ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<test_mq_timedsend_prototype>:
  endbr64
  sub    $0x18,%rsp
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  xor    %edi,%edi
  lea    0x4(%rsp),%r8
  call   <mq_timedsend@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_mq_timedsend_prototype(void) {
  mqd_t mqdes;
  struct timespec timeout;
  char *msgp;
  int err;
  size_t msg_len;
  unsigned msg_prio;
  mqdes = 0;
  msgp = NULL;
  msg_len = 0;
  msg_prio = 0;
  err = mq_timedsend(mqdes, msgp, msg_len, msg_prio, &timeout);
  (void)err;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movq   $0x0,-0x10(%rbp)
  cmpl   $0x2,-0x24(%rbp)
  je     L24
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L24:
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atol@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  shl    $0x3,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L34
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L34:
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movq   $0x0,-0x18(%rbp)
  jmp    L16
L8:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    -0x18(%rbp),%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rdx,%rcx
  cqto
  idivq  -0x10(%rbp)
  mov    %rdx,%rax
  mov    %rax,(%rcx)
  addq   $0x1,-0x18(%rbp)
L16:
  mov    -0x18(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jl     L8
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <getmax@plt>
  mov    %rax,%rsi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D33
type_infered: string
name: retrieve_data
arguments: data_label: D5
type_infered: string
name: retrieve_data
arguments: data_label: D28
type_infered: string
name: retrieve_data
arguments: data_label: D36
type_infered: string

tool: D33: usage: maxseq num

tool: D5: num = size of the array

tool: D28: Unable to allocate mem for an array of size %ld


tool: D36:  The maximum number in the array is: %ld


assistant: ```c
int main(int argc, char *argv[]) {
  long size = 0;
  long i;
  long *numbers;
  if (argc != 2) {
    printf("usage: maxseq num\n");
    printf("num = size of the array\n");
    exit(1);
  }
  size = atol(argv[1]);
  numbers = (long *)malloc(size * sizeof(long));
  if (!numbers) {
    printf("Unable to allocate mem for an array of size %ld\n", size);
    exit(1);
  }
  srand(time(NULL));
  for (i = 0; i < size; i++)
    numbers[i] = rand() % size;
  printf(" The maximum number in the array is: %ld\n", getmax(numbers, size));
  free(numbers);
  exit(0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbx
  cmp    $0x2,%edi
  jne    L1
L28:
  mov    0x8(%rbx),%rdi
  call   <atol@plt>
  mov    %rax,%rbx
  lea    0x0(,%rax,8),%rdi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L10
L3:
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %rbx,%rbx
  jle    L0
  mov    %r13,%r12
  mov    $0x0,%ebp
L51:
  mov    $0x0,%eax
  call   <rand@plt>
  cqto
  idiv   %rbx
  mov    %rdx,(%r12)
  add    $0x1,%rbp
  add    $0x8,%r12
  cmp    %rbp,%rbx
  jne    L51
L0:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <getmax@plt>
  mov    %rax,%rsi
  lea    D8(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <free@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L1:
  lea    D37(%rip),%rdi
  call   <puts@plt>
  lea    D30(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L28
L10:
  mov    %rbx,%rsi
  lea    D34(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L3
```

tool_calls: name: disassemble_data
arguments: label: D8
data_type: string
name: disassemble_data
arguments: label: D37
data_type: string
name: disassemble_data
arguments: label: D30
data_type: string
name: disassemble_data
arguments: label: D34
data_type: string

tool: D8:  The maximum number in the array is: %ld


tool: D37: usage: maxseq num

tool: D30: num = size of the array

tool: D34: Unable to allocate mem for an array of size %ld


assistant: ```c
int main(int argc, char *argv[]) {
  long size = 0;
  long i;
  long *numbers;
  if (argc != 2) {
    printf("usage: maxseq num\n");
    printf("num = size of the array\n");
    exit(1);
  }
  size = atol(argv[1]);
  numbers = (long *)malloc(size * sizeof(long));
  if (!numbers) {
    printf("Unable to allocate mem for an array of size %ld\n", size);
    exit(1);
  }
  srand(time(NULL));
  for (i = 0; i < size; i++)
    numbers[i] = rand() % size;
  printf(" The maximum number in the array is: %ld\n", getmax(numbers, size));
  free(numbers);
  exit(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   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L20
  lea    D22(%rip),%rdi
  call   <puts@plt>
  lea    D26(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L20:
  mov    0x8(%rbx),%rdi
  call   <atol@plt>
  lea    0x0(,%rax,8),%rdi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L30
L4:
  xor    %edi,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %rbp,%rbp
  jle    L47
  mov    %r13,%r12
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L45:
  xor    %eax,%eax
  add    $0x1,%rbx
  add    $0x8,%r12
  call   <rand@plt>
  cqto
  idiv   %rbp
  mov    %rdx,-0x8(%r12)
  cmp    %rbx,%rbp
  jne    L45
L47:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <getmax@plt>
  lea    D23(%rip),%rdi
  mov    %rax,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <free@plt>
  xor    %edi,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L30:
  mov    %rbp,%rsi
  lea    D21(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L4
```

