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

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x2,%edx
  mov    $0x3,%esi
  push   %r12
  push   %rbp
  sub    $0x3d0,%rsp
  movapd D16(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x3c8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  lea    0x50(%rsp),%r12
  mov    D11(%rip),%rax
  movaps %xmm0,(%rsp)
  movapd D2(%rip),%xmm0
  mov    %r13,%rdi
  lea    0xa0(%rsp),%rbp
  mov    %rax,0x40(%rsp)
  movaps %xmm0,0x10(%rsp)
  movapd D33(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  movapd D43(%rip),%xmm0
  movaps %xmm0,0x30(%rsp)
  call   <pm@plt>
  mov    %rbp,%rcx
  mov    $0x3,%edx
  mov    %r13,%rsi
  mov    $0x2,%r8d
  mov    %r12,%rdi
  call   <svdu1v@plt>
  mov    $0x5,%edx
  mov    $0x1,%esi
  mov    %r12,%rdi
  call   <pm@plt>
  mov    $0x2,%edx
  mov    $0x2,%esi
  mov    %rbp,%rdi
  call   <pm@plt>
  mov    0x3c8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x3d0,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D16
type: byte[16]
name: read_data
arguments: label: D11
type: byte[8]
name: read_data
arguments: label: D2
type: byte[16]
name: read_data
arguments: label: D33
type: byte[16]
name: read_data
arguments: label: D43
type: byte[16]

tool: {"D16": [0x00,0x00,0x00,0x00,0x00,0x00,0xf0,0x3f,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x40]}

tool: {"D11": [0x00,0x00,0x00,0x00,0x00,0x00,0x22,0x40]}

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x08,0x40,0x00,0x00,0x00,0x00,0x00,0x00,0x10,0x40]}

tool: {"D33": [0x00,0x00,0x00,0x00,0x00,0x00,0x14,0x40,0x00,0x00,0x00,0x00,0x00,0x00,0x18,0x40]}

tool: {"D43": [0x00,0x00,0x00,0x00,0x00,0x00,0x1c,0x40,0x00,0x00,0x00,0x00,0x00,0x00,0x20,0x40]}

assistant: ```c
int main(int argc, char **argv) {
  double ma[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
  double ou[100];
  double d[9];
  double ov[100];
  int m, n;
  m = 3;
  n = 2;
  pm(ma, m, n);
  svdu1v(d, ma, m, ov, n);
  pm(d, 1, 5);
  pm(ov, n, n);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<longinit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L16:
  mov    -0x10(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    %rdx,(%rax)
L4:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L16
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void longinit(long *in, size_t size, const long v) {
  long *pt = in, *fpt;
  fpt = in + size;
  while (pt < fpt)
    *pt++ = v;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<longinit>:
  endbr64
  lea    (%rdi,%rsi,8),%rax
  cmp    %rax,%rdi
  jae    L0
L1:
  add    $0x8,%rdi
  mov    %rdx,-0x8(%rdi)
  cmp    %rdi,%rax
  ja     L1
L0:
  ret
```

assistant: ```c
void longinit(long *in, size_t size, const long v) {
  long *pt = in, *fpt;
  fpt = in + size;
  while (pt < fpt)
    *pt++ = v;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<longinit>:
  endbr64
  lea    (%rdi,%rsi,8),%rax
  cmp    %rax,%rdi
  jae    L17
  nopl   (%rax)
L8:
  add    $0x8,%rdi
  mov    %rdx,-0x8(%rdi)
  cmp    %rdi,%rax
  ja     L8
L17:
  ret
```

