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

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x25,%eax
  movzbl %al,%edx
  test   $0x1,%al
  je     L2
  test   $0x4,%al
  je     L1
  test   $0x8,%al
  je     L9
L2:
  shl    $0x2,%edx
  and    $0x3c,%edx
  or     %edx,%eax
  xor    $0xffffff99,%eax
  add    $0x7c,%eax
  mov    %al,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  test   $0x2,%al
  je     L32
  mov    %eax,%ecx
  shr    $0x3,%cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%edx
  shl    $0x2,%edx
  and    $0x1c,%edx
  or     %edx,%eax
  xor    $0xffffff99,%eax
  add    $0x7c,%eax
  mov    %al,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L32:
  imul   %edx,%edx
L9:
  shl    $0x4,%edx
  or     %edx,%eax
  xor    $0xffffff99,%eax
  add    $0x7c,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = input[0UL] + (unsigned char)219;
    if (state[0UL] & (unsigned char)1) {
      if ((state[0UL] >> (unsigned char)2) & (unsigned char)1) {
        if ((state[0UL] >> (unsigned char)3) & (unsigned char)1) {
          state[0UL] |= (state[0UL] & (unsigned char)15) << 2UL;
        } else {
          state[0UL] |= (state[0UL] & (unsigned char)63) << 4UL;
        }
      } else if ((state[0UL] >> (unsigned char)1) & (unsigned char)1) {
        state[0UL] |= ((state[0UL] << (((state[0UL] >> (unsigned char)3) & (unsigned char)7) | 1UL)) & (unsigned char)7) << 2UL;
      } else {
        state[0UL] |= (state[0UL] * state[0UL] & (unsigned char)31) << 4UL;
      }
    } else {
      state[0UL] |= (state[0UL] & (unsigned char)15) << 2UL;
    }
    output[0UL] = (state[0UL] ^ 498160537UL) + (unsigned char)124;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<RandomFunc>:
  endbr64
  movzbl (%rdi),%eax
  sub    $0x25,%eax
  movzbl %al,%edx
  test   $0x1,%al
  je     L30
  test   $0x4,%al
  je     L20
  test   $0x8,%al
  je     L9
L30:
  shl    $0x2,%edx
  and    $0x3c,%edx
  or     %edx,%eax
  xor    $0xffffff99,%eax
  add    $0x7c,%eax
  mov    %al,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  test   $0x2,%al
  je     L15
  mov    %eax,%ecx
  shr    $0x3,%cl
  and    $0x6,%ecx
  or     $0x1,%ecx
  shl    %cl,%edx
  shl    $0x2,%edx
  and    $0x1c,%edx
  or     %edx,%eax
  xor    $0xffffff99,%eax
  add    $0x7c,%eax
  mov    %al,(%rsi)
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  imul   %edx,%edx
L9:
  shl    $0x4,%edx
  or     %edx,%eax
  xor    $0xffffff99,%eax
  add    $0x7c,%eax
  mov    %al,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = input[0UL] + (unsigned char)219;
    if (state[0UL] & (unsigned char)1) {
      if ((state[0UL] >> (unsigned char)2) & (unsigned char)1) {
        if ((state[0UL] >> (unsigned char)3) & (unsigned char)1) {
          state[0UL] |= (state[0UL] & (unsigned char)15) << 2UL;
        } else {
          state[0UL] |= (state[0UL] & (unsigned char)63) << 4UL;
        }
      } else if ((state[0UL] >> (unsigned char)1) & (unsigned char)1) {
        state[0UL] |= ((state[0UL] << (((state[0UL] >> (unsigned char)3) & (unsigned char)7) | 1UL)) & (unsigned char)7) << 2UL;
      } else {
        state[0UL] |= (state[0UL] * state[0UL] & (unsigned char)31) << 4UL;
      }
    } else {
      state[0UL] |= (state[0UL] & (unsigned char)15) << 2UL;
    }
    output[0UL] = (state[0UL] ^ 498160537UL) + (unsigned char)124;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<extract_min>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jg     L0
  mov    $0xffffffff,%eax
  jmp    L3
L0:
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%rdx
  add    $0x4,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0x1,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <siftdown@plt>
  mov    -0x4(%rbp),%eax
L3:
  leave
  ret
```

assistant: ```c
int extract_min(int *head, int *head_len) {
  int temp_len = *head_len;
  if (temp_len < 1) {
    return -1;
  }
  int res = head[1];
  head[1] = head[temp_len--];
  *head_len = temp_len;
  siftdown(head, *head_len, 1);
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<extract_min>:
  endbr64
  push   %rbx
  mov    (%rsi),%eax
  test   %eax,%eax
  jle    L15
  mov    0x4(%rdi),%ebx
  lea    -0x1(%rax),%ecx
  cltq
  mov    (%rdi,%rax,4),%eax
  mov    %eax,0x4(%rdi)
  mov    %ecx,(%rsi)
  mov    $0x1,%edx
  mov    %ecx,%esi
  call   <siftdown@plt>
L19:
  mov    %ebx,%eax
  pop    %rbx
  ret
L15:
  mov    $0xffffffff,%ebx
  jmp    L19
```

