ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<uu_decode_line>:
  endbr64
  mov    %edx,%r8d
  movsbl (%rsi),%edx
  sub    $0x20,%edx
  and    $0x3f,%edx
  cmp    $0x2d,%edx
  jg     L37
  mov    $0xaaaaaaab,%eax
  shl    $0x2,%edx
  push   %r14
  imul   %rax,%rdx
  push   %r13
  mov    $0x1,%eax
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  shr    $0x21,%rdx
  push   %rbx
  mov    %rsi,%rbx
  add    $0x8,%edx
  movslq %edx,%rdx
  nopw   0x0(%rax,%rax,1)
L34:
  movzbl (%rbx,%rax,1),%ecx
  mov    %eax,%r13d
  cmp    $0xd,%cl
  je     L57
  cmp    $0xa,%cl
  je     L57
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L34
  mov    %r8d,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  movsbl (%rbx),%r14d
  sub    $0x20,%r14d
  and    $0x3f,%r14d
L42:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L57:
  mov    %r8d,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  sub    $0x1,%r13d
  call   <memset@plt>
  movsbl (%rbx),%r14d
  sub    $0x20,%r14d
  and    $0x3f,%r14d
  test   %r13d,%r13d
  je     L42
  add    $0x1,%rbx
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L49:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x3,%r12d
  add    $0x3,%rbp
  call   <uu_decode_word24@plt>
  add    $0x4,%rbx
  cmp    %r12d,%r13d
  jg     L49
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L37:
  xor    %eax,%eax
  ret
```

assistant: ```c
char uu_decode_line(void *dest, char *src_uu, uint8_t bytes) {
  uint8_t i, j;
  char *dest_p = (char *)dest;
  char *src_p = (char *)src_uu;
  uint8_t len;
  uint8_t len_check;
  len = 0;
  len_check = (src_p[0] - ' ') & 0x3F;
  if (len_check > 45)
    return 0;
  for (i = 1; i < len_check * 4 / 3 + 8; i++) {
    if ((src_uu[i] == '\r') || (src_uu[i] == '\n')) {
      len = i - 1;
      break;
    }
  }
  memset(dest, 0, bytes);
  len_check = (src_p[0] - ' ') & 0x3F;
  if (((len_check * 4) <= (len * 3 - 12)) || (len_check * 4) > (len * 3)) {
    ;
  }
  j = 1;
  for (i = 0; i < len; i += 3) {
    uu_decode_word24(&dest_p[i], &src_p[j]);
    j += 4;
  }
  return len_check;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<uu_decode_line>:
  endbr64
  mov    %edx,%r8d
  movsbl (%rsi),%edx
  sub    $0x20,%edx
  and    $0x3f,%edx
  cmp    $0x2d,%edx
  jg     L29
  mov    $0xaaaaaaab,%eax
  shl    $0x2,%edx
  push   %r14
  imul   %rax,%rdx
  push   %r13
  mov    $0x1,%eax
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  shr    $0x21,%rdx
  push   %rbx
  mov    %rsi,%rbx
  add    $0x8,%edx
  movslq %edx,%rdx
  nopw   0x0(%rax,%rax,1)
L18:
  movzbl (%rbx,%rax,1),%ecx
  mov    %eax,%r13d
  cmp    $0xd,%cl
  je     L30
  cmp    $0xa,%cl
  je     L30
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L18
  mov    %r8d,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
  movsbl (%rbx),%r14d
  sub    $0x20,%r14d
  and    $0x3f,%r14d
L34:
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  mov    %r8d,%edx
  xor    %esi,%esi
  mov    %rbp,%rdi
  sub    $0x1,%r13d
  call   <memset@plt>
  movsbl (%rbx),%r14d
  sub    $0x20,%r14d
  and    $0x3f,%r14d
  test   %r13d,%r13d
  je     L34
  add    $0x1,%rbx
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L12:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  add    $0x3,%r12d
  add    $0x3,%rbp
  call   <uu_decode_word24@plt>
  add    $0x4,%rbx
  cmp    %r12d,%r13d
  jg     L12
  pop    %rbx
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L29:
  xor    %eax,%eax
  ret
```

