ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<swap32>:
  endbr64
  mov    %edi,%eax
  shl    $0x8,%edi
  shr    $0x8,%eax
  and    $0xff00ff00,%edi
  and    $0xff00ff,%eax
  or     %eax,%edi
  mov    %edi,%eax
  sar    $0x10,%edi
  shl    $0x10,%eax
  or     %edi,%eax
  ret
```

assistant: ```c
uint32_t swap32(uint32_t x) {
  x = ((x & 0x00ff00ff) << 8) | ((x & 0xff00ff00) >> 8);
  return (x << 16) | (x >> 16);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<swap32>:
  endbr64
  mov    %edi,%eax
  shl    $0x8,%edi
  shr    $0x8,%eax
  and    $0xff00ff00,%edi
  and    $0xff00ff,%eax
  or     %eax,%edi
  mov    %edi,%eax
  sar    $0x10,%edi
  shl    $0x10,%eax
  or     %edi,%eax
  ret
```

assistant: ```c
uint32_t swap32(uint32_t x) {
  x = ((x & 0x00ff00ff) << 8) | ((x & 0xff00ff00) >> 8);
  return (x << 16) | (x >> 16);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jle    L17
  mov    $0x0,%eax
  call   <print_test_data@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L11
L17:
  mov    $0x0,%eax
  call   <test@plt>
  mov    %eax,-0x4(%rbp)
L11:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int ret;
  if (argc > 1)
    ret = print_test_data();
  else
    ret = test();
  exit(ret);
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L18
  mov    $0x0,%eax
  call   <print_test_data@plt>
  mov    %eax,%edi
L16:
  call   <exit@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L18:
  mov    $0x0,%eax
  call   <test@plt>
  mov    %eax,%edi
  jmp    L16
```

assistant: ```c
int main(int argc, char *argv[]) {
  int ret;
  if (argc > 1)
    ret = print_test_data();
  else
    ret = test();
  exit(ret);
  return (0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L12
  call   <print_test_data@plt>
  mov    %eax,%edi
L9:
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L12:
  call   <test@plt>
  mov    %eax,%edi
  jmp    L9
```

assistant: ```c
int main(int argc, char *argv[]) {
  int ret;
  if (argc > 1)
    ret = print_test_data();
  else
    ret = test();
  exit(ret);
  return (0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L17
  call   <print_test_data@plt>
  mov    %eax,%edi
L7:
  call   <exit@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L17:
  call   <test@plt>
  mov    %eax,%edi
  jmp    L7
```

assistant: ```c
int main(int argc, char *argv[]) {
  int ret;
  if (argc > 1)
    ret = print_test_data();
  else
    ret = test();
  exit(ret);
  return (0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<heapBased1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movb   $0x61,(%rax)
  mov    -0x8(%rbp),%rax
  movzbl 0x1(%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movzbl -0x9(%rbp),%eax
  leave
  ret
```

assistant: ```c
char heapBased1() {
  char *buf = malloc(2);
  buf[0] = 'a';
  char result = buf[1];
  free(buf);
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<heapBased1>:
  endbr64
  push   %rbx
  mov    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  movb   $0x61,(%rax)
  movzbl 0x1(%rax),%ebx
  call   <free@plt>
  mov    %ebx,%eax
  pop    %rbx
  ret
```

assistant: ```c
char heapBased1() {
  char *buf = malloc(2);
  buf[0] = 'a';
  char result = buf[1];
  free(buf);
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<heapBased1>:
  endbr64
  push   %r12
  mov    $0x2,%edi
  call   <malloc@plt>
  movb   $0x61,(%rax)
  movzbl 0x1(%rax),%r12d
  mov    %rax,%rdi
  call   <free@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
char heapBased1() {
  char *buf = malloc(2);
  buf[0] = 'a';
  char result = buf[1];
  free(buf);
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<heapBased1>:
  endbr64
  push   %r12
  mov    $0x2,%edi
  call   <malloc@plt>
  movb   $0x61,(%rax)
  movzbl 0x1(%rax),%r12d
  mov    %rax,%rdi
  call   <free@plt>
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
char heapBased1() {
  char *buf = malloc(2);
  buf[0] = 'a';
  char result = buf[1];
  free(buf);
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<freeQueue>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x4,%edx
  mov    $0x4,%esi
  mov    %rax,%rdi
  call   <compressQueue@plt>
  mov    -0x10(%rbp),%rax
  movl   $0xffffffff,(%rax)
  mov    -0x18(%rbp),%rax
  movl   $0xffffffff,(%rax)
  mov    $0x2,%edi
  call   <showSucess@plt>
  nop
  leave
  ret
```

