ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<operation_sleep_notimer>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <ldo_power_off@plt>
  xor    %edx,%edx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <set_wakeup_timer@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <operation_sleep@plt>
```

assistant: ```c
void operation_sleep_notimer(void) {
  ldo_power_off();
  set_wakeup_timer(0, 0, 0);
  operation_sleep();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<init_curses>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <initscr@plt>
  mov    $0x0,%eax
  call   <cbreak@plt>
  mov    $0x0,%eax
  call   <noecho@plt>
  mov    $0x0,%eax
  call   <start_color@plt>
  mov    D5(%rip),%rax
  mov    (%rax),%edx
  mov    D3(%rip),%rax
  mov    (%rax),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <keypad@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void *init_curses() {
  initscr();
  cbreak();
  noecho();
  start_color();
  keypad(stdscr, TRUE);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<init_curses>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <initscr@plt>
  mov    $0x0,%eax
  call   <cbreak@plt>
  mov    $0x0,%eax
  call   <noecho@plt>
  mov    $0x0,%eax
  call   <start_color@plt>
  mov    D19(%rip),%rax
  mov    (%rax),%esi
  mov    D4(%rip),%rax
  mov    (%rax),%edi
  call   <keypad@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *init_curses() {
  initscr();
  cbreak();
  noecho();
  start_color();
  keypad(stdscr, TRUE);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<init_curses>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initscr@plt>
  xor    %eax,%eax
  call   <cbreak@plt>
  xor    %eax,%eax
  call   <noecho@plt>
  xor    %eax,%eax
  call   <start_color@plt>
  mov    D8(%rip),%rax
  mov    (%rax),%esi
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  call   <keypad@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *init_curses() {
  initscr();
  cbreak();
  noecho();
  start_color();
  keypad(stdscr, TRUE);
}
```
user: What is the c source code of the assembly code below?

```asm
<init_curses>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <initscr@plt>
  xor    %eax,%eax
  call   <cbreak@plt>
  xor    %eax,%eax
  call   <noecho@plt>
  xor    %eax,%eax
  call   <start_color@plt>
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <keypad@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void *init_curses() {
  initscr();
  cbreak();
  noecho();
  start_color();
  keypad(stdscr, TRUE);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L1
  movq   $0x0,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L37
L7:
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
  addq   $0x1,-0x10(%rbp)
  subq   $0x1,-0x8(%rbp)
L37:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L7
  jmp    L30
L1:
  nop
L30:
  leave
  ret
```

assistant: ```c
void reverse(int arr[], size_t len) {
  if (len == 0)
    return;
  size_t i, j;
  for (i = 0, j = len - 1; i < j; i++, j--)
    swap(arr, i, j);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<reverse>:
  endbr64
  test   %rsi,%rsi
  je     L5
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  sub    $0x1,%rbx
  je     L9
  mov    $0x0,%ebp
L26:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <swap@plt>
  add    $0x1,%rbp
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jb     L26
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  ret
```

assistant: ```c
void reverse(int arr[], size_t len) {
  if (len == 0)
    return;
  size_t i, j;
  for (i = 0, j = len - 1; i < j; i++, j--)
    swap(arr, i, j);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverse>:
  endbr64
  test   %rsi,%rsi
  je     L5
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x1,%rbx
  je     L1
  mov    %rdi,%r12
  xor    %ebp,%ebp
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%rbp
  call   <swap@plt>
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jb     L7
L1:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  ret
```

assistant: ```c
void reverse(int arr[], size_t len) {
  if (len == 0)
    return;
  size_t i, j;
  for (i = 0, j = len - 1; i < j; i++, j--)
    swap(arr, i, j);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<reverse>:
  endbr64
  test   %rsi,%rsi
  je     L13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x1,%rbx
  je     L11
  mov    %rdi,%r12
  xor    %ebp,%ebp
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%rbp
  call   <swap@plt>
  sub    $0x1,%rbx
  cmp    %rbx,%rbp
  jb     L7
L11:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  ret
```

assistant: ```c
void reverse(int arr[], size_t len) {
  if (len == 0)
    return;
  size_t i, j;
  for (i = 0, j = len - 1; i < j; i++, j--)
    swap(arr, i, j);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fl_fputs>:
  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   <strlen@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%rcx
  mov    -0x8(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fl_fwrite@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L17
  mov    -0x8(%rbp),%eax
  jmp    L8
L17:
  mov    -0x4(%rbp),%eax
L8:
  leave
  ret
```

assistant: ```c
int fl_fputs(const char *str, void *f) {
  int len = (int)strlen(str);
  int res = fl_fwrite(str, 1, len, f);
  if (res == len)
    return len;
  else
    return res;
}
```
user: What is the c source code of the assembly code below?

