ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  xor    %esi,%esi
  xor    %r12d,%r12d
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rdi
  call   <gettimeofday@plt>
  nop
L9:
  mov    %rbx,%rdi
  mov    $0x2,%esi
  add    $0x1,%rbx
  call   <pow@plt>
  add    %rax,%r12
  cmp    $0x65,%rbx
  jne    L9
  mov    $0x2,%esi
  mov    $0x13ba,%edi
  call   <pow@plt>
  mov    %r12,%rdx
  lea    D8(%rip),%rdi
  mov    %rax,%rcx
  mov    %rax,%rsi
  xor    %eax,%eax
  sub    %r12,%rcx
  call   <printf@plt>
  xor    %esi,%esi
  lea    0x8(%rsp),%rdi
  call   <gettimeofday@plt>
  mov    0x8(%rsp),%rsi
  xor    %eax,%eax
  sub    0x10(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
datatype: string
name: retrieve_data
arguments: data_tag: D3
datatype: string

tool: D8: %lu - %lu = %lu

tool: D3: 
took %lu usec


assistant: ```c
int main(void) {
  struct timeval stop, start;
  gettimeofday(&start, NULL);
  int i;
  long sum = 0, squared = 0, squaredsum = 0;
  for (i = 1; i <= 100; i++) {
    sum += i;
    squared += pow(i, 2);
  }
  squaredsum = pow(sum, 2);
  printf("%lu - %lu = %lu", squaredsum, squared, squaredsum - squared);
  gettimeofday(&stop, NULL);
  printf("\ntook %lu usec\n", stop.tv_usec - start.tv_usec);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cpucycles_x86cpuspeed>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  rdtsc
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long long cpucycles_x86cpuspeed(void) {
  long long result;
  asm volatile(".byte 15;.byte 49" : "=A"(result));
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cpucycles_x86cpuspeed>:
  endbr64
  rdtsc
  ret
```

assistant: ```c
long long cpucycles_x86cpuspeed(void) {
  long long result;
  asm volatile(".byte 15;.byte 49" : "=A"(result));
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cpucycles_x86cpuspeed>:
  endbr64
  rdtsc
  ret
```

assistant: ```c
long long cpucycles_x86cpuspeed(void) {
  long long result;
  asm volatile(".byte 15;.byte 49" : "=A"(result));
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<cpucycles_x86cpuspeed>:
  endbr64
  rdtsc
  ret
```

assistant: ```c
long long cpucycles_x86cpuspeed(void) {
  long long result;
  asm volatile(".byte 15;.byte 49" : "=A"(result));
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<gf2_matrix_times>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L28
L29:
  mov    -0x1c(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L14
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  xor    %eax,-0x4(%rbp)
L14:
  sarl   -0x1c(%rbp)
  addq   $0x4,-0x18(%rbp)
L28:
  cmpl   $0x0,-0x1c(%rbp)
  jne    L29
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t gf2_matrix_times(uint32_t *mat, uint32_t vec) {
  uint32_t sum;
  sum = 0;
  while (vec) {
    if (vec & 1)
      sum ^= *mat;
    vec >>= 1;
    mat++;
  }
  return (sum);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<gf2_matrix_times>:
  endbr64
  test   %esi,%esi
  je     L13
  mov    $0x0,%eax
  jmp    L7
L26:
  add    $0x4,%rdi
  sar    %esi
  je     L5
L7:
  test   $0x1,%sil
  je     L26
  xor    (%rdi),%eax
  jmp    L26
L5:
  ret
L13:
  mov    %esi,%eax
  ret
```