assistant: ```c
int extract_min(int *head, int *head_len) {
  int temp_len = *head_len;
  if (temp_len < 1) {
    return -1;
  }
  int res = head[1];
  head[1] = head[temp_len--];
  *head_len = temp_len;
  siftdown(head, *head_len, 1);
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<extract_min>:
  endbr64
  push   %r12
  movslq (%rsi),%rax
  test   %eax,%eax
  jle    L4
  lea    -0x1(%rax),%r8d
  mov    (%rdi,%rax,4),%eax
  mov    0x4(%rdi),%r12d
  mov    $0x1,%edx
  mov    %eax,0x4(%rdi)
  mov    %r8d,(%rsi)
  mov    %r8d,%esi
  call   <siftdown@plt>
L12:
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L4:
  mov    $0xffffffff,%r12d
  jmp    L12
```

assistant: ```c
int extract_min(int *head, int *head_len) {
  int temp_len = *head_len;
  if (temp_len < 1) {
    return -1;
  }
  int res = head[1];
  head[1] = head[temp_len--];
  *head_len = temp_len;
  siftdown(head, *head_len, 1);
  return res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<extract_min>:
  endbr64
  push   %r12
  movslq (%rsi),%rax
  test   %eax,%eax
  jle    L16
  lea    -0x1(%rax),%r8d
  mov    (%rdi,%rax,4),%eax
  mov    0x4(%rdi),%r12d
  mov    $0x1,%edx
  mov    %eax,0x4(%rdi)
  mov    %r8d,(%rsi)
  mov    %r8d,%esi
  call   <siftdown@plt>
L10:
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax)
L16:
  mov    $0xffffffff,%r12d
  jmp    L10
```

assistant: ```c
int extract_min(int *head, int *head_len) {
  int temp_len = *head_len;
  if (temp_len < 1) {
    return -1;
  }
  int res = head[1];
  head[1] = head[temp_len--];
  *head_len = temp_len;
  siftdown(head, *head_len, 1);
  return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x1,-0x4(%rbp)
  jmp    L16
L14:
  mov    -0x4(%rbp),%eax
  imul   %eax,%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L16:
  cmpl   $0x6,-0x4(%rbp)
  jle    L14
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D15
type_infered: string
name: retrieve_data
arguments: label: D9
type_infered: string

tool: D15:     n   n  cubed

tool: D9: %5d %5d


assistant: ```c
int main(void) {
  int num;
  printf("    n   n  cubed\n");
  for (num = 1; num <= 6; num++)
    printf("%5d %5d\n", num, num * num * num);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%ebx
  lea    D10(%rip),%rbp
L6:
  mov    %ebx,%edx
  imul   %ebx,%edx
  imul   %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x7,%ebx
  jne    L6
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D11
data_type: string
name: unpack_data
arguments: data_label: D10
data_type: string

tool: D11:     n   n  cubed

tool: D10: %5d %5d


assistant: ```c
int main(void) {
  int num;
  printf("    n   n  cubed\n");
  for (num = 1; num <= 6; num++)
    printf("%5d %5d\n", num, num * num * num);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D1(%rip),%rdi
  lea    D8(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  imul   %ebx,%edx
  imul   %ebx,%edx
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x7,%ebx
  jne    L3
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_guessed: string
name: unpack_data
arguments: data_label: D8
type_guessed: string

tool: D1:     n   n  cubed

tool: D8: %5d %5d


assistant: ```c
int main(void) {
  int num;
  printf("    n   n  cubed\n");
  for (num = 1; num <= 6; num++)
    printf("%5d %5d\n", num, num * num * num);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D18(%rip),%rdi
  lea    D6(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  call   <puts@plt>
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  imul   %ebx,%edx
  imul   %ebx,%edx
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0x7,%ebx
  jne    L0
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D18
type: string
name: parse_data
arguments: tag: D6
type: string