```asm
<fl_fputs>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  mov    %rbp,%rcx
  mov    %eax,%edx
  mov    $0x1,%esi
  mov    %rbx,%rdi
  call   <fl_fwrite@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int fl_fputs(const char *str, void *f) {
  int len = (int)strlen(str);
  int res = fl_fwrite(str, 1, len, f);
  if (res == len)
    return len;
  else
    return res;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fl_fputs>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %r12,%rcx
  mov    %rbp,%rdi
  mov    %eax,%edx
  pop    %rbp
  mov    $0x1,%esi
  pop    %r12
  jmp    1070 <fl_fwrite@plt>
```

assistant: ```c
int fl_fputs(const char *str, void *f) {
  int len = (int)strlen(str);
  int res = fl_fwrite(str, 1, len, f);
  if (res == len)
    return len;
  else
    return res;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fl_fputs>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  add    $0x8,%rsp
  mov    %r12,%rcx
  mov    %rbp,%rdi
  mov    %eax,%edx
  pop    %rbp
  mov    $0x1,%esi
  pop    %r12
  jmp    1070 <fl_fwrite@plt>
```

assistant: ```c
int fl_fputs(const char *str, void *f) {
  int len = (int)strlen(str);
  int res = fl_fwrite(str, 1, len, f);
  if (res == len)
    return len;
  else
    return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<now_usecs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x18(%rbp),%rax
  imul   $0xf4240,%rax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
long long now_usecs(void) {
  struct timeval tv;
  assert(!gettimeofday(&tv, NULL));
  return (long long)tv.tv_sec * 1000000 + tv.tv_usec;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<now_usecs>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    $0x0,%esi
  call   <gettimeofday@plt>
  test   %eax,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long now_usecs(void) {
  struct timeval tv;
  assert(!gettimeofday(&tv, NULL));
  return (long long)tv.tv_sec * 1000000 + tv.tv_usec;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<now_usecs>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long now_usecs(void) {
  struct timeval tv;
  assert(!gettimeofday(&tv, NULL));
  return (long long)tv.tv_sec * 1000000 + tv.tv_usec;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<now_usecs>:
  endbr64
  sub    $0x28,%rsp
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <gettimeofday@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  imul   $0xf4240,0x8(%rsp),%rax
  add    (%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long now_usecs(void) {
  struct timeval tv;
  assert(!gettimeofday(&tv, NULL));
  return (long long)tv.tv_sec * 1000000 + tv.tv_usec;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<close>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0xffffffff,%edi
  call   <_exit@plt>
  mov    $0xffffffff,%eax
  leave
  ret
```

assistant: ```c
int close(int fd) {
  _exit(-1);
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<close>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%edi
  call   <_exit@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int close(int fd) {
  _exit(-1);
  return -1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<close>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%edi
  call   <_exit@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int close(int fd) {
  _exit(-1);
  return -1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<close>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xffffffff,%edi
  call   <_exit@plt>
  mov    $0xffffffff,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int close(int fd) {
  _exit(-1);
  return -1;
}
```
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    $0xfc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0xfb8(%rbp)
  mov    $0x3e8,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x3e8,%esi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0xfac(%rbp)
  movl   $0x2,-0xfc0(%rbp)
  jmp    L41
L69:
  mov    -0xfc0(%rbp),%eax
  cltq
  movl   $0x1,-0xfb0(%rbp,%rax,4)
  addl   $0x1,-0xfc0(%rbp)
L41:
  cmpl   $0x3e8,-0xfc0(%rbp)
  jle    L69
  movl   $0x2,-0xfc0(%rbp)
  jmp    L4
L1:
  movl   $0x2,-0xfbc(%rbp)
  jmp    L52
L67:
  mov    -0xfc0(%rbp),%eax
  imul   -0xfbc(%rbp),%eax
  cltq
  movl   $0x0,-0xfb0(%rbp,%rax,4)
  addl   $0x1,-0xfbc(%rbp)
L52:
  mov    $0x3e8,%eax
  cltd
  idivl  -0xfc0(%rbp)
  cmp    %eax,-0xfbc(%rbp)
  jle    L67
  addl   $0x1,-0xfc0(%rbp)
L4:
  cmpl   $0x1f4,-0xfc0(%rbp)
  jle    L1
  movl   $0x1,-0xfc0(%rbp)
  jmp    L70
L50:
  mov    -0xfc0(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L38
  mov    -0xfc0(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0xfc0(%rbp),%eax
  cltq
  add    %rax,-0xfb8(%rbp)
L38:
  addl   $0x1,-0xfc0(%rbp)
L70:
  cmpl   $0x3e8,-0xfc0(%rbp)
  jle    L50
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0xfb8(%rbp),%rax
  mov    %rax,%rsi
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L80
  call   <__stack_chk_fail@plt>
L80:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D18
datatype: string
name: parse_data
arguments: data_label: D21
datatype: string
name: parse_data
arguments: data_label: D1
datatype: string
name: parse_data
arguments: data_label: D35
datatype: string

