ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<check_dnaflag>:
  endbr64
  test   %esi,%esi
  jle    L20
  push   %r15
  lea    D7(%rip),%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L56:
  add    $0x1,%ebp
L71:
  add    $0x1,%rbx
  cmp    %ebx,%r13d
  jl     L79
L18:
  movsbl (%r14,%rbx,1),%esi
  cmp    $0x2d,%sil
  je     L71
  add    $0x1,%r12d
  cmp    $0x4e,%sil
  je     L56
  mov    %r15,%rdi
  call   <res_index@plt>
  cmp    $0x80000000,%eax
  adc    $0x0,%ebp
  add    $0x1,%rbx
  cmp    %ebx,%r13d
  jge    L18
L79:
  test   %ebp,%ebp
  je     L62
  test   %r12d,%r12d
  je     L62
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %ebp,%xmm0
  cvtsi2ss %r12d,%xmm1
  divss  %xmm1,%xmm0
  cvtss2sd %xmm0,%xmm0
  comisd D9(%rip),%xmm0
  setae  %al
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L62:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L20:
  xor    %eax,%eax
  ret
```

tool_calls: name: parse_data
arguments: tag: D7
type: string
name: parse_data
arguments: tag: D9
type: f64

tool: D7: ACGTUN

tool: D9: 0.85

assistant: ```c
char check_dnaflag(char *seq, int slen) {
  int i, c, nresidues, nbases;
  float ratio;
  char *dna_codes = "ACGTUN";
  nresidues = nbases = 0;
  for (i = 1; i <= slen; i++) {
    if (seq[i] != '-') {
      nresidues++;
      if (seq[i] == 'N')
        nbases++;
      else {
        c = res_index(dna_codes, seq[i]);
        if (c >= 0)
          nbases++;
      }
    }
  }
  if ((nbases == 0) || (nresidues == 0))
    return 0;
  ratio = (float)nbases / (float)nresidues;
  if (ratio >= 0.85)
    return 1;
  else
    return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<check_dnaflag>:
  endbr64
  test   %esi,%esi
  jle    L76
  push   %r15
  lea    D7(%rip),%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  jmp    L67
  nopl   0x0(%rax,%rax,1)
L60:
  add    $0x1,%ebp
L6:
  add    $0x1,%rbx
  cmp    %ebx,%r13d
  jl     L34
L67:
  movsbl (%r14,%rbx,1),%esi
  cmp    $0x2d,%sil
  je     L6
  add    $0x1,%r12d
  cmp    $0x4e,%sil
  je     L60
  mov    %r15,%rdi
  call   <res_index@plt>
  cmp    $0x80000000,%eax
  adc    $0x0,%ebp
  add    $0x1,%rbx
  cmp    %ebx,%r13d
  jge    L67
L34:
  test   %ebp,%ebp
  je     L18
  test   %r12d,%r12d
  je     L18
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %ebp,%xmm0
  cvtsi2ss %r12d,%xmm1
  divss  %xmm1,%xmm0
  cvtss2sd %xmm0,%xmm0
  comisd D13(%rip),%xmm0
  setae  %al
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  xchg   %ax,%ax
L18:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L76:
  xor    %eax,%eax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type: string
name: extract_symbolic_data
arguments: data_label: D13
type: f64

tool: D7: ACGTUN

tool: D13: 0.85

assistant: ```c
char check_dnaflag(char *seq, int slen) {
  int i, c, nresidues, nbases;
  float ratio;
  char *dna_codes = "ACGTUN";
  nresidues = nbases = 0;
  for (i = 1; i <= slen; i++) {
    if (seq[i] != '-') {
      nresidues++;
      if (seq[i] == 'N')
        nbases++;
      else {
        c = res_index(dna_codes, seq[i]);
        if (c >= 0)
          nbases++;
      }
    }
  }
  if ((nbases == 0) || (nresidues == 0))
    return 0;
  ratio = (float)nbases / (float)nresidues;
  if (ratio >= 0.85)
    return 1;
  else
    return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x10(%rbp)
  movl   $0xa,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <interchange@plt>
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D12
type_infered: string
name: extract_symbolic_data
arguments: tag: D15
type_infered: string

tool: D12: Originally x = %d and y = %d.


