ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x0,-0x24(%rbp)
  jmp    L20
L18:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,%ecx
  movslq %ecx,%rax
  imul   $0x2aaaaaab,%rax,%rax
  shr    $0x20,%rax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  mov    %eax,%edx
  sub    %esi,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  movslq %edx,%rax
  mov    -0x20(%rbp,%rax,4),%eax
  lea    0x1(%rax),%ecx
  movslq %edx,%rax
  mov    %ecx,-0x20(%rbp,%rax,4)
  addl   $0x1,-0x24(%rbp)
L20:
  cmpl   $0x3e7,-0x24(%rbp)
  jle    L18
  movl   $0x0,-0x24(%rbp)
  jmp    L21
L42:
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x24(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x24(%rbp)
L21:
  cmpl   $0x5,-0x24(%rbp)
  jle    L42
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D3
type: string

tool: D3: %d times %d


assistant: ```c
int main(int argc, char **argv) {
  int counts[6] = {0};
  int i;
  srand(time(NULL));
  for (i = 0; i < (1000); i++) {
    counts[rand() % 6]++;
  }
  for (i = 0; i < 6; i++) {
    printf("%d times %d\n", counts[i], i + 1);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x3e8,%ebx
L20:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x2aaaaaab,%rdx,%rdx
  shr    $0x20,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,2),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  movslq %eax,%rdx
  addl   $0x1,(%rsp,%rdx,4)
  sub    $0x1,%ebx
  jne    L20
  mov    $0x1,%ebx
  lea    D0(%rip),%rbp
L0:
  mov    -0x4(%rsp,%rbx,4),%esi
  mov    %ebx,%edx
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    $0x7,%rbx
  jne    L0
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D0
type_guessed: string

tool: D0: %d times %d


assistant: ```c
int main(int argc, char **argv) {
  int counts[6] = {0};
  int i;
  srand(time(NULL));
  for (i = 0; i < (1000); i++) {
    counts[rand() % 6]++;
  }
  for (i = 0; i < 6; i++) {
    printf("%d times %d\n", counts[i], i + 1);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  xor    %edi,%edi
  push   %rbp
  push   %rbx
  mov    $0x3e8,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movq   $0x0,0x10(%rsp)
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
L9:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x2aaaaaab,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,2),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  movslq %eax,%rdx
  addl   $0x1,(%rsp,%rdx,4)
  sub    $0x1,%ebx
  jne    L9
  mov    $0x1,%ebx
  mov    %rsp,%r12
  lea    D0(%rip),%rbp
  xchg   %ax,%ax
L6:
  mov    -0x4(%r12,%rbx,4),%esi
  mov    %ebx,%edx
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x7,%rbx
  jne    L6
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
type: string

tool: D0: %d times %d


assistant: ```c
int main(int argc, char **argv) {
  int counts[6] = {0};
  int i;
  srand(time(NULL));
  for (i = 0; i < (1000); i++) {
    counts[rand() % 6]++;
  }
  for (i = 0; i < 6; i++) {
    printf("%d times %d\n", counts[i], i + 1);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  xor    %edi,%edi
  push   %rbp
  push   %rbx
  mov    $0x3e8,%ebx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  movq   $0x0,0x10(%rsp)
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
L1:
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x2aaaaaab,%rdx,%rdx
  sar    $0x1f,%ecx
  shr    $0x20,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,2),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  movslq %eax,%rdx
  addl   $0x1,(%rsp,%rdx,4)
  sub    $0x1,%ebx
  jne    L1
  mov    $0x1,%ebx
  mov    %rsp,%r12
  lea    D5(%rip),%rbp
  xchg   %ax,%ax
L21:
  mov    -0x4(%r12,%rbx,4),%esi
  mov    %ebx,%edx
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x7,%rbx
  jne    L21
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D5
type: string

tool: D5: %d times %d


assistant: ```c
int main(int argc, char **argv) {
  int counts[6] = {0};
  int i;
  srand(time(NULL));
  for (i = 0; i < (1000); i++) {
    counts[rand() % 6]++;
  }
  for (i = 0; i < 6; i++) {
    printf("%d times %d\n", counts[i], i + 1);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<KSRM16162742943295_br_strcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L16
  lea    D13(%rip),%rax
  mov    %rax,-0x28(%rbp)
L16:
  cmpq   $0x0,-0x30(%rbp)
  jne    L19
  lea    D13(%rip),%rax
  mov    %rax,-0x30(%rbp)
L19:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L0
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L0:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D13
type_infered: string

tool: 

