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

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  sub    $0x99f4bb,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L1
L22:
  mov    -0x10(%rbp),%eax
  mov    -0xc(%rbp,%rax,4),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,-0xc(%rbp,%rax,4)
  addl   $0x2,-0x10(%rbp)
L1:
  cmpl   $0x0,-0x10(%rbp)
  je     L22
  movl   $0x0,-0x10(%rbp)
  jmp    L41
L40:
  lea    -0xc(%rbp),%rax
  movzwl (%rax),%eax
  mov    %ax,-0x14(%rbp)
  lea    -0xc(%rbp),%rax
  add    $0x2,%rax
  lea    -0xc(%rbp),%rdx
  movzwl (%rax),%eax
  mov    %ax,(%rdx)
  lea    -0xc(%rbp),%rax
  add    $0x2,%rax
  movzwl -0x14(%rbp),%edx
  mov    %dx,(%rax)
  lea    -0xc(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x14(%rbp)
  lea    -0xc(%rbp),%rdx
  lea    -0xc(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rdx),%edx
  mov    %dx,(%rax)
  lea    -0xc(%rbp),%rax
  movzwl -0x14(%rbp),%edx
  mov    %dx,(%rax)
  lea    -0xc(%rbp),%rdx
  mov    -0x10(%rbp),%eax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    $0x2,%rax
  movzwl (%rax),%eax
  mov    %ax,-0x12(%rbp)
  lea    -0xc(%rbp),%rdx
  mov    -0x10(%rbp),%eax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    -0xc(%rbp),%rcx
  mov    -0x10(%rbp),%edx
  shl    $0x2,%rdx
  add    %rcx,%rdx
  add    $0x2,%rdx
  movzwl (%rax),%eax
  mov    %ax,(%rdx)
  lea    -0xc(%rbp),%rdx
  mov    -0x10(%rbp),%eax
  shl    $0x2,%rax
  add    %rax,%rdx
  movzwl -0x12(%rbp),%eax
  mov    %ax,(%rdx)
  addl   $0x1,-0x10(%rbp)
L41:
  cmpl   $0x0,-0x10(%rbp)
  je     L40
  mov    -0xc(%rbp),%eax
  lea    -0x3918917d(%rax),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy12;
  unsigned short copy13;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      state[local1] += state[local1];
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < 1UL) {
      copy12 = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = copy12;
      copy12 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = copy12;
      copy13 = *((unsigned short *)(&state[local1]) + 1);
      *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = copy13;
      local1++;
    }
    output[0UL] = state[0UL] - 957911421UL;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  lea    -0x133e976(%rax,%rax,1),%eax
  rol    $0x10,%eax
  lea    -0x3918917d(%rax),%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy12;
  unsigned short copy13;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      state[local1] += state[local1];
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < 1UL) {
      copy12 = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = copy12;
      copy12 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = copy12;
      copy13 = *((unsigned short *)(&state[local1]) + 1);
      *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = copy13;
      local1++;
    }
    output[0UL] = state[0UL] - 957911421UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  lea    -0x133e976(%rax,%rax,1),%eax
  rol    $0x10,%eax
  sub    $0x3918917d,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy12;
  unsigned short copy13;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      state[local1] += state[local1];
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < 1UL) {
      copy12 = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = copy12;
      copy12 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = copy12;
      copy13 = *((unsigned short *)(&state[local1]) + 1);
      *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = copy13;
      local1++;
    }
    output[0UL] = state[0UL] - 957911421UL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%eax
  lea    -0x133e976(%rax,%rax,1),%eax
  rol    $0x10,%eax
  sub    $0x3918917d,%eax
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned int input[1], unsigned int output[1]) {
  unsigned int state[1];
  unsigned int local1;
  unsigned short copy12;
  unsigned short copy13;
  {
    state[0UL] = input[0UL] + 4284877637U;
    local1 = 0UL;
    while (local1 < 1UL) {
      state[local1] += state[local1];
      local1 += 2UL;
    }
    local1 = 0UL;
    while (local1 < 1UL) {
      copy12 = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = copy12;
      copy12 = *((unsigned short *)(&state[0UL]) + 1);
      *((unsigned short *)(&state[0UL]) + 1) = *((unsigned short *)(&state[0UL]) + 0);
      *((unsigned short *)(&state[0UL]) + 0) = copy12;
      copy13 = *((unsigned short *)(&state[local1]) + 1);
      *((unsigned short *)(&state[local1]) + 1) = *((unsigned short *)(&state[local1]) + 0);
      *((unsigned short *)(&state[local1]) + 0) = copy13;
      local1++;
    }
    output[0UL] = state[0UL] - 957911421UL;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x1,-0x8(%rbp)
  jle    L6
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L7
L6:
  movl   $0x1,-0x8(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L7:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L17:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L17
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D32
type_guessed: string
name: read_data
arguments: label: D2
type_guessed: string
name: read_data
arguments: label: D7
type_guessed: string
name: read_data
arguments: label: D17
type_guessed: string