tool_calls: name: disassemble_data
arguments: tag: D22
datatype: string
name: disassemble_data
arguments: tag: D26
datatype: string
name: disassemble_data
arguments: tag: D23
datatype: string
name: disassemble_data
arguments: tag: D21
datatype: string

tool: D22: usage: maxseq num

tool: D26: num = size of the array

tool: D23:  The maximum number in the array is: %ld


tool: D21: Unable to allocate mem for an array of size %ld


assistant: ```c
int main(int argc, char *argv[]) {
  long size = 0;
  long i;
  long *numbers;
  if (argc != 2) {
    printf("usage: maxseq num\n");
    printf("num = size of the array\n");
    exit(1);
  }
  size = atol(argv[1]);
  numbers = (long *)malloc(size * sizeof(long));
  if (!numbers) {
    printf("Unable to allocate mem for an array of size %ld\n", size);
    exit(1);
  }
  srand(time(NULL));
  for (i = 0; i < size; i++)
    numbers[i] = rand() % size;
  printf(" The maximum number in the array is: %ld\n", getmax(numbers, size));
  free(numbers);
  exit(0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  cmp    $0x2,%edi
  je     L35
  lea    D2(%rip),%rdi
  call   <puts@plt>
  lea    D34(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L35:
  mov    0x8(%rbx),%rdi
  call   <atol@plt>
  lea    0x0(,%rax,8),%rdi
  mov    %rax,%rbp
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L10
L20:
  xor    %edi,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %rbp,%rbp
  jle    L1
  mov    %r13,%r12
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L14:
  xor    %eax,%eax
  add    $0x1,%rbx
  add    $0x8,%r12
  call   <rand@plt>
  cqto
  idiv   %rbp
  mov    %rdx,-0x8(%r12)
  cmp    %rbx,%rbp
  jne    L14
L1:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <getmax@plt>
  lea    D27(%rip),%rdi
  mov    %rax,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  call   <free@plt>
  xor    %edi,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  mov    %rbp,%rsi
  lea    D16(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L20
```

tool_calls: name: retrieve_data
arguments: tag: D2
type: string
name: retrieve_data
arguments: tag: D34
type: string
name: retrieve_data
arguments: tag: D27
type: string
name: retrieve_data
arguments: tag: D16
type: string

tool: D2: usage: maxseq num

tool: D34: num = size of the array

tool: D27:  The maximum number in the array is: %ld


tool: D16: Unable to allocate mem for an array of size %ld


assistant: ```c
int main(int argc, char *argv[]) {
  long size = 0;
  long i;
  long *numbers;
  if (argc != 2) {
    printf("usage: maxseq num\n");
    printf("num = size of the array\n");
    exit(1);
  }
  size = atol(argv[1]);
  numbers = (long *)malloc(size * sizeof(long));
  if (!numbers) {
    printf("Unable to allocate mem for an array of size %ld\n", size);
    exit(1);
  }
  srand(time(NULL));
  for (i = 0; i < size; i++)
    numbers[i] = rand() % size;
  printf(" The maximum number in the array is: %ld\n", getmax(numbers, size));
  free(numbers);
  exit(0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    $0x0,%eax
  call   <test@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int c, char **v, char **e) {
  test();
  getchar();
  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
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <test@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int c, char **v, char **e) {
  test();
  getchar();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main(int c, char **v, char **e) {
  test();
  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    $0x8,%rsp
  xor    %eax,%eax
  call   <test@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int c, char **v, char **e) {
  test();
  getchar();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<bsearch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %rcx,-0x40(%rbp)
  mov    %r8,-0x48(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %eax,-0x14(%rbp)
  jmp    L36
L8:
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  imul   -0x40(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    -0x48(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L24
  mov    -0x8(%rbp),%rax
  jmp    L13
L24:
  cmpl   $0x0,-0xc(%rbp)
  jns    L1
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L36
L1:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x18(%rbp)
L36:
  mov    -0x14(%rbp),%eax
  sub    -0x18(%rbp),%eax
  cmp    $0x1,%eax
  jg     L8
  mov    $0x0,%eax
L13:
  leave
  ret
```

