ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<trim>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  movzbl (%rbx),%edx
  mov    %rbx,%r8
  cmp    $0x3b,%dl
  ja     L17
  movabs $0xf7fffffefffffdff,%rcx
  nopw   0x0(%rax,%rax,1)
L19:
  bt     %rdx,%rcx
  jb     L17
  movzbl 0x1(%r8),%edx
  add    $0x1,%r8
  cmp    $0x3b,%dl
  jbe    L19
L17:
  cltq
  lea    -0x1(%rbx,%rax,1),%rax
  movzbl (%rax),%edx
  cmp    $0x3b,%dl
  ja     L36
  movabs $0xf7fffffefffffdff,%rcx
  nopl   (%rax)
L10:
  bt     %rdx,%rcx
  jb     L36
  movzbl -0x1(%rax),%edx
  sub    $0x1,%rax
  cmp    $0x3b,%dl
  jbe    L10
L36:
  movb   $0x0,0x1(%rax)
  mov    %r8,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *trim(char *str) {
  int len = strlen(str);
  char *c = str;
  while (*c == ' ' || *c == '\t' || *c == ';')
    c++;
  char *d = &str[len - 1];
  while (*d == ' ' || *d == '\t' || *d == ';')
    d--;
  *(d + 1) = 0;
  return c;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  nop
L9:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L9
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t strlen(const char *s) {
  const char *w = s;
  while (*w++)
    ;
  return w - s;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strlen>:
  endbr64
  mov    %rdi,%rax
L5:
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L5
  sub    %rdi,%rax
  ret
```

assistant: ```c
size_t strlen(const char *s) {
  const char *w = s;
  while (*w++)
    ;
  return w - s;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strlen>:
  endbr64
  mov    %rdi,%rax
  nopw   0x0(%rax,%rax,1)
L3:
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L3
  sub    %rdi,%rax
  ret
```

assistant: ```c
size_t strlen(const char *s) {
  const char *w = s;
  while (*w++)
    ;
  return w - s;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<strlen>:
  endbr64
  mov    %rdi,%rax
  nopw   0x0(%rax,%rax,1)
L7:
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L7
  sub    %rdi,%rax
  ret
```

assistant: ```c
size_t strlen(const char *s) {
  const char *w = s;
  while (*w++)
    ;
  return w - s;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<imat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jle    L16
  cmpl   $0x0,-0x18(%rbp)
  jg     L32
L16:
  mov    $0x0,%eax
  jmp    L13
L32:
  mov    -0x14(%rbp),%eax
  cltq
  mov    %eax,%edx
  mov    -0x18(%rbp),%eax
  cltq
  imul   %edx,%eax
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L24
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <fatalerr@plt>
L24:
  mov    -0x8(%rbp),%rax
L13:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D9
type_infered: string

tool: D9: integer matrix memory allocation error: n=%d,m=%d


assistant: ```c
extern int *imat(int n, int m) {
  int *p;
  if (n <= 0 || m <= 0)
    return NULL;
  if (!(p = (int *)malloc(sizeof(int) * n * m))) {
    fatalerr("integer matrix memory allocation error: n=%d,m=%d\n", n, m);
  }
  return p;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<imat>:
  endbr64
  test   %edi,%edi
  jle    L16
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    %esi,%ebp
  test   %esi,%esi
  jle    L8
  imul   %esi,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L9
L36:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  mov    %ebp,%edx
  mov    %ebx,%esi
  lea    D7(%rip),%rdi
  call   <fatalerr@plt>
  mov    $0x0,%eax
  jmp    L36
L8:
  mov    $0x0,%eax
  jmp    L36
L16:
  mov    $0x0,%eax
  ret
```

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

tool: D7: integer matrix memory allocation error: n=%d,m=%d


assistant: ```c
extern int *imat(int n, int m) {
  int *p;
  if (n <= 0 || m <= 0)
    return NULL;
  if (!(p = (int *)malloc(sizeof(int) * n * m))) {
    fatalerr("integer matrix memory allocation error: n=%d,m=%d\n", n, m);
  }
  return p;
}
```
user: What is the c source code of the assembly code below?