tool: D32: This program was called with %i parameters


tool: D2: This program was called with only 1 parameter

tool: D7: program = '%s'


tool: D17: parameter %i = '%s'


assistant: ```c
int main(int argc, char **argv) {
  int nb_params = argc - 1;
  if (nb_params > 1) {
    printf("This program was called with %i parameters\n", nb_params);
  } else if (nb_params = 1) {
    printf("This program was called with only 1 parameter\n");
  } else {
    printf("This program was called without any parameter\n");
  }
  printf("program = '%s'\n", argv[0]);
  int i;
  for (i = 0; i < nb_params; i++) {
    printf("parameter %i = '%s'\n", i, argv[i]);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%r13
  lea    -0x1(%rdi),%ebp
  cmp    $0x1,%ebp
  jle    L10
  mov    %ebp,%esi
  lea    D15(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L5:
  mov    0x0(%r13),%rsi
  lea    D35(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  lea    D18(%rip),%r12
L4:
  mov    0x0(%r13,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,%ebp
  jg     L4
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L10:
  lea    D26(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%ebp
  jmp    L5
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D15
datatype: string
name: extract_symbolic_data
arguments: data_tag: D35
datatype: string
name: extract_symbolic_data
arguments: data_tag: D18
datatype: string
name: extract_symbolic_data
arguments: data_tag: D26
datatype: string

tool: D15: This program was called with %i parameters


tool: D35: program = '%s'


tool: D18: parameter %i = '%s'


tool: D26: This program was called with only 1 parameter

assistant: ```c
int main(int argc, char **argv) {
  int nb_params = argc - 1;
  if (nb_params > 1) {
    printf("This program was called with %i parameters\n", nb_params);
  } else if (nb_params = 1) {
    printf("This program was called with only 1 parameter\n");
  } else {
    printf("This program was called without any parameter\n");
  }
  printf("program = '%s'\n", argv[0]);
  int i;
  for (i = 0; i < nb_params; i++) {
    printf("parameter %i = '%s'\n", i, argv[i]);
  }
  return 0;
}
```
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
  mov    %rsi,%r12
  push   %rbp
  lea    -0x1(%rdi),%ebp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%ebp
  jle    L17
  mov    %ebp,%esi
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L2:
  mov    (%r12),%rsi
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  movslq %ebp,%rbp
  xor    %ebx,%ebx
  lea    D35(%rip),%r13
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L23:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbp,%rbx
  jne    L23
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  lea    D0(%rip),%rdi
  mov    $0x1,%ebp
  call   <puts@plt>
  jmp    L2
```

tool_calls: name: retrieve_data
arguments: data_tag: D18
type_infered: string
name: retrieve_data
arguments: data_tag: D1
type_infered: string
name: retrieve_data
arguments: data_tag: D35
type_infered: string
name: retrieve_data
arguments: data_tag: D0
type_infered: string

tool: D18: This program was called with %i parameters


tool: D1: program = '%s'


tool: D35: parameter %i = '%s'


tool: D0: This program was called with only 1 parameter

assistant: ```c
int main(int argc, char **argv) {
  int nb_params = argc - 1;
  if (nb_params > 1) {
    printf("This program was called with %i parameters\n", nb_params);
  } else if (nb_params = 1) {
    printf("This program was called with only 1 parameter\n");
  } else {
    printf("This program was called without any parameter\n");
  }
  printf("program = '%s'\n", argv[0]);
  int i;
  for (i = 0; i < nb_params; i++) {
    printf("parameter %i = '%s'\n", i, argv[i]);
  }
  return 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
  mov    %rsi,%r12
  push   %rbp
  lea    -0x1(%rdi),%ebp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%ebp
  jle    L18
  mov    %ebp,%esi
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L14:
  mov    (%r12),%rsi
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  movslq %ebp,%rbp
  xor    %ebx,%ebx
  lea    D16(%rip),%r13
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L5:
  mov    (%r12,%rbx,8),%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L5
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  lea    D36(%rip),%rdi
  mov    $0x1,%ebp
  call   <puts@plt>
  jmp    L14
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
data_type: string
name: extract_symbolic_data
arguments: tag: D37
data_type: string
name: extract_symbolic_data
arguments: tag: D16
data_type: string
name: extract_symbolic_data
arguments: tag: D36
data_type: string