assistant: ```c
void longinit(long *in, size_t size, const long v) {
  long *pt = in, *fpt;
  fpt = in + size;
  while (pt < fpt)
    *pt++ = v;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<longinit>:
  endbr64
  shl    $0x3,%rsi
  mov    %rdi,%rcx
  lea    (%rdi,%rsi,1),%rax
  cmp    %rax,%rdi
  jae    L28
  sub    $0x1,%rsi
  mov    %rdi,%rax
  mov    %rsi,%rdi
  shr    $0x3,%rdi
  add    $0x1,%rdi
  cmp    $0x7,%rsi
  jbe    L30
  mov    %rdi,%rsi
  movq   %rdx,%xmm0
  shr    %rsi
  punpcklqdq %xmm0,%xmm0
  shl    $0x4,%rsi
  add    %rcx,%rsi
  nopw   0x0(%rax,%rax,1)
L5:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rax,%rsi
  jne    L5
  mov    %rdi,%rax
  and    $0xfffffffffffffffe,%rax
  lea    (%rcx,%rax,8),%rcx
  cmp    %rax,%rdi
  je     L28
L30:
  mov    %rdx,(%rcx)
L28:
  ret
```

assistant: ```c
void longinit(long *in, size_t size, const long v) {
  long *pt = in, *fpt;
  fpt = in + size;
  while (pt < fpt)
    *pt++ = v;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sizeu3>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdx
  lea    D27(%rip),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    D28(%rip),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D27
type_guessed: string

tool: D27: %zu

assistant: ```c
char *sizeu3(size_t num) {
  static char sizeu3_buf[28];
  sprintf(sizeu3_buf,
          "%"
          "zu",
          num);
  return sizeu3_buf;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sizeu3>:
  endbr64
  push   %rbx
  mov    %rdi,%rdx
  lea    D9(%rip),%rsi
  lea    D5(%rip),%rbx
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: data_label: D9
datatype: string

tool: D9: %zu

assistant: ```c
char *sizeu3(size_t num) {
  static char sizeu3_buf[28];
  sprintf(sizeu3_buf,
          "%"
          "zu",
          num);
  return sizeu3_buf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sizeu3>:
  endbr64
  push   %r12
  lea    D14(%rip),%r12
  mov    %rdi,%rdx
  lea    D2(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
datatype: string

tool: D2: %zu

assistant: ```c
char *sizeu3(size_t num) {
  static char sizeu3_buf[28];
  sprintf(sizeu3_buf,
          "%"
          "zu",
          num);
  return sizeu3_buf;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sizeu3>:
  endbr64
  push   %r12
  lea    D19(%rip),%r12
  mov    %rdi,%rdx
  lea    D4(%rip),%rsi
  mov    %r12,%rdi
  call   <sprintf@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D4
type_guessed: string

tool: D4: %zu

assistant: ```c
char *sizeu3(size_t num) {
  static char sizeu3_buf[28];
  sprintf(sizeu3_buf,
          "%"
          "zu",
          num);
  return sizeu3_buf;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<fact_recur>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L1
  mov    $0x1,%eax
  jmp    L8
L1:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fact_recur>
  imul   -0x4(%rbp),%eax
L8:
  leave
  ret
```

assistant: ```c
int fact_recur(int n) {
  if (n == 0)
    return 1;
  return n * fact_recur(n - 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<fact_recur>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  jne    L1
  ret
L1:
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <fact_recur>
  imul   %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int fact_recur(int n) {
  if (n == 0)
    return 1;
  return n * fact_recur(n - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fact_recur>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L18
  nopl   (%rax)
L7:
  imul   %edi,%eax
  sub    $0x1,%edi
  jne    L7
L18:
  ret
```

assistant: ```c
int fact_recur(int n) {
  if (n == 0)
    return 1;
  return n * fact_recur(n - 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fact_recur>:
  endbr64
  mov    $0x1,%eax
  test   %edi,%edi
  je     L18
  nopl   (%rax)
L12:
  imul   %edi,%eax
  sub    $0x1,%edi
  jne    L12
L18:
  ret
```

assistant: ```c
int fact_recur(int n) {
  if (n == 0)
    return 1;
  return n * fact_recur(n - 1);
}
```
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    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_bool@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L15
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
L15:
  cmpl   $0x0,-0xc(%rbp)
  je     L9
  mov    -0x8(%rbp),%rax
  movl   $0x29a,(%rax)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L9:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  bool selector = __VERIFIER_nondet_bool();
  int *ptr;
  if (selector) {
    ptr = malloc(sizeof(int));
  }
  if (selector) {
    *ptr = 666;
    free(ptr);
  }
  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
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_bool@plt>
  test   %eax,%eax
  jne    L13
L6:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L13:
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  movl   $0x29a,(%rax)
  call   <free@plt>
  jmp    L6
```