assistant: ```c
uint32_t gf2_matrix_times(uint32_t *mat, uint32_t vec) {
  uint32_t sum;
  sum = 0;
  while (vec) {
    if (vec & 1)
      sum ^= *mat;
    vec >>= 1;
    mat++;
  }
  return (sum);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gf2_matrix_times>:
  endbr64
  xor    %eax,%eax
  test   %esi,%esi
  je     L9
  nopw   0x0(%rax,%rax,1)
L28:
  test   $0x1,%sil
  je     L29
  xor    (%rdi),%eax
L29:
  add    $0x4,%rdi
  sar    %esi
  jne    L28
  ret
  nopl   0x0(%rax)
L9:
  ret
```

assistant: ```c
uint32_t gf2_matrix_times(uint32_t *mat, uint32_t vec) {
  uint32_t sum;
  sum = 0;
  while (vec) {
    if (vec & 1)
      sum ^= *mat;
    vec >>= 1;
    mat++;
  }
  return (sum);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<gf2_matrix_times>:
  endbr64
  xor    %eax,%eax
  test   %esi,%esi
  je     L19
  nopw   0x0(%rax,%rax,1)
L6:
  test   $0x1,%sil
  je     L21
  xor    (%rdi),%eax
L21:
  add    $0x4,%rdi
  sar    %esi
  jne    L6
  ret
  nopl   0x0(%rax)
L19:
  ret
```

assistant: ```c
uint32_t gf2_matrix_times(uint32_t *mat, uint32_t vec) {
  uint32_t sum;
  sum = 0;
  while (vec) {
    if (vec & 1)
      sum ^= *mat;
    vec >>= 1;
    mat++;
  }
  return (sum);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <ParseOptions@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  ParseOptions(argc, argv);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <ParseOptions@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  ParseOptions(argc, argv);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <ParseOptions@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  ParseOptions(argc, argv);
  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
  call   <ParseOptions@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  ParseOptions(argc, argv);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x4d2,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D2
type_guessed: string

tool: D2: %zu

assistant: ```c
int main(int argc, char **argv) {
  size_t i;
  i = 1234;
  printf("%zu", i);
  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
  mov    $0x4d2,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type_infered: string

tool: D8: %zu

assistant: ```c
int main(int argc, char **argv) {
  size_t i;
  i = 1234;
  printf("%zu", i);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4d2,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D7: %zu

assistant: ```c
int main(int argc, char **argv) {
  size_t i;
  i = 1234;
  printf("%zu", i);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4d2,%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
data_type: string

tool: D2: %zu

assistant: ```c
int main(int argc, char **argv) {
  size_t i;
  i = 1234;
  printf("%zu", i);
  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
<string_iconv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  jne    L19
  mov    $0x0,%eax
  jmp    L10
L19:
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
L10:
  leave
  ret
```

assistant: ```c
char *string_iconv(int from_utf8, const char *from_code, const char *to_code, const char *string) {
  char *outbuf;
  if (!string)
    return NULL;
  (void)from_utf8;
  (void)from_code;
  (void)to_code;
  outbuf = strdup(string);
  return outbuf;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<string_iconv>:
  endbr64
  mov    %rcx,%rdi
  test   %rcx,%rcx
  je     L1
  sub    $0x8,%rsp
  call   <strdup@plt>
  add    $0x8,%rsp
  ret
L1:
  mov    %rcx,%rax
  ret
```