tool: D18: Program that prints the sum of all prime numbers
 below %d and also prints thier sum


tool: D21: 

Prime Numbers below %d: 

tool: D1: %4d

tool: D35: 
The sum of all the prime numbers is: %ld


assistant: ```c
int main(void) {
  long int sum = 0;
  int i, j, a[1000 + 1];
  printf("Program that prints the sum of all prime numbers\n "
         "below %d and also prints thier sum\n",
         1000);
  printf("\n\nPrime Numbers below %d: ", 1000);
  for (a[1] = 0, i = 2; i <= 1000; i++)
    a[i] = 1;
  for (i = 2; i <= 1000 / 2; i++)
    for (j = 2; j <= 1000 / i; j++)
      a[i * j] = 0;
  for (i = 1; i <= 1000; i++) {
    if (a[i]) {
      printf("%4d", i);
      sum += i;
    }
  }
  printf("\n");
  printf("\nThe sum of all the prime numbers is: %ld\n", sum);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    $0x3e8,%esi
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x3e8,%esi
  lea    D10(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,0x4(%rsp)
  lea    0x8(%rsp),%rax
  lea    0xfa4(%rsp),%rdx
L63:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L63
  lea    0x10(%rsp),%r9
  mov    $0x8,%esi
  mov    $0x2,%r8d
  mov    $0x3e8,%r10d
  jmp    L99
L25:
  movl   $0x0,(%rcx)
  add    $0x1,%edx
  add    %rsi,%rcx
  cmp    %edi,%edx
  jne    L25
L3:
  add    $0x1,%r8d
  add    $0x8,%r9
  add    $0x4,%rsi
  cmp    $0x1f5,%r8d
  je     L47
L99:
  mov    %r10d,%eax
  cltd
  idiv   %r8d
  lea    0x1(%rax),%edi
  mov    %r9,%rcx
  mov    $0x2,%edx
  cmp    $0x1,%eax
  jg     L25
  jmp    L3
L47:
  mov    $0x1,%ebx
  mov    $0x0,%ebp
  lea    D29(%rip),%r12
  jmp    L18
L57:
  add    $0x1,%rbx
  cmp    $0x3e9,%rbx
  je     L11
L18:
  mov    %ebx,%esi
  cmpl   $0x0,(%rsp,%rbx,4)
  je     L57
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    %rbx,%rbp
  jmp    L57
L11:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %rbp,%rsi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L80
  mov    $0x0,%eax
  add    $0xfb0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L80:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
type_infered: string
name: extract_symbolic_data
arguments: tag: D10
type_infered: string
name: extract_symbolic_data
arguments: tag: D29
type_infered: string
name: extract_symbolic_data
arguments: tag: D18
type_infered: string

tool: D7: Program that prints the sum of all prime numbers
 below %d and also prints thier sum


tool: D10: 