assistant: ```c
char *KSRM16162742943295_br_strcat(const char *str1, const char *str2) {
  char *result;
  size_t len1, len2;
  if (str1 == ((char *)0))
    str1 = "";
  if (str2 == ((char *)0))
    str2 = "";
  len1 = strlen(str1);
  len2 = strlen(str2);
  result = (char *)malloc(len1 + len2 + 1);
  if (result) {
    memcpy(result, str1, len1);
    memcpy(result + len1, str2, len2);
    result[len1 + len2] = '\0';
  }
  return result;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<KSRM16162742943295_br_strcat>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  test   %rdi,%rdi
  lea    D3(%rip),%rax
  cmove  %rax,%rbp
  test   %rsi,%rsi
  cmove  %rax,%rbx
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%r13
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %rax,%r14
  lea    0x0(%r13,%rax,1),%r15
  lea    0x1(%r15),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L17
L8:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L17:
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  lea    (%r12,%r13,1),%rdi
  mov    %r14,%rdx
  mov    %rbx,%rsi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%r15,1)
  jmp    L8
```

tool_calls: name: unpack_data
arguments: tag: D3
type: string

tool: 

assistant: ```c
char *KSRM16162742943295_br_strcat(const char *str1, const char *str2) {
  char *result;
  size_t len1, len2;
  if (str1 == ((char *)0))
    str1 = "";
  if (str2 == ((char *)0))
    str2 = "";
  len1 = strlen(str1);
  len2 = strlen(str2);
  result = (char *)malloc(len1 + len2 + 1);
  if (result) {
    memcpy(result, str1, len1);
    memcpy(result + len1, str2, len2);
    result[len1 + len2] = '\0';
  }
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<KSRM16162742943295_br_strcat>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    D1(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L29
  mov    %rdi,%r13
  call   <strlen@plt>
  mov    %rax,%rbx
L29:
  test   %rbp,%rbp
  je     L10
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%r14
  lea    (%rbx,%rax,1),%r15
L44:
  lea    0x1(%r15),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L21
L3:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L21:
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  lea    (%r12,%rbx,1),%rdi
  mov    %r14,%rdx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%r15,1)
  jmp    L3
  nopl   0x0(%rax)
L10:
  mov    %rbx,%r15
  xor    %r14d,%r14d
  lea    D1(%rip),%rbp
  jmp    L44
```

tool_calls: name: disassemble_data
arguments: tag: D1
data_type: string

tool: 

assistant: ```c
char *KSRM16162742943295_br_strcat(const char *str1, const char *str2) {
  char *result;
  size_t len1, len2;
  if (str1 == ((char *)0))
    str1 = "";
  if (str2 == ((char *)0))
    str2 = "";
  len1 = strlen(str1);
  len2 = strlen(str2);
  result = (char *)malloc(len1 + len2 + 1);
  if (result) {
    memcpy(result, str1, len1);
    memcpy(result + len1, str2, len2);
    result[len1 + len2] = '\0';
  }
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<KSRM16162742943295_br_strcat>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    D14(%rip),%r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L26
  mov    %rdi,%r13
  call   <strlen@plt>
  mov    %rax,%rbx
L26:
  test   %rbp,%rbp
  je     L36
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rax,%r14
  lea    (%rax,%rbx,1),%r15
L19:
  lea    0x1(%r15),%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L3
L8:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L3:
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  lea    (%r12,%rbx,1),%rdi
  mov    %r14,%rdx
  mov    %rbp,%rsi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%r15,1)
  jmp    L8
  nopl   0x0(%rax)
L36:
  mov    %rbx,%r15
  xor    %r14d,%r14d
  lea    D14(%rip),%rbp
  jmp    L19
```

tool_calls: name: read_data
arguments: tag: D14
type: string

tool: 

assistant: ```c
char *KSRM16162742943295_br_strcat(const char *str1, const char *str2) {
  char *result;
  size_t len1, len2;
  if (str1 == ((char *)0))
    str1 = "";
  if (str2 == ((char *)0))
    str2 = "";
  len1 = strlen(str1);
  len2 = strlen(str2);
  result = (char *)malloc(len1 + len2 + 1);
  if (result) {
    memcpy(result, str1, len1);
    memcpy(result + len1, str2, len2);
    result[len1 + len2] = '\0';
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<_quick_rec_first>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L14
L49:
  addl   $0x1,-0x10(%rbp)
L22:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jg     L49
  jmp    L68
L76:
  subl   $0x1,-0xc(%rbp)
L68:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jl     L76
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L14
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x10(%rbp)
  subl   $0x1,-0xc(%rbp)
L14:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L22
  mov    -0x1c(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jge    L44
  mov    -0xc(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <_quick_rec_first>
L44:
  mov    -0x10(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L59
  mov    -0x20(%rbp),%edx
  mov    -0x10(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <_quick_rec_first>
L59:
  nop
  leave
  ret
```

