ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<print_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L19:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L19
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D5
datatype: string

tool: D5: %d 

assistant: ```c
void print_array(int *array, int size) {
  for (int i = 0; i < size; i++) {
    printf("%d ", array[i]);
  }
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<print_array>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L5
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%r12
  lea    D1(%rip),%rbp
L7:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L7
L5:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
data_type: string

tool: D1: %d 

assistant: ```c
void print_array(int *array, int size) {
  for (int i = 0; i < size; i++) {
    printf("%d ", array[i]);
  }
  printf("\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<print_array>:
  endbr64
  test   %esi,%esi
  jle    L0
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L15:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L15
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L0:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
type_infered: string

tool: D5: %d 

assistant: ```c
void print_array(int *array, int size) {
  for (int i = 0; i < size; i++) {
    printf("%d ", array[i]);
  }
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<print_array>:
  endbr64
  test   %esi,%esi
  jle    L15
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L0:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L0
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L15:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
datatype: string

tool: D7: %d 

assistant: ```c
void print_array(int *array, int size) {
  for (int i = 0; i < size; i++) {
    printf("%d ", array[i]);
  }
  printf("\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mem_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  add    $0x20,%rax
  add    $0x3,%eax
  and    $0xfffffffc,%eax
  cltq
  mov    %rax,%rdx
  mov    -0x20(%rbp),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    $0x8,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L9
L4:
  mov    -0x20(%rbp),%eax
  add    $0x1,%eax
  imul   -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %eax,-0xc(%rbp)
  mov    -0x1c(%rbp),%eax
  imul   -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  lea    -0xc(%rbp),%rax
  mov    $0x4,%edx
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
L9:
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  test   %eax,%eax
  jne    L4
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
void mem_init(void *mem_pool, uint16_t mem_size, uint16_t mem_count, void **mem_head) {
  *mem_head = mem_pool;
  *((uint16_t *)(((uint32_t)((uint8_t *)mem_pool + sizeof(mem_pool)) + 3) & (~((uint32_t)3)))) = mem_count;
  memset(((uint8_t *)mem_pool + (mem_size * (--mem_count))), 0, sizeof(mem_pool));
  while (mem_count--) {
    uint32_t next;
    next = (uint32_t)((uint8_t *)mem_pool + (mem_size * (mem_count + 1)));
    memcpy(((uint8_t *)mem_pool + (mem_size * mem_count)), (void *)&next, sizeof(next));
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<mem_init>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %esi,%r14d
  mov    %edx,%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rdi,(%rcx)
  lea    0x23(%rdi),%eax
  and    $0xfffffffc,%eax
  cltq
  mov    %edx,(%rax)
  lea    -0x1(%rdx),%r13d
  mov    %r13d,%r12d
  imul   %esi,%r12d
  movslq %r12d,%rax
  lea    (%rdi,%rax,4),%rdi
  mov    $0x8,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  sub    $0x2,%ebp
  test   %r13d,%r13d
  je     L4
  mov    %r14d,%r13d
  neg    %r13d
  movslq %r13d,%r13
  shl    $0x2,%r13
  mov    %r12d,%eax
  sub    %r14d,%eax
  movslq %eax,%rdx
  lea    (%rbx,%rdx,4),%rbx
  sub    %eax,%r12d
  shl    $0x2,%r12d
  lea    0x4(%rsp),%r14
L16:
  lea    (%r12,%rbx,1),%eax
  mov    %eax,0x4(%rsp)
  mov    $0x4,%edx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    %ebp,%eax
  sub    $0x1,%ebp
  add    %r13,%rbx
  test   %eax,%eax
  jne    L16
L4:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void mem_init(void *mem_pool, uint16_t mem_size, uint16_t mem_count, void **mem_head) {
  *mem_head = mem_pool;
  *((uint16_t *)(((uint32_t)((uint8_t *)mem_pool + sizeof(mem_pool)) + 3) & (~((uint32_t)3)))) = mem_count;