assistant: ```c
char uu_decode_line(void *dest, char *src_uu, uint8_t bytes) {
  uint8_t i, j;
  char *dest_p = (char *)dest;
  char *src_p = (char *)src_uu;
  uint8_t len;
  uint8_t len_check;
  len = 0;
  len_check = (src_p[0] - ' ') & 0x3F;
  if (len_check > 45)
    return 0;
  for (i = 1; i < len_check * 4 / 3 + 8; i++) {
    if ((src_uu[i] == '\r') || (src_uu[i] == '\n')) {
      len = i - 1;
      break;
    }
  }
  memset(dest, 0, bytes);
  len_check = (src_p[0] - ' ') & 0x3F;
  if (((len_check * 4) <= (len * 3 - 12)) || (len_check * 4) > (len * 3)) {
    ;
  }
  j = 1;
  for (i = 0; i < len; i += 3) {
    uu_decode_word24(&dest_p[i], &src_p[j]);
    j += 4;
  }
  return len_check;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<putdata2d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L14
L45:
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L44:
  mov    -0x14(%rbp),%eax
  imul   $0x1be5,%eax,%eax
  lea    0xd5f5(%rax),%edx
  movslq %edx,%rax
  imul   $0xffffffff81742e05,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x11,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x3f480,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  D5(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L2:
  mov    -0x28(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L44
  addl   $0x1,-0x10(%rbp)
L14:
  mov    -0x24(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jg     L45
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
type_guessed: byte[8]

tool: {"D5": [0x71,0x8b,0x89,0xc0,0x85,0x2e,0xd0,0x3e]}

assistant: ```c
void putdata2d(int n1, int n2, double **a) {
  int j1, j2, seed = 0;
  for (j1 = 0; j1 <= n1 - 1; j1++) {
    for (j2 = 0; j2 <= n2 - 1; j2++) {
      a[j1][j2] = ((*(&seed) = (*(&seed) * 7141 + 54773) % 259200) * (1.0 / 259200));
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<putdata2d>:
  endbr64
  mov    %esi,%r8d
  test   %edi,%edi
  jle    L56
  mov    %rdx,%rsi
  lea    -0x1(%rdi),%eax
  lea    0x8(%rdx,%rax,8),%r9
  lea    -0x1(%r8),%eax
  lea    0x8(,%rax,8),%rdi
  mov    $0x0,%eax
  movsd  D6(%rip),%xmm1
  jmp    L34
L17:
  imul   $0x1be5,%eax,%eax
  lea    0xd5f5(%rax),%edx
  movslq %edx,%rax
  imul   $0xffffffff81742e05,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x11,%eax
  mov    %edx,%r10d
  sar    $0x1f,%r10d
  sub    %r10d,%eax
  imul   $0x3f480,%eax,%r10d
  mov    %edx,%eax
  sub    %r10d,%eax
  mov    (%rsi),%rdx
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rdx,%rcx,1)
  add    $0x8,%rcx
  cmp    %rcx,%rdi
  jne    L17
L29:
  add    $0x8,%rsi
  cmp    %r9,%rsi
  je     L56
L34:
  mov    $0x0,%ecx
  test   %r8d,%r8d
  jg     L17
  jmp    L29
L56:
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type: byte[8]