tool: D2: This program was called with %i parameters


tool: D37: program = '%s'


tool: D16: parameter %i = '%s'


tool: D36: This program was called with only 1 parameter

assistant: ```c
int main(int argc, char **argv) {
  int nb_params = argc - 1;
  if (nb_params > 1) {
    printf("This program was called with %i parameters\n", nb_params);
  } else if (nb_params = 1) {
    printf("This program was called with only 1 parameter\n");
  } else {
    printf("This program was called without any parameter\n");
  }
  printf("program = '%s'\n", argv[0]);
  int i;
  for (i = 0; i < nb_params; i++) {
    printf("parameter %i = '%s'\n", i, argv[i]);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strstr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    -0x4(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jns    L27
  mov    $0x0,%eax
  jmp    L39
L27:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x4(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L26
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  jmp    L39
L26:
  subl   $0x1,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jns    L27
  mov    $0x0,%eax
L39:
  leave
  ret
```

assistant: ```c
char *strstr(s1, s2)
char *s1, *s2;
{
  int x, y;
  y = strlen(s2);
  x = strlen(s1) - y;
  if (x < 0)
    return ((char *)0L);
  do {
    if (strncmp(s1 + x, s2, y) == 0)
      return (s1 + x);
  } while (--x >= 0);
  return ((char *)0L);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strstr>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  mov    %eax,%r12d
  mov    %r13,%rdi
  call   <strlen@plt>
  sub    %ebx,%eax
  js     L9
  movslq %eax,%rdx
  lea    0x0(%r13,%rdx,1),%rbx
  lea    -0x1(%r13,%rdx,1),%r13
  mov    %eax,%eax
  sub    %rax,%r13
  movslq %r12d,%r12
L16:
  mov    %rbx,%r14
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  je     L26
  sub    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L16
  mov    $0x0,%r14d
L26:
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L9:
  mov    $0x0,%r14d
  jmp    L26
```

assistant: ```c
char *strstr(s1, s2)
char *s1, *s2;
{
  int x, y;
  y = strlen(s2);
  x = strlen(s1) - y;
  if (x < 0)
    return ((char *)0L);
  do {
    if (strncmp(s1 + x, s2, y) == 0)
      return (s1 + x);
  } while (--x >= 0);
  return ((char *)0L);
}
```
user: Translate the below assembly code into C programming language:

```asm
<strstr>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%r12
  movslq %eax,%rbp
  call   <strlen@plt>
  sub    %r12d,%eax
  js     L1
  movslq %eax,%rdx
  mov    %eax,%eax
  lea    0x0(%r13,%rdx,1),%r12
  lea    -0x1(%r13,%rdx,1),%r13
  sub    %rax,%r13
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L8:
  sub    $0x1,%r12
  cmp    %r13,%r12
  je     L1
L31:
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L8
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L1:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strstr(s1, s2)
char *s1, *s2;
{
  int x, y;
  y = strlen(s2);
  x = strlen(s1) - y;
  if (x < 0)
    return ((char *)0L);
  do {
    if (strncmp(s1 + x, s2, y) == 0)
      return (s1 + x);
  } while (--x >= 0);
  return ((char *)0L);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strstr>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %r13,%rdi
  mov    %rax,%r12
  movslq %eax,%rbp
  call   <strlen@plt>
  sub    %r12d,%eax
  js     L2
  movslq %eax,%rdx
  mov    %eax,%eax
  lea    0x0(%r13,%rdx,1),%r12
  lea    -0x1(%r13,%rdx,1),%r13
  sub    %rax,%r13
  jmp    L37
  nopl   0x0(%rax,%rax,1)
L18:
  sub    $0x1,%r12
  cmp    %r13,%r12
  je     L2
L37:
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <strncmp@plt>
  test   %eax,%eax
  jne    L18
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L2:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *strstr(s1, s2)
char *s1, *s2;
{
  int x, y;
  y = strlen(s2);
  x = strlen(s1) - y;
  if (x < 0)
    return ((char *)0L);
  do {
    if (strncmp(s1 + x, s2, y) == 0)
      return (s1 + x);
  } while (--x >= 0);
  return ((char *)0L);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<longestConsec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  movl   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpl   $0x0,-0x3c(%rbp)
  je     L42
  mov    -0x40(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jg     L42
  cmpl   $0x0,-0x40(%rbp)
  jg     L108
L42:
  lea    D5(%rip),%rax
  jmp    L60
L108:
  movl   $0x0,-0x24(%rbp)
  jmp    L98
L57:
  movl   $0x0,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L24
L94:
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    %eax,-0x1c(%rbp)
  addl   $0x1,-0x20(%rbp)
L24:
  mov    -0x20(%rbp),%eax
  cmp    -0x40(%rbp),%eax
  jl     L94
  cmpq   $0x0,-0x10(%rbp)
  je     L70
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L70:
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L100
L110:
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  addl   $0x1,-0x20(%rbp)
L100:
  mov    -0x20(%rbp),%eax
  cmp    -0x40(%rbp),%eax
  jl     L110
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L41
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L41:
  addl   $0x1,-0x24(%rbp)
L98:
  mov    -0x24(%rbp),%edx
  mov    -0x40(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x3c(%rbp)
  jge    L57
  mov    -0x8(%rbp),%rax
L60:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D5
type_infered: string