tool: D15: Now x = %d and y = %d.


assistant: ```c
int main(void) {
  int x = 5, y = 10;
  printf("Originally x = %d and y = %d.\n", x, y);
  interchange(&x, &y);
  printf("Now x = %d and y = %d.\n", x, y);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x5,(%rsp)
  movl   $0xa,0x4(%rsp)
  mov    $0xa,%edx
  mov    $0x5,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rsp,%rdi
  call   <interchange@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D3
datatype: string
name: read_data
arguments: label: D5
datatype: string

tool: D3: Originally x = %d and y = %d.


tool: D5: Now x = %d and y = %d.


assistant: ```c
int main(void) {
  int x = 5, y = 10;
  printf("Originally x = %d and y = %d.\n", x, y);
  interchange(&x, &y);
  printf("Now x = %d and y = %d.\n", x, y);
  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    $0xa,%edx
  mov    $0x5,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  movl   $0x5,(%rsp)
  movl   $0xa,0x4(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rsp,%rdi
  call   <interchange@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
data_type: string
name: extract_symbolic_data
arguments: data_label: D18
data_type: string

tool: D1: Originally x = %d and y = %d.


tool: D18: Now x = %d and y = %d.


assistant: ```c
int main(void) {
  int x = 5, y = 10;
  printf("Originally x = %d and y = %d.\n", x, y);
  interchange(&x, &y);
  printf("Now x = %d and y = %d.\n", x, y);
  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    $0xa,%edx
  mov    $0x5,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  movl   $0x5,(%rsp)
  movl   $0xa,0x4(%rsp)
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %rsp,%rdi
  call   <interchange@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%esi
  lea    D15(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type_guessed: string
name: extract_symbolic_data
arguments: tag: D15
type_guessed: string

tool: D6: Originally x = %d and y = %d.


tool: D15: Now x = %d and y = %d.


assistant: ```c
int main(void) {
  int x = 5, y = 10;
  printf("Originally x = %d and y = %d.\n", x, y);
  interchange(&x, &y);
  printf("Now x = %d and y = %d.\n", x, y);
  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
<ft_printf_X>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L17
  addl   $0x1,-0x8(%rbp)
  jmp    L17
L16:
  mov    -0x14(%rbp),%eax
  shr    $0x4,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x8(%rbp)
L17:
  cmpl   $0x0,-0x14(%rbp)
  jne    L16
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <ft_putnbr_X@plt>
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int ft_printf_X(unsigned int nbr) {
  int ret;
  int copy;
  copy = nbr;
  ret = 0;
  if (nbr == 0)
    ret++;
  while (nbr >= 1) {
    nbr = nbr / 16;
    ret++;
  }
  ft_putnbr_X(copy);
  return (ret);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_printf_X>:
  endbr64
  push   %rbx
  mov    %edi,%edx
  test   %edi,%edi
  sete   %bl
  movzbl %bl,%ebx
  je     L18
L9:
  mov    %edi,%eax
  shr    $0x4,%edi
  add    $0x1,%ebx
  cmp    $0xf,%eax
  ja     L9
L18:
  mov    %edx,%edi
  call   <ft_putnbr_X@plt>
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int ft_printf_X(unsigned int nbr) {
  int ret;
  int copy;
  copy = nbr;
  ret = 0;
  if (nbr == 0)
    ret++;
  while (nbr >= 1) {
    nbr = nbr / 16;
    ret++;
  }
  ft_putnbr_X(copy);
  return (ret);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ft_printf_X>:
  endbr64
  push   %r12
  mov    $0x1,%r12d
  test   %edi,%edi
  je     L18
  mov    %edi,%eax
  xor    %r12d,%r12d
  nopl   (%rax)
L0:
  add    $0x1,%r12d
  shr    $0x4,%eax
  jne    L0
L18:
  call   <ft_putnbr_X@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int ft_printf_X(unsigned int nbr) {
  int ret;
  int copy;