tool: D18:     n   n  cubed

tool: D6: %5d %5d


assistant: ```c
int main(void) {
  int num;
  printf("    n   n  cubed\n");
  for (num = 1; num <= 6; num++)
    printf("%5d %5d\n", num, num * num * num);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<rtrim>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%ebx
  jmp    L9
L1:
  sub    $0x1,%ebx
L9:
  movslq %ebx,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L1
  movslq %ebx,%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *rtrim(char *str) {
  register int p;
  p = strlen(str) - 1;
  while (isspace(str[p]))
    p--;
  str[p + 1] = '\0';
  return str;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rtrim>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    -0x1(%rax),%ebx
  movslq %ebx,%rbx
L8:
  movsbl 0x0(%rbp,%rbx,1),%edi
  call   <isspace@plt>
  mov    %rbx,%rdx
  sub    $0x1,%rbx
  test   %rax,%rax
  jne    L8
  movb   $0x0,0x1(%rbp,%rdx,1)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *rtrim(char *str) {
  register int p;
  p = strlen(str) - 1;
  while (isspace(str[p]))
    p--;
  str[p + 1] = '\0';
  return str;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rtrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    -0x1(%rax),%ebx
  movslq %ebx,%rbx
  nopl   0x0(%rax)
L0:
  movsbl (%r12,%rbx,1),%edi
  call   <isspace@plt>
  mov    %rbx,%rdx
  sub    $0x1,%rbx
  test   %rax,%rax
  jne    L0
  movb   $0x0,0x1(%r12,%rdx,1)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *rtrim(char *str) {
  register int p;
  p = strlen(str) - 1;
  while (isspace(str[p]))
    p--;
  str[p + 1] = '\0';
  return str;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<rtrim>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    -0x1(%rax),%ebx
  movslq %ebx,%rbx
  nopl   0x0(%rax)
L9:
  movsbl (%r12,%rbx,1),%edi
  call   <isspace@plt>
  mov    %rbx,%rdx
  sub    $0x1,%rbx
  test   %rax,%rax
  jne    L9
  movb   $0x0,0x1(%r12,%rdx,1)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
char *rtrim(char *str) {
  register int p;
  p = strlen(str) - 1;
  while (isspace(str[p]))
    p--;
  str[p + 1] = '\0';
  return str;
}
```
user: What is the c source code of the assembly code below?

```asm
<shell_copy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  lea    D6(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <popen@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L37
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x0,%eax
  jmp    L34
L19:
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fread@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  sub    %rax,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  add    %rax,-0x28(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x1c(%rbp)
L37:
  cmpq   $0x0,-0x18(%rbp)
  je     L27
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L27
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <ferror@plt>
  test   %eax,%eax
  je     L19
L27:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <pclose@plt>
  mov    -0x1c(%rbp),%eax
L34:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
data_type: string
name: unpack_data
arguments: data_label: D10
data_type: string

tool: D6: r

tool: D10: popen

assistant: ```c
int shell_copy(char *buf, size_t n, char *cmd) {
  int total = 0;
  FILE *pf;
  size_t ret, remaining = n;
  pf = popen(cmd, "r");
  if (pf == NULL) {
    perror("popen");
    return 0;
  }
  while (remaining > 0 && !feof(pf) && !ferror(pf)) {
    ret = fread(buf, 1, remaining, pf);
    remaining -= ret;
    buf += ret;
    total += ret;
  }
  pclose(pf);
  return total;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<shell_copy>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %rdx,%rdi
  lea    D13(%rip),%rsi
  call   <popen@plt>
  test   %rax,%rax
  je     L22
  mov    %rax,%rbp
  mov    $0x0,%r13d
  test   %rbx,%rbx
  je     L16
L50:
  mov    %rbp,%rdi
  call   <feof@plt>
  test   %eax,%eax
  jne    L16
  mov    %rbp,%rdi
  call   <ferror@plt>
  test   %eax,%eax
  jne    L16
  mov    %rbp,%rcx
  mov    %rbx,%rdx
  mov    $0x1,%esi
  mov    %r12,%rdi
  call   <fread@plt>
  add    %rax,%r12
  add    %eax,%r13d
  sub    %rax,%rbx
  jne    L50
L16:
  mov    %rbp,%rdi
  call   <pclose@plt>
L46:
  mov    %r13d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  lea    D17(%rip),%rdi
  call   <perror@plt>
  mov    $0x0,%r13d
  jmp    L46
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
type_infered: string
name: retrieve_data
arguments: data_tag: D17
type_infered: string