assistant: ```c
void freeQueue(int *queue, int *begin, int *end) {
  compressQueue(queue, (5 - 1), (5 - 1));
  *begin = -1;
  *end = -1;
  showSucess(2);
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<freeQueue>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    $0x4,%edx
  mov    $0x4,%esi
  call   <compressQueue@plt>
  movl   $0xffffffff,0x0(%rbp)
  movl   $0xffffffff,(%rbx)
  mov    $0x2,%edi
  call   <showSucess@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void freeQueue(int *queue, int *begin, int *end) {
  compressQueue(queue, (5 - 1), (5 - 1));
  *begin = -1;
  *end = -1;
  showSucess(2);
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<freeQueue>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x4,%esi
  push   %rbx
  mov    %rdx,%rbx
  mov    $0x4,%edx
  sub    $0x8,%rsp
  call   <compressQueue@plt>
  movl   $0xffffffff,0x0(%rbp)
  mov    $0x2,%edi
  movl   $0xffffffff,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  jmp    1070 <showSucess@plt>
```

assistant: ```c
void freeQueue(int *queue, int *begin, int *end) {
  compressQueue(queue, (5 - 1), (5 - 1));
  *begin = -1;
  *end = -1;
  showSucess(2);
  return;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<freeQueue>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x4,%esi
  push   %rbx
  mov    %rdx,%rbx
  mov    $0x4,%edx
  sub    $0x8,%rsp
  call   <compressQueue@plt>
  movl   $0xffffffff,0x0(%rbp)
  mov    $0x2,%edi
  movl   $0xffffffff,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  jmp    1070 <showSucess@plt>
```

assistant: ```c
void freeQueue(int *queue, int *begin, int *end) {
  compressQueue(queue, (5 - 1), (5 - 1));
  *begin = -1;
  *end = -1;
  showSucess(2);
  return;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<divide_media_token>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x70,%rsp
  mov    %rsi,%r14
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    D10(%rip),%rsi
  call   <strstr@plt>
  mov    $0x0,%r12d
  test   %rax,%rax
  je     L37
  mov    %rax,%rbx
  lea    D10(%rip),%rdi
  call   <strlen@plt>
  cltq
  lea    (%rbx,%rax,1),%rsi
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <strcpy@plt>
  movb   $0x0,0x63(%rsp)
  movb   $0x0,0x62(%rsp)
  movb   $0x0,0x61(%rsp)
  mov    $0x20,%esi
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L4
  movb   $0x0,(%rax)
L4:
  mov    %rsp,%rdi
  mov    $0x9,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L73
  movb   $0x0,(%rax)
L73:
  mov    %rsp,%rdi
  mov    $0xa,%esi
  call   <strchr@plt>
  test   %rax,%rax
  je     L25
  movb   $0x0,(%rax)
L25:
  mov    %r14,%r13
  mov    $0x0,%r12d
  mov    %rsp,%rbp
  jmp    L59
L26:
  movb   $0x0,(%rax)
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <strcpy@plt>
  add    $0x1,%r12d
  lea    0x1(%rbx),%rbp
  add    $0x1e,%r13
  cmp    $0x5,%r12d
  je     L37
L59:
  cmpb   $0x0,0x0(%rbp)
  je     L37
  mov    $0x2c,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L26
  movslq %r12d,%rdx
  mov    %rdx,%rax
  shl    $0x4,%rax
  sub    %rdx,%rax
  lea    (%r14,%rax,2),%rdi
  mov    %rbp,%rsi
  call   <strcpy@plt>
  add    $0x1,%r12d
L37:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  mov    %r12d,%eax
  add    $0x70,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D10
type_guessed: string

