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

```asm
<init_map>:
  endbr64
  push   %r14
  mov    $0x2e,%r8d
  mov    $0x40,%ecx
  xor    %edx,%edx
  push   %r13
  mov    $0x7f,%esi
  xor    %edi,%edi
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0xc8,%ebx
  call   <draw_region@plt>
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  call   <rand@plt>
  mov    $0x6c,%esi
  mov    $0x1,%edi
  mov    %eax,%edx
  mov    %eax,%r12d
  call   <rr@plt>
  mov    %r12d,%edx
  lea    0x13(%rax),%esi
  lea    0x5(%rax),%edi
  mov    %eax,%ebp
  call   <rr@plt>
  mov    %r12d,%edx
  mov    $0x38,%esi
  mov    $0x1,%edi
  mov    %eax,%r14d
  call   <rr@plt>
  mov    %r12d,%edx
  lea    0x7(%rax),%esi
  lea    0x3(%rax),%edi
  mov    %eax,%r13d
  call   <rr@plt>
  mov    $0x20,%r8d
  mov    %r13d,%edx
  mov    %ebp,%edi
  mov    %eax,%ecx
  mov    %r14d,%esi
  call   <draw_region@plt>
  sub    $0x1,%ebx
  jne    L17
  mov    $0x5,%esi
  mov    $0x40,%edi
  mov    $0xa,%ebx
  call   <rand_place@plt>
  nop
L12:
  mov    $0x2,%esi
  mov    $0x6d,%edi
  call   <rand_place@plt>
  sub    $0x1,%ebx
  jne    L12
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void init_map(void) {
  int i, x1, x2, y1, y2;
  draw_region(0, 128 - 1, 0, 64, '.');
  for (i = 0; i < 200; i++) {
    int r = rand();
    x1 = rr(1, 128 - 20, r);
    x2 = rr(x1 + 5, x1 + 19, r);
    y1 = rr(1, 64 - 8, r);
    y2 = rr(y1 + 3, y1 + 7, r);
    draw_region(x1, x2, y1, y2, ' ');
  }
  rand_place('@', 5);
  for (i = 0; i < 10; i++) {
    rand_place('m', 2);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<init_map>:
  endbr64
  push   %r14
  mov    $0x2e,%r8d
  mov    $0x40,%ecx
  xor    %edx,%edx
  push   %r13
  mov    $0x7f,%esi
  xor    %edi,%edi
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0xc8,%ebx
  call   <draw_region@plt>
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  call   <rand@plt>
  mov    $0x6c,%esi
  mov    $0x1,%edi
  mov    %eax,%edx
  mov    %eax,%r12d
  call   <rr@plt>
  mov    %r12d,%edx
  lea    0x13(%rax),%esi
  lea    0x5(%rax),%edi
  mov    %eax,%ebp
  call   <rr@plt>
  mov    %r12d,%edx
  mov    $0x38,%esi
  mov    $0x1,%edi
  mov    %eax,%r14d
  call   <rr@plt>
  mov    %r12d,%edx
  lea    0x7(%rax),%esi
  lea    0x3(%rax),%edi
  mov    %eax,%r13d
  call   <rr@plt>
  mov    $0x20,%r8d
  mov    %r13d,%edx
  mov    %ebp,%edi
  mov    %eax,%ecx
  mov    %r14d,%esi
  call   <draw_region@plt>
  sub    $0x1,%ebx
  jne    L17
  mov    $0x5,%esi
  mov    $0x40,%edi
  mov    $0xa,%ebx
  call   <rand_place@plt>
  nop
L8:
  mov    $0x2,%esi
  mov    $0x6d,%edi
  call   <rand_place@plt>
  sub    $0x1,%ebx
  jne    L8
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void init_map(void) {
  int i, x1, x2, y1, y2;
  draw_region(0, 128 - 1, 0, 64, '.');
  for (i = 0; i < 200; i++) {
    int r = rand();
    x1 = rr(1, 128 - 20, r);
    x2 = rr(x1 + 5, x1 + 19, r);
    y1 = rr(1, 64 - 8, r);
    y2 = rr(y1 + 3, y1 + 7, r);
    draw_region(x1, x2, y1, y2, ' ');
  }
  rand_place('@', 5);
  for (i = 0; i < 10; i++) {
    rand_place('m', 2);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_get_from_sar_array>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  movabs $0x300003000,%rax
  mov    (%rax),%rax
  mov    %rax,-0x10(%rbp)
  movabs $0x300003008,%rax
  mov    %rax,-0x18(%rbp)
  jmp    L18
L14:
  mov    -0x18(%rbp),%rax
  add    $0x40,%rax
  mov    (%rax),%rax
  mov    -0x18(%rbp),%rdx
  sub    $0xffffffffffffff80,%rdx
  mov    (%rdx),%rdx
  sub    %rdx,%rax
  cmp    %rax,-0x28(%rbp)
  jg     L37
  mov    -0x18(%rbp),%rax
  mov    0x80(%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  sub    $0xffffffffffffff80,%rax
  mov    (%rax),%rcx
  mov    -0x18(%rbp),%rax
  sub    $0xffffffffffffff80,%rax
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  jmp    L13
L37:
  addq   $0xc0,-0x18(%rbp)
L18:
  mov    -0x18(%rbp),%rax
  cmp    %rax,-0x10(%rbp)
  jge    L14
  mov    $0x0,%eax
L13:
  pop    %rbp
  ret
```