  copy = nbr;
  ret = 0;
  if (nbr == 0)
    ret++;
  while (nbr >= 1) {
    nbr = nbr / 16;
    ret++;
  }
  ft_putnbr_X(copy);
  return (ret);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_printf_X>:
  endbr64
  push   %r12
  mov    $0x1,%r12d
  test   %edi,%edi
  je     L5
  mov    %edi,%eax
  xor    %r12d,%r12d
  nopl   (%rax)
L8:
  add    $0x1,%r12d
  shr    $0x4,%eax
  jne    L8
L5:
  call   <ft_putnbr_X@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int ft_printf_X(unsigned int nbr) {
  int ret;
  int copy;
  copy = nbr;
  ret = 0;
  if (nbr == 0)
    ret++;
  while (nbr >= 1) {
    nbr = nbr / 16;
    ret++;
  }
  ft_putnbr_X(copy);
  return (ret);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<nth_digit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <log10l@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L17
L0:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jg     L0
  mov    -0x8(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%eax
  leave
  ret
```

assistant: ```c
int nth_digit(int n, int i) {
  int m = n;
  int j = log10l(n);
  while (j > i) {
    m /= 10;
    j--;
  }
  return m % 10;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<nth_digit>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  call   <log10l@plt>
  cmp    %ebp,%eax
  jle    L0
L7:
  movslq %ebx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %ebx,%ecx
  sar    $0x1f,%ecx
  mov    %edx,%ebx
  sub    %ecx,%ebx
  sub    $0x1,%eax
  cmp    %eax,%ebp
  jne    L7
L0:
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int nth_digit(int n, int i) {
  int m = n;
  int j = log10l(n);
  while (j > i) {
    m /= 10;
    j--;
  }
  return m % 10;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<nth_digit>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <log10l@plt>
  cmp    %ebp,%eax
  jle    L7
  nopw   0x0(%rax,%rax,1)
L16:
  movslq %ebx,%rdx
  mov    %ebx,%ecx
  sub    $0x1,%eax
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  mov    %edx,%ebx
  sub    %ecx,%ebx
  cmp    %eax,%ebp
  jne    L16
L7:
  movslq %ebx,%rax
  mov    %ebx,%edx
  add    $0x8,%rsp
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  add    %edx,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int nth_digit(int n, int i) {
  int m = n;
  int j = log10l(n);
  while (j > i) {
    m /= 10;
    j--;
  }
  return m % 10;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<nth_digit>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  call   <log10l@plt>
  cmp    %ebp,%eax
  jle    L6
  nopw   0x0(%rax,%rax,1)
L13:
  movslq %ebx,%rdx
  mov    %ebx,%ecx
  sub    $0x1,%eax
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  mov    %edx,%ebx
  sub    %ecx,%ebx
  cmp    %eax,%ebp
  jne    L13
L6:
  movslq %ebx,%rax
  mov    %ebx,%edx
  add    $0x8,%rsp
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  add    %edx,%edx
  sub    %edx,%eax
  ret
```

assistant: ```c
int nth_digit(int n, int i) {
  int m = n;
  int j = log10l(n);
  while (j > i) {
    m /= 10;
    j--;
  }
  return m % 10;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xac(%rbp)
  jmp    L50
L30:
  lea    -0x70(%rbp),%rcx
  mov    -0xac(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  lea    (%rcx,%rax,1),%rdx
  mov    -0xac(%rbp),%eax
  cltq
  mov    %rdx,-0xa0(%rbp,%rax,8)
  addl   $0x1,-0xac(%rbp)
L50:
  cmpl   $0x4,-0xac(%rbp)
  jle    L30
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0xac(%rbp)
  jmp    L21
L49:
  mov    -0xac(%rbp),%eax
  cltq
  mov    -0xa0(%rbp,%rax,8),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0xac(%rbp)
L21:
  cmpl   $0x4,-0xac(%rbp)
  jle    L49
  lea    -0xa0(%rbp),%rax
  mov    %rax,-0xa8(%rbp)
  mov    -0xa8(%rbp),%rax
  mov    %rax,%rdi
  call   <sort@plt>