tool: D10: media=

assistant: ```c
int divide_media_token(char *input, char output[5][30]) {
  char media_command[100];
  char *media, *p, *pp;
  int i;
  p = strstr(input, "media=");
  if (p) {
    strcpy(media_command, p + strlen("media="));
  } else {
    return 0;
  }
  media_command[sizeof(media_command) - 1] = 0;
  media_command[sizeof(media_command) - 2] = 0;
  media_command[sizeof(media_command) - 3] = 0;
  p = strchr(media_command, ' ');
  if (p)
    *p = 0;
  p = strchr(media_command, '\t');
  if (p)
    *p = 0;
  p = strchr(media_command, '\n');
  if (p)
    *p = 0;
  p = media_command;
  for (i = 0; i < 5;) {
    if (*p == 0)
      break;
    pp = strchr(p, ',');
    if (pp)
      *pp = 0;
    strcpy(output[i], p);
    i++;
    if (pp == NULL)
      break;
    p = pp + 1;
  }
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<divide_media_token>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  mov    %rbp,%rsi
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  call   <strstr@plt>
  test   %rax,%rax
  je     L83
  mov    %rbp,%rdi
  mov    %rax,%rbx
  mov    %rsp,%rbp
  call   <strlen@plt>
  mov    %rbp,%rdi
  cltq
  lea    (%rbx,%rax,1),%rsi
  call   <strcpy@plt>
  xor    %eax,%eax
  mov    $0x20,%esi
  mov    %rbp,%rdi
  movb   $0x0,0x61(%rsp)
  mov    %ax,0x62(%rsp)
  call   <strchr@plt>
  test   %rax,%rax
  je     L82
  movb   $0x0,(%rax)
L82:
  mov    $0x9,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L34
  movb   $0x0,(%rax)
L34:
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L41
  movb   $0x0,(%rax)
L41:
  xor    %r13d,%r13d
L74:
  cmpb   $0x0,0x0(%rbp)
  je     L83
  mov    $0x2c,%esi
  mov    %rbp,%rdi
  add    $0x1,%r13d
  call   <strchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L5
  movb   $0x0,(%rax)
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    0x1(%rbx),%rbp
  add    $0x1e,%r12
  call   <strcpy@plt>
  cmp    $0x5,%r13d
  jne    L74
L83:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x78,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L5:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  jmp    L83
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D7
data_type: string

