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

```asm
<search_cmd_in_hist>:
  endbr64
  push   %r14
  xor    %eax,%eax
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <reset_actual_ptr@plt>
  xor    %eax,%eax
  call   <is_empty_hist@plt>
  test   %eax,%eax
  je     L23
L3:
  xor    %eax,%eax
  call   <reset_actual_ptr@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L32:
  cmpb   $0x0,(%rdi)
  je     L63
  mov    %r12,%rsi
  call   <strstr@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffff,%ebx
L23:
  xor    %eax,%eax
  call   <get_prev_cmd@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L32
L63:
  lea    0x0(,%rbx,8),%edi
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,%r13
  xor    %eax,%eax
  call   <reset_actual_ptr@plt>
  jmp    L61
  nopl   (%rax)
L27:
  cmpb   $0x0,0x0(%rbp)
  je     L64
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L43
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,0x0(%r13,%rbx,8)
L43:
  add    $0x1,%rbx
L61:
  xor    %eax,%eax
  movslq %ebx,%r14
  call   <get_prev_cmd@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L27
L64:
  movq   $0x0,0x0(%r13,%r14,8)
  jmp    L3
```

assistant: ```c
char **search_cmd_in_hist(char *cmd_in) {
  int nb_cmd;
  char *my_cmd;
  char **tab_cmd = NULL;
  reset_actual_ptr();
  if (!is_empty_hist()) {
    while ((my_cmd = get_prev_cmd())) {
      if (my_cmd[0] != '\0') {
        if (strstr(my_cmd, cmd_in) != NULL)
          nb_cmd++;
      } else
        break;
    }
    tab_cmd = malloc((nb_cmd + 1) * sizeof(char *));
    nb_cmd = 0;
    reset_actual_ptr();
    while ((my_cmd = get_prev_cmd())) {
      if (my_cmd[0] != '\0') {
        if (strstr(my_cmd, cmd_in) != NULL)
          tab_cmd[nb_cmd] = strdup(my_cmd);
        nb_cmd++;
      } else
        break;
    }
    tab_cmd[nb_cmd] = NULL;
  }
  reset_actual_ptr();
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<search_cmd_in_hist>:
  endbr64
  push   %r14
  xor    %eax,%eax
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    $0x1,%ebx
  call   <reset_actual_ptr@plt>
  xor    %eax,%eax
  call   <is_empty_hist@plt>
  test   %eax,%eax
  je     L30
L69:
  xor    %eax,%eax
  call   <reset_actual_ptr@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  cmpb   $0x0,(%rdi)
  je     L78
  mov    %r12,%rsi
  call   <strstr@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffff,%ebx
L30:
  xor    %eax,%eax
  call   <get_prev_cmd@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L6
L78:
  lea    0x0(,%rbx,8),%edi
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,%r13
  xor    %eax,%eax
  call   <reset_actual_ptr@plt>
  jmp    L29
  nopl   (%rax)
L36:
  cmpb   $0x0,0x0(%rbp)
  je     L23
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L24
  mov    %rbp,%rdi
  call   <strdup@plt>
  mov    %rax,0x0(%r13,%rbx,8)
L24:
  add    $0x1,%rbx
L29:
  xor    %eax,%eax
  movslq %ebx,%r14
  call   <get_prev_cmd@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L36
L23:
  movq   $0x0,0x0(%r13,%r14,8)
  jmp    L69
```

assistant: ```c
char **search_cmd_in_hist(char *cmd_in) {
  int nb_cmd;
  char *my_cmd;
  char **tab_cmd = NULL;
  reset_actual_ptr();
  if (!is_empty_hist()) {
    while ((my_cmd = get_prev_cmd())) {
      if (my_cmd[0] != '\0') {
        if (strstr(my_cmd, cmd_in) != NULL)
          nb_cmd++;
      } else
        break;
    }
    tab_cmd = malloc((nb_cmd + 1) * sizeof(char *));
    nb_cmd = 0;
    reset_actual_ptr();
    while ((my_cmd = get_prev_cmd())) {
      if (my_cmd[0] != '\0') {
        if (strstr(my_cmd, cmd_in) != NULL)
          tab_cmd[nb_cmd] = strdup(my_cmd);
        nb_cmd++;
      } else
        break;
    }
    tab_cmd[nb_cmd] = NULL;
  }
  reset_actual_ptr();
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GREEDY_ALLOC_ROUND_UP>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x2,-0x18(%rbp)
  ja     L30
  mov    $0x2,%eax
  jmp    L39
L30:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <ALIGN_POWER2@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L6
  mov    -0x18(%rbp),%rax
  jmp    L39
L6:
  mov    -0x8(%rbp),%rax
L39:
  leave
  ret
```

