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
<escape_space>:
  endbr64
  push   %r12
  lea    D5(%rip),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L8:
  movb   $0x0,(%rax)
  lea    0x1(%rax),%rbp
  mov    %rbx,%rdi
  mov    %rbp,%rsi
  call   <strcat@plt>
L9:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <strstr@plt>
  test   %rax,%rax
  jne    L8
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_label: D5
type_infered: string

tool: D5: \ 

assistant: ```c
void escape_space(char *buf) {
  char *p1, *p2;
  p1 = p2 = buf;
  while ((p1 = strstr(p1, "\\ ")) != NULL) {
    *p1++ = 0;
    strcat(buf, p1);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<h2o_mem_alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L7
  lea    D9(%rip),%rax
  mov    %rax,%rdx
  mov    $0x196,%esi
  lea    D15(%rip),%rax
  mov    %rax,%rdi
  call   <h2o__fatal@plt>
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type_guessed: string
name: retrieve_data
arguments: data_tag: D15
type_guessed: string

tool: D9: no memory

tool: D15: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/devnexen/h2o/refs/heads/master/include/h2o/memory.h

assistant: ```c
void *h2o_mem_alloc(size_t sz) {
  void *p = malloc(sz);
  if (p == NULL)
    h2o__fatal("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/devnexen/h2o/refs/heads/master/include/h2o/memory.h", 406, "no memory");
  return p;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<h2o_mem_alloc>:
  endbr64
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L9
L7:
  mov    %rbx,%rax
  pop    %rbx
  ret
L9:
  lea    D2(%rip),%rdx
  mov    $0x196,%esi
  lea    D16(%rip),%rdi
  call   <h2o__fatal@plt>
  jmp    L7
```

tool_calls: name: unpack_data
arguments: data_label: D2
data_type: string
name: unpack_data
arguments: data_label: D16
data_type: string

tool: D2: no memory

tool: D16: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/devnexen/h2o/refs/heads/master/include/h2o/memory.h

assistant: ```c
void *h2o_mem_alloc(size_t sz) {
  void *p = malloc(sz);
  if (p == NULL)
    h2o__fatal("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/devnexen/h2o/refs/heads/master/include/h2o/memory.h", 406, "no memory");
  return p;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<h2o_mem_alloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L17
L19:
  mov    %r12,%rax
  pop    %r12
  ret
L17:
  lea    D6(%rip),%rdx
  mov    $0x196,%esi
  lea    D1(%rip),%rdi
  call   <h2o__fatal@plt>
  jmp    L19
```

tool_calls: name: parse_data
arguments: tag: D6
datatype: string
name: parse_data
arguments: tag: D1
datatype: string

tool: D6: no memory

tool: D1: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/devnexen/h2o/refs/heads/master/include/h2o/memory.h

assistant: ```c
void *h2o_mem_alloc(size_t sz) {
  void *p = malloc(sz);
  if (p == NULL)
    h2o__fatal("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/devnexen/h2o/refs/heads/master/include/h2o/memory.h", 406, "no memory");
  return p;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<h2o_mem_alloc>:
  endbr64
  push   %r12
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L14
L15:
  mov    %r12,%rax
  pop    %r12
  ret
L14:
  lea    D1(%rip),%rdx
  mov    $0x196,%esi
  lea    D9(%rip),%rdi
  call   <h2o__fatal@plt>
  jmp    L15
```

tool_calls: name: unpack_data
arguments: data_label: D1
data_type: string
name: unpack_data
arguments: data_label: D9
data_type: string

tool: D1: no memory

tool: D9: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/devnexen/h2o/refs/heads/master/include/h2o/memory.h

assistant: ```c
void *h2o_mem_alloc(size_t sz) {
  void *p = malloc(sz);
  if (p == NULL)
    h2o__fatal("/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/devnexen/h2o/refs/heads/master/include/h2o/memory.h", 406, "no memory");
  return p;
}
```
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   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xa,-0x8(%rbp)
  movl   $0x3,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D8
type: string

tool: D8: Contenuto di n: %d
%d

assistant: ```c
int main() {
  int n = 10;
  int i = 3;
  printf("Contenuto di n: %d\n%d", n, i);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%edx
  mov    $0xa,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
type_guessed: string

