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

```asm
<option_alloc>:
  endbr64
  push   %r15
  lea    -0x4(%rdx),%eax
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  and    $0xfffffffb,%eax
  je     L33
  lea    -0x1(%rdx),%eax
  cmp    $0x1,%eax
  ja     L24
L33:
  cmp    $0x7,%r14d
  ja     L24
  xor    %edi,%edi
  mov    0x0(%rbp),%r13
  call   <CMSG_LEN@plt>
  xor    %edi,%edi
  sub    %eax,%r13d
  sete   %dil
  call   <__glibc_unlikely@plt>
  test   %rax,%rax
  je     L29
  addq   $0x4,0x0(%rbp)
  mov    $0x4,%r13d
L29:
  lea    -0x1(%rbx),%eax
  mov    %rbp,%rdi
  and    %eax,%r13d
  sub    %r13d,%ebx
  and    %eax,%ebx
  lea    (%rbx,%r14,1),%esi
  call   <add_pad@plt>
  mov    %rbp,%rdi
  call   <CMSG_DATA@plt>
  xor    %edi,%edi
  mov    0x0(%rbp),%r15
  mov    %rax,%r13
  call   <CMSG_LEN@plt>
  add    0x0(%rbp),%r12
  xor    %edi,%edi
  mov    %r12,0x0(%rbp)
  mov    %rax,%r14
  call   <CMSG_LEN@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  sub    %r12d,%esi
  and    $0x7,%esi
  call   <add_pad@plt>
  xor    %edi,%edi
  mov    0x0(%rbp),%rbx
  call   <CMSG_LEN@plt>
  xor    %edi,%edi
  sub    %rax,%rbx
  and    $0x7,%ebx
  sete   %dil
  call   <assert@plt>
  mov    0x0(%rbp),%rbx
  xor    %edi,%edi
  call   <CMSG_LEN@plt>
  mov    %rax,%r8
  mov    %rbx,%rax
  sub    %r8,%rax
  lea    0x7(%rax),%rbx
  cmovns %rax,%rbx
  sar    $0x3,%rbx
  sub    $0x1,%ebx
  cmp    $0xff,%ebx
  jg     L24
  mov    %rbp,%rdi
  add    %r15,%r13
  call   <CMSG_DATA@plt>
  sub    %r14,%r13
  mov    %ebx,(%rax)
  jmp    L9
  nopl   0x0(%rax)
L24:
  xor    %r13d,%r13d
L9:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
uint8_t *option_alloc(struct cmsghdr *cmsg, int datalen, int multx, int plusy) {
  if ((multx != 1 && multx != 2 && multx != 4 && multx != 8) || !(plusy >= 0 && plusy <= 7))
    return NULL;
  int dsize = cmsg->cmsg_len - CMSG_LEN(0);
  if (__glibc_unlikely(dsize == 0)) {
    cmsg->cmsg_len += sizeof(struct ip6_ext);
    dsize = sizeof(struct ip6_ext);
  }
  add_pad(cmsg, ((multx - (dsize & (multx - 1))) & (multx - 1)) + plusy);
  uint8_t *result = CMSG_DATA(cmsg) + cmsg->cmsg_len - CMSG_LEN(0);
  cmsg->cmsg_len += datalen;
  dsize = cmsg->cmsg_len - CMSG_LEN(0);
  add_pad(cmsg, (8 - (dsize & (8 - 1))) & (8 - 1));
  assert(((cmsg->cmsg_len - CMSG_LEN(0)) % 8) == 0);
  int len8b = (cmsg->cmsg_len - CMSG_LEN(0)) / 8 - 1;
  if (len8b >= 256)
    return NULL;
  struct ip6_ext *ie = (void *)CMSG_DATA(cmsg);
  ie->ip6e_len = len8b;
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rop3_228_0_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  movzbl -0x20(%rbp),%eax
  xor    -0x1c(%rbp),%al
  mov    %al,-0x3(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x3(%rbp),%al
  mov    %al,-0x2(%rbp)
  movzbl -0x1c(%rbp),%eax
  xor    -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  and    -0x1c(%rbp),%al
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  not    %eax
  mov    %eax,%edx
  movzbl -0x1(%rbp),%eax
  and    %edx,%eax
  or     %eax,%ecx
  mov    %ecx,%edx
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %eax,%ecx
  movzbl -0x20(%rbp),%eax
  and    %eax,%ecx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    %ecx,%eax
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_228_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = T ^ S;
  stk2 = *D & stk3;
  stk1 = S ^ stk2;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rop3_228_0_1>:
  endbr64
  mov    %edx,%ecx
  xor    %esi,%ecx
  and    (%rdi),%cl
  xor    %esi,%ecx
  mov    %edx,%eax
  not    %eax
  or     %esi,%eax
  and    %ecx,%eax
  not    %esi
  and    %edx,%esi
  and    (%rdi),%sil
  or     %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_228_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = T ^ S;
  stk2 = *D & stk3;
  stk1 = S ^ stk2;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<rop3_228_0_1>:
  endbr64
  mov    %edx,%ecx
  mov    %edx,%eax
  xor    %esi,%ecx
  not    %eax
  and    (%rdi),%cl
  xor    %esi,%ecx
  or     %esi,%eax
  not    %esi
  and    %ecx,%eax
  and    %edx,%esi
  and    (%rdi),%sil
  or     %esi,%eax
  mov    %al,(%rdi)
  ret
```

