67,0x20,0x77,0x69,0x74,0x68,0x20,0x73,0x70,0x61,0x63,0x65,0x73]}

tool: D28: %s


assistant: ```c
int main(int argc, char **args) {
  char test1[] = "wonderful";
  char test2[] = "racecar";
  char test3[] = "1";
  char test4[] = "extremelylong string with spaces";
  rev_string(test1, strlen(test1));
  rev_string(test2, strlen(test2));
  rev_string(test3, strlen(test3));
  rev_string(test4, strlen(test4));
  printf("%s\n", test1);
  printf("%s\n", test2);
  printf("%s\n", test3);
  printf("%s\n", test4);
}
```