assistant: ```c
char *string_iconv(int from_utf8, const char *from_code, const char *to_code, const char *string) {
  char *outbuf;
  if (!string)
    return NULL;
  (void)from_utf8;
  (void)from_code;
  (void)to_code;
  outbuf = strdup(string);
  return outbuf;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<string_iconv>:
  endbr64
  mov    %rcx,%rdi
  test   %rcx,%rcx
  je     L7
  jmp    1050 <strdup@plt>
  nopl   0x0(%rax)
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *string_iconv(int from_utf8, const char *from_code, const char *to_code, const char *string) {
  char *outbuf;
  if (!string)
    return NULL;
  (void)from_utf8;
  (void)from_code;
  (void)to_code;
  outbuf = strdup(string);
  return outbuf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<string_iconv>:
  endbr64
  mov    %rcx,%rdi
  test   %rcx,%rcx
  je     L1
  jmp    1050 <strdup@plt>
  nopl   0x0(%rax)
L1:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *string_iconv(int from_utf8, const char *from_code, const char *to_code, const char *string) {
  char *outbuf;
  if (!string)
    return NULL;
  (void)from_utf8;
  (void)from_code;
  (void)to_code;
  outbuf = strdup(string);
  return outbuf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_se>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    (%rax),%esi
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <se@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  add    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    -0x10(%rbp),%ecx
  sub    %ecx,%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int read_se(const unsigned char *buffer, int *bits_offset, int *total_bits) {
  int val, bits;
  val = se(buffer, *bits_offset, *total_bits, &bits);
  *bits_offset += bits;
  *total_bits -= bits;
  return val;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<read_se>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rcx
  mov    (%rdx),%edx
  mov    (%rsi),%esi
  call   <se@plt>
  mov    0x4(%rsp),%edx
  add    %edx,0x0(%rbp)
  sub    %edx,(%rbx)
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_se(const unsigned char *buffer, int *bits_offset, int *total_bits) {
  int val, bits;
  val = se(buffer, *bits_offset, *total_bits, &bits);
  *bits_offset += bits;
  *total_bits -= bits;
  return val;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_se>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  mov    (%rdx),%edx
  mov    (%rsi),%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rcx
  call   <se@plt>
  mov    0x4(%rsp),%edx
  add    %edx,0x0(%rbp)
  sub    %edx,(%rbx)
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_se(const unsigned char *buffer, int *bits_offset, int *total_bits) {
  int val, bits;
  val = se(buffer, *bits_offset, *total_bits, &bits);
  *bits_offset += bits;
  *total_bits -= bits;
  return val;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_se>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdx,%rbx
  sub    $0x18,%rsp
  mov    (%rdx),%edx
  mov    (%rsi),%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rcx
  call   <se@plt>
  mov    0x4(%rsp),%edx
  add    %edx,0x0(%rbp)
  sub    %edx,(%rbx)
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int read_se(const unsigned char *buffer, int *bits_offset, int *total_bits) {
  int val, bits;
  val = se(buffer, *bits_offset, *total_bits, &bits);
  *bits_offset += bits;
  *total_bits -= bits;
  return val;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_bigendian>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D9(%rip),%eax
  cmp    $0x1020304,%eax
  sete   %al
  movzbl %al,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D9
type_infered: byte[4]

tool: {"D9": [0x01,0x00,0x00,0x00]}

assistant: ```c
int is_bigendian() {
  static const union {
    uint8_t u8[4];
    uint32_t u32;
  } byte_order __attribute__((aligned(4))) = {.u8 = {1, 2, 3, 4}};
  return byte_order.u32 == 0x01020304;
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
int is_bigendian() {
  static const union {
    uint8_t u8[4];
    uint32_t u32;
  } byte_order __attribute__((aligned(4))) = {.u8 = {1, 2, 3, 4}};
  return byte_order.u32 == 0x01020304;
}
```
user: 请将以下的汇编代码转换成C语言代码:

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

assistant: ```c
int is_bigendian() {
  static const union {
    uint8_t u8[4];
    uint32_t u32;
  } byte_order __attribute__((aligned(4))) = {.u8 = {1, 2, 3, 4}};
  return byte_order.u32 == 0x01020304;
}
```
user: What is the c source code of the assembly code below?