tool: D8: Contenuto di n: %d
%d

assistant: ```c
int main() {
  int n = 10;
  int i = 3;
  printf("Contenuto di n: %d\n%d", n, i);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%edx
  mov    $0xa,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D5
type_infered: string

tool: D5: Contenuto di n: %d
%d

assistant: ```c
int main() {
  int n = 10;
  int i = 3;
  printf("Contenuto di n: %d\n%d", n, i);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%edx
  mov    $0xa,%esi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D5
type_infered: string

tool: D5: Contenuto di n: %d
%d

assistant: ```c
int main() {
  int n = 10;
  int i = 3;
  printf("Contenuto di n: %d\n%d", n, i);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<print_time>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x0,%esi
  test   %edi,%edi
  js     L47
L25:
  movslq %edi,%rdx
  imul   $0xffffffff91a2b3c5,%rdx,%rdx
  shr    $0x20,%rdx
  add    %edi,%edx
  sar    $0xb,%edx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%edx
  imul   $0xe10,%edx,%eax
  sub    %eax,%edi
  movslq %edi,%rax
  imul   $0xffffffff88888889,%rax,%rax
  shr    $0x20,%rax
  add    %edi,%eax
  sar    $0x5,%eax
  mov    %edi,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x3c,%eax,%ecx
  sub    %ecx,%edi
  movslq %edi,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  mov    %edi,%r8d
  sar    $0x1f,%r8d
  sub    %r8d,%ecx
  lea    (%rcx,%rcx,4),%r8d
  add    %r8d,%r8d
  sub    %r8d,%edi
  add    $0x30,%edi
  mov    %dil,(%rsp)
  add    $0x30,%ecx
  mov    %cl,0x1(%rsp)
  movb   $0x3a,0x2(%rsp)
  movslq %eax,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  mov    %eax,%edi
  sar    $0x1f,%edi
  sub    %edi,%ecx
  lea    (%rcx,%rcx,4),%edi
  add    %edi,%edi
  sub    %edi,%eax
  add    $0x30,%eax
  mov    %al,0x3(%rsp)
  add    $0x30,%ecx
  mov    %cl,0x4(%rsp)
  mov    $0x5,%r8d
  test   %edx,%edx
  je     L15
  movb   $0x3a,0x5(%rsp)
  mov    $0x7,%ecx
  lea    -0x1(%rsp),%r9
L8:
  mov    %ecx,%r8d
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%edi
  sar    $0x1f,%edi
  sub    %edi,%eax
  lea    (%rax,%rax,4),%edi
  add    %edi,%edi
  sub    %edi,%edx
  add    $0x30,%edx
  mov    %dl,(%r9,%rcx,1)
  mov    %eax,%edx
  add    $0x1,%rcx
  test   %eax,%eax
  jne    L8
L15:
  test   %esi,%esi
  je     L7
  movslq %r8d,%rax
  movb   $0x2d,(%rsp,%rax,1)
  lea    0x1(%r8),%r8d
L7:
  mov    %rsp,%rdi
  mov    %r8d,%esi
  call   <stack_print@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x38,%rsp
  ret
L47:
  neg    %edi
  mov    $0x1,%esi
  jmp    L25
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void print_time(int t) {
  int h, m, s;
  char stack[32];
  int neg = 0;
  int p = 0;
  if (t < 0) {
    neg = 1;
    t *= -1;
  }
  h = t / 3600;
  t = t % 3600;
  m = t / 60;
  s = t % 60;
  stack[p++] = s % 10 + '0';
  stack[p++] = s / 10 + '0';
  stack[p++] = ':';
  stack[p++] = m % 10 + '0';
  stack[p++] = m / 10 + '0';
  if (h) {
    stack[p++] = ':';
    do {
      stack[p++] = h % 10 + '0';
      h /= 10;
    } while (h);
  }
  if (neg)
    stack[p++] = '-';
  stack_print(stack, p);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<print_time>:
  endbr64
  sub    $0x38,%rsp
  xor    %r8d,%r8d
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jns    L37
  neg    %edi
  mov    $0x1,%r8d
L37:
  mov    %edi,%eax
  mov    $0x91a2b3c5,%edx
  mov    $0x88888889,%ecx
  mov    $0xcccccccd,%r9d
  imul   %rdx,%rax
  movb   $0x3a,0x2(%rsp)
  mov    $0x5,%r10d
  shr    $0x2b,%rax
  imul   $0xe10,%eax,%edx
  sub    %edx,%edi
  mov    %edi,%edx
  imul   %rcx,%rdx
  shr    $0x25,%rdx
  imul   $0x3c,%edx,%ecx
  sub    %ecx,%edi
  mov    %edi,%ecx
  imul   %r9,%rcx
  shr    $0x23,%rcx
  lea    (%rcx,%rcx,4),%esi
  add    $0x30,%ecx
  mov    %cl,0x1(%rsp)
  mov    %edx,%ecx
  add    %esi,%esi
  imul   %r9,%rcx
  sub    %esi,%edi
  add    $0x30,%edi
  mov    %dil,(%rsp)
  shr    $0x23,%rcx
  lea    (%rcx,%rcx,4),%esi
  add    $0x30,%ecx
  add    %esi,%esi
  mov    %cl,0x4(%rsp)
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,0x3(%rsp)
  test   %eax,%eax
  je     L3
  movb   $0x3a,0x5(%rsp)
  mov    $0x7,%ecx
  lea    -0x1(%rsp),%rdi
  nopl   0x0(%rax)
L17:
  mov    %eax,%edx
  mov    %ecx,%r10d
  imul   %r9,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%esi
  add    %esi,%esi
  sub    %esi,%eax
  add    $0x30,%eax