assistant: ```c
void _quick_rec_first(int *vet, int begin, int end) {
  int i = begin, j = end;
  int tmp;
  int pivot = vet[begin];
  while (i <= j) {
    while (vet[i] < pivot)
      i++;
    while (vet[j] > pivot)
      j--;
    if (i <= j) {
      tmp = vet[i];
      vet[i] = vet[j];
      vet[j] = tmp;
      i++;
      j--;
    }
  }
  if (begin < j) {
    _quick_rec_first(vet, begin, j);
  }
  if (i < end) {
    _quick_rec_first(vet, i, end);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<_quick_rec_first>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r10d
  mov    %edx,%r12d
  movslq %esi,%rax
  mov    (%rdi,%rax,4),%esi
  mov    %r10d,%ebx
  jmp    L115
L19:
  mov    %edi,(%r8)
  mov    %ecx,(%r9)
  add    $0x1,%ebx
  sub    $0x1,%edx
  jmp    L115
L82:
  cmp    %r10d,%edx
  jg     L84
L71:
  cmp    %r12d,%ebx
  jl     L13
L21:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L84:
  mov    %r10d,%esi
  mov    %rbp,%rdi
  call   <_quick_rec_first>
  jmp    L71
L13:
  mov    %r12d,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <_quick_rec_first>
  jmp    L21
L17:
  movslq %edx,%rax
  lea    0x0(%rbp,%rax,4),%r9
  mov    (%r9),%edi
  cmp    %edi,%esi
  jge    L19
L80:
  movslq %edx,%rax
  lea    -0x4(%rbp,%rax,4),%rax
L100:
  sub    $0x1,%edx
  mov    %rax,%r9
  sub    $0x4,%rax
  mov    0x4(%rax),%edi
  cmp    %esi,%edi
  jg     L100
L113:
  cmp    %ebx,%edx
  jge    L19
L115:
  cmp    %edx,%ebx
  jg     L82
  movslq %ebx,%rax
  shl    $0x2,%rax
  lea    0x0(%rbp,%rax,1),%r8
  mov    (%r8),%ecx
  lea    0x4(%rbp,%rax,1),%rax
  cmp    %ecx,%esi
  jle    L17
L95:
  add    $0x1,%ebx
  mov    %rax,%r8
  add    $0x4,%rax
  mov    -0x4(%rax),%ecx
  cmp    %esi,%ecx
  jl     L95
  movslq %edx,%rax
  lea    0x0(%rbp,%rax,4),%r9
  mov    (%r9),%edi
  cmp    %edi,%esi
  jl     L80
  jmp    L113
```

assistant: ```c
void _quick_rec_first(int *vet, int begin, int end) {
  int i = begin, j = end;
  int tmp;
  int pivot = vet[begin];
  while (i <= j) {
    while (vet[i] < pivot)
      i++;
    while (vet[j] > pivot)
      j--;
    if (i <= j) {
      tmp = vet[i];
      vet[i] = vet[j];
      vet[j] = tmp;
      i++;
      j--;
    }
  }
  if (begin < j) {
    _quick_rec_first(vet, begin, j);
  }
  if (i < end) {
    _quick_rec_first(vet, i, end);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<_quick_rec_first>:
  endbr64
  push   %r12
  mov    %esi,%r10d
  mov    %rdi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
L8:
  movslq %r10d,%rax
  mov    %ebp,%edx
  mov    %r10d,%ebx
  mov    (%r12,%rax,4),%edi
  nopl   0x0(%rax)
L27:
  cmp    %edx,%ebx
  jg     L93
L10:
  movslq %ebx,%rax
  movslq %edx,%r11
  shl    $0x2,%rax
  lea    (%r12,%r11,4),%r9
  lea    (%r12,%rax,1),%rcx
  mov    (%r9),%r8d
  lea    0x4(%r12,%rax,1),%rax
  mov    (%rcx),%esi
  cmp    %edi,%esi
  jge    L110
  nopl   0x0(%rax)
L25:
  mov    %rax,%rcx
  mov    (%rax),%esi
  lea    0x4(%rax),%rax
  add    $0x1,%ebx
  cmp    %edi,%esi
  jl     L25
  cmp    %r8d,%edi
  jge    L126
L70:
  lea    -0x4(%r12,%r11,4),%rax
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %rax,%r9
  mov    (%rax),%r8d
  sub    $0x4,%rax
  sub    $0x1,%edx
  cmp    %edi,%r8d
  jg     L1
L126:
  cmp    %ebx,%edx
  jl     L27
  mov    %r8d,(%rcx)
  add    $0x1,%ebx
  sub    $0x1,%edx
  mov    %esi,(%r9)
L23:
  cmp    %edx,%ebx
  jle    L10
L93:
  cmp    %r10d,%edx
  jg     L17
L58:
  cmp    %ebp,%ebx
  jge    L127
  mov    %ebx,%r10d
  jmp    L8
  nop
L110:
  cmp    %r8d,%edi
  jl     L70
  mov    %r8d,(%rcx)
  add    $0x1,%ebx
  sub    $0x1,%edx
  mov    %esi,(%r9)
  jmp    L23
L127:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  mov    %r10d,%esi
  mov    %r12,%rdi
  call   <_quick_rec_first>
  jmp    L58
```