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

assistant: ```c
int is_bigendian() {
  static const union {
    uint8_t u8[4];
    uint32_t u32;
  } byte_order __attribute__((aligned(4))) = {.u8 = {1, 2, 3, 4}};
  return byte_order.u32 == 0x01020304;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<isr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    $0x0,%eax
  call   <irq_pending@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <irq_getmask@plt>
  and    %ebx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x8,%eax
  test   %eax,%eax
  je     L22
  mov    $0x0,%eax
  call   <uart_isr@plt>
L22:
  mov    -0x14(%rbp),%eax
  and    $0x20,%eax
  test   %eax,%eax
  je     L2
  mov    $0x0,%eax
  call   <timer0_isr@plt>
L2:
  mov    -0x14(%rbp),%eax
  and    $0x40,%eax
  test   %eax,%eax
  je     L24
  mov    $0x0,%eax
  call   <timer1_isr@plt>
L24:
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void isr(void) {
  unsigned int irqs;
  irqs = irq_pending() & irq_getmask();
  if (irqs & (0x00000008))
    uart_isr();
  if (irqs & (0x00000020))
    timer0_isr();
  if (irqs & (0x00000040))
    timer1_isr();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<isr>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <irq_pending@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <irq_getmask@plt>
  and    %eax,%ebx
  test   $0x8,%bl
  jne    L9
L44:
  test   $0x20,%bl
  jne    L58
L19:
  test   $0x40,%bl
  jne    L26
L47:
  pop    %rbx
  ret
L9:
  mov    $0x0,%eax
  call   <uart_isr@plt>
  jmp    L44
L58:
  mov    $0x0,%eax
  call   <timer0_isr@plt>
  jmp    L19
L26:
  mov    $0x0,%eax
  call   <timer1_isr@plt>
  jmp    L47
```

assistant: ```c
void isr(void) {
  unsigned int irqs;
  irqs = irq_pending() & irq_getmask();
  if (irqs & (0x00000008))
    uart_isr();
  if (irqs & (0x00000020))
    timer0_isr();
  if (irqs & (0x00000040))
    timer1_isr();
}
```
user: What is the c source code of the assembly code below?

```asm
<isr>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <irq_pending@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <irq_getmask@plt>
  and    %eax,%ebx
  test   $0x8,%bl
  jne    L28
  test   $0x20,%bl
  jne    L5
L8:
  and    $0x40,%ebx
  jne    L20
L2:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L28:
  xor    %eax,%eax
  call   <uart_isr@plt>
  test   $0x20,%bl
  je     L8
L5:
  xor    %eax,%eax
  call   <timer0_isr@plt>
  and    $0x40,%ebx
  je     L2
L20:
  xor    %eax,%eax
  pop    %rbx
  jmp    10a0 <timer1_isr@plt>
```

assistant: ```c
void isr(void) {
  unsigned int irqs;
  irqs = irq_pending() & irq_getmask();
  if (irqs & (0x00000008))
    uart_isr();
  if (irqs & (0x00000020))
    timer0_isr();
  if (irqs & (0x00000040))
    timer1_isr();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<isr>:
  endbr64
  push   %rbx
  xor    %eax,%eax
  call   <irq_pending@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <irq_getmask@plt>
  and    %eax,%ebx
  test   $0x8,%bl
  jne    L27
  test   $0x20,%bl
  jne    L41
L28:
  and    $0x40,%ebx
  jne    L24
L15:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L27:
  xor    %eax,%eax
  call   <uart_isr@plt>
  test   $0x20,%bl
  je     L28
L41:
  xor    %eax,%eax
  call   <timer0_isr@plt>
  and    $0x40,%ebx
  je     L15
L24:
  xor    %eax,%eax
  pop    %rbx
  jmp    10a0 <timer1_isr@plt>
```

assistant: ```c
void isr(void) {
  unsigned int irqs;
  irqs = irq_pending() & irq_getmask();
  if (irqs & (0x00000008))
    uart_isr();
  if (irqs & (0x00000020))
    timer0_isr();
  if (irqs & (0x00000040))
    timer1_isr();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<lauto_free>:
  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
  mov    (%rax),%rax
  test   %rax,%rax
  je     L1
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  movq   $0x0,(%rax)
L1:
  nop
  leave
  ret
```

assistant: ```c
void lauto_free(void *ptr) {
  void **location = ptr;
  if (*location) {
    free(*location);
    *location = NULL;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<lauto_free>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L8
  call   <free@plt>
  movq   $0x0,(%rbx)
L8:
  pop    %rbx
  ret
```

assistant: ```c
void lauto_free(void *ptr) {
  void **location = ptr;
  if (*location) {
    free(*location);
    *location = NULL;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<lauto_free>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L8
  call   <free@plt>
  movq   $0x0,(%rbx)
L8:
  pop    %rbx
  ret
```