  mov    %al,(%rdi,%rcx,1)
  mov    %edx,%eax
  add    $0x1,%rcx
  test   %edx,%edx
  jne    L17
L3:
  test   %r8d,%r8d
  je     L14
  movslq %r10d,%rax
  add    $0x1,%r10d
  movb   $0x2d,(%rsp,%rax,1)
L14:
  mov    %rsp,%rdi
  mov    %r10d,%esi
  call   <stack_print@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L32
  add    $0x38,%rsp
  ret
L32:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void print_time(int t) {
  int h, m, s;
  char stack[32];
  int neg = 0;
  int p = 0;
  if (t < 0) {
    neg = 1;
    t *= -1;
  }
  h = t / 3600;
  t = t % 3600;
  m = t / 60;
  s = t % 60;
  stack[p++] = s % 10 + '0';
  stack[p++] = s / 10 + '0';
  stack[p++] = ':';
  stack[p++] = m % 10 + '0';
  stack[p++] = m / 10 + '0';
  if (h) {
    stack[p++] = ':';
    do {
      stack[p++] = h % 10 + '0';
      h /= 10;
    } while (h);
  }
  if (neg)
    stack[p++] = '-';
  stack_print(stack, p);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_time>:
  endbr64
  push   %rbx
  xor    %r8d,%r8d
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  test   %edi,%edi
  jns    L24
  neg    %edi
  mov    $0x1,%r8d
L24:
  mov    %edi,%eax
  mov    $0x91a2b3c5,%edx
  mov    $0x88888889,%ecx
  xor    %ebx,%ebx
  imul   %rdx,%rax
  mov    $0xcccccccd,%r9d
  movb   $0x3a,0x2(%rsp)
  mov    $0x5,%r10d
  shr    $0x2b,%rax
  imul   $0xe10,%eax,%edx
  sub    %edx,%edi
  mov    %edi,%edx
  imul   %rcx,%rdx
  shr    $0x25,%rdx
  imul   $0x3c,%edx,%ecx
  sub    %ecx,%edi
  mov    %edi,%ecx
  imul   %r9,%rcx
  shr    $0x23,%rcx
  lea    (%rcx,%rcx,4),%esi
  add    $0x30,%ecx
  add    %esi,%esi
  sub    %esi,%edi
  add    $0x30,%edi
  mov    %dil,%bl
  mov    %cl,%bh
  mov    %edx,%ecx
  imul   %r9,%rcx
  mov    %bx,(%rsp)
  shr    $0x23,%rcx
  lea    (%rcx,%rcx,4),%esi
  add    $0x30,%ecx
  add    %esi,%esi
  mov    %cl,0x4(%rsp)
  sub    %esi,%edx
  add    $0x30,%edx
  mov    %dl,0x3(%rsp)
  test   %eax,%eax
  je     L26
  movb   $0x3a,0x5(%rsp)
  mov    $0x7,%ecx
  lea    -0x1(%rsp),%rdi
  nopl   0x0(%rax,%rax,1)
L40:
  mov    %eax,%edx
  mov    %ecx,%r10d
  imul   %r9,%rdx
  shr    $0x23,%rdx
  lea    (%rdx,%rdx,4),%esi
  add    %esi,%esi
  sub    %esi,%eax
  add    $0x30,%eax
  mov    %al,(%rdi,%rcx,1)
  mov    %edx,%eax
  add    $0x1,%rcx
  test   %edx,%edx
  jne    L40
L26:
  test   %r8d,%r8d
  je     L14
  movslq %r10d,%rax
  add    $0x1,%r10d
  movb   $0x2d,(%rsp,%rax,1)
L14:
  mov    %rsp,%rdi
  mov    %r10d,%esi
  call   <stack_print@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L45
  add    $0x30,%rsp
  pop    %rbx
  ret
L45:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void print_time(int t) {
  int h, m, s;
  char stack[32];
  int neg = 0;
  int p = 0;
  if (t < 0) {
    neg = 1;
    t *= -1;
  }
  h = t / 3600;
  t = t % 3600;
  m = t / 60;
  s = t % 60;
  stack[p++] = s % 10 + '0';
  stack[p++] = s / 10 + '0';
  stack[p++] = ':';
  stack[p++] = m % 10 + '0';
  stack[p++] = m / 10 + '0';
  if (h) {
    stack[p++] = ':';
    do {
      stack[p++] = h % 10 + '0';
      h /= 10;
    } while (h);
  }
  if (neg)
    stack[p++] = '-';
  stack_print(stack, p);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x4(%rbp)
  jmp    L25
L36:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L38
  mov    $0x0,%eax
  jmp    L12
L38:
  addl   $0x1,-0x4(%rbp)
L25:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L36
  mov    $0x1,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int prime(int n) {
  int a = 2;
  while (a < n) {
    if (n % a == 0)
      return 0;
    a++;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L23
  test   $0x1,%dil
  je     L43
  mov    $0x2,%ecx
L18:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  je     L46
  mov    %edi,%eax
  cltd
  idiv   %ecx
  mov    %edx,%eax
  test   %edx,%edx
  jne    L18
L26:
  ret
L46:
  mov    $0x1,%eax
  ret
L23:
  mov    $0x1,%eax
  ret
L43:
  mov    $0x0,%eax
  jmp    L26
```

assistant: ```c
int prime(int n) {
  int a = 2;
  while (a < n) {
    if (n % a == 0)
      return 0;
    a++;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L39
  test   $0x1,%dil
  je     L3
  mov    $0x2,%ecx
  jmp    L38
  cs nopw 0x0(%rax,%rax,1)
L27:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L29
L38:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L27
L39:
  mov    $0x1,%edx
L29:
  mov    %edx,%eax
  ret
L3:
  xor    %edx,%edx
  jmp    L29
```