assistant: ```c
void _quick_rec_first(int *vet, int begin, int end) {
  int i = begin, j = end;
  int tmp;
  int pivot = vet[begin];
  while (i <= j) {
    while (vet[i] < pivot)
      i++;
    while (vet[j] > pivot)
      j--;
    if (i <= j) {
      tmp = vet[i];
      vet[i] = vet[j];
      vet[j] = tmp;
      i++;
      j--;
    }
  }
  if (begin < j) {
    _quick_rec_first(vet, begin, j);
  }
  if (i < end) {
    _quick_rec_first(vet, i, end);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  movl   $0x64,-0x28(%rbp)
  movsd  D7(%rip),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movq   $0x40000000,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %eax,%esi
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x28(%rbp),%xmm0
  movsd  -0x20(%rbp),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%esi
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm1,%xmm1
  cvtsi2sdq -0x18(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%esi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movb   $0x0,-0x29(%rbp)
  movsbl -0x29(%rbp),%eax
  mov    -0x28(%rbp),%edx
  imul   %edx,%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type: f64
name: retrieve_data
arguments: data_tag: D29
type: string
name: retrieve_data
arguments: data_tag: D52
type: string
name: retrieve_data
arguments: data_tag: D50
type: string
name: retrieve_data
arguments: data_tag: D24
type: string
name: retrieve_data
arguments: data_tag: D23
type: string

tool: D7: 1.2

tool: D29: You have %d bugs at the imaginary rate of %f.


tool: D52: The entire universe has %ld bugs.


tool: D50: You are expected to have %f bugs.


tool: D24: That is only a %e portion of the universe.


tool: D23: Which means you should care %d%%.


assistant: ```c
int main(int argc, const char *argv[]) {
  int bugs = 100;
  double bugRate = 1.2;
  printf("You have %d bugs at the imaginary rate of %f.\n", bugs, bugRate);
  long universeOfDefects = 1L * 1024L * 1024L * 1024L;
  printf("The entire universe has %ld bugs.\n", universeOfDefects);
  double expectedBugs = bugs * bugRate;
  printf("You are expected to have %f bugs.\n", expectedBugs);
  double partOfTheUniverse = expectedBugs / universeOfDefects;
  printf("That is only a %e portion of the universe.\n", partOfTheUniverse);
  char nulByte = '\0';
  int carePercentage = bugs * nulByte;
  printf("Which means you should care %d%%.\n", carePercentage);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  movsd  D35(%rip),%xmm0
  mov    $0x64,%esi
  lea    D27(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x40000000,%esi
  lea    D41(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x78,%esi
  lea    D33(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%esi
  lea    D31(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%esi
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D35
data_type: f64
name: read_data
arguments: data_label: D27
data_type: string
name: read_data
arguments: data_label: D41
data_type: string
name: read_data
arguments: data_label: D33
data_type: string
name: read_data
arguments: data_label: D31
data_type: string
name: read_data
arguments: data_label: D38
data_type: string

tool: D35: 1.2

tool: D27: You have %d bugs at the imaginary rate of %f.


tool: D41: The entire universe has %ld bugs.


tool: D33: You are expected to have %f bugs.


tool: D31: That is only a %e portion of the universe.


tool: D38: Which means you should care %d%%.


assistant: ```c
int main(int argc, const char *argv[]) {
  int bugs = 100;
  double bugRate = 1.2;
  printf("You have %d bugs at the imaginary rate of %f.\n", bugs, bugRate);
  long universeOfDefects = 1L * 1024L * 1024L * 1024L;
  printf("The entire universe has %ld bugs.\n", universeOfDefects);
  double expectedBugs = bugs * bugRate;
  printf("You are expected to have %f bugs.\n", expectedBugs);
  double partOfTheUniverse = expectedBugs / universeOfDefects;
  printf("That is only a %e portion of the universe.\n", partOfTheUniverse);
  char nulByte = '\0';
  int carePercentage = bugs * nulByte;
  printf("Which means you should care %d%%.\n", carePercentage);
  return 0;
}
```