  memset(((uint8_t *)mem_pool + (mem_size * (--mem_count))), 0, sizeof(mem_pool));
  while (mem_count--) {
    uint32_t next;
    next = (uint32_t)((uint8_t *)mem_pool + (mem_size * (mem_count + 1)));
    memcpy(((uint8_t *)mem_pool + (mem_size * mem_count)), (void *)&next, sizeof(next));
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mem_init>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  lea    -0x1(%rdx),%r13d
  push   %r12
  mov    %r13d,%r12d
  push   %rbp
  imul   %esi,%r12d
  mov    %edx,%ebp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x2,%ebp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x23(%rdi),%eax
  mov    %rdi,(%rcx)
  and    $0xfffffffc,%eax
  cltq
  mov    %edx,(%rax)
  movslq %r12d,%rax
  mov    $0x8,%edx
  lea    (%rdi,%rax,4),%rdi
  call   <memset@plt>
  test   %r13d,%r13d
  je     L21
  mov    %r14d,%r13d
  mov    %r12d,%eax
  sub    %r14d,%eax
  neg    %r13d
  lea    0x4(%rsp),%r14
  movslq %r13d,%r13
  movslq %eax,%rdx
  sub    %eax,%r12d
  shl    $0x2,%r13
  lea    (%rbx,%rdx,4),%rbx
  shl    $0x2,%r12d
  nopl   0x0(%rax)
L25:
  lea    (%r12,%rbx,1),%eax
  mov    %rbx,%rdi
  mov    $0x4,%edx
  mov    %r14,%rsi
  sub    $0x1,%ebp
  mov    %eax,0x4(%rsp)
  add    %r13,%rbx
  call   <memcpy@plt>
  cmp    $0xffffffff,%ebp
  jne    L25
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void mem_init(void *mem_pool, uint16_t mem_size, uint16_t mem_count, void **mem_head) {
  *mem_head = mem_pool;
  *((uint16_t *)(((uint32_t)((uint8_t *)mem_pool + sizeof(mem_pool)) + 3) & (~((uint32_t)3)))) = mem_count;
  memset(((uint8_t *)mem_pool + (mem_size * (--mem_count))), 0, sizeof(mem_pool));
  while (mem_count--) {
    uint32_t next;
    next = (uint32_t)((uint8_t *)mem_pool + (mem_size * (mem_count + 1)));
    memcpy(((uint8_t *)mem_pool + (mem_size * mem_count)), (void *)&next, sizeof(next));
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<mem_init>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  lea    -0x1(%rdx),%r13d
  push   %r12
  mov    %r13d,%r12d
  push   %rbp
  imul   %esi,%r12d
  mov    %edx,%ebp
  xor    %esi,%esi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x2,%ebp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x23(%rdi),%eax
  mov    %rdi,(%rcx)
  and    $0xfffffffc,%eax
  cltq
  mov    %edx,(%rax)
  movslq %r12d,%rax
  mov    $0x8,%edx
  lea    (%rdi,%rax,4),%rdi
  call   <memset@plt>
  test   %r13d,%r13d
  je     L21
  mov    %r14d,%r13d
  mov    %r12d,%eax
  sub    %r14d,%eax
  neg    %r13d
  lea    0x4(%rsp),%r14
  movslq %r13d,%r13
  movslq %eax,%rdx
  sub    %eax,%r12d
  shl    $0x2,%r13
  lea    (%rbx,%rdx,4),%rbx
  shl    $0x2,%r12d
  nopl   0x0(%rax)
L15:
  lea    (%r12,%rbx,1),%eax
  mov    %rbx,%rdi
  mov    $0x4,%edx
  mov    %r14,%rsi
  sub    $0x1,%ebp
  mov    %eax,0x4(%rsp)
  add    %r13,%rbx
  call   <memcpy@plt>
  cmp    $0xffffffff,%ebp
  jne    L15
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void mem_init(void *mem_pool, uint16_t mem_size, uint16_t mem_count, void **mem_head) {
  *mem_head = mem_pool;
  *((uint16_t *)(((uint32_t)((uint8_t *)mem_pool + sizeof(mem_pool)) + 3) & (~((uint32_t)3)))) = mem_count;
  memset(((uint8_t *)mem_pool + (mem_size * (--mem_count))), 0, sizeof(mem_pool));
  while (mem_count--) {
    uint32_t next;
    next = (uint32_t)((uint8_t *)mem_pool + (mem_size * (mem_count + 1)));
    memcpy(((uint8_t *)mem_pool + (mem_size * mem_count)), (void *)&next, sizeof(next));
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<printArray>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L8:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
type_infered: string
name: retrieve_data
arguments: tag: D8
type_infered: string

tool: D7: %d 

tool: D8: 


assistant: ```c
void printArray(int *array, int n) {
  int i = 0;
  for (i = 0; i < n; i++) {
    printf("%d ", array[i]);
  }
  printf("\n\n");
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<printArray>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L13
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%r12
  lea    D7(%rip),%rbp
L10:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L10
L13:
  lea    D3(%rip),%rdi
  call   <puts@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type_infered: string
name: extract_symbolic_data
arguments: label: D3
type_infered: string

tool: D7: %d 

tool: D3: 