assistant: ```c
int prime(int n) {
  int a = 2;
  while (a < n) {
    if (n % a == 0)
      return 0;
    a++;
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L58
  test   $0x1,%dil
  je     L41
  mov    $0x2,%ecx
  jmp    L22
  cs nopw 0x0(%rax,%rax,1)
L34:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L1
L22:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L34
L58:
  mov    $0x1,%edx
L1:
  mov    %edx,%eax
  ret
L41:
  xor    %edx,%edx
  jmp    L1
```

assistant: ```c
int prime(int n) {
  int a = 2;
  while (a < n) {
    if (n % a == 0)
      return 0;
    a++;
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<file_exist>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <stat@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int file_exist(char *filename) {
  struct stat buffer;
  return (stat(filename, &buffer) == 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<file_exist>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %rax,%rax
  sete   %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int file_exist(char *filename) {
  struct stat buffer;
  return (stat(filename, &buffer) == 0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<file_exist>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %rax,%rax
  sete   %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int file_exist(char *filename) {
  struct stat buffer;
  return (stat(filename, &buffer) == 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<file_exist>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %rax,%rax
  sete   %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int file_exist(char *filename) {
  struct stat buffer;
  return (stat(filename, &buffer) == 0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<for_loop_exits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L16:
  addl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x9,-0x4(%rbp)
  jle    L16
  mov    $0x0,%eax
  call   <clang_analyzer_warnIfReached@plt>
  nop
  leave
  ret
```