assistant: ```c
char *_get_from_sar_array(long size) {
  long i;
  long last_element;
  long *sar_array_element;
  long old_value;
  last_element = *(long *)(0x300003000);
  for (sar_array_element = (long *)(0x300003008); (long)sar_array_element <= last_element; sar_array_element = sar_array_element + 24) {
    if (((*(sar_array_element + 8)) - (*(sar_array_element + 16))) >= size) {
      old_value = *(sar_array_element + 16);
      *(sar_array_element + 16) = *(sar_array_element + 16) + size;
      return (char *)old_value;
    }
  }
  return (char *)0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_get_from_sar_array>:
  endbr64
  movabs 0x300003000,%rax
  movabs $0x300003007,%rdx
  cmp    %rdx,%rax
  jle    L7
  lea    0x1(%rdx),%rdx
L3:
  mov    0x80(%rdx),%rsi
  mov    0x40(%rdx),%rcx
  sub    %rsi,%rcx
  cmp    %rdi,%rcx
  jge    L23
  add    $0xc0,%rdx
  cmp    %rax,%rdx
  jle    L3
  mov    $0x0,%eax
  ret
L23:
  add    %rsi,%rdi
  mov    %rdi,0x80(%rdx)
  mov    %rsi,%rax
  ret
L7:
  mov    $0x0,%eax
  ret
```

assistant: ```c
char *_get_from_sar_array(long size) {
  long i;
  long last_element;
  long *sar_array_element;
  long old_value;
  last_element = *(long *)(0x300003000);
  for (sar_array_element = (long *)(0x300003008); (long)sar_array_element <= last_element; sar_array_element = sar_array_element + 24) {
    if (((*(sar_array_element + 8)) - (*(sar_array_element + 16))) >= size) {
      old_value = *(sar_array_element + 16);
      *(sar_array_element + 16) = *(sar_array_element + 16) + size;
      return (char *)old_value;
    }
  }
  return (char *)0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_get_from_sar_array>:
  endbr64
  movabs 0x300003000,%rax
  movabs $0x300003007,%rdx
  cmp    %rdx,%rax
  jle    L11
  add    $0x1,%rdx
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L20:
  add    $0xc0,%rdx
  cmp    %rax,%rdx
  jg     L11
L31:
  mov    0x80(%rdx),%rsi
  mov    0x40(%rdx),%rcx
  sub    %rsi,%rcx
  cmp    %rdi,%rcx
  jl     L20
  add    %rsi,%rdi
  mov    %rsi,%rax
  mov    %rdi,0x80(%rdx)
  ret
  nopl   (%rax)
L11:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *_get_from_sar_array(long size) {
  long i;
  long last_element;
  long *sar_array_element;
  long old_value;
  last_element = *(long *)(0x300003000);
  for (sar_array_element = (long *)(0x300003008); (long)sar_array_element <= last_element; sar_array_element = sar_array_element + 24) {
    if (((*(sar_array_element + 8)) - (*(sar_array_element + 16))) >= size) {
      old_value = *(sar_array_element + 16);
      *(sar_array_element + 16) = *(sar_array_element + 16) + size;
      return (char *)old_value;
    }
  }
  return (char *)0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<_get_from_sar_array>:
  endbr64
  movabs 0x300003000,%rax
  movabs $0x300003007,%rdx
  cmp    %rdx,%rax
  jle    L8
  add    $0x1,%rdx
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L5:
  add    $0xc0,%rdx
  cmp    %rax,%rdx
  jg     L8
L9:
  mov    0x80(%rdx),%rsi
  mov    0x40(%rdx),%rcx
  sub    %rsi,%rcx
  cmp    %rdi,%rcx
  jl     L5
  add    %rsi,%rdi
  mov    %rsi,%rax
  mov    %rdi,0x80(%rdx)
  ret
  nopl   (%rax)
L8:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *_get_from_sar_array(long size) {
  long i;
  long last_element;
  long *sar_array_element;
  long old_value;
  last_element = *(long *)(0x300003000);
  for (sar_array_element = (long *)(0x300003008); (long)sar_array_element <= last_element; sar_array_element = sar_array_element + 24) {
    if (((*(sar_array_element + 8)) - (*(sar_array_element + 16))) >= size) {
      old_value = *(sar_array_element + 16);
      *(sar_array_element + 16) = *(sar_array_element + 16) + size;
      return (char *)old_value;
    }
  }
  return (char *)0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<copy_if_changed>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  je     L8
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movl   $0x1,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
bool copy_if_changed(void *dst, const void *src, const unsigned int length) {
  bool changed = false;
  if (memcmp(dst, src, length)) {
    memcpy(dst, src, length);
    changed = true;
  }
  return changed;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<copy_if_changed>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %edx,%r12d
  call   <memcmp@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L10
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  mov    %r12d,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    $0x1,%eax
  jmp    L17
```