assistant: ```c
void *bsearch(const void *searchKey, const void *base, size_t num, size_t size, int (*compare)(const void *, const void *)) {
  int low = 0;
  int high = num;
  while (high - low > 1) {
    int mid = (low + high) / 2;
    void *midKey = (char *)base + mid * size;
    int compVal = (*compare)(searchKey, midKey);
    if (compVal == 0)
      return midKey;
    else if (compVal < 0)
      high = mid;
    else
      low = mid;
  }
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<bsearch>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  mov    %rsi,0x8(%rsp)
  cmp    $0x1,%edx
  jle    L18
  mov    %rcx,%r14
  mov    %r8,%r15
  mov    %edx,%r12d
  mov    $0x0,%r13d
L23:
  lea    0x0(%r13,%r12,1),%eax
  mov    %eax,%ebx
  shr    $0x1f,%ebx
  add    %eax,%ebx
  sar    %ebx
  movslq %ebx,%rbp
  imul   %r14,%rbp
  add    0x8(%rsp),%rbp
  mov    %rbp,%rsi
  mov    (%rsp),%rdi
  call   *%r15
  test   %eax,%eax
  je     L6
  cmovns %ebx,%r13d
  cmovs  %ebx,%r12d
  mov    %r12d,%eax
  sub    %r13d,%eax
  cmp    $0x1,%eax
  jg     L23
  mov    $0x0,%ebp
L6:
  mov    %rbp,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L18:
  mov    $0x0,%ebp
  jmp    L6
```

assistant: ```c
void *bsearch(const void *searchKey, const void *base, size_t num, size_t size, int (*compare)(const void *, const void *)) {
  int low = 0;
  int high = num;
  while (high - low > 1) {
    int mid = (low + high) / 2;
    void *midKey = (char *)base + mid * size;
    int compVal = (*compare)(searchKey, midKey);
    if (compVal == 0)
      return midKey;
    else if (compVal < 0)
      high = mid;
    else
      low = mid;
  }
  return NULL;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bsearch>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,(%rsp)
  mov    %rcx,0x8(%rsp)
  cmp    $0x1,%edx
  jle    L34
  mov    %rdi,%rbp
  mov    %r8,%r14
  mov    %edx,%r12d
  xor    %r13d,%r13d
  jmp    L45
  xchg   %ax,%ax
L13:
  cmovs  %ebx,%r12d
  cmovns %ebx,%r13d
  mov    %r12d,%eax
  sub    %r13d,%eax
  cmp    $0x1,%eax
  jle    L34
L45:
  lea    0x0(%r13,%r12,1),%eax
  mov    %rbp,%rdi
  mov    %eax,%ebx
  shr    $0x1f,%ebx
  add    %eax,%ebx
  sar    %ebx
  movslq %ebx,%r15
  imul   0x8(%rsp),%r15
  add    (%rsp),%r15
  mov    %r15,%rsi
  call   *%r14
  test   %eax,%eax
  jne    L13
L15:
  add    $0x18,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L34:
  xor    %r15d,%r15d
  jmp    L15
```