assistant: ```c
int main() {
  bool selector = __VERIFIER_nondet_bool();
  int *ptr;
  if (selector) {
    ptr = malloc(sizeof(int));
  }
  if (selector) {
    *ptr = 666;
    free(ptr);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <__VERIFIER_nondet_bool@plt>
  test   %eax,%eax
  jne    L5
L2:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L5:
  mov    $0x4,%edi
  call   <malloc@plt>
  movl   $0x29a,(%rax)
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L2
```

assistant: ```c
int main() {
  bool selector = __VERIFIER_nondet_bool();
  int *ptr;
  if (selector) {
    ptr = malloc(sizeof(int));
  }
  if (selector) {
    *ptr = 666;
    free(ptr);
  }
  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
  xor    %eax,%eax
  call   <__VERIFIER_nondet_bool@plt>
  test   %eax,%eax
  jne    L3
L9:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L3:
  mov    $0x4,%edi
  call   <malloc@plt>
  movl   $0x29a,(%rax)
  mov    %rax,%rdi
  call   <free@plt>
  jmp    L9
```

assistant: ```c
int main() {
  bool selector = __VERIFIER_nondet_bool();
  int *ptr;
  if (selector) {
    ptr = malloc(sizeof(int));
  }
  if (selector) {
    *ptr = 666;
    free(ptr);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copyout>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int copyout(void *src, void *dst, unsigned long size) {
  memcpy(dst, src, size);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<copyout>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rax
  mov    %rsi,%rdi
  mov    %rax,%rsi
  call   <memcpy@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int copyout(void *src, void *dst, unsigned long size) {
  memcpy(dst, src, size);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<copyout>:
  endbr64
  mov    %rdi,%r8
  sub    $0x8,%rsp
  mov    %rsi,%rdi
  mov    %r8,%rsi
  call   <memcpy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int copyout(void *src, void *dst, unsigned long size) {
  memcpy(dst, src, size);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<copyout>:
  endbr64
  mov    %rdi,%r8
  sub    $0x8,%rsp
  mov    %rsi,%rdi
  mov    %r8,%rsi
  call   <memcpy@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int copyout(void *src, void *dst, unsigned long size) {
  memcpy(dst, src, size);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    $0x0,%eax
  call   <getuid@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D7
type_infered: string
name: unpack_data
arguments: label: D8
type_infered: string

tool: D7: static int builder = %d;


tool: D8: static uint64_t builtat = %d;


assistant: ```c
int main(int argc, char *argv[]) {
  struct timeval tv;
  int uid;
  gettimeofday(&tv, NULL);
  uid = getuid();
  printf("static int builder = %d;\n", uid);
  printf("static uint64_t builtat = %d;\n", tv.tv_sec);
  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
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  mov    $0x0,%eax
  call   <getuid@plt>
  mov    %eax,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D8
data_type: string
name: disassemble_data
arguments: tag: D18
data_type: string

tool: D8: static int builder = %d;


tool: D18: static uint64_t builtat = %d;


assistant: ```c
int main(int argc, char *argv[]) {
  struct timeval tv;
  int uid;
  gettimeofday(&tv, NULL);
  uid = getuid();
  printf("static int builder = %d;\n", uid);
  printf("static uint64_t builtat = %d;\n", tv.tv_sec);
  exit(0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <gettimeofday@plt>
  xor    %eax,%eax
  call   <getuid@plt>
  lea    D8(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D8
type_guessed: string
name: read_data
arguments: data_label: D7
type_guessed: string

tool: D8: static int builder = %d;


tool: D7: static uint64_t builtat = %d;


assistant: ```c
int main(int argc, char *argv[]) {
  struct timeval tv;
  int uid;
  gettimeofday(&tv, NULL);
  uid = getuid();
  printf("static int builder = %d;\n", uid);
  printf("static uint64_t builtat = %d;\n", tv.tv_sec);
  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
  sub    $0x18,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <gettimeofday@plt>
  xor    %eax,%eax
  call   <getuid@plt>
  lea    D11(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x4(%rsp),%esi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D11
type_infered: string
name: retrieve_data
arguments: data_label: D12
type_infered: string

tool: D11: static int builder = %d;


tool: D12: static uint64_t builtat = %d;