assistant: ```c
size_t GREEDY_ALLOC_ROUND_UP(size_t l) {
  size_t m;
  if (l <= 2)
    return 2;
  m = ALIGN_POWER2(l);
  if (m == 0)
    return l;
  return m;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<GREEDY_ALLOC_ROUND_UP>:
  endbr64
  mov    $0x2,%eax
  cmp    $0x2,%rdi
  jbe    L6
  push   %rbx
  mov    %rdi,%rbx
  call   <ALIGN_POWER2@plt>
  test   %rax,%rax
  cmove  %rbx,%rax
  pop    %rbx
  ret
L6:
  ret
```

assistant: ```c
size_t GREEDY_ALLOC_ROUND_UP(size_t l) {
  size_t m;
  if (l <= 2)
    return 2;
  m = ALIGN_POWER2(l);
  if (m == 0)
    return l;
  return m;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<GREEDY_ALLOC_ROUND_UP>:
  endbr64
  cmp    $0x2,%rdi
  jbe    L8
  push   %rbx
  mov    %rdi,%rbx
  call   <ALIGN_POWER2@plt>
  test   %rax,%rax
  cmove  %rbx,%rax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L8:
  mov    $0x2,%eax
  ret
```

assistant: ```c
size_t GREEDY_ALLOC_ROUND_UP(size_t l) {
  size_t m;
  if (l <= 2)
    return 2;
  m = ALIGN_POWER2(l);
  if (m == 0)
    return l;
  return m;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<GREEDY_ALLOC_ROUND_UP>:
  endbr64
  cmp    $0x2,%rdi
  jbe    L3
  push   %rbx
  mov    %rdi,%rbx
  call   <ALIGN_POWER2@plt>
  test   %rax,%rax
  cmove  %rbx,%rax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L3:
  mov    $0x2,%eax
  ret
```

assistant: ```c
size_t GREEDY_ALLOC_ROUND_UP(size_t l) {
  size_t m;
  if (l <= 2)
    return 2;
  m = ALIGN_POWER2(l);
  if (m == 0)
    return l;
  return m;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_hash_subg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x2c(%rbp),%eax
  and    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rcx
  lea    -0x10(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rax
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  shr    $0x3,%rdx
  mov    %edx,%esi
  mov    -0x14(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rdx,%rcx
  mov    %rax,%rdx
  mov    %rcx,%rdi
  call   <hash@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
unsigned long get_hash_subg(int *subg, int m) {
  int off = m % (sizeof(unsigned long) / sizeof(int));
  unsigned long hash0;
  memcpy(&hash0, subg, off * sizeof(int));
  return hash((unsigned long *)&subg[off], (int)(m * sizeof(int) / sizeof(unsigned long)), hash0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_hash_subg>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %esi,%edx
  and    $0x1,%edx
  mov    %esi,%r12d
  and    $0x1,%r12d
  shl    $0x2,%edx
  mov    %rsp,%rdi
  mov    %rbp,%rsi
  call   <memcpy@plt>
  movslq %ebx,%rsi
  shl    $0x1f,%rsi
  lea    0x0(%rbp,%r12,4),%rdi
  mov    (%rsp),%rdx
  shr    $0x20,%rsi
  call   <hash@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long get_hash_subg(int *subg, int m) {
  int off = m % (sizeof(unsigned long) / sizeof(int));
  unsigned long hash0;
  memcpy(&hash0, subg, off * sizeof(int));
  return hash((unsigned long *)&subg[off], (int)(m * sizeof(int) / sizeof(unsigned long)), hash0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_hash_subg>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  and    $0x1,%r12d
  mov    %rdi,%rbp
  push   %rbx
  lea    0x0(,%r12,4),%edx
  mov    %esi,%ebx
  mov    %rbp,%rsi
  movslq %r12d,%r12
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rsi
  mov    (%rsp),%rdx
  lea    0x0(%rbp,%r12,4),%rdi
  shl    $0x1f,%rsi
  shr    $0x20,%rsi
  call   <hash@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned long get_hash_subg(int *subg, int m) {
  int off = m % (sizeof(unsigned long) / sizeof(int));
  unsigned long hash0;
  memcpy(&hash0, subg, off * sizeof(int));
  return hash((unsigned long *)&subg[off], (int)(m * sizeof(int) / sizeof(unsigned long)), hash0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x40,-0x1c(%rbp)
  movl   $0x4,-0x14(%rbp)
  cmpl   $0x1,-0x24(%rbp)
  jle    L4
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atol@plt>
  mov    %eax,-0x1c(%rbp)
L4:
  mov    -0x14(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L20
L42:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x18(%rbp)
L20:
  mov    -0x18(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L42
  movl   $0x0,-0x18(%rbp)
  jmp    L6
L25:
  mov    $0x0,%eax
  call   <rand@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  pxor   %xmm2,%xmm2
  cvtsi2sd %eax,%xmm2
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x18(%rbp)
L6:
  mov    -0x18(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L25
  mov    D3(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x1c(%rbp),%edx
  mov    -0x14(%rbp),%esi
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <fwrite@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  int *array;
  int N = 64;
  int i, j, tmp, intSize = sizeof(int);
  if (argc > 1)
    N = atol(argv[1]);
  array = (int *)malloc(intSize * N);
  for (i = 0; i < N; i++)
    array[i] = i + 1;
  for (i = 0; i < N; i++) {
    j = (rand() * 1. / RAND_MAX) * (N - 1);
    tmp = array[i];
    array[i] = array[j];
    array[j] = tmp;
  }
  fwrite(array, intSize, N, stdout);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jg     L31
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x40,%r12d
L45:
  mov    %rbp,%rdx
  mov    $0x0,%eax
L28:
  add    $0x1,%eax
  mov    %eax,(%rdx)
  add    $0x4,%rdx
  cmp    %r12d,%eax
  jl     L28
  mov    $0x0,%ebx
  lea    -0x1(%r12),%r13d
L3:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    0x0(%rbp,%rbx,4),%edx
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mov    D17(%rip),%rax
  pxor   %xmm1,%xmm1
  cvtsi2sdl (%rax),%xmm1
  divsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %r13d,%xmm1
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  cltq
  lea    0x0(%rbp,%rax,4),%rax
  mov    (%rax),%ecx
  mov    %ecx,0x0(%rbp,%rbx,4)
  mov    %edx,(%rax)
  add    $0x1,%rbx
  cmp    %ebx,%r12d
  jg     L3
L22:
  mov    D0(%rip),%rax
  mov    (%rax),%ecx
  mov    %r12d,%edx
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <fwrite@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L31:
  mov    0x8(%rsi),%rdi
  call   <atol@plt>
  mov    %eax,%r12d
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r12d,%r12d
  jg     L45
  jmp    L22
```