tool: D7: media=

assistant: ```c
int divide_media_token(char *input, char output[5][30]) {
  char media_command[100];
  char *media, *p, *pp;
  int i;
  p = strstr(input, "media=");
  if (p) {
    strcpy(media_command, p + strlen("media="));
  } else {
    return 0;
  }
  media_command[sizeof(media_command) - 1] = 0;
  media_command[sizeof(media_command) - 2] = 0;
  media_command[sizeof(media_command) - 3] = 0;
  p = strchr(media_command, ' ');
  if (p)
    *p = 0;
  p = strchr(media_command, '\t');
  if (p)
    *p = 0;
  p = strchr(media_command, '\n');
  if (p)
    *p = 0;
  p = media_command;
  for (i = 0; i < 5;) {
    if (*p == 0)
      break;
    pp = strchr(p, ',');
    if (pp)
      *pp = 0;
    strcpy(output[i], p);
    i++;
    if (pp == NULL)
      break;
    p = pp + 1;
  }
  return i;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<divide_media_token>:
  endbr64
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  mov    %rbp,%rsi
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  call   <strstr@plt>
  test   %rax,%rax
  je     L38
  mov    %rbp,%rdi
  mov    %rax,%rbx
  mov    %rsp,%rbp
  call   <strlen@plt>
  mov    %rbp,%rdi
  cltq
  lea    (%rbx,%rax,1),%rsi
  call   <strcpy@plt>
  xor    %eax,%eax
  mov    $0x20,%esi
  mov    %rbp,%rdi
  movb   $0x0,0x61(%rsp)
  mov    %ax,0x62(%rsp)
  call   <strchr@plt>
  test   %rax,%rax
  je     L49
  movb   $0x0,(%rax)
L49:
  mov    $0x9,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L57
  movb   $0x0,(%rax)
L57:
  mov    $0xa,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L61
  movb   $0x0,(%rax)
L61:
  xor    %r13d,%r13d
L76:
  cmpb   $0x0,0x0(%rbp)
  je     L38
  mov    $0x2c,%esi
  mov    %rbp,%rdi
  add    $0x1,%r13d
  call   <strchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L47
  movb   $0x0,(%rax)
  mov    %rbp,%rsi
  mov    %r12,%rdi
  lea    0x1(%rbx),%rbp
  add    $0x1e,%r12
  call   <strcpy@plt>
  cmp    $0x5,%r13d
  jne    L76
L38:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0x78,%rsp
  mov    %r13d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L47:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  jmp    L38
L45:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D7
type_guessed: string

tool: D7: media=

assistant: ```c
int divide_media_token(char *input, char output[5][30]) {
  char media_command[100];
  char *media, *p, *pp;
  int i;
  p = strstr(input, "media=");
  if (p) {
    strcpy(media_command, p + strlen("media="));
  } else {
    return 0;
  }
  media_command[sizeof(media_command) - 1] = 0;
  media_command[sizeof(media_command) - 2] = 0;
  media_command[sizeof(media_command) - 3] = 0;
  p = strchr(media_command, ' ');
  if (p)
    *p = 0;
  p = strchr(media_command, '\t');
  if (p)
    *p = 0;
  p = strchr(media_command, '\n');
  if (p)
    *p = 0;
  p = media_command;
  for (i = 0; i < 5;) {
    if (*p == 0)
      break;
    pp = strchr(p, ',');
    if (pp)
      *pp = 0;
    strcpy(output[i], p);
    i++;
    if (pp == NULL)
      break;
    p = pp + 1;
  }
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x2c(%rbp),%rax
  mov    %rax,%rsi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  jmp    L5
L19:
  lea    -0x20(%rbp),%rdx
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x28(%rbp),%rax
  mov    -0x20(%rbp),%rcx
  cqto
  idiv   %rcx
  mov    %rdx,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L5:
  mov    -0x2c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x2c(%rbp)
  test   %eax,%eax
  jg     L19
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D20
type_guessed: string
name: retrieve_data
arguments: tag: D9
type_guessed: string

tool: D20: %d


tool: D9: %lld %lld


assistant: ```c
int main() {
  int T;
  long long x, k;
  long long p, q;
  scanf("%d\n", &T);
  while (T-- > 0) {
    scanf("%lld %lld\n", &x, &k);
    q = x % k;
    p = k - q;
    printf("%lld %lld\n", p, q);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D16(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jle    L7
  lea    D11(%rip),%rbx
L4:
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x10(%rsp),%rsi
  mov    0x8(%rsp),%rax
  cqto
  idiv   %rsi
  sub    %rdx,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jg     L4
L7:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D16
type_guessed: string
name: disassemble_data
arguments: data_label: D11
type_guessed: string