assistant: ```c
void rop3_228_0_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  stk3 = T ^ S;
  stk2 = *D & stk3;
  stk1 = S ^ stk2;
  *D = (stk1 & S) | (stk1 & (~T)) | (T & (~S) & *D);
}
```
user: What is the c source code of the assembly code below?

```asm
<call4029>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  call   *%rax
  nop
  leave
  ret
```

assistant: ```c
void call4029(void *arg) {
  void (*fn)(void) = arg;
  fn();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<call4029>:
  endbr64
  sub    $0x8,%rsp
  call   *%rdi
  add    $0x8,%rsp
  ret
```

assistant: ```c
void call4029(void *arg) {
  void (*fn)(void) = arg;
  fn();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<call4029>:
  endbr64
  jmp    *%rdi
```

assistant: ```c
void call4029(void *arg) {
  void (*fn)(void) = arg;
  fn();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<call4029>:
  endbr64
  jmp    *%rdi
```

assistant: ```c
void call4029(void *arg) {
  void (*fn)(void) = arg;
  fn();
}
```
user: What is the c source code of the assembly code below:

```asm
<countAccountId>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x28(%rbp)
  cmpq   $0x0,-0x38(%rbp)
  jne    L25
  mov    $0xffffffff,%eax
  jmp    L3
L25:
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    $0x40,%eax
  je     L5
  mov    $0xffffffff,%eax
  jmp    L3
L5:
  mov    -0x40(%rbp),%rax
  lea    D3(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L66
  mov    $0xffffffff,%eax
  jmp    L3
L86:
  mov    -0x20(%rbp),%rax
  mov    -0x38(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L66
  addl   $0x1,-0x28(%rbp)
L66:
  mov    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rcx
  lea    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <getline@plt>
  mov    %eax,-0x24(%rbp)
  cmpl   $0xffffffff,-0x24(%rbp)
  jne    L86
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    -0x28(%rbp),%eax
L3:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_infered: string

tool: D3: r

assistant: ```c
int countAccountId(const char *pAccountId, const char *pAccounts) {
  char *line = NULL;
  size_t len = 0;
  ssize_t read;
  FILE *fp;
  int i = 0;
  if (pAccountId == NULL) {
    return -1;
  }
  if (strlen(pAccountId) != 64) {
    return -1;
  }
  fp = fopen(pAccounts, "r");
  if (fp == NULL) {
    return -1;
  }
  while ((read = getline(&line, &len, fp)) != -1) {
    if (strstr(line, pAccountId) != NULL) {
      i++;
    }
  }
  free(line);
  fclose(fp);
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<countAccountId>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  test   %rdi,%rdi
  je     L34
  mov    %rdi,%r12
  mov    %rsi,%rbx
  call   <strlen@plt>
  cmp    $0x40,%eax
  jne    L43
  lea    D4(%rip),%rsi
  mov    %rbx,%rdi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L2
  mov    $0x0,%ebx
  lea    0x10(%rsp),%r13
  jmp    L25
L27:
  mov    %r12,%rsi
  mov    0x8(%rsp),%rdi
  call   <strstr@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffff,%ebx
L25:
  lea    0x8(%rsp),%rdi
  mov    %rbp,%rdx
  mov    %r13,%rsi
  call   <getline@plt>
  cmp    $0xffffffff,%eax
  jne    L27
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
L14:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L84
  mov    %ebx,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L34:
  mov    $0xffffffff,%ebx
  jmp    L14
L43:
  mov    $0xffffffff,%ebx
  jmp    L14
L2:
  mov    $0xffffffff,%ebx
  jmp    L14
L84:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D4
type_guessed: string