assistant: ```c
int main(int argc, char **argv) {
  int *array;
  int N = 64;
  int i, j, tmp, intSize = sizeof(int);
  if (argc > 1)
    N = atol(argv[1]);
  array = (int *)malloc(intSize * N);
  for (i = 0; i < N; i++)
    array[i] = i + 1;
  for (i = 0; i < N; i++) {
    j = (rand() * 1. / RAND_MAX) * (N - 1);
    tmp = array[i];
    array[i] = array[j];
    array[j] = tmp;
  }
  fwrite(array, intSize, N, stdout);
}
```
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
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  cmp    $0x1,%edi
  jg     L36
  mov    $0x100,%edi
  mov    $0x40,%r12d
  call   <malloc@plt>
  mov    %rax,%rbp
L11:
  mov    %rbp,%r14
  mov    %rbp,%rdx
  xor    %eax,%eax
  nopl   0x0(%rax)
L9:
  add    $0x1,%eax
  add    $0x4,%rdx
  mov    %eax,-0x4(%rdx)
  cmp    %r12d,%eax
  jl     L9
  lea    -0x1(%r12),%edx
  pxor   %xmm3,%xmm3
  mov    D14(%rip),%rbx
  cvtsi2sd %edx,%xmm3
  lea    0x4(%rbp,%rdx,4),%r13
  movsd  %xmm3,0x8(%rsp)
  cs nopw 0x0(%rax,%rax,1)
L18:
  xor    %eax,%eax
  add    $0x4,%r14
  call   <rand@plt>
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  mov    -0x4(%r14),%edx
  cvtsi2sdl (%rbx),%xmm1
  cvtsi2sd %eax,%xmm0
  divsd  %xmm1,%xmm0
  mulsd  0x8(%rsp),%xmm0
  cvttsd2si %xmm0,%eax
  cltq
  lea    0x0(%rbp,%rax,4),%rax
  mov    (%rax),%ecx
  mov    %ecx,-0x4(%r14)
  mov    %edx,(%rax)
  cmp    %r13,%r14
  jne    L18
L19:
  mov    D8(%rip),%rax
  mov    %r12d,%edx
  mov    %rbp,%rdi
  mov    $0x4,%esi
  mov    (%rax),%ecx
  call   <fwrite@plt>
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L36:
  mov    0x8(%rsi),%rdi
  call   <atol@plt>
  lea    0x0(,%rax,4),%edi
  mov    %eax,%r12d
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r12d,%r12d
  jg     L11
  jmp    L19
```