tool: D13: r

tool: D17: popen

assistant: ```c
int shell_copy(char *buf, size_t n, char *cmd) {
  int total = 0;
  FILE *pf;
  size_t ret, remaining = n;
  pf = popen(cmd, "r");
  if (pf == NULL) {
    perror("popen");
    return 0;
  }
  while (remaining > 0 && !feof(pf) && !ferror(pf)) {
    ret = fread(buf, 1, remaining, pf);
    remaining -= ret;
    buf += ret;
    total += ret;
  }
  pclose(pf);
  return total;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<shell_copy>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    %rdx,%rdi
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  lea    D16(%rip),%rsi
  sub    $0x8,%rsp
  call   <popen@plt>
  test   %rax,%rax
  je     L29
  mov    %rax,%rbp
  xor    %r13d,%r13d
  test   %rbx,%rbx
  jne    L34
  jmp    L44
  nopl   (%rax)
L40:
  mov    %rbp,%rdi
  call   <ferror@plt>
  test   %eax,%eax
  jne    L44
  mov    %r12,%rdi
  mov    %rbp,%rcx
  mov    %rbx,%rdx
  mov    $0x1,%esi
  call   <fread@plt>
  add    %rax,%r12
  add    %eax,%r13d
  sub    %rax,%rbx
  je     L44
L34:
  mov    %rbp,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L40
L44:
  mov    %rbp,%rdi
  call   <pclose@plt>
L12:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L29:
  lea    D2(%rip),%rdi
  xor    %r13d,%r13d
  call   <perror@plt>
  jmp    L12
```

tool_calls: name: unpack_data
arguments: data_tag: D16
data_type: string
name: unpack_data
arguments: data_tag: D2
data_type: string

tool: D16: r

tool: D2: popen

assistant: ```c
int shell_copy(char *buf, size_t n, char *cmd) {
  int total = 0;
  FILE *pf;
  size_t ret, remaining = n;
  pf = popen(cmd, "r");
  if (pf == NULL) {
    perror("popen");
    return 0;
  }
  while (remaining > 0 && !feof(pf) && !ferror(pf)) {
    ret = fread(buf, 1, remaining, pf);
    remaining -= ret;
    buf += ret;
    total += ret;
  }
  pclose(pf);
  return total;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<shell_copy>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  mov    %rdx,%rdi
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  lea    D9(%rip),%rsi
  sub    $0x8,%rsp
  call   <popen@plt>
  test   %rax,%rax
  je     L49
  mov    %rax,%rbp
  xor    %r13d,%r13d
  test   %rbx,%rbx
  jne    L9
  jmp    L65
  nopl   (%rax)
L34:
  mov    %rbp,%rdi
  call   <ferror@plt>
  test   %eax,%eax
  jne    L65
  mov    %r12,%rdi
  mov    %rbp,%rcx
  mov    %rbx,%rdx
  mov    $0x1,%esi
  call   <fread@plt>
  add    %rax,%r12
  add    %eax,%r13d
  sub    %rax,%rbx
  je     L65
L9:
  mov    %rbp,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L34
L65:
  mov    %rbp,%rdi
  call   <pclose@plt>
L68:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L49:
  lea    D5(%rip),%rdi
  xor    %r13d,%r13d
  call   <perror@plt>
  jmp    L68
```

tool_calls: name: read_data
arguments: label: D9
type_guessed: string
name: read_data
arguments: label: D5
type_guessed: string

tool: D9: r

tool: D5: popen

assistant: ```c
int shell_copy(char *buf, size_t n, char *cmd) {
  int total = 0;
  FILE *pf;
  size_t ret, remaining = n;
  pf = popen(cmd, "r");
  if (pf == NULL) {
    perror("popen");
    return 0;
  }
  while (remaining > 0 && !feof(pf) && !ferror(pf)) {
    ret = fread(buf, 1, remaining, pf);
    remaining -= ret;
    buf += ret;
    total += ret;
  }
  pclose(pf);
  return total;
}
```
user: What is the c source code of the assembly code below?

