ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<my_lltoa>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %ecx,%r12d
  lea    -0x1(%rdx),%r13d
  cmp    %ecx,%r13d
  jl     L34
L24:
  mov    %rbp,%rdi
  mov    %rbp,%r8
  mov    $0x0,%esi
  movabs $0xcccccccccccccccd,%r9
L108:
  mov    %esi,%ecx
  add    $0x1,%esi
  mov    %rbx,%rax
  mul    %r9
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%r10
  add    %r10,%r10
  mov    %rbx,%rax
  sub    %r10,%rax
  add    $0x30,%eax
  mov    %al,(%r8)
  mov    %rbx,%rax
  mov    %rdx,%rbx
  add    $0x1,%r8
  cmp    %esi,%r13d
  jle    L65
  cmp    $0x9,%rax
  ja     L108
L65:
  movslq %esi,%r10
  lea    0x0(%rbp,%r10,1),%r8
  movb   $0x0,(%r8)
  test   %ecx,%ecx
  jle    L107
  movslq %ecx,%rax
  mov    $0x0,%edx
L13:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  movzbl 0x0(%rbp,%rax,1),%r9d
  mov    %r9b,0x0(%rbp,%rdx,1)
  mov    %cl,0x0(%rbp,%rax,1)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %eax,%edx
  jl     L13
L107:
  cmp    %esi,%r12d
  jle    L59
  sub    %esi,%r12d
  test   %esi,%esi
  js     L103
  mov    %r8,%rax
  lea    -0x1(%rbp,%r10,1),%rcx
  mov    %esi,%esi
  sub    %rsi,%rcx
  movslq %r12d,%rsi
L83:
  movzbl (%rax),%edx
  mov    %dl,(%rax,%rsi,1)
  sub    $0x1,%rax
  cmp    %rcx,%rax
  jne    L83
L103:
  test   %r12d,%r12d
  jle    L59
  lea    -0x1(%r12),%eax
  lea    0x1(%rbp,%rax,1),%rax
L1:
  movb   $0x30,(%rdi)
  add    $0x1,%rdi
  cmp    %rdi,%rax
  jne    L1
L59:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L34:
  mov    %r13d,%edx
  mov    %ecx,%esi
  lea    D8(%rip),%rdi
  call   <debuga@plt>
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L24
```

tool_calls: name: retrieve_data
arguments: tag: D8
type_guessed: string

tool: D8: The requested number of digits passed to my_lltoa (%d) is bigger than the output buffer size (%d)


assistant: ```c
void my_lltoa(unsigned long long int n, char *s, int ssize, int len) {
  int i;
  int slen = 0;
  int j;
  char c;
  ssize--;
  if (len > ssize) {
    debuga(("The requested number of digits passed to my_lltoa (%d) is bigger than the output buffer size (%d)\n"), len, ssize);
    abort();
  }
  do {
    s[slen++] = (n % 10) + '0';
  } while ((n /= 10) > 0 && slen < ssize);
  s[slen] = '\0';
  for (i = 0, j = slen - 1; i < j; i++, j--) {
    c = s[i];
    s[i] = s[j];
    s[j] = c;
  }
  if (len > slen) {
    i = len - slen;
    for (j = slen; j >= 0; j--)
      s[j + i] = s[j];
    for (j = 0; j < i; j++)
      s[j] = '0';
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_lltoa>:
  endbr64
  push   %r13
  lea    -0x1(%rdx),%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cmp    %ecx,%r13d
  jl     L30
L0:
  movabs $0xcccccccccccccccd,%r9
  mov    %rbp,%r8
  xor    %esi,%esi
  jmp    L2
  nopl   (%rax)
L69:
  cmp    $0x9,%rax
  jbe    L44
L2:
  mov    %rbx,%rax
  mov    %esi,%ecx
  add    $0x1,%esi
  add    $0x1,%r8
  mul    %r9
  mov    %rbx,%rax
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rdi
  add    %rdi,%rdi
  sub    %rdi,%rax
  add    $0x30,%eax
  mov    %al,-0x1(%r8)
  mov    %rbx,%rax
  mov    %rdx,%rbx
  cmp    %esi,%r13d
  jg     L69
L44:
  movslq %esi,%r9
  lea    0x0(%rbp,%r9,1),%rdi
  movb   $0x0,(%rdi)
  test   %ecx,%ecx
  je     L64
  movslq %ecx,%rax
  xor    %edx,%edx
  nopl   0x0(%rax)
L27:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  movzbl 0x0(%rbp,%rax,1),%r8d
  mov    %r8b,0x0(%rbp,%rdx,1)
  add    $0x1,%rdx
  mov    %cl,0x0(%rbp,%rax,1)
  sub    $0x1,%rax
  cmp    %edx,%eax
  jg     L27
L64:
  cmp    %esi,%r12d
  jle    L61
  sub    %esi,%r12d
  lea    -0x1(%rbp,%r9,1),%rsi