assistant: ```c
bool copy_if_changed(void *dst, const void *src, const unsigned int length) {
  bool changed = false;
  if (memcmp(dst, src, length)) {
    memcpy(dst, src, length);
    changed = true;
  }
  return changed;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<copy_if_changed>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <memcmp@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L0
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %r13d,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  pop    %rbp
  mov    $0x1,%eax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
bool copy_if_changed(void *dst, const void *src, const unsigned int length) {
  bool changed = false;
  if (memcmp(dst, src, length)) {
    memcpy(dst, src, length);
    changed = true;
  }
  return changed;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<copy_if_changed>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  call   <memcmp@plt>
  mov    %rax,%r8
  xor    %eax,%eax
  test   %r8,%r8
  jne    L9
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  mov    %r13d,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  pop    %rbp
  mov    $0x1,%eax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
bool copy_if_changed(void *dst, const void *src, const unsigned int length) {
  bool changed = false;
  if (memcmp(dst, src, length)) {
    memcpy(dst, src, length);
    changed = true;
  }
  return changed;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    -0x2c(%rbp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    -0x2c(%rbp),%edx
  test   %edx,%edx
  js     L105
  movslq %edx,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L70:
  cmp    %rcx,%rsp
  je     L98
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L70
L105:
  lea    D6(%rip),%rdi
  call   <puts@plt>
  jmp    L82
L98:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L62
  orq    $0x0,-0x8(%rsp,%rax,1)
L62:
  mov    %rsp,%r13
  test   %edx,%edx
  jle    L67
  mov    $0x0,%r12d
  mov    $0x2,%ebx
  jmp    L41
L42:
  add    $0x1,%ebx
  cmp    %r12d,-0x2c(%rbp)
  jle    L67
L41:
  mov    %ebx,%edi
  call   <isPrime@plt>
  test   %rax,%rax
  je     L42
  movslq %r12d,%rax
  mov    %ebx,0x0(%r13,%rax,4)
  lea    0x1(%r12),%r12d
  jmp    L42
L67:
  lea    D34(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x0,-0x2c(%rbp)
  jle    L82
  mov    $0x0,%ebx
  lea    D42(%rip),%r12
L57:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,-0x2c(%rbp)
  jg     L57
L82:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    $0x0,%eax
  lea    -0x18(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D3
type_infered: string
name: unpack_data
arguments: data_label: D24
type_infered: string
name: unpack_data
arguments: data_label: D6
type_infered: string
name: unpack_data
arguments: data_label: D34
type_infered: string
name: unpack_data
arguments: data_label: D42
type_infered: string

tool: D3: How many prime numbers? 

tool: D24: %i

tool: D6: 

  A minimum of one prime number is required.