```asm
<utoa>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%r12
  mov    %edx,%esi
  mov    %edx,%ebp
  lea    -0x2(%rdx),%edx
  mov    $0x0,%eax
  cmp    $0x22,%edx
  ja     L6
  mov    %edi,%ebx
  call   <digits@plt>
  mov    %eax,%edi
  mov    %eax,%ecx
  sub    $0x1,%ecx
  js     L18
  movslq %ecx,%rcx
  add    %r12,%rcx
  cltq
  lea    -0x2(%r12,%rax,1),%rsi
  lea    -0x1(%rdi),%eax
  sub    %rax,%rsi
  lea    D6(%rip),%r8
L27:
  mov    %ebx,%eax
  mov    $0x0,%edx
  div    %ebp
  mov    %eax,%ebx
  movzbl (%r8,%rdx,1),%eax
  mov    %al,(%rcx)
  sub    $0x1,%rcx
  cmp    %rsi,%rcx
  jne    L27
L18:
  movslq %edi,%rdi
  movb   $0x0,(%r12,%rdi,1)
  mov    %r12,%rax
L6:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D6
data_type: string

tool: D6: 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ

assistant: ```c
char *utoa(unsigned int value, char *str, int base) {
  if (base < 2 || base > 36)
    return NULL;
  static const char *alphabet = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
  int len = digits(value, base);
  for (int i = len - 1; i >= 0; --i) {
    str[i] = alphabet[value % base];
    value /= base;
  }
  str[len] = 0;
  return str;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<utoa>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x2,%edx
  cmp    $0x22,%edx
  ja     L15
  mov    %rsi,%rbp
  mov    %ebx,%esi
  mov    %edi,%r12d
  call   <digits@plt>
  mov    %eax,%edx
  movslq %eax,%r8
  sub    $0x1,%edx
  js     L12
  movslq %eax,%r8
  movslq %edx,%rcx
  lea    D9(%rip),%rdi
  mov    %edx,%edx
  lea    -0x2(%rbp,%r8,1),%rsi
  add    %rbp,%rcx
  sub    %rdx,%rsi
  nopl   (%rax)
L16:
  mov    %r12d,%eax
  xor    %edx,%edx
  sub    $0x1,%rcx
  div    %ebx
  mov    %eax,%r12d
  movzbl (%rdi,%rdx,1),%eax
  mov    %al,0x1(%rcx)
  cmp    %rsi,%rcx
  jne    L16
L12:
  movb   $0x0,0x0(%rbp,%r8,1)
  mov    %rbp,%rax
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
data_type: string

tool: D9: 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ

assistant: ```c
char *utoa(unsigned int value, char *str, int base) {
  if (base < 2 || base > 36)
    return NULL;
  static const char *alphabet = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
  int len = digits(value, base);
  for (int i = len - 1; i >= 0; --i) {
    str[i] = alphabet[value % base];
    value /= base;
  }
  str[len] = 0;
  return str;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<utoa>:
  endbr64
  push   %r12
  xor    %eax,%eax
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x2,%edx
  cmp    $0x22,%edx
  ja     L11
  mov    %rsi,%rbp
  mov    %ebx,%esi
  mov    %edi,%r12d
  call   <digits@plt>
  mov    %eax,%edx
  movslq %eax,%r8
  sub    $0x1,%edx
  js     L2
  movslq %eax,%r8
  movslq %edx,%rcx
  lea    D7(%rip),%rdi
  mov    %edx,%edx
  lea    -0x2(%rbp,%r8,1),%rsi
  add    %rbp,%rcx
  sub    %rdx,%rsi
  nopl   (%rax)
L28:
  mov    %r12d,%eax
  xor    %edx,%edx
  sub    $0x1,%rcx
  div    %ebx
  mov    %eax,%r12d
  movzbl (%rdi,%rdx,1),%eax
  mov    %al,0x1(%rcx)
  cmp    %rcx,%rsi
  jne    L28
L2:
  movb   $0x0,0x0(%rbp,%r8,1)
  mov    %rbp,%rax
L11:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
data_type: string

tool: D7: 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ

assistant: ```c
char *utoa(unsigned int value, char *str, int base) {
  if (base < 2 || base > 36)
    return NULL;
  static const char *alphabet = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
  int len = digits(value, base);
  for (int i = len - 1; i >= 0; --i) {
    str[i] = alphabet[value % base];
    value /= base;
  }
  str[len] = 0;
  return str;
}
```