  mov    %rdi,%rax
  sub    %r9,%rsi
  movslq %r12d,%rdx
  xchg   %ax,%ax
L82:
  movzbl (%rax),%ecx
  sub    $0x1,%rax
  mov    %cl,0x1(%rax,%rdx,1)
  cmp    %rax,%rsi
  jne    L82
  add    $0x8,%rsp
  mov    %rbp,%rdi
  mov    $0x30,%esi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1080 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L61:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L30:
  mov    %r13d,%edx
  mov    %ecx,%esi
  lea    D5(%rip),%rdi
  call   <debuga@plt>
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L0
```

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

tool: D5: The requested number of digits passed to my_lltoa (%d) is bigger than the output buffer size (%d)


assistant: ```c
void my_lltoa(unsigned long long int n, char *s, int ssize, int len) {
  int i;
  int slen = 0;
  int j;
  char c;
  ssize--;
  if (len > ssize) {
    debuga(("The requested number of digits passed to my_lltoa (%d) is bigger than the output buffer size (%d)\n"), len, ssize);
    abort();
  }
  do {
    s[slen++] = (n % 10) + '0';
  } while ((n /= 10) > 0 && slen < ssize);
  s[slen] = '\0';
  for (i = 0, j = slen - 1; i < j; i++, j--) {
    c = s[i];
    s[i] = s[j];
    s[j] = c;
  }
  if (len > slen) {
    i = len - slen;
    for (j = slen; j >= 0; j--)
      s[j + i] = s[j];
    for (j = 0; j < i; j++)
      s[j] = '0';
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<my_lltoa>:
  endbr64
  push   %r13
  lea    -0x1(%rdx),%r13d
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cmp    %ecx,%r13d
  jl     L63
L12:
  movabs $0xcccccccccccccccd,%r9
  mov    %rbp,%r8
  xor    %esi,%esi
  jmp    L76
  nopl   (%rax)
L39:
  cmp    $0x9,%rax
  jbe    L62
L76:
  mov    %rbx,%rax
  mov    %esi,%ecx
  add    $0x1,%esi
  add    $0x1,%r8
  mul    %r9
  mov    %rbx,%rax
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rdi
  add    %rdi,%rdi
  sub    %rdi,%rax
  add    $0x30,%eax
  mov    %al,-0x1(%r8)
  mov    %rbx,%rax
  mov    %rdx,%rbx
  cmp    %esi,%r13d
  jg     L39
L62:
  movslq %esi,%r9
  lea    0x0(%rbp,%r9,1),%rdi
  movb   $0x0,(%rdi)
  test   %ecx,%ecx
  je     L75
  movslq %ecx,%rax
  xor    %edx,%edx
  nopl   0x0(%rax)
L6:
  movzbl 0x0(%rbp,%rdx,1),%ecx
  movzbl 0x0(%rbp,%rax,1),%r8d
  mov    %r8b,0x0(%rbp,%rdx,1)
  add    $0x1,%rdx
  mov    %cl,0x0(%rbp,%rax,1)
  sub    $0x1,%rax
  cmp    %edx,%eax
  jg     L6
L75:
  cmp    %esi,%r12d
  jle    L49
  sub    %esi,%r12d
  lea    -0x1(%rbp,%r9,1),%rsi
  mov    %rdi,%rax
  sub    %r9,%rsi
  movslq %r12d,%rdx
  xchg   %ax,%ax
L34:
  movzbl (%rax),%ecx
  sub    $0x1,%rax
  mov    %cl,0x1(%rax,%rdx,1)
  cmp    %rsi,%rax
  jne    L34
  add    $0x8,%rsp
  mov    %rbp,%rdi
  mov    $0x30,%esi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1080 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L49:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L63:
  mov    %r13d,%edx
  mov    %ecx,%esi
  lea    D5(%rip),%rdi
  call   <debuga@plt>
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L12
```

tool_calls: name: retrieve_data
arguments: label: D5
data_type: string

tool: D5: The requested number of digits passed to my_lltoa (%d) is bigger than the output buffer size (%d)


assistant: ```c
void my_lltoa(unsigned long long int n, char *s, int ssize, int len) {
  int i;
  int slen = 0;
  int j;
  char c;
  ssize--;
  if (len > ssize) {
    debuga(("The requested number of digits passed to my_lltoa (%d) is bigger than the output buffer size (%d)\n"), len, ssize);
    abort();
  }
  do {
    s[slen++] = (n % 10) + '0';
  } while ((n /= 10) > 0 && slen < ssize);
  s[slen] = '\0';
  for (i = 0, j = slen - 1; i < j; i++, j--) {