tool: D4: r

assistant: ```c
int countAccountId(const char *pAccountId, const char *pAccounts) {
  char *line = NULL;
  size_t len = 0;
  ssize_t read;
  FILE *fp;
  int i = 0;
  if (pAccountId == NULL) {
    return -1;
  }
  if (strlen(pAccountId) != 64) {
    return -1;
  }
  fp = fopen(pAccounts, "r");
  if (fp == NULL) {
    return -1;
  }
  while ((read = getline(&line, &len, fp)) != -1) {
    if (strstr(line, pAccountId) != NULL) {
      i++;
    }
  }
  free(line);
  fclose(fp);
  return i;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<countAccountId>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  test   %rdi,%rdi
  je     L28
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  cmp    $0x40,%eax
  jne    L28
  mov    %rbp,%rdi
  lea    D3(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L28
  xor    %r12d,%r12d
  lea    0x10(%rsp),%r14
  lea    0x8(%rsp),%r13
  jmp    L42
  nopl   (%rax)
L60:
  mov    0x8(%rsp),%rdi
  mov    %rbx,%rsi
  call   <strstr@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffff,%r12d
L42:
  mov    %rbp,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <getline@plt>
  cmp    $0xffffffff,%eax
  jne    L60
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
L24:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x20,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L28:
  mov    $0xffffffff,%r12d
  jmp    L24
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D3
type_infered: string

tool: D3: r

assistant: ```c
int countAccountId(const char *pAccountId, const char *pAccounts) {
  char *line = NULL;
  size_t len = 0;
  ssize_t read;
  FILE *fp;
  int i = 0;
  if (pAccountId == NULL) {
    return -1;
  }
  if (strlen(pAccountId) != 64) {
    return -1;
  }
  fp = fopen(pAccounts, "r");
  if (fp == NULL) {
    return -1;
  }
  while ((read = getline(&line, &len, fp)) != -1) {
    if (strstr(line, pAccountId) != NULL) {
      i++;
    }
  }
  free(line);
  fclose(fp);
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<countAccountId>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  test   %rdi,%rdi
  je     L67
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  cmp    $0x40,%eax
  jne    L67
  mov    %rbp,%rdi
  lea    D6(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L67
  xor    %r12d,%r12d
  lea    0x10(%rsp),%r14
  lea    0x8(%rsp),%r13
  jmp    L31
  nopl   (%rax)
L61:
  mov    0x8(%rsp),%rdi
  mov    %rbx,%rsi
  call   <strstr@plt>
  cmp    $0x1,%rax
  sbb    $0xffffffff,%r12d
L31:
  mov    %rbp,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <getline@plt>
  cmp    $0xffffffff,%eax
  jne    L61
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
L38:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x20,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L67:
  mov    $0xffffffff,%r12d
  jmp    L38
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D6
type_infered: string

tool: D6: r

assistant: ```c
int countAccountId(const char *pAccountId, const char *pAccounts) {
  char *line = NULL;
  size_t len = 0;
  ssize_t read;
  FILE *fp;
  int i = 0;
  if (pAccountId == NULL) {
    return -1;
  }
  if (strlen(pAccountId) != 64) {
    return -1;
  }
  fp = fopen(pAccounts, "r");
  if (fp == NULL) {
    return -1;
  }
  while ((read = getline(&line, &len, fp)) != -1) {
    if (strstr(line, pAccountId) != NULL) {
      i++;
    }
  }
  free(line);
  fclose(fp);
  return i;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D8
data_type: string
name: retrieve_data
arguments: label: D28
data_type: string
name: retrieve_data
arguments: label: D15
data_type: string

tool: D8: Usage: ucwiki [OPTIONS]


tool: D28:        -i	: input directory


tool: D15:        -o	: output directory


assistant: ```c
void usage() {
  printf("Usage: ucwiki [OPTIONS]\n");
  printf("       -i\t: input directory\n");
  printf("       -o\t: output directory\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    D27(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
type: string
name: parse_data
arguments: data_tag: D25
type: string
name: parse_data
arguments: data_tag: D27
type: string

tool: D2: Usage: ucwiki [OPTIONS]


tool: D25:        -i	: input directory


tool: D27:        -o	: output directory


assistant: ```c
void usage() {
  printf("Usage: ucwiki [OPTIONS]\n");
  printf("       -i\t: input directory\n");
  printf("       -o\t: output directory\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D29(%rip),%rdi
  call   <printf@plt>
  lea    D14(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D29
datatype: string
name: disassemble_data
arguments: data_tag: D14
datatype: string

tool: D24: Usage: ucwiki [OPTIONS]


tool: D29:        -i	: input directory


tool: D14:        -o	: output directory


assistant: ```c
void usage() {
  printf("Usage: ucwiki [OPTIONS]\n");
  printf("       -i\t: input directory\n");
  printf("       -o\t: output directory\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D0(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: tag: D8
type_guessed: string
name: parse_data
arguments: tag: D0
type_guessed: string
name: parse_data
arguments: tag: D11
type_guessed: string