assistant: ```c
int main(int argc, char *argv[]) {
  struct timeval tv;
  int uid;
  gettimeofday(&tv, NULL);
  uid = getuid();
  printf("static int builder = %d;\n", uid);
  printf("static uint64_t builtat = %d;\n", tv.tv_sec);
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rand_text>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x3,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  lea    0x0(,%rax,4),%ecx
  add    %ecx,%eax
  mov    %edx,%ecx
  sub    %eax,%ecx
  lea    -0x1(%rcx),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    $0x0,%eax
  jmp    L37
L16:
  movl   $0x0,-0x10(%rbp)
  jmp    L12
L11:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  lea    0x61(%rax),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x10(%rbp)
L12:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jb     L11
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L37:
  leave
  ret
```

assistant: ```c
char *rand_text() {
  srand(time(NULL));
  unsigned length = (rand() % 25) - 1;
  char *t = (char *)malloc(length * sizeof(char));
  if (!t)
    return NULL;
  unsigned i = 0;
  for (; i < length; ++i) {
    t[i] = (char)((rand() % 26) + 'a');
  }
  t[i + 1] = '\0';
  return t;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rand_text>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rbp
  imul   $0x51eb851f,%rbp,%rbp
  sar    $0x23,%rbp
  cltd
  sub    %edx,%ebp
  lea    0x0(%rbp,%rbp,4),%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  mov    %eax,%ebp
  lea    -0x1(%rax),%r13d
  mov    %r13d,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L27
  mov    %rax,%r12
  mov    %rax,%r14
  test   %r13d,%r13d
  je     L36
  mov    %rax,%rbx
  mov    %ebp,%r13d
  lea    -0x2(%rbp),%eax
  lea    0x1(%r12,%rax,1),%rbp
L10:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x23,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x61,%eax
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L10
  sub    $0x1,%r13d
L36:
  lea    0x1(%r13),%eax
  movb   $0x0,(%r12,%rax,1)
L19:
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  mov    $0x0,%r14d
  jmp    L19
```

assistant: ```c
char *rand_text() {
  srand(time(NULL));
  unsigned length = (rand() % 25) - 1;
  char *t = (char *)malloc(length * sizeof(char));
  if (!t)
    return NULL;
  unsigned i = 0;
  for (; i < length; ++i) {
    t[i] = (char)((rand() % 26) + 'a');
  }
  t[i + 1] = '\0';
  return t;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<rand_text>:
  endbr64
  push   %r13
  xor    %edi,%edi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rbx
  cltd
  imul   $0x51eb851f,%rbx,%rbx
  sar    $0x23,%rbx
  sub    %edx,%ebx
  lea    (%rbx,%rbx,4),%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  lea    -0x1(%rbx),%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L21
  test   %ebp,%ebp
  je     L7
  mov    %rax,%r13
  lea    -0x2(%rbx),%eax
  lea    0x1(%r12,%rax,1),%rbp
  nop
L39:
  xor    %eax,%eax
  add    $0x1,%r13
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x61,%eax
  mov    %al,-0x1(%r13)
  cmp    %rbp,%r13
  jne    L39
L27:
  movb   $0x0,(%r12,%rbx,1)
L21:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L7:
  mov    $0x1,%ebx
  jmp    L27
```

assistant: ```c
char *rand_text() {
  srand(time(NULL));
  unsigned length = (rand() % 25) - 1;
  char *t = (char *)malloc(length * sizeof(char));
  if (!t)
    return NULL;
  unsigned i = 0;
  for (; i < length; ++i) {
    t[i] = (char)((rand() % 26) + 'a');
  }
  t[i + 1] = '\0';
  return t;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<rand_text>:
  endbr64
  push   %r13
  xor    %edi,%edi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rbx
  cltd
  imul   $0x51eb851f,%rbx,%rbx
  sar    $0x23,%rbx
  sub    %edx,%ebx
  lea    (%rbx,%rbx,4),%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  lea    -0x1(%rbx),%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L22
  test   %ebp,%ebp
  je     L25
  mov    %rax,%r13
  lea    -0x2(%rbx),%eax
  lea    0x1(%r12,%rax,1),%rbp
  nop
L36:
  xor    %eax,%eax
  add    $0x1,%r13
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x61,%eax
  mov    %al,-0x1(%r13)
  cmp    %rbp,%r13
  jne    L36
L23:
  movb   $0x0,(%r12,%rbx,1)
L22:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L25:
  mov    $0x1,%ebx
  jmp    L23
```

assistant: ```c
char *rand_text() {
  srand(time(NULL));
  unsigned length = (rand() % 25) - 1;
  char *t = (char *)malloc(length * sizeof(char));
  if (!t)
    return NULL;
  unsigned i = 0;
  for (; i < length; ++i) {
    t[i] = (char)((rand() % 26) + 'a');
  }
  t[i + 1] = '\0';
  return t;
}
```