tool: D34: 


tool: D42: %i 

assistant: ```c
int main(void) {
  int i, n, count;
  printf("How many prime numbers? ");
  scanf("%i", &n);
  if (n < 0) {
    printf("\n\n  A minimum of one prime number is required.\n");
    return 0;
  }
  count = 0;
  int prime[n];
  for (i = 2; count < n; ++i) {
    if (isPrime(i))
      prime[count++] = i;
  }
  printf("\n\n");
  for (i = 0; i < n; ++i)
    printf("%i ", prime[i]);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rdi
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x2c(%rbp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    -0x2c(%rbp),%edx
  test   %edx,%edx
  js     L33
  movslq %edx,%rax
  mov    %rsp,%rsi
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  and    $0xfffffffffffffff0,%rcx
  sub    %rax,%rsi
L36:
  cmp    %rsi,%rsp
  je     L41
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L36
L41:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L108
L76:
  mov    %rsp,%r13
  test   %edx,%edx
  je     L70
  xor    %r12d,%r12d
  mov    $0x2,%ebx
  nopl   (%rax)
L31:
  mov    %ebx,%edi
  call   <isPrime@plt>
  test   %rax,%rax
  je     L79
  movslq %r12d,%rax
  add    $0x1,%r12d
  mov    %ebx,0x0(%r13,%rax,4)
L79:
  add    $0x1,%ebx
  cmp    %r12d,-0x2c(%rbp)
  jg     L31
L70:
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    -0x2c(%rbp),%eax
  test   %eax,%eax
  jle    L106
  xor    %ebx,%ebx
  lea    D22(%rip),%r12
  nopl   (%rax)
L52:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,-0x2c(%rbp)
  jg     L52
L106:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  lea    -0x18(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
L108:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L76
L33:
  lea    D33(%rip),%rdi
  call   <puts@plt>
  jmp    L106
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D4
datatype: string
name: parse_data
arguments: label: D3
datatype: string
name: parse_data
arguments: label: D19
datatype: string
name: parse_data
arguments: label: D22
datatype: string
name: parse_data
arguments: label: D33
datatype: string

tool: D4: How many prime numbers? 

tool: D3: %i

tool: D19: 


tool: D22: %i 

tool: D33: 

  A minimum of one prime number is required.

assistant: ```c
int main(void) {
  int i, n, count;
  printf("How many prime numbers? ");
  scanf("%i", &n);
  if (n < 0) {
    printf("\n\n  A minimum of one prime number is required.\n");
    return 0;
  }
  count = 0;
  int prime[n];
  for (i = 2; count < n; ++i) {
    if (isPrime(i))
      prime[count++] = i;
  }
  printf("\n\n");
  for (i = 0; i < n; ++i)
    printf("%i ", prime[i]);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D16(%rip),%rdi
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    -0x2c(%rbp),%rsi
  lea    D31(%rip),%rdi
  call   <scanf@plt>
  mov    -0x2c(%rbp),%edx
  test   %edx,%edx
  js     L25
  movslq %edx,%rax
  mov    %rsp,%rsi
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  and    $0xfffffffffffffff0,%rcx
  sub    %rax,%rsi
L44:
  cmp    %rsi,%rsp
  je     L38
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L44
L38:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L139
L79:
  mov    %rsp,%r13
  test   %edx,%edx
  je     L4
  xor    %r12d,%r12d
  mov    $0x2,%ebx
  nopl   (%rax)
L20:
  mov    %ebx,%edi
  call   <isPrime@plt>
  test   %rax,%rax
  je     L98
L52:
  mov    %ebx,0x0(%r13,%r12,4)
  lea    0x1(%r12),%eax
  add    $0x1,%ebx
  cmp    %eax,-0x2c(%rbp)
  jle    L4
  mov    %ebx,%edi
  movslq %eax,%r12
  call   <isPrime@plt>
  test   %rax,%rax
  jne    L52
L98:
  add    $0x1,%ebx
  cmp    %r12d,-0x2c(%rbp)
  jg     L20
L4:
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    -0x2c(%rbp),%eax
  test   %eax,%eax
  jle    L14
  xor    %ebx,%ebx
  lea    D17(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L95:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,-0x2c(%rbp)
  jg     L95
L14:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L102
  lea    -0x18(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
L139:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L79
L25:
  lea    D21(%rip),%rdi
  call   <puts@plt>
  jmp    L14
L102:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D16
type_infered: string
name: unpack_data
arguments: data_label: D31
type_infered: string
name: unpack_data
arguments: data_label: D11
type_infered: string
name: unpack_data
arguments: data_label: D17
type_infered: string
name: unpack_data
arguments: data_label: D21
type_infered: string

tool: D16: How many prime numbers? 

tool: D31: %i

tool: D11: 


tool: D17: %i 

tool: D21: 

  A minimum of one prime number is required.

assistant: ```c
int main(void) {
  int i, n, count;
  printf("How many prime numbers? ");
  scanf("%i", &n);
  if (n < 0) {
    printf("\n\n  A minimum of one prime number is required.\n");
    return 0;
  }
  count = 0;
  int prime[n];
  for (i = 2; count < n; ++i) {
    if (isPrime(i))
      prime[count++] = i;
  }
  printf("\n\n");
  for (i = 0; i < n; ++i)
    printf("%i ", prime[i]);
  return 0;
}
```
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    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <doit@plt>
  mov    $0x0,%eax
  leave
  ret
```