tool: {"D6": [0x71,0x8b,0x89,0xc0,0x85,0x2e,0xd0,0x3e]}

assistant: ```c
void putdata2d(int n1, int n2, double **a) {
  int j1, j2, seed = 0;
  for (j1 = 0; j1 <= n1 - 1; j1++) {
    for (j2 = 0; j2 <= n2 - 1; j2++) {
      a[j1][j2] = ((*(&seed) = (*(&seed) * 7141 + 54773) % 259200) * (1.0 / 259200));
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<putdata2d>:
  endbr64
  mov    %esi,%r9d
  test   %edi,%edi
  jle    L18
  lea    -0x1(%rdi),%eax
  mov    %rdx,%r8
  movsd  D4(%rip),%xmm1
  lea    0x8(%rdx,%rax,8),%r10
  lea    -0x1(%rsi),%edx
  xor    %eax,%eax
  lea    0x8(,%rdx,8),%r11
  nopl   0x0(%rax,%rax,1)
L2:
  test   %r9d,%r9d
  jle    L17
  mov    (%r8),%rdx
  lea    (%r11,%rdx,1),%rdi
  nopl   0x0(%rax)
L33:
  imul   $0x1be5,%eax,%eax
  pxor   %xmm0,%xmm0
  add    $0x8,%rdx
  lea    0xd5f5(%rax),%ecx
  mov    %ecx,%eax
  shr    $0x7,%eax
  imul   $0x205d0b9,%rax,%rax
  shr    $0x24,%rax
  imul   $0x3f480,%eax,%esi
  mov    %ecx,%eax
  sub    %esi,%eax
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rdx)
  cmp    %rdx,%rdi
  jne    L33
L17:
  add    $0x8,%r8
  cmp    %r10,%r8
  jne    L2
L18:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D4
type_guessed: byte[8]

tool: {"D4": [0x71,0x8b,0x89,0xc0,0x85,0x2e,0xd0,0x3e]}

assistant: ```c
void putdata2d(int n1, int n2, double **a) {
  int j1, j2, seed = 0;
  for (j1 = 0; j1 <= n1 - 1; j1++) {
    for (j2 = 0; j2 <= n2 - 1; j2++) {
      a[j1][j2] = ((*(&seed) = (*(&seed) * 7141 + 54773) % 259200) * (1.0 / 259200));
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<putdata2d>:
  endbr64
  test   %edi,%edi
  jle    L24
  test   %esi,%esi
  jle    L24
  lea    -0x1(%rdi),%eax
  mov    %esi,%esi
  movsd  D7(%rip),%xmm1
  mov    %rdx,%r8
  lea    0x8(%rdx,%rax,8),%r10
  lea    0x0(,%rsi,8),%r9
  xor    %eax,%eax
  nop
L4:
  mov    (%r8),%rdx
  lea    (%r9,%rdx,1),%rdi
  nopw   0x0(%rax,%rax,1)
L18:
  imul   $0x1be5,%eax,%eax
  pxor   %xmm0,%xmm0
  add    $0x8,%rdx
  lea    0xd5f5(%rax),%ecx
  mov    %ecx,%eax
  shr    $0x7,%eax
  imul   $0x205d0b9,%rax,%rax
  shr    $0x24,%rax
  imul   $0x3f480,%eax,%esi
  mov    %ecx,%eax
  sub    %esi,%eax
  cvtsi2sd %eax,%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rdx)
  cmp    %rdx,%rdi
  jne    L18
  add    $0x8,%r8
  cmp    %r8,%r10
  jne    L4
L24:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type_guessed: byte[8]

tool: {"D7": [0x71,0x8b,0x89,0xc0,0x85,0x2e,0xd0,0x3e]}

assistant: ```c
void putdata2d(int n1, int n2, double **a) {
  int j1, j2, seed = 0;
  for (j1 = 0; j1 <= n1 - 1; j1++) {
    for (j2 = 0; j2 <= n2 - 1; j2++) {
      a[j1][j2] = ((*(&seed) = (*(&seed) * 7141 + 54773) % 259200) * (1.0 / 259200));
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x7,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <f@plt>
  leave
  ret
```

assistant: ```c
int main() {
  int a = 7;
  return f(a);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7,%edi
  call   <f@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int a = 7;
  return f(a);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  mov    $0x7,%edi
  jmp    1050 <f@plt>
```