assistant: ```c
void lauto_free(void *ptr) {
  void **location = ptr;
  if (*location) {
    free(*location);
    *location = NULL;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<lauto_free>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%rdi
  test   %rdi,%rdi
  je     L4
  call   <free@plt>
  movq   $0x0,(%rbx)
L4:
  pop    %rbx
  ret
```

assistant: ```c
void lauto_free(void *ptr) {
  void **location = ptr;
  if (*location) {
    free(*location);
    *location = NULL;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<dec_bin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movq   $0x1,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L19
L10:
  mov    -0x28(%rbp),%rax
  cqto
  shr    $0x3f,%rdx
  add    %rdx,%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %rax,-0x28(%rbp)
  mov    -0x8(%rbp),%rax
  imul   -0x18(%rbp),%rax
  add    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,-0x18(%rbp)
L19:
  cmpq   $0x0,-0x28(%rbp)
  jne    L10
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long dec_bin(long n) {
  long rem;
  long i = 1;
  long binary = 0;
  while (n != 0) {
    rem = n % 2;
    n /= 2;
    binary += rem * i;
    i *= 10;
  }
  return binary;
}
```
user: Translate the below assembly code into C programming language:

```asm
<dec_bin>:
  endbr64
  test   %rdi,%rdi
  je     L29
  mov    $0x0,%esi
  mov    $0x1,%ecx
L1:
  mov    %rdi,%rdx
  shr    $0x3f,%rdx
  lea    (%rdi,%rdx,1),%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  mov    %rdi,%rdx
  shr    $0x3f,%rdx
  add    %rdi,%rdx
  sar    %rdx
  mov    %rdx,%rdi
  imul   %rcx,%rax
  add    %rax,%rsi
  lea    (%rcx,%rcx,4),%rcx
  add    %rcx,%rcx
  test   %rdx,%rdx
  jne    L1
L5:
  mov    %rsi,%rax
  ret
L29:
  mov    %rdi,%rsi
  jmp    L5
```

assistant: ```c
long dec_bin(long n) {
  long rem;
  long i = 1;
  long binary = 0;
  while (n != 0) {
    rem = n % 2;
    n /= 2;
    binary += rem * i;
    i *= 10;
  }
  return binary;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dec_bin>:
  endbr64
  xor    %r8d,%r8d
  test   %rdi,%rdi
  je     L6
  mov    $0x1,%edx
  nopl   0x0(%rax)
L19:
  mov    %rdi,%rcx
  shr    $0x3f,%rcx
  lea    (%rdi,%rcx,1),%rax
  and    $0x1,%eax
  sub    %rcx,%rax
  mov    %rdi,%rcx
  imul   %rdx,%rax
  shr    $0x3f,%rcx
  lea    (%rdx,%rdx,4),%rdx
  add    %rcx,%rdi
  add    %rdx,%rdx
  add    %rax,%r8
  sar    %rdi
  jne    L19
L6:
  mov    %r8,%rax
  ret
```

assistant: ```c
long dec_bin(long n) {
  long rem;
  long i = 1;
  long binary = 0;
  while (n != 0) {
    rem = n % 2;
    n /= 2;
    binary += rem * i;
    i *= 10;
  }
  return binary;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dec_bin>:
  endbr64
  xor    %r8d,%r8d
  test   %rdi,%rdi
  je     L12
  mov    $0x1,%edx
  nopl   0x0(%rax)
L1:
  mov    %rdi,%rcx
  shr    $0x3f,%rcx
  lea    (%rdi,%rcx,1),%rax
  and    $0x1,%eax
  sub    %rcx,%rax
  mov    %rdi,%rcx
  imul   %rdx,%rax
  shr    $0x3f,%rcx
  lea    (%rdx,%rdx,4),%rdx
  add    %rcx,%rdi
  add    %rdx,%rdx
  add    %rax,%r8
  sar    %rdi
  jne    L1
L12:
  mov    %r8,%rax
  ret
```

assistant: ```c
long dec_bin(long n) {
  long rem;
  long i = 1;
  long binary = 0;
  while (n != 0) {
    rem = n % 2;
    n /= 2;
    binary += rem * i;
    i *= 10;
  }
  return binary;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<testfutimes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <futimes@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L17
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  jmp    L10
L17:
  nop
L10:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_guessed: string

tool: D6: futimes

assistant: ```c
int testfutimes(int fd, const struct timeval tv[2]) {
  int ret;
  ret = futimes(fd, tv);
  if (ret) {
    perror("futimes");
    goto out;
  }
out:
  return ret;
}
```