tool: D16: %d


tool: D11: %lld %lld


assistant: ```c
int main() {
  int T;
  long long x, k;
  long long p, q;
  scanf("%d\n", &T);
  while (T-- > 0) {
    scanf("%lld %lld\n", &x, &k);
    q = x % k;
    p = k - q;
    printf("%lld %lld\n", p, q);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D19(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jle    L26
  lea    0x10(%rsp),%r12
  lea    0x8(%rsp),%rbp
  lea    D9(%rip),%rbx
  nopl   (%rax)
L21:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  mov    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  cqto
  idiv   %rsi
  sub    %rdx,%rsi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jg     L21
L26:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D19
type: string
name: disassemble_data
arguments: tag: D9
type: string

tool: D19: %d


tool: D9: %lld %lld


assistant: ```c
int main() {
  int T;
  long long x, k;
  long long p, q;
  scanf("%d\n", &T);
  while (T-- > 0) {
    scanf("%lld %lld\n", &x, &k);
    q = x % k;
    p = k - q;
    printf("%lld %lld\n", p, q);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  lea    D13(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jle    L1
  lea    0x10(%rsp),%r12
  lea    0x8(%rsp),%rbp
  lea    D17(%rip),%rbx
  nopl   (%rax)
L3:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  mov    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  cqto
  idiv   %rsi
  sub    %rdx,%rsi
  call   <printf@plt>
  mov    0x4(%rsp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,0x4(%rsp)
  test   %eax,%eax
  jg     L3
L1:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D13
type_infered: string
name: parse_data
arguments: data_label: D17
type_infered: string

tool: D13: %d


tool: D17: %lld %lld


assistant: ```c
int main() {
  int T;
  long long x, k;
  long long p, q;
  scanf("%d\n", &T);
  while (T-- > 0) {
    scanf("%lld %lld\n", &x, &k);
    q = x % k;
    p = k - q;
    printf("%lld %lld\n", p, q);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dagutil_lockallstreams>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L16
L6:
  mov    -0x4(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <dagutil_lockstream@plt>
  test   %rax,%rax
  je     L37
  mov    $0xffffffff,%eax
  jmp    L36
L37:
  addl   $0x1,-0x4(%rbp)
L16:
  cmpl   $0x3f,-0x4(%rbp)
  jle    L6
  mov    $0x0,%eax
L36:
  leave
  ret
```

assistant: ```c
int dagutil_lockallstreams(int dagfd) {
  int stream;
  for (stream = 0; stream < 64; stream++) {
    if (0 != dagutil_lockstream(dagfd, stream)) {
      return -1;
    }
  }
  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
<dagutil_lockallstreams>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    $0x0,%ebx
L2:
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <dagutil_lockstream@plt>
  test   %rax,%rax
  jne    L19
  add    $0x1,%ebx
  cmp    $0x40,%ebx
  jne    L2
  jmp    L18
L19:
  mov    $0xffffffff,%eax
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int dagutil_lockallstreams(int dagfd) {
  int stream;
  for (stream = 0; stream < 64; stream++) {
    if (0 != dagutil_lockstream(dagfd, stream)) {
      return -1;
    }
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dagutil_lockallstreams>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L18
L16:
  add    $0x1,%ebx
  cmp    $0x40,%ebx
  je     L6
L18:
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <dagutil_lockstream@plt>
  test   %rax,%rax
  je     L16
  mov    $0xffffffff,%eax
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int dagutil_lockallstreams(int dagfd) {
  int stream;
  for (stream = 0; stream < 64; stream++) {
    if (0 != dagutil_lockstream(dagfd, stream)) {
      return -1;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dagutil_lockallstreams>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  jmp    L10
L17:
  add    $0x1,%ebx
  cmp    $0x40,%ebx
  je     L18
L10:
  mov    %ebx,%esi
  mov    %ebp,%edi
  call   <dagutil_lockstream@plt>
  test   %rax,%rax
  je     L17
  mov    $0xffffffff,%eax
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int dagutil_lockallstreams(int dagfd) {
  int stream;
  for (stream = 0; stream < 64; stream++) {
    if (0 != dagutil_lockstream(dagfd, stream)) {
      return -1;
    }
  }
  return 0;
}
```