```asm
<imat>:
  endbr64
  test   %edi,%edi
  jle    L22
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L11
  mov    %edi,%ebp
  imul   %esi,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L6
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
L6:
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D2(%rip),%rdi
  call   <fatalerr@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L22:
  xor    %eax,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: label: D2
data_type: string

tool: D2: integer matrix memory allocation error: n=%d,m=%d


assistant: ```c
extern int *imat(int n, int m) {
  int *p;
  if (n <= 0 || m <= 0)
    return NULL;
  if (!(p = (int *)malloc(sizeof(int) * n * m))) {
    fatalerr("integer matrix memory allocation error: n=%d,m=%d\n", n, m);
  }
  return p;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<imat>:
  endbr64
  test   %edi,%edi
  jle    L6
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %esi,%esi
  jle    L7
  mov    %edi,%ebp
  imul   %esi,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L21
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
L21:
  mov    %r12d,%edx
  mov    %ebp,%esi
  lea    D1(%rip),%rdi
  call   <fatalerr@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D1
datatype: string

tool: D1: integer matrix memory allocation error: n=%d,m=%d


assistant: ```c
extern int *imat(int n, int m) {
  int *p;
  if (n <= 0 || m <= 0)
    return NULL;
  if (!(p = (int *)malloc(sizeof(int) * n * m))) {
    fatalerr("integer matrix memory allocation error: n=%d,m=%d\n", n, m);
  }
  return p;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  add    $0x45,%eax
  mov    %al,-0x9(%rbp)
  movb   $0x0,-0xa(%rbp)
  jmp    L39
L9:
  movzbl -0x9(%rbp),%eax
  cmp    %al,-0xa(%rbp)
  jae    L21
  movzbl -0xa(%rbp),%eax
  cltq
  movzbl -0x9(%rbp,%rax,1),%eax
  movzbl -0x9(%rbp),%edx
  sub    %edx,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  jmp    L41
L21:
  movzbl -0xa(%rbp),%eax
  cltq
  movzbl -0x9(%rbp,%rax,1),%edx
  movzbl -0xa(%rbp),%eax
  add    %edx,%edx
  cltq
  mov    %dl,-0x9(%rbp,%rax,1)
  movzbl -0x9(%rbp),%ecx
  movzbl -0xa(%rbp),%eax
  cltq
  movzbl -0x9(%rbp,%rax,1),%edx
  movzbl -0xa(%rbp),%eax
  add    %ecx,%edx
  cltq
  mov    %dl,-0x9(%rbp,%rax,1)
L41:
  addb   $0x2,-0xa(%rbp)
L39:
  cmpb   $0x0,-0xa(%rbp)
  je     L9
  movzbl -0x9(%rbp),%eax
  lea    0x10(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)69;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        state[0UL] = state[local1] - state[0UL];
        state[0UL] += state[0UL];
      } else {
        state[local1] += state[local1];
        state[local1] = state[0UL] + state[local1];
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + (unsigned char)16;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<RandomFunc>:
  endbr64
  movb   $0x10,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)69;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        state[0UL] = state[local1] - state[0UL];
        state[0UL] += state[0UL];
      } else {
        state[local1] += state[local1];
        state[local1] = state[0UL] + state[local1];
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + (unsigned char)16;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<RandomFunc>:
  endbr64
  movb   $0x10,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)69;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        state[0UL] = state[local1] - state[0UL];
        state[0UL] += state[0UL];
      } else {
        state[local1] += state[local1];
        state[local1] = state[0UL] + state[local1];
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + (unsigned char)16;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<RandomFunc>:
  endbr64
  movb   $0x10,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  unsigned char local1;
  {
    state[0UL] = input[0UL] + (unsigned char)69;
    local1 = 0UL;
    while (local1 < 1UL) {
      if (state[0UL] > local1) {
        state[0UL] = state[local1] - state[0UL];
        state[0UL] += state[0UL];
      } else {
        state[local1] += state[local1];
        state[local1] = state[0UL] + state[local1];
      }
      local1 += 2UL;
    }
    output[0UL] = state[0UL] + (unsigned char)16;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<c_vector3_cross_product>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x30(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  mov    -0x38(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mulss  %xmm1,%xmm0
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm2
  mov    -0x38(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  mulss  %xmm2,%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x14(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  mov    -0x38(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  %xmm1,%xmm0
  mov    -0x30(%rbp),%rax
  movss  (%rax),%xmm2
  mov    -0x38(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  mulss  %xmm2,%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  movss  (%rax),%xmm1
  mov    -0x38(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mulss  %xmm1,%xmm0
  mov    -0x30(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm2
  mov    -0x38(%rbp),%rax
  movss  (%rax),%xmm1
  mulss  %xmm2,%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  movss  -0x14(%rbp),%xmm0
  mov    -0x28(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x28(%rbp),%rax
  add    $0x4,%rax
  movss  -0x10(%rbp),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  movss  -0xc(%rbp),%xmm0
  movss  %xmm0,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
void c_vector3_cross_product(float *result, const float *a, const float *b) {
  float tmp[3];
  tmp[0] = a[1] * b[2] - a[2] * b[1];
  tmp[1] = a[2] * b[0] - a[0] * b[2];
  tmp[2] = a[0] * b[1] - a[1] * b[0];
  result[0] = tmp[0];
  result[1] = tmp[1];
  result[2] = tmp[2];
}
```
user: What is the c source code of the assembly code below?