Prime Numbers below %d: 

tool: D29: %4d

tool: D18: 
The sum of all the prime numbers is: %ld


assistant: ```c
int main(void) {
  long int sum = 0;
  int i, j, a[1000 + 1];
  printf("Program that prints the sum of all prime numbers\n "
         "below %d and also prints thier sum\n",
         1000);
  printf("\n\nPrime Numbers below %d: ", 1000);
  for (a[1] = 0, i = 2; i <= 1000; i++)
    a[i] = 1;
  for (i = 2; i <= 1000 / 2; i++)
    for (j = 2; j <= 1000 / i; j++)
      a[i * j] = 0;
  for (i = 1; i <= 1000; i++) {
    if (a[i]) {
      printf("%4d", i);
      sum += i;
    }
  }
  printf("\n");
  printf("\nThe sum of all the prime numbers is: %ld\n", sum);
  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
  push   %r12
  mov    $0x3e8,%esi
  lea    D16(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x3e8,%esi
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rax
  lea    0xfa4(%rsp),%rdx
  nopl   0x0(%rax,%rax,1)
L18:
  movl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L18
  lea    0x10(%rsp),%r8
  mov    $0x8,%esi
  mov    $0x2,%edi
  mov    $0x3e8,%r9d
  nopl   0x0(%rax)
L59:
  mov    %r9d,%eax
  mov    %r8,%rcx
  cltd
  idiv   %edi
  mov    $0x2,%edx
  add    $0x1,%eax
  nopl   0x0(%rax)
L30:
  add    $0x1,%edx
  movl   $0x0,(%rcx)
  add    %rsi,%rcx
  cmp    %eax,%edx
  jne    L30
  add    $0x1,%edi
  add    $0x8,%r8
  add    $0x4,%rsi
  cmp    $0x1f5,%edi
  jne    L59
  mov    $0x2,%ebx
  xor    %ebp,%ebp
  lea    D23(%rip),%r12
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L67:
  mov    (%rsp,%rbx,4),%eax
  test   %eax,%eax
  je     L31
  mov    %r12,%rdi
  xor    %eax,%eax
  add    %rbx,%rbp
  call   <printf@plt>
L31:
  add    $0x1,%rbx
L32:
  mov    %ebx,%esi
  cmp    $0x3e9,%rbx
  jne    L67
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0xfb0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D16
datatype: string
name: unpack_data
arguments: tag: D30
datatype: string
name: unpack_data
arguments: tag: D23
datatype: string
name: unpack_data
arguments: tag: D28
datatype: string

tool: D16: Program that prints the sum of all prime numbers
 below %d and also prints thier sum


tool: D30: 

Prime Numbers below %d: 