assistant: ```c
int main() {
  int a = 7;
  return f(a);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  mov    $0x7,%edi
  jmp    1050 <f@plt>
```

assistant: ```c
int main() {
  int a = 7;
  return f(a);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<r>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edi
  call   <g@plt>
  mov    $0x1,%edi
  call   <g@plt>
  mov    $0x0,%edi
  call   <a@plt>
  mov    $0x0,%edi
  call   <a@plt>
  mov    $0x0,%edi
  call   <a@plt>
  mov    $0x1,%edi
  call   <a@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void r() {
  g(0);
  g(1);
  a(0);
  a(0);
  a(0);
  a(1);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<r>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <g@plt>
  mov    $0x1,%edi
  call   <g@plt>
  mov    $0x0,%edi
  call   <a@plt>
  mov    $0x0,%edi
  call   <a@plt>
  mov    $0x0,%edi
  call   <a@plt>
  mov    $0x1,%edi
  call   <a@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void r() {
  g(0);
  g(1);
  a(0);
  a(0);
  a(0);
  a(1);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<r>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <g@plt>
  mov    $0x1,%edi
  call   <g@plt>
  xor    %edi,%edi
  call   <a@plt>
  xor    %edi,%edi
  call   <a@plt>
  xor    %edi,%edi
  call   <a@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <a@plt>
```

assistant: ```c
void r() {
  g(0);
  g(1);
  a(0);
  a(0);
  a(0);
  a(1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<r>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <g@plt>
  mov    $0x1,%edi
  call   <g@plt>
  xor    %edi,%edi
  call   <a@plt>
  xor    %edi,%edi
  call   <a@plt>
  xor    %edi,%edi
  call   <a@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <a@plt>
```

assistant: ```c
void r() {
  g(0);
  g(1);
  a(0);
  a(0);
  a(0);
  a(1);
}
```
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   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D6(%rip),%rax
  mov    %rax,-0x48(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
L11:
  mov    -0x48(%rbp),%rax
  mov    %rax,%rdi
  call   <opendir@plt>
  mov    %rax,-0x40(%rbp)
  jmp    L26
L14:
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x5,%eax
  jle    L26
  mov    -0x48(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcat@plt>
  lea    -0x30(%rbp),%rax
  lea    D8(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L9
  lea    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <file_read_play@plt>
L9:
  lea    -0x30(%rbp),%rax
  mov    $0x20,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
L26:
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <readdir@plt>
  mov    %rax,-0x38(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  jne    L14
  mov    -0x40(%rbp),%rax
  mov    %rax,%rdi
  call   <closedir@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L11
```

tool_calls: name: read_data
arguments: data_label: D6
type: string
name: read_data
arguments: data_label: D8
type: string

tool: D6: ../../data/media/

tool: D8: media_

assistant: ```c
int main(void) {
  struct dirent *entry;
  DIR *dir;
  char *path = "../../data/media/";
  char fname[32] = {0};
  while (1) {
    dir = opendir(path);
    while ((entry = readdir(dir)) != NULL) {
      if (strlen(entry->d_name) > 5) {
        strcat(fname, path);
        strcat(fname, entry->d_name);
        if (strstr(fname, "media_"))
          file_read_play(fname);
        memset(fname, 0, sizeof(fname));
      }
    }
    closedir(dir);
    sleep(1);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  lea    D1(%rip),%r12
  mov    %rsp,%r13
  lea    D11(%rip),%r14
  jmp    L28
L56:
  mov    %r13,%rdi
  call   <file_read_play@plt>
  jmp    L25
L15:
  mov    %rbp,%rdi
  call   <closedir@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
L28:
  mov    %r12,%rdi
  call   <opendir@plt>
  mov    %rax,%rbp
L23:
  mov    %rbp,%rdi
  call   <readdir@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L15
  mov    (%rbx),%rdi
  call   <strlen@plt>
  cmp    $0x5,%eax
  jle    L23
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <strcat@plt>
  mov    (%rbx),%rsi
  mov    %r13,%rdi
  call   <strcat@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L56
L25:
  mov    $0x20,%edx
  mov    $0x0,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  jmp    L23
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type: string
name: retrieve_data
arguments: data_tag: D11
type: string

tool: D1: ../../data/media/

tool: D11: media_

assistant: ```c
int main(void) {
  struct dirent *entry;
  DIR *dir;
  char *path = "../../data/media/";
  char fname[32] = {0};
  while (1) {
    dir = opendir(path);
    while ((entry = readdir(dir)) != NULL) {
      if (strlen(entry->d_name) > 5) {
        strcat(fname, path);
        strcat(fname, entry->d_name);
        if (strstr(fname, "media_"))
          file_read_play(fname);
        memset(fname, 0, sizeof(fname));
      }
    }
    closedir(dir);
    sleep(1);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r14
  pxor   %xmm0,%xmm0
  lea    D13(%rip),%r14
  push   %r13
  push   %r12
  lea    D11(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  xchg   %ax,%ax
L19:
  mov    %r12,%rdi
  call   <opendir@plt>
  mov    %rax,%rbp
L18:
  mov    %rbp,%rdi
  call   <readdir@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L58
L51:
  mov    (%rbx),%rdi
  call   <strlen@plt>
  cmp    $0x5,%eax
  jle    L18
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <strcat@plt>
  mov    (%rbx),%rsi
  mov    %r13,%rdi
  call   <strcat@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L48
L27:
  mov    $0x20,%edx
  xor    %esi,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  mov    %rbp,%rdi
  call   <readdir@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L51
  nopl   (%rax)
L58:
  mov    %rbp,%rdi
  call   <closedir@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L48:
  mov    %r13,%rdi
  call   <file_read_play@plt>
  jmp    L27
```