tool: 

assistant: ```c
char *longestConsec(char *arr[], int n, int k) {
  int i, j, len, strl, max = 0;
  char *str = 0, *longest = 0;
  if (n == 0 || k > n || k <= 0)
    return "";
  for (i = 0; i + k <= n; i++) {
    for (len = j = 0; j < k; j++)
      len += strlen(arr[i + j]);
    if (str)
      free(str);
    str = calloc(len + 1, sizeof(char));
    for (j = 0; j < k; j++)
      strcat(str, arr[i + j]);
    if ((strl = strlen(str)) > max)
      max = strl, strcpy(longest = malloc(strl + 1), str);
  }
  return longest;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<longestConsec>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,0x14(%rsp)
  mov    %edx,%r12d
  test   %esi,%esi
  sete   %al
  cmp    %edx,%esi
  setl   %dl
  or     %dl,%al
  jne    L90
  test   %r12d,%r12d
  jle    L90
  mov    %rdi,%r14
  movq   $0x0,0x18(%rsp)
  mov    $0x0,%r13d
  movl   $0x0,0x10(%rsp)
  movl   $0x0,0xc(%rsp)
  jmp    L4
L57:
  lea    0x1(%rbp),%edi
  mov    $0x1,%esi
  call   <calloc@plt>
  mov    %rax,%r13
  mov    $0x0,%ebx
L40:
  mov    (%r15,%rbx,8),%rsi
  mov    %r13,%rdi
  call   <strcat@plt>
  add    $0x1,%rbx
  cmp    %ebx,%r12d
  jg     L40
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,%ebx
  cmp    0x10(%rsp),%eax
  jg     L38
L76:
  addl   $0x1,0xc(%rsp)
  mov    0xc(%rsp),%eax
  add    $0x8,%r14
  add    %r12d,%eax
  cmp    %eax,0x14(%rsp)
  jl     L24
L4:
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  mov    %r14,%r15
L46:
  mov    (%r14,%rbx,8),%rdi
  call   <strlen@plt>
  add    %eax,%ebp
  add    $0x1,%rbx
  cmp    %ebx,%r12d
  jg     L46
  test   %r13,%r13
  je     L57
  mov    %r13,%rdi
  call   <free@plt>
  jmp    L57
L38:
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %ebx,0x10(%rsp)
  jmp    L76
L90:
  lea    D8(%rip),%rax
  mov    %rax,0x18(%rsp)
L24:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D8
type_infered: string

tool: 

assistant: ```c
char *longestConsec(char *arr[], int n, int k) {
  int i, j, len, strl, max = 0;
  char *str = 0, *longest = 0;
  if (n == 0 || k > n || k <= 0)
    return "";
  for (i = 0; i + k <= n; i++) {
    for (len = j = 0; j < k; j++)
      len += strlen(arr[i + j]);
    if (str)
      free(str);
    str = calloc(len + 1, sizeof(char));
    for (j = 0; j < k; j++)
      strcat(str, arr[i + j]);
    if ((strl = strlen(str)) > max)
      max = strl, strcpy(longest = malloc(strl + 1), str);
  }
  return longest;
}
```