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

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D7(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D32(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <espar@plt>
  test   %rax,%rax
  jne    L24
  mov    $0x1,%ebx
  cmpl   $0x1,0x4(%rsp)
  jg     L63
  jmp    L0
L24:
  mov    0x4(%rsp),%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L0:
  lea    D20(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L18:
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jle    L0
L63:
  mov    %ebx,%edi
  call   <esprimo@plt>
  test   %rax,%rax
  je     L18
  mov    0x4(%rsp),%esi
  mov    %ebx,%edi
  call   <goldbach@plt>
  jmp    L18
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D7
type_guessed: string
name: disassemble_data
arguments: data_label: D32
type_guessed: string
name: disassemble_data
arguments: data_label: D16
type_guessed: string
name: disassemble_data
arguments: data_label: D20
type_guessed: string

tool: D7: Introduce un numero par 

tool: D32: %d

tool: D16: El numero %d no es par 


tool: D20: PAUSE

assistant: ```c
int main(void) {
  int num;
  int i = 1;
  printf("Introduce un numero par \n");
  scanf("%d", &num);
  if (espar(num)) {
    printf("El numero %d no es par \n", num);
  } else {
    while (i < num) {
      if (esprimo(i)) {
        goldbach(i, num);
      }
      i++;
    }
  }
  system("PAUSE");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbx
  lea    D0(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D25(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <espar@plt>
  test   %rax,%rax
  jne    L20
  cmpl   $0x1,0x4(%rsp)
  mov    $0x1,%ebx
  jg     L24
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L58:
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jle    L34
L24:
  mov    %ebx,%edi
  call   <esprimo@plt>
  test   %rax,%rax
  je     L58
  mov    0x4(%rsp),%esi
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <goldbach@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L24
  nopl   0x0(%rax)
L34:
  lea    D22(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L40
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
L20:
  mov    0x4(%rsp),%esi
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L34
L40:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D0
type_guessed: string
name: unpack_data
arguments: label: D25
type_guessed: string
name: unpack_data
arguments: label: D22
type_guessed: string
name: unpack_data
arguments: label: D5
type_guessed: string

tool: D0: Introduce un numero par 

tool: D25: %d

tool: D22: PAUSE

tool: D5: El numero %d no es par 


assistant: ```c
int main(void) {
  int num;
  int i = 1;
  printf("Introduce un numero par \n");
  scanf("%d", &num);
  if (espar(num)) {
    printf("El numero %d no es par \n", num);
  } else {
    while (i < num) {
      if (esprimo(i)) {
        goldbach(i, num);
      }
      i++;
    }
  }
  system("PAUSE");
  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   %rbx
  lea    D23(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D30(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <espar@plt>
  test   %rax,%rax
  jne    L73
  cmpl   $0x1,0x4(%rsp)
  mov    $0x1,%ebx
  jle    L64
L40:
  mov    %ebx,%edi
  call   <esprimo@plt>
  test   %rax,%rax
  jne    L59
  add    $0x1,%ebx
  cmp    %ebx,0x4(%rsp)
  jg     L40
L64:
  lea    D1(%rip),%rdi
  call   <system@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L59:
  mov    0x4(%rsp),%esi
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <goldbach@plt>
  cmp    0x4(%rsp),%ebx
  jl     L40
  jmp    L64
L73:
  mov    0x4(%rsp),%esi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L64
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D23
datatype: string
name: parse_data
arguments: label: D30
datatype: string
name: parse_data
arguments: label: D1
datatype: string
name: parse_data
arguments: label: D13
datatype: string