    c = s[i];
    s[i] = s[j];
    s[j] = c;
  }
  if (len > slen) {
    i = len - slen;
    for (j = slen; j >= 0; j--)
      s[j + i] = s[j];
    for (j = 0; j < i; j++)
      s[j] = '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   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D69(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D58(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x10(%rbp),%xmm1
  movss  D73(%rip),%xmm0
  mulss  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtss2sd %xmm0,%xmm1
  movsd  D49(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm2,%xmm2
  cvtss2sd -0xc(%rbp),%xmm2
  movq   %xmm2,%rax
  movq   %rax,%xmm0
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D86(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D80(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D69
data_type: string
name: parse_data
arguments: tag: D33
data_type: string
name: parse_data
arguments: tag: D10
data_type: string
name: parse_data
arguments: tag: D58
data_type: string
name: parse_data
arguments: tag: D73
data_type: byte[4]
name: parse_data
arguments: tag: D49
data_type: f64
name: parse_data
arguments: tag: D29
data_type: string
name: parse_data
arguments: tag: D86
data_type: string
name: parse_data
arguments: tag: D80
data_type: string

tool: D69: Are you worth you weight in rhodium?

tool: D33: Let's check it out.

tool: D10: Please enter your weight in pounds:

tool: D58: %f

tool: {"D73": [0x00,0x80,0x40,0x44]}

tool: D49: 14.5833

tool: D29: Your weight in rhodium is $%.2f.


tool: D86: You are easily worth that! If rhodium prices drop,

tool: D80: eat more to maintain in your value.

assistant: ```c
int main(void) {
  float weight;
  float value;
  printf("Are you worth you weight in rhodium?\n");
  printf("Let's check it out.\n");
  printf("Please enter your weight in pounds:");
  scanf("%f", &weight);
  value = 770 * weight * 14.5833;
  printf("Your weight in rhodium is $%.2f.\n", value);
  printf("You are easily worth that! If rhodium prices drop,\n");
  printf("eat more to maintain in your value.\n");
  getchar();
  getchar();
  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
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D76(%rip),%rdi
  call   <puts@plt>
  lea    D30(%rip),%rdi
  call   <puts@plt>
  lea    D86(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D35(%rip),%rdi
  call   <scanf@plt>
  movss  D66(%rip),%xmm0
  mulss  0x4(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  D85(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  lea    D25(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D70(%rip),%rdi
  call   <puts@plt>
  lea    D14(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D76
data_type: string
name: extract_symbolic_data
arguments: tag: D30
data_type: string
name: extract_symbolic_data
arguments: tag: D86
data_type: string
name: extract_symbolic_data
arguments: tag: D35
data_type: string
name: extract_symbolic_data
arguments: tag: D66
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D85
data_type: f64
name: extract_symbolic_data
arguments: tag: D25
data_type: string
name: extract_symbolic_data
arguments: tag: D70
data_type: string
name: extract_symbolic_data
arguments: tag: D14
data_type: string

tool: D76: Are you worth you weight in rhodium?

tool: D30: Let's check it out.

tool: D86: Please enter your weight in pounds:

tool: D35: %f

tool: {"D66": [0x00,0x80,0x40,0x44]}

tool: D85: 14.5833

tool: D25: Your weight in rhodium is $%.2f.


tool: D70: You are easily worth that! If rhodium prices drop,

tool: D14: eat more to maintain in your value.