assistant: ```c
void printArray(int *array, int n) {
  int i = 0;
  for (i = 0; i < n; i++) {
    printf("%d ", array[i]);
  }
  printf("\n\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<printArray>:
  endbr64
  test   %esi,%esi
  jle    L10
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D15(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L16:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L16
  pop    %rbx
  lea    D9(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <puts@plt>
  nopl   (%rax)
L10:
  lea    D9(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D15
datatype: string
name: disassemble_data
arguments: data_tag: D9
datatype: string

tool: D15: %d 

tool: D9: 


assistant: ```c
void printArray(int *array, int n) {
  int i = 0;
  for (i = 0; i < n; i++) {
    printf("%d ", array[i]);
  }
  printf("\n\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<printArray>:
  endbr64
  test   %esi,%esi
  jle    L3
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L10:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L10
  pop    %rbx
  lea    D25(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <puts@plt>
  nopl   (%rax)
L3:
  lea    D25(%rip),%rdi
  jmp    1060 <puts@plt>
```

tool_calls: name: parse_data
arguments: tag: D2
datatype: string
name: parse_data
arguments: tag: D25
datatype: string

tool: D2: %d 

tool: D25: 


assistant: ```c
void printArray(int *array, int n) {
  int i = 0;
  for (i = 0; i < n; i++) {
    printf("%d ", array[i]);
  }
  printf("\n\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<sleep_ms>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x24(%rbp),%eax
  mov    %eax,%eax
  imul   $0x10624dd3,%rax,%rax
  shr    $0x20,%rax
  shr    $0x6,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  imul   $0x3e8,%eax,%eax
  sub    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  imul   $0xf4240,%eax,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L23
L18:
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
L23:
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <nanosleep@plt>
  test   %rax,%rax
  jne    L18
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
void sleep_ms(unsigned int ms) {
  struct timespec req, rem;
  req.tv_sec = ms / 1000;
  ms -= req.tv_sec * 1000;
  req.tv_nsec = ms * 1000000;
  while (nanosleep(&req, &rem))
    req = rem;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<sleep_ms>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  imul   $0x10624dd3,%rax,%rax
  shr    $0x26,%rax
  mov    %eax,0x8(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  mov    %edi,0xc(%rsp)
  lea    0x10(%rsp),%rbx
  jmp    L12
L20:
  mov    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
L12:
  lea    0x8(%rsp),%rdi
  mov    %rbx,%rsi
  call   <nanosleep@plt>
  test   %rax,%rax
  jne    L20
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x20,%rsp
  pop    %rbx
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sleep_ms(unsigned int ms) {
  struct timespec req, rem;
  req.tv_sec = ms / 1000;
  ms -= req.tv_sec * 1000;
  req.tv_nsec = ms * 1000000;
  while (nanosleep(&req, &rem))
    req = rem;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sleep_ms>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  lea    0x10(%rsp),%rbp
  lea    0x8(%rsp),%rbx
  imul   $0x10624dd3,%rax,%rax
  shr    $0x26,%rax
  mov    %eax,0x8(%rsp)
  imul   $0x3e8,%eax,%eax
  sub    %eax,%edi
  imul   $0xf4240,%edi,%edi
  mov    %edi,0xc(%rsp)
  jmp    L14
  nopl   0x0(%rax)
L10:
  mov    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
L14:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <nanosleep@plt>
  test   %rax,%rax
  jne    L10
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sleep_ms(unsigned int ms) {
  struct timespec req, rem;
  req.tv_sec = ms / 1000;
  ms -= req.tv_sec * 1000;
  req.tv_nsec = ms * 1000000;
  while (nanosleep(&req, &rem))
    req = rem;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sleep_ms>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %edi,%eax
  lea    0x10(%rsp),%rbp
  lea    0x8(%rsp),%rbx
  imul   $0x10624dd3,%rax,%rax
  shr    $0x26,%rax
  imul   $0x3e8,%eax,%edx
  movd   %eax,%xmm0
  sub    %edx,%edi
  imul   $0xf4240,%edi,%edi
  movd   %edi,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,0x8(%rsp)
  jmp    L7
  nopl   0x0(%rax,%rax,1)
L23:
  mov    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
L7:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <nanosleep@plt>
  test   %rax,%rax
  jne    L23
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void sleep_ms(unsigned int ms) {
  struct timespec req, rem;
  req.tv_sec = ms / 1000;
  ms -= req.tv_sec * 1000;
  req.tv_nsec = ms * 1000000;
  while (nanosleep(&req, &rem))
    req = rem;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<pali>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x9,-0x4(%rbp)
  jle    L16
  cmpl   $0x63,-0x4(%rbp)
  jg     L33
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %edx,%esi
  sub    %eax,%esi
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  cmp    %edx,%esi
  je     L16
L33:
  cmpl   $0x64,-0x4(%rbp)
  jle    L34
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x51eb851f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x5,%edx
  sar    $0x1f,%eax
  mov    %edx,%esi
  sub    %eax,%esi
  mov    -0x4(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  cmp    %edx,%esi
  jne    L34
L16:
  mov    $0x1,%eax
  jmp    L37
L34:
  mov    $0x0,%eax
L37:
  pop    %rbp
  ret
```