tool_calls: name: unpack_data
arguments: data_label: D13
datatype: string
name: unpack_data
arguments: data_label: D11
datatype: string

tool: D13: media_

tool: D11: ../../data/media/

assistant: ```c
int main(void) {
  struct dirent *entry;
  DIR *dir;
  char *path = "../../data/media/";
  char fname[32] = {0};
  while (1) {
    dir = opendir(path);
    while ((entry = readdir(dir)) != NULL) {
      if (strlen(entry->d_name) > 5) {
        strcat(fname, path);
        strcat(fname, entry->d_name);
        if (strstr(fname, "media_"))
          file_read_play(fname);
        memset(fname, 0, sizeof(fname));
      }
    }
    closedir(dir);
    sleep(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   %r14
  pxor   %xmm0,%xmm0
  lea    D9(%rip),%r14
  push   %r13
  push   %r12
  lea    D16(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  xchg   %ax,%ax
L11:
  mov    %r12,%rdi
  call   <opendir@plt>
  mov    %rax,%rbp
L3:
  mov    %rbp,%rdi
  call   <readdir@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L36
L0:
  mov    (%rbx),%rdi
  call   <strlen@plt>
  cmp    $0x5,%eax
  jle    L3
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <strcat@plt>
  mov    (%rbx),%rsi
  mov    %r13,%rdi
  call   <strcat@plt>
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L23
L59:
  mov    $0x20,%edx
  xor    %esi,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  mov    %rbp,%rdi
  call   <readdir@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L0
  nopl   (%rax)
L36:
  mov    %rbp,%rdi
  call   <closedir@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  jmp    L11
  nopw   0x0(%rax,%rax,1)
L23:
  mov    %r13,%rdi
  call   <file_read_play@plt>
  jmp    L59
```

tool_calls: name: unpack_data
arguments: label: D9
type_guessed: string
name: unpack_data
arguments: label: D16
type_guessed: string

tool: D9: media_

tool: D16: ../../data/media/

assistant: ```c
int main(void) {
  struct dirent *entry;
  DIR *dir;
  char *path = "../../data/media/";
  char fname[32] = {0};
  while (1) {
    dir = opendir(path);
    while ((entry = readdir(dir)) != NULL) {
      if (strlen(entry->d_name) > 5) {
        strcat(fname, path);
        strcat(fname, entry->d_name);
        if (strstr(fname, "media_"))
          file_read_play(fname);
        memset(fname, 0, sizeof(fname));
      }
    }
    closedir(dir);
    sleep(1);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<divs100>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x1f,%eax
  and    $0x63,%eax
  add    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sar    %eax
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0x3,%eax
  add    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0x6,%eax