assistant: ```c
int main(int argc, char **argv) {
  int *array;
  int N = 64;
  int i, j, tmp, intSize = sizeof(int);
  if (argc > 1)
    N = atol(argv[1]);
  array = (int *)malloc(intSize * N);
  for (i = 0; i < N; i++)
    array[i] = i + 1;
  for (i = 0; i < N; i++) {
    j = (rand() * 1. / RAND_MAX) * (N - 1);
    tmp = array[i];
    array[i] = array[j];
    array[j] = tmp;
  }
  fwrite(array, intSize, N, stdout);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x4c(%rbp)
  movl   $0x0,-0x48(%rbp)
  jmp    L16
L37:
  movl   $0x0,-0x44(%rbp)
  jmp    L31
L46:
  mov    -0x48(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x48(%rbp)
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x4c(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  add    %rcx,%rdx
  mov    %eax,-0x40(%rbp,%rdx,4)
  addl   $0x1,-0x44(%rbp)
L31:
  cmpl   $0x3,-0x44(%rbp)
  jle    L46
  addl   $0x1,-0x4c(%rbp)
L16:
  cmpl   $0x2,-0x4c(%rbp)
  jle    L37
  lea    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <matriz_ponteiro@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int main() {
  int mtr[3][4];
  for (int lin = 0, cont = 0; lin < 3; lin++)
    for (int col = 0; col < 4; col++)
      mtr[lin][col] = cont++;
  matriz_ponteiro(mtr);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x0,(%rsp)
  movl   $0x1,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0x3,0xc(%rsp)
  movl   $0x4,0x10(%rsp)
  movl   $0x5,0x14(%rsp)
  movl   $0x6,0x18(%rsp)
  movl   $0x7,0x1c(%rsp)
  movl   $0x8,0x20(%rsp)
  movl   $0x9,0x24(%rsp)
  movl   $0xa,0x28(%rsp)
  movl   $0xb,0x2c(%rsp)
  mov    %rsp,%rdi
  call   <matriz_ponteiro@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x48,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int mtr[3][4];
  for (int lin = 0, cont = 0; lin < 3; lin++)
    for (int col = 0; col < 4; col++)
      mtr[lin][col] = cont++;
  matriz_ponteiro(mtr);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x100000000,%rax
  mov    %rsp,%rdi
  mov    %rax,(%rsp)
  movabs $0x300000002,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x500000004,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x700000006,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x900000008,%rax
  mov    %rax,0x20(%rsp)
  movabs $0xb0000000a,%rax
  mov    %rax,0x28(%rsp)
  call   <matriz_ponteiro@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x48,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int mtr[3][4];
  for (int lin = 0, cont = 0; lin < 3; lin++)
    for (int col = 0; col < 4; col++)
      mtr[lin][col] = cont++;
  matriz_ponteiro(mtr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  movdqa D8(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movaps %xmm0,(%rsp)
  movdqa D23(%rip),%xmm0
  movaps %xmm0,0x10(%rsp)
  movdqa D11(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  call   <matriz_ponteiro@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x48,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D8
type_guessed: byte[16]
name: unpack_data
arguments: label: D23
type_guessed: byte[16]
name: unpack_data
arguments: label: D11
type_guessed: byte[16]

tool: {"D8": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D23": [0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00,0x07,0x00,0x00,0x00]}