assistant: ```c
bool pali(int x) {
  if ((x < 10) || (x < 100 && (x / 10 == x % 10)) || (x > 100 && (x / 100 == x % 10))) {
    return true;
  }
  return false;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<pali>:
  endbr64
  mov    %edi,%eax
  mov    $0x1,%edx
  cmp    $0x9,%edi
  jle    L30
  cmp    $0x63,%edi
  jg     L0
  movslq %edi,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %edi,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  cmp    %eax,%edx
  sete   %dl
  movzbl %dl,%edx
L30:
  mov    %edx,%eax
  ret
L0:
  cmp    $0x64,%edi
  jle    L38
  movslq %edi,%rdx
  imul   $0x51eb851f,%rdx,%rcx
  sar    $0x25,%rcx
  mov    %edi,%esi
  sar    $0x1f,%esi
  sub    %esi,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cmp    %eax,%ecx
  sete   %dl
  movzbl %dl,%edx
  jmp    L30
L38:
  mov    $0x0,%edx
  jmp    L30
```

assistant: ```c
bool pali(int x) {
  if ((x < 10) || (x < 100 && (x / 10 == x % 10)) || (x > 100 && (x / 100 == x % 10))) {
    return true;
  }
  return false;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<pali>:
  endbr64
  mov    %edi,%eax
  mov    $0x1,%r8d
  cmp    $0x9,%edi
  jle    L23
  cmp    $0x63,%edi
  jle    L8
  cmp    $0x64,%edi
  je     L15
  movslq %edi,%rdx
  mov    %edi,%esi
  xor    %r8d,%r8d
  imul   $0x51eb851f,%rdx,%rcx
  sar    $0x1f,%esi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x25,%rcx
  sar    $0x22,%rdx
  sub    %esi,%ecx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cmp    %eax,%ecx
  sete   %r8b
L23:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L8:
  movslq %edi,%rdx
  mov    %edi,%ecx
  xor    %r8d,%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  cmp    %eax,%edx
  sete   %r8b
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L15:
  xor    %r8d,%r8d
  jmp    L23
```

assistant: ```c
bool pali(int x) {
  if ((x < 10) || (x < 100 && (x / 10 == x % 10)) || (x > 100 && (x / 100 == x % 10))) {
    return true;
  }
  return false;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<pali>:
  endbr64
  mov    %edi,%eax
  mov    $0x1,%r8d
  cmp    $0x9,%edi
  jle    L2
  cmp    $0x63,%edi
  jle    L26
  cmp    $0x64,%edi
  je     L0
  movslq %edi,%rdx
  mov    %edi,%esi
  xor    %r8d,%r8d
  imul   $0x51eb851f,%rdx,%rcx
  sar    $0x1f,%esi
  imul   $0x66666667,%rdx,%rdx
  sar    $0x25,%rcx
  sar    $0x22,%rdx
  sub    %esi,%ecx
  sub    %esi,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  cmp    %eax,%ecx
  sete   %r8b
L2:
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L26:
  movslq %edi,%rdx
  mov    %edi,%ecx
  xor    %r8d,%r8d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  cmp    %eax,%edx
  sete   %r8b
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L0:
  xor    %r8d,%r8d
  jmp    L2
```

assistant: ```c
bool pali(int x) {
  if ((x < 10) || (x < 100 && (x / 10 == x % 10)) || (x > 100 && (x / 100 == x % 10))) {
    return true;
  }
  return false;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<nvram_set_boot>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <nvram_get_option@plt>
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  test   %eax,%eax
  jns    L50
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  jmp    L49
L50:
  mov    -0x10(%rbp),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L22
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <nvram_get@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L22
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L22
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  jmp    L49
L22:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <_nvram_set@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
L49:
  leave
  ret
```

assistant: ```c
int nvram_set_boot(const char *name, const char *value) {
  unsigned int opt = nvram_get_option(name);
  int ret = 0;
  if (opt & 0x80000000) {
    return EINVAL;
  }
  if (opt & 0x02) {
    char *exist = nvram_get(name);
    if (exist && *exist) {
      return EACCES;
    }
  }
  ret = _nvram_set(name, value);
  return ret;
}
```