tool: D23: Introduce un numero par 

tool: D30: %d

tool: D1: PAUSE

tool: D13: El numero %d no es par 


assistant: ```c
int main(void) {
  int num;
  int i = 1;
  printf("Introduce un numero par \n");
  scanf("%d", &num);
  if (espar(num)) {
    printf("El numero %d no es par \n", num);
  } else {
    while (i < num) {
      if (esprimo(i)) {
        goldbach(i, num);
      }
      i++;
    }
  }
  system("PAUSE");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memmove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jae    L9
  mov    -0x8(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x10(%rbp)
  jae    L9
  mov    -0x28(%rbp),%rax
  add    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  add    %rax,-0x10(%rbp)
  jmp    L21
L30:
  subq   $0x1,-0x8(%rbp)
  subq   $0x1,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
L21:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L30
  mov    -0x18(%rbp),%rax
  jmp    L40
L9:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <do_memcpy@plt>
L40:
  leave
  ret
```

assistant: ```c
void *memmove(void *dst, const void *src, size_t len) {
  char *d = dst;
  const char *s = src;
  if (s < d && d < s + len) {
    s += len;
    d += len;
    while (len--)
      *--d = *--s;
    return dst;
  } else
    return do_memcpy(dst, src, len);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memmove>:
  endbr64
  cmp    %rsi,%rdi
  jbe    L22
  lea    (%rsi,%rdx,1),%rax
  cmp    %rax,%rdi
  jae    L22
  lea    -0x1(%rdx),%rcx
  test   %rdx,%rdx
  je     L31
L12:
  movzbl (%rsi,%rcx,1),%eax
  mov    %al,(%rdi,%rcx,1)
  sub    $0x1,%rcx
  cmp    $0xffffffffffffffff,%rcx
  jne    L12
  mov    %rdi,%rax
  ret
L22:
  sub    $0x8,%rsp
  call   <do_memcpy@plt>
  add    $0x8,%rsp
  ret
L31:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memmove(void *dst, const void *src, size_t len) {
  char *d = dst;
  const char *s = src;
  if (s < d && d < s + len) {
    s += len;
    d += len;
    while (len--)
      *--d = *--s;
    return dst;
  } else
    return do_memcpy(dst, src, len);
}
```
user: Translate the below assembly code into C programming language:

```asm
<memmove>:
  endbr64
  cmp    %rsi,%rdi
  jbe    L9
  lea    (%rsi,%rdx,1),%rax
  cmp    %rax,%rdi
  jae    L9
  lea    -0x1(%rdx),%rax
  test   %rdx,%rdx
  je     L16
  nopl   0x0(%rax,%rax,1)
L2:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  sub    $0x1,%rax
  jae    L2
L16:
  mov    %rdi,%rax
  ret
  nopl   0x0(%rax)
L9:
  jmp    1050 <do_memcpy@plt>
```

assistant: ```c
void *memmove(void *dst, const void *src, size_t len) {
  char *d = dst;
  const char *s = src;
  if (s < d && d < s + len) {
    s += len;
    d += len;
    while (len--)
      *--d = *--s;
    return dst;
  } else
    return do_memcpy(dst, src, len);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memmove>:
  endbr64
  cmp    %rsi,%rdi
  jbe    L13
  lea    (%rsi,%rdx,1),%rax
  cmp    %rax,%rdi
  jae    L13
  lea    -0x1(%rdx),%rax
  test   %rdx,%rdx
  je     L21
  nopl   0x0(%rax,%rax,1)
L39:
  movzbl (%rsi,%rax,1),%edx
  mov    %dl,(%rdi,%rax,1)
  sub    $0x1,%rax
  jae    L39
L21:
  mov    %rdi,%rax
  ret
  nopl   0x0(%rax)
L13:
  jmp    1050 <do_memcpy@plt>
```

assistant: ```c
void *memmove(void *dst, const void *src, size_t len) {
  char *d = dst;
  const char *s = src;
  if (s < d && d < s + len) {
    s += len;
    d += len;
    while (len--)
      *--d = *--s;
    return dst;
  } else
    return do_memcpy(dst, src, len);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parport_init>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x1,%edx
  mov    $0x2,%esi
  mov    $0x378,%edi
  call   <ioperm@plt>
  test   %rax,%rax
  je     L0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L0:
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D9
datatype: string

tool: D9: Can't get I/O privileges for parallel port

assistant: ```c
void parport_init() {
  if (ioperm(0x378, 0x379 - 0x378 + 1, 1)) {
    perror("Can't get I/O privileges for parallel port");
    exit(1);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<parport_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  mov    $0x2,%esi
  mov    $0x378,%edi
  call   <ioperm@plt>
  test   %rax,%rax
  jne    L7
L11:
  add    $0x8,%rsp
  ret
L7:
  lea    D9(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L11
```

tool_calls: name: parse_data
arguments: label: D9
type_guessed: string

tool: D9: Can't get I/O privileges for parallel port

assistant: ```c
void parport_init() {
  if (ioperm(0x378, 0x379 - 0x378 + 1, 1)) {
    perror("Can't get I/O privileges for parallel port");
    exit(1);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<parport_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  mov    $0x2,%esi
  mov    $0x378,%edi
  call   <ioperm@plt>
  test   %rax,%rax
  jne    L2
  add    $0x8,%rsp
  ret
  cs nopw 0x0(%rax,%rax,1)
L2:
  lea    D6(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D6
type_infered: string