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

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

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

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

assistant: ```c
int main(int argc, char **argv) {
  doit(argc);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<trim_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L31
  mov    -0x18(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <reverse_line@plt>
  jmp    L61
L62:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x4(%rbp)
L61:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L62
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x9,%al
  je     L62
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L62
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L62
  subl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0xa,(%rax)
  subl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <reverse_line@plt>
  jmp    L16
L31:
  nop
L16:
  leave
  ret
```

assistant: ```c
void trim_line(char line[]) {
  int i = 0;
  if (line[1] == '\0')
    return;
  reverse_line(line, 1000);
  while (line[i] == ' ' || line[i] == '\t' || line[i] == '\n' || line[i] == '\0') {
    line[i] = '\0';
    i++;
  }
  i--;
  line[i] = '\n';
  i--;
  line[i] = '\0';
  reverse_line(line, 1000);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<trim_line>:
  endbr64
  cmpb   $0x0,0x1(%rdi)
  jne    L39
  ret
L39:
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x3e8,%esi
  call   <reverse_line@plt>
  movzbl (%rbx),%eax
  lea    -0x9(%rax),%edx
  cmp    $0x1,%dl
  jbe    L47
  test   $0xdf,%al
  jne    L44
L47:
  mov    %rbx,%rax
  mov    $0x1,%edi
  sub    %ebx,%edi
L22:
  movb   $0x0,(%rax)
  lea    (%rdi,%rax,1),%esi
  add    $0x1,%rax
  movzbl (%rax),%edx
  lea    -0x9(%rdx),%ecx
  cmp    $0x1,%cl
  jbe    L22
  test   $0xdf,%dl
  je     L22
L0:
  lea    -0x1(%rsi),%eax
  cltq
  movb   $0xa,(%rbx,%rax,1)
  sub    $0x2,%esi
  movslq %esi,%rsi
  movb   $0x0,(%rbx,%rsi,1)
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <reverse_line@plt>
  pop    %rbx
  ret
L44:
  mov    $0x0,%esi
  jmp    L0
```

assistant: ```c
void trim_line(char line[]) {
  int i = 0;
  if (line[1] == '\0')
    return;
  reverse_line(line, 1000);
  while (line[i] == ' ' || line[i] == '\t' || line[i] == '\n' || line[i] == '\0') {
    line[i] = '\0';
    i++;
  }
  i--;
  line[i] = '\n';
  i--;
  line[i] = '\0';
  reverse_line(line, 1000);
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<trim_line>:
  endbr64
  cmpb   $0x0,0x1(%rdi)
  jne    L26
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  push   %rbp
  mov    $0x3e8,%esi
  mov    %rdi,%rbp
  call   <reverse_line@plt>
  movzbl 0x0(%rbp),%eax
  lea    -0x9(%rax),%edx
  cmp    $0x1,%dl
  jbe    L9
  test   $0xdf,%al
  jne    L22
L9:
  mov    %rbp,%rax
  xor    %edx,%edx
  nopl   0x0(%rax,%rax,1)
L20:
  movb   $0x0,(%rax)
  movzbl 0x1(%rax),%ecx
  mov    %edx,%r8d
  add    $0x1,%rax
  add    $0x1,%edx
  lea    -0x9(%rcx),%esi
  cmp    $0x1,%sil
  jbe    L20
  and    $0xdf,%ecx
  je     L20
  movslq %r8d,%rax
  sub    $0x1,%r8d
  movslq %r8d,%r8
L25:
  movb   $0xa,0x0(%rbp,%rax,1)
  mov    %rbp,%rdi
  mov    $0x3e8,%esi
  movb   $0x0,0x0(%rbp,%r8,1)
  pop    %rbp
  jmp    1050 <reverse_line@plt>
L22:
  mov    $0xfffffffffffffffe,%r8
  mov    $0xffffffffffffffff,%rax
  jmp    L25
```

assistant: ```c
void trim_line(char line[]) {
  int i = 0;
  if (line[1] == '\0')
    return;
  reverse_line(line, 1000);
  while (line[i] == ' ' || line[i] == '\t' || line[i] == '\n' || line[i] == '\0') {
    line[i] = '\0';
    i++;
  }
  i--;
  line[i] = '\n';
  i--;
  line[i] = '\0';
  reverse_line(line, 1000);
  return;
}
```