tool: {"D11": [0x08,0x00,0x00,0x00,0x09,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x0b,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int mtr[3][4];
  for (int lin = 0, cont = 0; lin < 3; lin++)
    for (int col = 0; col < 4; col++)
      mtr[lin][col] = cont++;
  matriz_ponteiro(mtr);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movq   $0x0,-0x10(%rbp)
  lea    D111(%rip),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x10(%rbp)
  mov    D67(%rip),%rdx
  mov    -0x10(%rbp),%rax
  movq   %rdx,%xmm0
  mov    $0x1,%ecx
  lea    D64(%rip),%rdx
  lea    D32(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  mov    D84(%rip),%rdx
  mov    -0x10(%rbp),%rax
  movq   %rdx,%xmm0
  mov    $0x2,%ecx
  lea    D106(%rip),%rdx
  lea    D32(%rip),%rsi
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  mov    -0x10(%rbp),%rax
  lea    D50(%rip),%rdx
  lea    D5(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  mov    D86(%rip),%rdx
  mov    -0x10(%rbp),%rax
  movq   %rdx,%xmm0
  lea    D77(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  movq   $0x0,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D111
type_guessed: string
name: parse_data
arguments: data_tag: D9
type_guessed: string
name: parse_data
arguments: data_tag: D67
type_guessed: f64
name: parse_data
arguments: data_tag: D64
type_guessed: string
name: parse_data
arguments: data_tag: D32
type_guessed: string
name: parse_data
arguments: data_tag: D84
type_guessed: f64
name: parse_data
arguments: data_tag: D106
type_guessed: string
name: parse_data
arguments: data_tag: D50
type_guessed: string
name: parse_data
arguments: data_tag: D5
type_guessed: string
name: parse_data
arguments: data_tag: D86
type_guessed: f64
name: parse_data
arguments: data_tag: D77
type_guessed: string

tool: D111: a+

tool: D9: /tmp/cpu_memory_measurement.txt

tool: D67: 2.1

tool: D64: apple

tool: D32: %s : %d : %f


tool: D84: 2.7

tool: D106: banana

tool: D50: fox 1

tool: D5: %s


tool: D86: 1.2

tool: D77: %%CPU: %f


assistant: ```c
void test1() {
  FILE *fp = NULL;
  fp = fopen("/tmp/cpu_memory_measurement.txt", "a+");
  fprintf(fp, "%s : %d : %f\n", "apple", 1, 2.1);
  fprintf(fp, "%s : %d : %f\n", "banana", 2, 2.7);
  fprintf(fp, "%s\n", "fox 1");
  fprintf(fp, "%%CPU: %f\n", 1.2);
  fputc('\n', fp);
  fputs("\n", fp);
  fclose(fp);
  int *pi = NULL;
  free(pi);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<test1>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D16(%rip),%rsi
  lea    D100(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%rbx
  movsd  D76(%rip),%xmm0
  mov    $0x1,%ecx
  lea    D101(%rip),%rdx
  lea    D91(%rip),%rbp
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  movsd  D45(%rip),%xmm0
  mov    $0x2,%ecx
  lea    D13(%rip),%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  lea    D64(%rip),%rdx
  lea    D35(%rip),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <fprintf@plt>
  movsd  D33(%rip),%xmm0
  lea    D69(%rip),%rsi
  mov    %rbx,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  mov    %rbx,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  mov    %rbx,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  mov    %rbx,%rdi
  call   <fclose@plt>
  mov    $0x0,%edi
  call   <free@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D16
type_infered: string
name: retrieve_data
arguments: data_tag: D100
type_infered: string
name: retrieve_data
arguments: data_tag: D76
type_infered: f64
name: retrieve_data
arguments: data_tag: D101
type_infered: string
name: retrieve_data
arguments: data_tag: D91
type_infered: string
name: retrieve_data
arguments: data_tag: D45
type_infered: f64
name: retrieve_data
arguments: data_tag: D13
type_infered: string
name: retrieve_data
arguments: data_tag: D64
type_infered: string
name: retrieve_data
arguments: data_tag: D35
type_infered: string
name: retrieve_data
arguments: data_tag: D33
type_infered: f64
name: retrieve_data
arguments: data_tag: D69
type_infered: string

tool: D16: a+

tool: D100: /tmp/cpu_memory_measurement.txt

tool: D76: 2.1

tool: D101: apple

tool: D91: %s : %d : %f


tool: D45: 2.7

tool: D13: banana

tool: D64: fox 1

tool: D35: %s


tool: D33: 1.2

tool: D69: %%CPU: %f


assistant: ```c
void test1() {
  FILE *fp = NULL;
  fp = fopen("/tmp/cpu_memory_measurement.txt", "a+");
  fprintf(fp, "%s : %d : %f\n", "apple", 1, 2.1);
  fprintf(fp, "%s : %d : %f\n", "banana", 2, 2.7);
  fprintf(fp, "%s\n", "fox 1");
  fprintf(fp, "%%CPU: %f\n", 1.2);
  fputc('\n', fp);
  fputs("\n", fp);
  fclose(fp);
  int *pi = NULL;
  free(pi);
}
```