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0xac(%rbp)
  jmp    L23
L43:
  mov    -0xac(%rbp),%eax
  cltq
  mov    -0xa0(%rbp,%rax,8),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addl   $0x1,-0xac(%rbp)
L23:
  cmpl   $0x4,-0xac(%rbp)
  jle    L43
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L68
  call   <__stack_chk_fail@plt>
L68:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D9
type_guessed: string
name: parse_data
arguments: data_label: D7
type_guessed: string
name: parse_data
arguments: data_label: D1
type_guessed: string

tool: D9: input 5 strings:

tool: D7: %s

tool: D1: strings sorted:

assistant: ```c
int main() {
  int i;
  char **p1, *p2[5], str[5][20];
  for (i = 0; i < 5; i++)
    p2[i] = str[i];
  printf("input 5 strings:\n");
  for (i = 0; i < 5; i++)
    scanf("%s", p2[i]);
  p1 = p2;
  sort(p1);
  printf("strings sorted:\n");
  for (i = 0; i < 5; i++)
    printf("%s\n", p2[i]);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rax
  mov    %rax,(%rsp)
  lea    0x44(%rsp),%rax
  mov    %rax,0x8(%rsp)
  lea    0x58(%rsp),%rax
  mov    %rax,0x10(%rsp)
  lea    0x6c(%rsp),%rax
  mov    %rax,0x18(%rsp)
  lea    0x80(%rsp),%rax
  mov    %rax,0x20(%rsp)
  lea    D0(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rbx
  lea    0x28(%rsp),%r12
  mov    %rbx,%rbp
  lea    D18(%rip),%r13
L28:
  mov    0x0(%rbp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  add    $0x8,%rbp
  cmp    %rbp,%r12
  jne    L28
  mov    %rsp,%rdi
  call   <sort@plt>
  lea    D8(%rip),%rdi
  call   <puts@plt>
L19:
  mov    (%rbx),%rdi
  call   <puts@plt>
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L19
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  mov    $0x0,%eax
  add    $0xa8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
datatype: string
name: parse_data
arguments: data_tag: D18
datatype: string
name: parse_data
arguments: data_tag: D8
datatype: string

tool: D0: input 5 strings:

tool: D18: %s

tool: D8: strings sorted:

assistant: ```c
int main() {
  int i;
  char **p1, *p2[5], str[5][20];
  for (i = 0; i < 5; i++)
    p2[i] = str[i];
  printf("input 5 strings:\n");
  for (i = 0; i < 5; i++)
    scanf("%s", p2[i]);
  p1 = p2;
  sort(p1);
  printf("strings sorted:\n");
  for (i = 0; i < 5; i++)
    printf("%s\n", p2[i]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  lea    D21(%rip),%rdi
  push   %r13
  lea    D27(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rax
  mov    %rsp,%r14
  lea    0x28(%rsp),%r12
  mov    %rax,(%rsp)
  lea    0x44(%rsp),%rax
  mov    %r14,%rbx
  mov    %r14,%rbp
  mov    %rax,0x8(%rsp)
  lea    0x58(%rsp),%rax
  mov    %rax,0x10(%rsp)
  lea    0x6c(%rsp),%rax
  mov    %rax,0x18(%rsp)
  lea    0x80(%rsp),%rax
  mov    %rax,0x20(%rsp)
  call   <puts@plt>
L25:
  mov    0x0(%rbp),%rsi
  mov    %r13,%rdi
  add    $0x8,%rbp
  call   <scanf@plt>
  cmp    %r12,%rbp
  jne    L25
  mov    %r14,%rdi
  call   <sort@plt>
  lea    D29(%rip),%rdi
  call   <puts@plt>
L20:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <puts@plt>
  cmp    %r12,%rbx
  jne    L20
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D21
datatype: string
name: disassemble_data
arguments: tag: D27
datatype: string
name: disassemble_data
arguments: tag: D29
datatype: string

tool: D21: input 5 strings:

tool: D27: %s

tool: D29: strings sorted:

assistant: ```c
int main() {
  int i;
  char **p1, *p2[5], str[5][20];
  for (i = 0; i < 5; i++)
    p2[i] = str[i];
  printf("input 5 strings:\n");
  for (i = 0; i < 5; i++)
    scanf("%s", p2[i]);
  p1 = p2;
  sort(p1);
  printf("strings sorted:\n");
  for (i = 0; i < 5; i++)
    printf("%s\n", p2[i]);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D4(%rip),%rdi
  push   %r13