assistant: ```c
int main(void) {
  float weight;
  float value;
  printf("Are you worth you weight in rhodium?\n");
  printf("Let's check it out.\n");
  printf("Please enter your weight in pounds:");
  scanf("%f", &weight);
  value = 770 * weight * 14.5833;
  printf("Your weight in rhodium is $%.2f.\n", value);
  printf("You are easily worth that! If rhodium prices drop,\n");
  printf("eat more to maintain in your value.\n");
  getchar();
  getchar();
  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
  sub    $0x18,%rsp
  lea    D53(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D40(%rip),%rdi
  call   <puts@plt>
  lea    D10(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D88(%rip),%rdi
  call   <scanf@plt>
  movss  D68(%rip),%xmm0
  mulss  0x4(%rsp),%xmm0
  lea    D20(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  mulsd  D50(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  lea    D69(%rip),%rdi
  call   <puts@plt>
  lea    D75(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D53
type_guessed: string
name: read_data
arguments: tag: D40
type_guessed: string
name: read_data
arguments: tag: D10
type_guessed: string
name: read_data
arguments: tag: D88
type_guessed: string
name: read_data
arguments: tag: D68
type_guessed: byte[4]
name: read_data
arguments: tag: D20
type_guessed: string
name: read_data
arguments: tag: D50
type_guessed: f64
name: read_data
arguments: tag: D69
type_guessed: string
name: read_data
arguments: tag: D75
type_guessed: string

tool: D53: Are you worth you weight in rhodium?

tool: D40: Let's check it out.

tool: D10: Please enter your weight in pounds:

tool: D88: %f

tool: {"D68": [0x00,0x80,0x40,0x44]}

tool: D20: Your weight in rhodium is $%.2f.


tool: D50: 14.5833

tool: D69: You are easily worth that! If rhodium prices drop,

tool: D75: eat more to maintain in your value.

assistant: ```c
int main(void) {
  float weight;
  float value;
  printf("Are you worth you weight in rhodium?\n");
  printf("Let's check it out.\n");
  printf("Please enter your weight in pounds:");
  scanf("%f", &weight);
  value = 770 * weight * 14.5833;
  printf("Your weight in rhodium is $%.2f.\n", value);
  printf("You are easily worth that! If rhodium prices drop,\n");
  printf("eat more to maintain in your value.\n");
  getchar();
  getchar();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D33(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D28(%rip),%rdi
  call   <puts@plt>
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D68(%rip),%rdi
  call   <scanf@plt>
  movss  D66(%rip),%xmm0
  mulss  0x4(%rsp),%xmm0
  lea    D84(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm0,%xmm0
  mulsd  D74(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  lea    D77(%rip),%rdi
  call   <puts@plt>
  lea    D10(%rip),%rdi
  call   <puts@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D33
type_guessed: string
name: parse_data
arguments: label: D28
type_guessed: string
name: parse_data
arguments: label: D37
type_guessed: string
name: parse_data
arguments: label: D68
type_guessed: string
name: parse_data
arguments: label: D66
type_guessed: byte[4]
name: parse_data
arguments: label: D84
type_guessed: string
name: parse_data
arguments: label: D74
type_guessed: f64
name: parse_data
arguments: label: D77
type_guessed: string
name: parse_data
arguments: label: D10
type_guessed: string

tool: D33: Are you worth you weight in rhodium?

tool: D28: Let's check it out.

tool: D37: Please enter your weight in pounds:

tool: D68: %f

tool: {"D66": [0x00,0x80,0x40,0x44]}

tool: D84: Your weight in rhodium is $%.2f.


tool: D74: 14.5833

tool: D77: You are easily worth that! If rhodium prices drop,

tool: D10: eat more to maintain in your value.