tool: D8: Usage: ucwiki [OPTIONS]


tool: D0:        -i	: input directory


tool: D11:        -o	: output directory


assistant: ```c
void usage() {
  printf("Usage: ucwiki [OPTIONS]\n");
  printf("       -i\t: input directory\n");
  printf("       -o\t: output directory\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getent>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int getent(char *cp, char *name) {
  (void)cp;
  (void)name;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getent>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int getent(char *cp, char *name) {
  (void)cp;
  (void)name;
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<getent>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int getent(char *cp, char *name) {
  (void)cp;
  (void)name;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getent>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int getent(char *cp, char *name) {
  (void)cp;
  (void)name;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L16
L22:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  lea    -0xc(%rbp),%rcx
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rsi
  mov    %eax,%edi
  call   <decomposeIntoSumThreeSquare@plt>
  test   %rax,%rax
  je     L28
  mov    -0xc(%rbp),%ecx
  mov    -0x10(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L26
L28:
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L26:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
L16:
  mov    -0x1c(%rbp),%eax
  test   %eax,%eax
  jne    L22
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L44
  call   <__stack_chk_fail@plt>
L44:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D12
type_infered: string
name: disassemble_data
arguments: label: D20
type_infered: string
name: disassemble_data
arguments: label: D4
type_infered: string

tool: D12: %d

tool: D20: %d %d %d


tool: D4: -1

assistant: ```c
int main(void) {
  int T, N, A, B, C;
  scanf("%d", &T);
  while (T) {
    scanf("%d", &N);
    if (decomposeIntoSumThreeSquare(N, &A, &B, &C))
      printf("%d %d %d\n", A, B, C);
    else
      printf("-1\n");
    T--;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D13(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x4(%rsp)
  je     L31
  lea    0x8(%rsp),%rbp
  lea    D13(%rip),%rbx
  lea    D0(%rip),%r13
  lea    D39(%rip),%r12
  jmp    L30
L13:
  mov    %r13,%rdi
  call   <puts@plt>
L19:
  mov    0x4(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x4(%rsp)
  je     L31
L30:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    0xc(%rsp),%rsi
  mov    0x8(%rsp),%edi
  call   <decomposeIntoSumThreeSquare@plt>
  test   %rax,%rax
  je     L13
  mov    0x14(%rsp),%ecx
  mov    0x10(%rsp),%edx
  mov    0xc(%rsp),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L19
L31:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L33:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D13
type_infered: string
name: unpack_data
arguments: label: D0
type_infered: string
name: unpack_data
arguments: label: D39
type_infered: string

tool: D13: %d

tool: D0: -1

tool: D39: %d %d %d


assistant: ```c
int main(void) {
  int T, N, A, B, C;
  scanf("%d", &T);
  while (T) {
    scanf("%d", &N);
    if (decomposeIntoSumThreeSquare(N, &A, &B, &C))
      printf("%d %d %d\n", A, B, C);
    else
      printf("-1\n");
    T--;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D17(%rip),%rbx
  mov    %rbx,%rdi
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  je     L1
  lea    0x8(%rsp),%r14
  lea    0x14(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    0xc(%rsp),%rbp
  lea    D13(%rip),%r15
  jmp    L5
  nopl   0x0(%rax,%rax,1)
L20:
  mov    0x14(%rsp),%ecx
  mov    0x10(%rsp),%edx
  xor    %eax,%eax
  lea    D26(%rip),%rdi
  mov    0xc(%rsp),%esi
  call   <printf@plt>
  subl   $0x1,0x4(%rsp)
  je     L1
L5:
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%edi
  mov    %r13,%rcx
  mov    %r12,%rdx
  mov    %rbp,%rsi
  call   <decomposeIntoSumThreeSquare@plt>
  test   %rax,%rax
  jne    L20
  mov    %r15,%rdi
  call   <puts@plt>
  subl   $0x1,0x4(%rsp)
  jne    L5
L1:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D17
datatype: string
name: extract_symbolic_data
arguments: label: D13
datatype: string
name: extract_symbolic_data
arguments: label: D26
datatype: string

tool: D17: %d

tool: D13: -1

tool: D26: %d %d %d


assistant: ```c
int main(void) {
  int T, N, A, B, C;
  scanf("%d", &T);
  while (T) {
    scanf("%d", &N);
    if (decomposeIntoSumThreeSquare(N, &A, &B, &C))
      printf("%d %d %d\n", A, B, C);
    else
      printf("-1\n");
    T--;
  }
  return 0;
}
```