assistant: ```c
void for_loop_exits() {
  int i;
  for (i = 0; i < 10; ++i) {
  }
  clang_analyzer_warnIfReached();
}
```
user: 请将以下的汇编代码转换成C语言代码：

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

assistant: ```c
void for_loop_exits() {
  int i;
  for (i = 0; i < 10; ++i) {
  }
  clang_analyzer_warnIfReached();
}
```
user: What is the c source code of the assembly code below:

```asm
<for_loop_exits>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <clang_analyzer_warnIfReached@plt>
```

assistant: ```c
void for_loop_exits() {
  int i;
  for (i = 0; i < 10; ++i) {
  }
  clang_analyzer_warnIfReached();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<for_loop_exits>:
  endbr64
  xor    %eax,%eax
  jmp    1050 <clang_analyzer_warnIfReached@plt>
```

assistant: ```c
void for_loop_exits() {
  int i;
  for (i = 0; i < 10; ++i) {
  }
  clang_analyzer_warnIfReached();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<divide>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movsd  -0x18(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jp     L26
  pxor   %xmm1,%xmm1
  ucomisd %xmm1,%xmm0
  jne    L26
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L8
L26:
  movsd  -0x20(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  movsd  -0x10(%rbp),%xmm1
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
L8:
  lea    D35(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L25
  call   <__stack_chk_fail@plt>
L25:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D9
datatype: string
name: retrieve_data
arguments: tag: D33
datatype: string
name: retrieve_data
arguments: tag: D1
datatype: string
name: retrieve_data
arguments: tag: D35
datatype: string
name: retrieve_data
arguments: tag: D25
datatype: string

tool: D9: Division!
 please input two numbers:

tool: D33: %lf %lf

tool: D1: Error:divisor can not be zero!

tool: D35: ===========================================

tool: D25: %lf / %lf = %lf


assistant: ```c
int divide() {
  double num1, num2, div;
  printf("Division!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  if (num2 == 0) {
    printf("Error:divisor can not be zero!\n");
    printf("===========================================\n");
  } else {
    div = num1 / num2;
    printf("%lf / %lf = %lf\n", num1, num2, div);
  }
  printf("===========================================\n");
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<divide>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D52(%rip),%rdi
  call   <scanf@plt>
  movsd  0x10(%rsp),%xmm1
  ucomisd D13(%rip),%xmm1
  jp     L19
  jne    L19
  lea    D45(%rip),%rdi
  call   <puts@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
L23:
  lea    D1(%rip),%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L19:
  movsd  0x8(%rsp),%xmm0
  movapd %xmm0,%xmm2
  divsd  %xmm1,%xmm2
  lea    D39(%rip),%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L23
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
data_type: string
name: extract_symbolic_data
arguments: data_label: D52
data_type: string
name: extract_symbolic_data
arguments: data_label: D13
data_type: byte[8]
name: extract_symbolic_data
arguments: data_label: D45
data_type: string
name: extract_symbolic_data
arguments: data_label: D1
data_type: string
name: extract_symbolic_data
arguments: data_label: D39
data_type: string

tool: D8: Division!
 please input two numbers:

tool: D52: %lf %lf

tool: {"D13": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D45: Error:divisor can not be zero!

tool: D1: ===========================================

tool: D39: %lf / %lf = %lf


assistant: ```c
int divide() {
  double num1, num2, div;
  printf("Division!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  if (num2 == 0) {
    printf("Error:divisor can not be zero!\n");
    printf("===========================================\n");
  } else {
    div = num1 / num2;
    printf("%lf / %lf = %lf\n", num1, num2, div);
  }
  printf("===========================================\n");
  return 0;
}
```