assistant: ```c
int main(void) {
  float weight;
  float value;
  printf("Are you worth you weight in rhodium?\n");
  printf("Let's check it out.\n");
  printf("Please enter your weight in pounds:");
  scanf("%f", &weight);
  value = 770 * weight * 14.5833;
  printf("Your weight in rhodium is $%.2f.\n", value);
  printf("You are easily worth that! If rhodium prices drop,\n");
  printf("eat more to maintain in your value.\n");
  getchar();
  getchar();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<update_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L6
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  movq   $0x0,(%rax)
  mov    $0x0,%eax
  jmp    L9
L6:
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <replace_string@plt>
L9:
  leave
  ret
```

assistant: ```c
int update_string(char **str, const char to[]) {
  if (to == NULL) {
    free(*str);
    *str = NULL;
    return 0;
  }
  return replace_string(str, to);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<update_string>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rsi,%rsi
  je     L17
  call   <replace_string@plt>
L4:
  pop    %rbx
  ret
L17:
  mov    (%rdi),%rdi
  call   <free@plt>
  movq   $0x0,(%rbx)
  mov    $0x0,%eax
  jmp    L4
```

assistant: ```c
int update_string(char **str, const char to[]) {
  if (to == NULL) {
    free(*str);
    *str = NULL;
    return 0;
  }
  return replace_string(str, to);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<update_string>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  test   %rsi,%rsi
  je     L8
  pop    %rbp
  jmp    1070 <replace_string@plt>
  nopl   0x0(%rax,%rax,1)
L8:
  mov    (%rdi),%rdi
  call   <free@plt>
  movq   $0x0,0x0(%rbp)
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int update_string(char **str, const char to[]) {
  if (to == NULL) {
    free(*str);
    *str = NULL;
    return 0;
  }
  return replace_string(str, to);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<update_string>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  test   %rsi,%rsi
  je     L7
  pop    %rbp
  jmp    1070 <replace_string@plt>
  nopl   0x0(%rax,%rax,1)
L7:
  mov    (%rdi),%rdi
  call   <free@plt>
  movq   $0x0,0x0(%rbp)
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int update_string(char **str, const char to[]) {
  if (to == NULL) {
    free(*str);
    *str = NULL;
    return 0;
  }
  return replace_string(str, to);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ungReadi>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x64(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %rdx,-0x78(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x78(%rbp),%rax
  mov    %eax,%edx
  mov    -0x70(%rbp),%rcx
  mov    -0x64(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x54(%rbp)
  mov    -0x54(%rbp),%eax
  cltq
  cmp    %rax,-0x78(%rbp)
  jbe    L1
  mov    -0x54(%rbp),%eax
  jmp    L0
L1:
  mov    -0x54(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x70(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L6
  nop
L35:
  lea    -0x50(%rbp),%rcx
  mov    -0x64(%rbp),%eax
  mov    $0x40,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  cmp    $0x3f,%eax
  jg     L35
L6:
  mov    -0x54(%rbp),%eax
L0:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  leave
  ret
```

assistant: ```c
int ungReadi(int fd, char *buff, size_t size) {
  char dump[64];
  int i = read(fd, buff, size);
  if (i < size)
    return i;
  else {
    if (buff[i - 1] != 0xa)
      while (read(fd, dump, 64) >= 64)
        ;
  }
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ungReadi>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %edi,%ebx
  mov    %rsi,%r13
  mov    %rdx,%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  call   <read@plt>
  mov    %eax,%ebp
  cltq
  cmp    %r12,%rax
  jb     L25
  cmpb   $0xa,-0x1(%r13,%rax,1)
  je     L25
  mov    %rsp,%r12
L34:
  mov    $0x40,%edx
  mov    %r12,%rsi
  mov    %ebx,%edi
  call   <read@plt>
  cmp    $0x3f,%eax
  jg     L34
L25:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    %ebp,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ungReadi(int fd, char *buff, size_t size) {
  char dump[64];
  int i = read(fd, buff, size);
  if (i < size)
    return i;
  else {
    if (buff[i - 1] != 0xa)
      while (read(fd, dump, 64) >= 64)
        ;
  }
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<ungReadi>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  call   <read@plt>
  mov    %eax,%r12d
  cltq
  cmp    %rbp,%rax
  jb     L24
  cmpb   $0xa,-0x1(%r13,%rax,1)
  je     L24
  mov    %rsp,%rbp
L16:
  mov    $0x40,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <read@plt>
  cmp    $0x3f,%eax
  jg     L16
L24:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x58,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ungReadi(int fd, char *buff, size_t size) {
  char dump[64];
  int i = read(fd, buff, size);
  if (i < size)
    return i;
  else {
    if (buff[i - 1] != 0xa)
      while (read(fd, dump, 64) >= 64)
        ;
  }
  return i;
}
```
user: What is the c source code of the assembly code below?

```asm
<ungReadi>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  call   <read@plt>
  mov    %eax,%r12d
  cltq
  cmp    %rbp,%rax
  jb     L3
  cmpb   $0xa,-0x1(%r13,%rax,1)
  je     L3
  mov    %rsp,%rbp
L19:
  mov    $0x40,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <read@plt>
  cmp    $0x3f,%eax
  jg     L19
L3:
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x58,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int ungReadi(int fd, char *buff, size_t size) {
  char dump[64];
  int i = read(fd, buff, size);
  if (i < size)
    return i;
  else {
    if (buff[i - 1] != 0xa)
      while (read(fd, dump, 64) >= 64)
        ;
  }
  return i;
}
```