assistant: ```c
void *bsearch(const void *searchKey, const void *base, size_t num, size_t size, int (*compare)(const void *, const void *)) {
  int low = 0;
  int high = num;
  while (high - low > 1) {
    int mid = (low + high) / 2;
    void *midKey = (char *)base + mid * size;
    int compVal = (*compare)(searchKey, midKey);
    if (compVal == 0)
      return midKey;
    else if (compVal < 0)
      high = mid;
    else
      low = mid;
  }
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<bsearch>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,(%rsp)
  mov    %rcx,0x8(%rsp)
  cmp    $0x1,%edx
  jle    L44
  mov    %rdi,%r12
  mov    %r8,%r15
  mov    %edx,%ebp
  xor    %r13d,%r13d
  nopl   0x0(%rax,%rax,1)
L48:
  lea    0x0(%r13,%rbp,1),%eax
  mov    %r12,%rdi
  mov    %eax,%ebx
  shr    $0x1f,%ebx
  add    %eax,%ebx
  mov    (%rsp),%rax
  sar    %ebx
  movslq %ebx,%r8
  imul   0x8(%rsp),%r8
  lea    (%rax,%r8,1),%r14
  mov    %r14,%rsi
  call   *%r15
  test   %eax,%eax
  je     L30
  js     L55
  mov    %ebp,%eax
  sub    %ebx,%eax
  cmp    $0x1,%eax
  jle    L44
  mov    %ebx,%r13d
  jmp    L48
  nopl   0x0(%rax)
L55:
  mov    %ebx,%eax
  sub    %r13d,%eax
  cmp    $0x1,%eax
  jle    L44
  mov    %ebx,%ebp
  jmp    L48
  xchg   %ax,%ax
L44:
  xor    %r14d,%r14d
L30:
  add    $0x18,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void *bsearch(const void *searchKey, const void *base, size_t num, size_t size, int (*compare)(const void *, const void *)) {
  int low = 0;
  int high = num;
  while (high - low > 1) {
    int mid = (low + high) / 2;
    void *midKey = (char *)base + mid * size;
    int compVal = (*compare)(searchKey, midKey);
    if (compVal == 0)
      return midKey;
    else if (compVal < 0)
      high = mid;
    else
      low = mid;
  }
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x20(%rbp)
  movl   $0x1,-0x1c(%rbp)
  movq   $0x40000000,-0x10(%rbp)
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L23
L10:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x18(%rbp),%rax
  cmp    %rax,-0x10(%rbp)
  jge    L49
  addl   $0x1,-0x1c(%rbp)
  jmp    L45
L49:
  movl   $0x1,-0x1c(%rbp)
L45:
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  cmp    %eax,%edx
  cmovge %edx,%eax
  mov    %eax,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
L23:
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  test   %eax,%eax
  jne    L10
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L22
  call   <__stack_chk_fail@plt>
L22:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D20
datatype: string
name: disassemble_data
arguments: tag: D6
datatype: string