```asm
<c_vector3_cross_product>:
  endbr64
  movss  0x4(%rsi),%xmm2
  movss  0x8(%rdx),%xmm3
  movss  0x8(%rsi),%xmm1
  movss  0x4(%rdx),%xmm0
  movss  (%rdx),%xmm5
  movss  (%rsi),%xmm6
  movaps %xmm2,%xmm4
  mulss  %xmm3,%xmm4
  movaps %xmm1,%xmm7
  mulss  %xmm0,%xmm7
  subss  %xmm7,%xmm4
  movss  %xmm4,(%rdi)
  mulss  %xmm5,%xmm1
  mulss  %xmm6,%xmm3
  subss  %xmm3,%xmm1
  movss  %xmm1,0x4(%rdi)
  mulss  %xmm6,%xmm0
  mulss  %xmm5,%xmm2
  subss  %xmm2,%xmm0
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void c_vector3_cross_product(float *result, const float *a, const float *b) {
  float tmp[3];
  tmp[0] = a[1] * b[2] - a[2] * b[1];
  tmp[1] = a[2] * b[0] - a[0] * b[2];
  tmp[2] = a[0] * b[1] - a[1] * b[0];
  result[0] = tmp[0];
  result[1] = tmp[1];
  result[2] = tmp[2];
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<c_vector3_cross_product>:
  endbr64
  movss  0x4(%rsi),%xmm2
  movss  0x8(%rsi),%xmm1
  movss  0x8(%rdx),%xmm3
  movss  0x4(%rdx),%xmm0
  movss  (%rdx),%xmm5
  movss  (%rsi),%xmm6
  movaps %xmm2,%xmm4
  movaps %xmm1,%xmm7
  mulss  %xmm3,%xmm4
  mulss  %xmm0,%xmm7
  mulss  %xmm5,%xmm1
  mulss  %xmm6,%xmm3
  mulss  %xmm6,%xmm0
  mulss  %xmm5,%xmm2
  subss  %xmm7,%xmm4
  subss  %xmm3,%xmm1
  movss  %xmm4,(%rdi)
  subss  %xmm2,%xmm0
  movss  %xmm1,0x4(%rdi)
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void c_vector3_cross_product(float *result, const float *a, const float *b) {
  float tmp[3];
  tmp[0] = a[1] * b[2] - a[2] * b[1];
  tmp[1] = a[2] * b[0] - a[0] * b[2];
  tmp[2] = a[0] * b[1] - a[1] * b[0];
  result[0] = tmp[0];
  result[1] = tmp[1];
  result[2] = tmp[2];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<c_vector3_cross_product>:
  endbr64
  movss  (%rdx),%xmm4
  movss  (%rsi),%xmm5
  movss  0x8(%rdx),%xmm2
  movss  0x8(%rsi),%xmm3
  movq   0x4(%rsi),%xmm1
  movq   0x4(%rdx),%xmm0
  unpcklps %xmm4,%xmm2
  unpcklps %xmm5,%xmm3
  mulps  %xmm1,%xmm2
  mulps  %xmm0,%xmm3
  mulss  %xmm4,%xmm1
  mulss  %xmm5,%xmm0
  subps  %xmm3,%xmm2
  subss  %xmm1,%xmm0
  movlps %xmm2,(%rdi)
  movss  %xmm0,0x8(%rdi)
  ret
```

assistant: ```c
void c_vector3_cross_product(float *result, const float *a, const float *b) {
  float tmp[3];
  tmp[0] = a[1] * b[2] - a[2] * b[1];
  tmp[1] = a[2] * b[0] - a[0] * b[2];
  tmp[2] = a[0] * b[1] - a[1] * b[0];
  result[0] = tmp[0];
  result[1] = tmp[1];
  result[2] = tmp[2];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0xc8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0xcc(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0xc8(%rbp)
  jmp    L10
L25:
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0xc8(%rbp),%eax
  cltq
  mov    %rdx,-0xc0(%rbp,%rax,8)
  movl   $0x0,-0xc4(%rbp)
  nop
L29:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0xcd(%rbp)
  cmpb   $0xa,-0xcd(%rbp)
  je     L29
  mov    -0xc8(%rbp),%eax
  cltq
  mov    -0xc0(%rbp,%rax,8),%rcx
  mov    -0xc4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc4(%rbp)