tool: D23: %4d

tool: D28: 
The sum of all the prime numbers is: %ld


assistant: ```c
int main(void) {
  long int sum = 0;
  int i, j, a[1000 + 1];
  printf("Program that prints the sum of all prime numbers\n "
         "below %d and also prints thier sum\n",
         1000);
  printf("\n\nPrime Numbers below %d: ", 1000);
  for (a[1] = 0, i = 2; i <= 1000; i++)
    a[i] = 1;
  for (i = 2; i <= 1000 / 2; i++)
    for (j = 2; j <= 1000 / i; j++)
      a[i * j] = 0;
  for (i = 1; i <= 1000; i++) {
    if (a[i]) {
      printf("%4d", i);
      sum += i;
    }
  }
  printf("\n");
  printf("\nThe sum of all the prime numbers is: %ld\n", sum);
  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   %r13
  mov    $0x3e8,%esi
  lea    D8(%rip),%rdi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x3e8,%esi
  lea    D31(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movdqa D2(%rip),%xmm0
  lea    0x8(%rsp),%rax
  lea    0xf98(%rsp),%rdx
  nopl   (%rax)
L41:
  movups %xmm0,(%rax)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L41
  lea    0x10(%rsp),%r8
  mov    $0x8,%esi
  mov    $0x2,%edi
  mov    D2(%rip),%rax
  movl   $0x1,0xfa0(%rsp)
  mov    $0x3e8,%r9d
  mov    %rax,0xf98(%rsp)
L39:
  mov    %r9d,%eax
  mov    %r8,%rcx
  cltd
  idiv   %edi
  mov    $0x2,%edx
  add    $0x1,%eax
  nopl   0x0(%rax)
L59:
  add    $0x1,%edx
  movl   $0x0,(%rcx)
  add    %rsi,%rcx
  cmp    %eax,%edx
  jne    L59
  add    $0x1,%edi
  add    $0x8,%r8
  add    $0x4,%rsi
  cmp    $0x1f5,%edi
  jne    L39
  mov    $0x2,%ebx
  xor    %r12d,%r12d
  mov    %rsp,%rbp
  lea    D37(%rip),%r13
  nopl   (%rax)
L57:
  mov    0x0(%rbp,%rbx,4),%eax
  mov    %ebx,%esi
  test   %eax,%eax
  je     L34
  mov    %r13,%rdi
  xor    %eax,%eax
  add    %rbx,%r12
  call   <printf@plt>
L34:
  add    $0x1,%rbx
  cmp    $0x3e9,%rbx
  jne    L57
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  mov    %r12,%rsi
  lea    D40(%rip),%rdi
  call   <printf@plt>
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0xfb8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D8
type: string
name: read_data
arguments: data_tag: D31
type: string
name: read_data
arguments: data_tag: D2
type: byte[16]
name: read_data
arguments: data_tag: D37
type: string
name: read_data
arguments: data_tag: D40
type: string

tool: D8: Program that prints the sum of all prime numbers
 below %d and also prints thier sum


tool: D31: 

Prime Numbers below %d: 

tool: {"D2": [0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x01,0x00,0x00,0x00]}

tool: D37: %4d

tool: D40: 
The sum of all the prime numbers is: %ld


assistant: ```c
int main(void) {
  long int sum = 0;
  int i, j, a[1000 + 1];
  printf("Program that prints the sum of all prime numbers\n "
         "below %d and also prints thier sum\n",
         1000);
  printf("\n\nPrime Numbers below %d: ", 1000);
  for (a[1] = 0, i = 2; i <= 1000; i++)
    a[i] = 1;
  for (i = 2; i <= 1000 / 2; i++)
    for (j = 2; j <= 1000 / i; j++)
      a[i * j] = 0;
  for (i = 1; i <= 1000; i++) {
    if (a[i]) {
      printf("%4d", i);
      sum += i;
    }
  }
  printf("\n");
  printf("\nThe sum of all the prime numbers is: %ld\n", sum);
  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
<modDecimal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  ucomisd -0x20(%rbp),%xmm0
  jp     L33
  pxor   %xmm0,%xmm0
  ucomisd -0x20(%rbp),%xmm0
  jne    L33
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  jmp    L29
L33:
  pxor   %xmm0,%xmm0
  comisd -0x20(%rbp),%xmm0
  jb     L16
  movsd  -0x20(%rbp),%xmm0
  movq   D2(%rip),%xmm1
  xorpd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L80
L16:
  movsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  jmp    L80
L77:
  movsd  -0x10(%rbp),%xmm0
  addsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
L80:
  pxor   %xmm0,%xmm0
  comisd -0x10(%rbp),%xmm0
  ja     L77
  jmp    L34
L1:
  movsd  -0x10(%rbp),%xmm0
  subsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
L34:
  movsd  -0x10(%rbp),%xmm0
  comisd -0x8(%rbp),%xmm0
  ja     L1
L29:
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
data_type: byte[8]

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80]}

assistant: ```c
double modDecimal(double x, double divisor) {
  double temp = x;
  double d = divisor;
  if (divisor == 0) {
    temp = 0;
  } else {
    if (divisor <= 0) {
      d = -divisor;
    } else {
      d = divisor;
    }
    while (temp < 0) {
      temp = temp + d;
    }
    while (temp > d) {
      temp = temp - d;
    }
  }
  return temp;
}
```