tool: D20: %d

tool: D6: %I64d

assistant: ```c
int main() {
  int n, m = 1, c = 1;
  long long p = 1 << 30, a;
  scanf("%d", &n);
  while (n--) {
    scanf("%I64d", &a);
    if (a > p)
      ++c;
    else
      c = 1;
    m = m > c ? m : c;
    p = a;
  }
  printf("%d", m);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L35
  mov    $0x40000000,%r12d
  mov    $0x1,%ebx
  mov    $0x1,%ebp
  lea    D18(%rip),%r13
L2:
  lea    0x10(%rsp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    %r12,%rax
  mov    0x10(%rsp),%r12
  add    $0x1,%ebx
  cmp    %rax,%r12
  mov    $0x1,%eax
  cmovle %eax,%ebx
  cmp    %ebx,%ebp
  cmovl  %ebx,%ebp
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  jne    L2
L29:
  mov    %ebp,%esi
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L35:
  mov    $0x1,%ebp
  jmp    L29
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D24
datatype: string
name: read_data
arguments: data_label: D18
datatype: string

tool: D24: %d

tool: D18: %I64d

assistant: ```c
int main() {
  int n, m = 1, c = 1;
  long long p = 1 << 30, a;
  scanf("%d", &n);
  while (n--) {
    scanf("%I64d", &a);
    if (a > p)
      ++c;
    else
      c = 1;
    m = m > c ? m : c;
    p = a;
  }
  printf("%d", m);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D7(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L15
  mov    $0x40000000,%r14d
  mov    $0x1,%ebx
  mov    $0x1,%r15d
  mov    $0x1,%ebp
  lea    0x10(%rsp),%r13
  lea    D24(%rip),%r12
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <scanf@plt>
  mov    %r14,%rax
  mov    0x10(%rsp),%r14
  cmp    %rax,%r14
  mov    0xc(%rsp),%eax
  cmovle %ebp,%ebx
  lea    -0x1(%rax),%ecx
  cmp    %ebx,%r15d
  mov    %ecx,0xc(%rsp)
  cmovl  %ebx,%r15d
  test   %eax,%eax
  jne    L3
L0:
  mov    %r15d,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  mov    $0x1,%r15d
  jmp    L0
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
data_type: string
name: retrieve_data
arguments: data_tag: D24
data_type: string

tool: D7: %d

tool: D24: %I64d

assistant: ```c
int main() {
  int n, m = 1, c = 1;
  long long p = 1 << 30, a;
  scanf("%d", &n);
  while (n--) {
    scanf("%I64d", &a);
    if (a > p)
      ++c;
    else
      c = 1;
    m = m > c ? m : c;
    p = a;
  }
  printf("%d", m);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  lea    D12(%rip),%r14
  push   %r13
  mov    %r14,%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0xc(%rsp)
  test   %eax,%eax
  je     L37
  mov    $0x40000000,%ebx
  mov    $0x1,%r15d
  mov    $0x1,%ebp
  lea    0x10(%rsp),%r13
  lea    D0(%rip),%r12
  jmp    L29
  nopl   0x0(%rax,%rax,1)
L61:
  mov    0xc(%rsp),%edx
  add    $0x1,%r15d
  cmp    %r15d,%ebp
  lea    -0x1(%rdx),%ecx
  cmovl  %r15d,%ebp
  mov    %ecx,0xc(%rsp)
  test   %edx,%edx
  je     L12
L17:
  mov    %rax,%rbx
L29:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rax
  cmp    %rbx,%rax
  jg     L61
  mov    0xc(%rsp),%edx
  lea    -0x1(%rdx),%ecx
  mov    %ecx,0xc(%rsp)
  test   %edx,%edx
  je     L12
  mov    $0x1,%r15d
  jmp    L17
L37:
  mov    $0x1,%ebp
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
datatype: string
name: extract_symbolic_data
arguments: label: D0
datatype: string

tool: D12: %d

tool: D0: %I64d

assistant: ```c
int main() {
  int n, m = 1, c = 1;
  long long p = 1 << 30, a;
  scanf("%d", &n);
  while (n--) {
    scanf("%I64d", &a);
    if (a > p)
      ++c;
    else
      c = 1;
    m = m > c ? m : c;
    p = a;
  }
  printf("%d", m);
}
```