  lea    D29(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xa0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x98(%rsp)
  xor    %eax,%eax
  lea    0x30(%rsp),%rax
  mov    %rsp,%r14
  lea    0x28(%rsp),%r12
  movq   %rax,%xmm0
  lea    0x44(%rsp),%rax
  mov    %r14,%rbx
  mov    %r14,%rbp
  movq   %rax,%xmm1
  lea    0x58(%rsp),%rax
  punpcklqdq %xmm1,%xmm0
  movaps %xmm0,(%rsp)
  movq   %rax,%xmm0
  lea    0x6c(%rsp),%rax
  movq   %rax,%xmm2
  lea    0x80(%rsp),%rax
  punpcklqdq %xmm2,%xmm0
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <puts@plt>
L10:
  mov    0x0(%rbp),%rsi
  mov    %r13,%rdi
  add    $0x8,%rbp
  call   <scanf@plt>
  cmp    %rbp,%r12
  jne    L10
  mov    %r14,%rdi
  call   <sort@plt>
  lea    D8(%rip),%rdi
  call   <puts@plt>
L25:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <puts@plt>
  cmp    %rbx,%r12
  jne    L25
  mov    0x98(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0xa0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D4
type_infered: string
name: disassemble_data
arguments: tag: D29
type_infered: string
name: disassemble_data
arguments: tag: D8
type_infered: string

tool: D4: input 5 strings:

tool: D29: %s

tool: D8: strings sorted:

assistant: ```c
int main() {
  int i;
  char **p1, *p2[5], str[5][20];
  for (i = 0; i < 5; i++)
    p2[i] = str[i];
  printf("input 5 strings:\n");
  for (i = 0; i < 5; i++)
    scanf("%s", p2[i]);
  p1 = p2;
  sort(p1);
  printf("strings sorted:\n");
  for (i = 0; i < 5; i++)
    printf("%s\n", p2[i]);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<setattr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  movl   $0x3,-0x8(%rbp)
  jmp    L31
L48:
  subl   $0x1,-0x8(%rbp)
  mov    D2(%rip),%rax
  mov    (%rax),%rdx
  mov    D1(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L0
  nop
L31:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <tcsetattr@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xffffffff,-0x4(%rbp)
  jne    L22
  cmpl   $0x0,-0x8(%rbp)
  jg     L48
  jmp    L22
L0:
  nop
L22:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int setattr(int fd, int opt, const struct termios *arg) {
  int result, retries = 3;
  while (-1 == (result = tcsetattr(fd, opt, arg)) && retries > 0) {
    retries--;
    if (EINTR == errno)
      continue;
    break;
  }
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<setattr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  mov    %esi,%ebp
  mov    %rdx,%rbx
  mov    $0x3,%r13d
L15:
  mov    %rbx,%rdx
  mov    %ebp,%esi
  mov    %r12d,%edi
  call   <tcsetattr@plt>
  test   %r13d,%r13d
  jle    L22
  cmp    $0xffffffff,%eax
  jne    L22
  sub    $0x1,%r13d
  mov    D1(%rip),%rdx
  mov    D14(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L15
  mov    $0xffffffff,%eax
L22:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int setattr(int fd, int opt, const struct termios *arg) {
  int result, retries = 3;
  while (-1 == (result = tcsetattr(fd, opt, arg)) && retries > 0) {
    retries--;
    if (EINTR == errno)
      continue;
    break;
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<setattr>:
  endbr64
  push   %r13
  mov    %edi,%r13d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    $0x4,%ebx
  sub    $0x8,%rsp
L18:
  mov    %rbp,%rdx
  mov    %r12d,%esi
  mov    %r13d,%edi
  call   <tcsetattr@plt>
  cmp    $0xffffffff,%eax
  jne    L21
  sub    $0x1,%ebx
  je     L21
  mov    D15(%rip),%rdx
  mov    D1(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L18
L21:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int setattr(int fd, int opt, const struct termios *arg) {
  int result, retries = 3;
  while (-1 == (result = tcsetattr(fd, opt, arg)) && retries > 0) {
    retries--;
    if (EINTR == errno)
      continue;
    break;
  }
  return result;
}
```