  add    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0xa,%eax
  sub    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0xc,%eax
  add    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0xd,%eax
  add    %eax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  sar    $0x14,%eax
  add    %eax,-0x8(%rbp)
  sarl   $0x6,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  imul   $0xffffff9c,%eax,%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    $0x1c,%eax
  sar    $0x7,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int divs100(int n) {
  int q, r;
  n = n + (n >> 31 & 99);
  q = (n >> 1) + (n >> 3) + (n >> 6) - (n >> 10) + (n >> 12) + (n >> 13) - (n >> 16);
  q = q + (q >> 20);
  q = q >> 6;
  r = n - q * 100;
  return q + ((r + 28) >> 7);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<divs100>:
  endbr64
  mov    %edi,%eax
  sar    $0x1f,%edi
  and    $0x63,%edi
  add    %eax,%edi
  mov    %edi,%eax
  sar    %eax
  mov    %edi,%edx
  sar    $0x3,%edx
  add    %edx,%eax
  mov    %edi,%edx
  sar    $0x6,%edx
  add    %edx,%eax
  mov    %edi,%edx
  sar    $0xa,%edx
  sub    %edx,%eax
  mov    %edi,%edx
  sar    $0xc,%edx
  add    %edx,%eax
  mov    %edi,%edx
  sar    $0xd,%edx
  add    %edx,%eax
  mov    %edi,%edx
  sar    $0x10,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  sar    $0x14,%edx
  add    %eax,%edx
  sar    $0x6,%edx
  imul   $0xffffff9c,%edx,%eax
  lea    0x1c(%rdi,%rax,1),%eax
  sar    $0x7,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int divs100(int n) {
  int q, r;
  n = n + (n >> 31 & 99);
  q = (n >> 1) + (n >> 3) + (n >> 6) - (n >> 10) + (n >> 12) + (n >> 13) - (n >> 16);
  q = q + (q >> 20);
  q = q >> 6;
  r = n - q * 100;
  return q + ((r + 28) >> 7);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<divs100>:
  endbr64
  mov    %edi,%eax
  sar    $0x1f,%edi
  and    $0x63,%edi
  add    %eax,%edi
  mov    %edi,%eax
  mov    %edi,%edx
  sar    $0x3,%edx
  sar    %eax
  add    %edx,%eax
  mov    %edi,%edx
  sar    $0x6,%edx
  add    %edx,%eax
  mov    %edi,%edx
  sar    $0xa,%edx
  sub    %edx,%eax
  mov    %edi,%edx
  sar    $0xc,%edx
  add    %edx,%eax
  mov    %edi,%edx
  sar    $0xd,%edx
  add    %edx,%eax
  mov    %edi,%edx
  sar    $0x10,%edx
  sub    %edx,%eax
  mov    %eax,%edx
  sar    $0x14,%edx
  add    %eax,%edx
  sar    $0x6,%edx
  imul   $0xffffff9c,%edx,%eax
  lea    0x1c(%rdi,%rax,1),%eax
  sar    $0x7,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int divs100(int n) {
  int q, r;
  n = n + (n >> 31 & 99);
  q = (n >> 1) + (n >> 3) + (n >> 6) - (n >> 10) + (n >> 12) + (n >> 13) - (n >> 16);
  q = q + (q >> 20);
  q = q >> 6;
  r = n - q * 100;
  return q + ((r + 28) >> 7);
}
```