  cltq
  lea    (%rcx,%rax,1),%rdx
  movzbl -0xcd(%rbp),%eax
  mov    %al,(%rdx)
  jmp    L24
L58:
  addl   $0x1,-0xc4(%rbp)
L24:
  mov    -0xc8(%rbp),%eax
  cltq
  mov    -0xc0(%rbp,%rax,8),%rdx
  mov    -0xc4(%rbp),%eax
  cltq
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  movzbl (%rbx),%eax
  cmp    $0xa,%al
  jne    L58
  mov    -0xc8(%rbp),%eax
  cltq
  mov    -0xc0(%rbp,%rax,8),%rdx
  mov    -0xc4(%rbp),%eax
  cltq
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0xc8(%rbp)
L10:
  mov    -0xcc(%rbp),%eax
  cmp    %eax,-0xc8(%rbp)
  jl     L25
  mov    -0xc8(%rbp),%eax
  cltq
  movq   $0x0,-0xc0(%rbp,%rax,8)
  mov    -0xcc(%rbp),%eax
  lea    0x1(%rax),%edx
  lea    -0xc0(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <char_array_sort@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0xc8(%rbp)
  jmp    L32
L23:
  mov    -0xc8(%rbp),%eax
  cltq
  mov    -0xc0(%rbp,%rax,8),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addl   $0x1,-0xc8(%rbp)
L32:
  mov    -0xc8(%rbp),%eax
  cltq
  mov    -0xc0(%rbp,%rax,8),%rax
  test   %rax,%rax
  jne    L23
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D5
data_type: string
name: unpack_data
arguments: data_label: D14
data_type: string

tool: D5: input sting number:


tool: D14: %d

assistant: ```c
int main(void) {
  char *p[20];
  int i, j, string_count;
  char ch;
  printf("input sting number:\n");
  scanf("%d", &string_count);
  for (i = 0; i < string_count; i++) {
    p[i] = (char *)malloc(sizeof(char));
    j = 0;
    while ((ch = getchar()) == '\n')
      ;
    p[i][j++] = ch;
    while ((p[i][j] = getchar()) != '\n')
      j++;
    p[i][j] = '\0';
  }
  p[i] = NULL;
  char_array_sort(p, string_count + 1);
  putchar('\n');
  for (i = 0; p[i] != NULL; i++)
    puts(p[i]);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xb8(%rsp)
  xor    %eax,%eax
  lea    D16(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L19
  lea    0x10(%rsp),%r13
  mov    $0x0,%r14d
L49:
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %r13,%r12
  mov    %rax,0x0(%r13)
L36:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  je     L36
  mov    (%r12),%rdx
  mov    %al,(%rdx)
  mov    $0x1,%ebx
L14:
  mov    %rbx,%rbp
  add    (%r12),%rbp
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,0x0(%rbp)
  mov    %rbx,%rdx
  add    $0x1,%rbx
  cmp    $0xa,%al
  jne    L14
  mov    (%r12),%rax
  movb   $0x0,(%rax,%rdx,1)
  add    $0x1,%r14d
  mov    0xc(%rsp),%eax
  add    $0x8,%r13
  cmp    %r14d,%eax
  jg     L49
L32:
  movslq %r14d,%r14
  movq   $0x0,0x10(%rsp,%r14,8)
  lea    0x1(%rax),%esi
  lea    0x10(%rsp),%rdi
  call   <char_array_sort@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x10(%rsp),%rdi
  test   %rdi,%rdi
  je     L67
  lea    0x18(%rsp),%rbx
L30:
  call   <puts@plt>
  add    $0x8,%rbx
  mov    -0x8(%rbx),%rdi
  test   %rdi,%rdi
  jne    L30
L67:
  mov    0xb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L72
  mov    $0x0,%eax
  add    $0xc0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  mov    $0x0,%r14d
  jmp    L32
L72:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D16
type_guessed: string
name: parse_data
arguments: data_label: D18
type_guessed: string

tool: D16: input sting number:


tool: D18: %d

assistant: ```c
int main(void) {
  char *p[20];
  int i, j, string_count;
  char ch;
  printf("input sting number:\n");
  scanf("%d", &string_count);
  for (i = 0; i < string_count; i++) {
    p[i] = (char *)malloc(sizeof(char));
    j = 0;
    while ((ch = getchar()) == '\n')
      ;
    p[i][j++] = ch;
    while ((p[i][j] = getchar()) != '\n')
      j++;
    p[i][j] = '\0';
  }
  p[i] = NULL;
  char_array_sort(p, string_count + 1);
  putchar('\n');
  for (i = 0; p[i] != NULL; i++)
    puts(p[i]);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D4(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xb8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L42
  lea    0x10(%rsp),%r13
  xor    %r12d,%r12d
  mov    %r13,%rbp
  nopw   0x0(%rax,%rax,1)
L8:
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  xchg   %ax,%ax
L33:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  je     L33
  mov    0x0(%rbp),%rdx
  mov    $0x1,%ebx
  mov    %al,(%rdx)
  cs nopw 0x0(%rax,%rax,1)
L21:
  mov    0x0(%rbp),%r14
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %rbx,%rdx
  add    %rbx,%r14
  add    $0x1,%rbx
  mov    %al,(%r14)
  cmp    $0xa,%al
  jne    L21
  mov    0x0(%rbp),%rax
  add    $0x1,%r12d
  add    $0x8,%rbp
  movb   $0x0,(%rax,%rdx,1)
  mov    0xc(%rsp),%eax
  cmp    %r12d,%eax
  jg     L8
L3:
  lea    0x1(%rax),%esi
  mov    %r13,%rdi
  movslq %r12d,%r12
  movq   $0x0,0x10(%rsp,%r12,8)
  call   <char_array_sort@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x10(%rsp),%rdi
  test   %rdi,%rdi
  je     L73
  lea    0x18(%rsp),%rbx
  nopl   (%rax)
L28:
  call   <puts@plt>
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  test   %rdi,%rdi
  jne    L28
L73:
  mov    0xb8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0xc0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L42:
  xor    %r12d,%r12d
  lea    0x10(%rsp),%r13
  jmp    L3
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D4
type_guessed: string
name: unpack_data
arguments: tag: D2
type_guessed: string

tool: D4: input sting number:


tool: D2: %d

assistant: ```c
int main(void) {
  char *p[20];
  int i, j, string_count;
  char ch;
  printf("input sting number:\n");
  scanf("%d", &string_count);
  for (i = 0; i < string_count; i++) {
    p[i] = (char *)malloc(sizeof(char));
    j = 0;
    while ((ch = getchar()) == '\n')
      ;
    p[i][j++] = ch;
    while ((p[i][j] = getchar()) != '\n')
      j++;
    p[i][j] = '\0';
  }
  p[i] = NULL;
  char_array_sort(p, string_count + 1);
  putchar('\n');
  for (i = 0; p[i] != NULL; i++)
    puts(p[i]);
  return 0;
}
```