tool: D6: Can't get I/O privileges for parallel port

assistant: ```c
void parport_init() {
  if (ioperm(0x378, 0x379 - 0x378 + 1, 1)) {
    perror("Can't get I/O privileges for parallel port");
    exit(1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<parport_init>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edx
  mov    $0x2,%esi
  mov    $0x378,%edi
  call   <ioperm@plt>
  test   %rax,%rax
  jne    L9
  add    $0x8,%rsp
  ret
  cs nopw 0x0(%rax,%rax,1)
L9:
  lea    D0(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1090 <exit@plt>
```

tool_calls: name: unpack_data
arguments: label: D0
type: string

tool: D0: Can't get I/O privileges for parallel port

assistant: ```c
void parport_init() {
  if (ioperm(0x378, 0x379 - 0x378 + 1, 1)) {
    perror("Can't get I/O privileges for parallel port");
    exit(1);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcpyArray>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <assert@plt>
  movl   $0x0,-0x1c(%rbp)
  nop
L14:
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L14
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L5
L15:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x18(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  addl   $0x1,-0x1c(%rbp)
L5:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L15
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movq   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char **strcpyArray(char **strings) {
  assert(strings);
  int count = 0;
  while (strings[count++])
    ;
  char **newArray = malloc(sizeof(char *) * count);
  count = 0;
  while (strings[count]) {
    newArray[count] = malloc(sizeof(char) * (1 + strlen(strings[count])));
    strcpy(newArray[count], strings[count]);
    ++count;
  }
  newArray[count] = 0;
  return newArray;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcpyArray>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <assert@plt>
  mov    $0x1,%eax
L23:
  mov    %eax,%edi
  add    $0x1,%rax
  cmpq   $0x0,-0x10(%rbp,%rax,8)
  jne    L23
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    0x0(%rbp),%rdi
  test   %rdi,%rdi
  je     L17
  mov    $0x0,%ebx
L20:
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,(%r12,%rbx,1)
  mov    0x0(%rbp,%rbx,1),%rsi
  call   <strcpy@plt>
  add    $0x8,%rbx
  mov    0x0(%rbp,%rbx,1),%rdi
  test   %rdi,%rdi
  jne    L20
L6:
  movq   $0x0,(%r12,%rbx,1)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L17:
  mov    $0x0,%ebx
  jmp    L6
```

assistant: ```c
char **strcpyArray(char **strings) {
  assert(strings);
  int count = 0;
  while (strings[count++])
    ;
  char **newArray = malloc(sizeof(char *) * count);
  count = 0;
  while (strings[count]) {
    newArray[count] = malloc(sizeof(char) * (1 + strlen(strings[count])));
    strcpy(newArray[count], strings[count]);
    ++count;
  }
  newArray[count] = 0;
  return newArray;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcpyArray>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <assert@plt>
  mov    $0x1,%eax
  nopl   (%rax)
L23:
  mov    %eax,%edi
  add    $0x1,%rax
  cmpq   $0x0,-0x10(%rbp,%rax,8)
  jne    L23
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    0x0(%rbp),%rdi
  mov    %rax,%r12
  test   %rdi,%rdi
  je     L14
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L18:
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,(%r12,%rbx,1)
  mov    0x0(%rbp,%rbx,1),%rsi
  mov    %rax,%rdi
  add    $0x8,%rbx
  call   <strcpy@plt>
  mov    0x0(%rbp,%rbx,1),%rdi
  test   %rdi,%rdi
  jne    L18
  add    %r12,%rbx
  mov    %r12,%rax
  movq   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L14:
  mov    %rax,%rbx
  mov    %r12,%rax
  movq   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char **strcpyArray(char **strings) {
  assert(strings);
  int count = 0;
  while (strings[count++])
    ;
  char **newArray = malloc(sizeof(char *) * count);
  count = 0;
  while (strings[count]) {
    newArray[count] = malloc(sizeof(char) * (1 + strlen(strings[count])));
    strcpy(newArray[count], strings[count]);
    ++count;
  }
  newArray[count] = 0;
  return newArray;
}
```
user: What is the c source code of the assembly code below?

```asm
<strcpyArray>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  call   <assert@plt>
  mov    $0x1,%eax
  nopl   (%rax)
L23:
  mov    %eax,%edi
  add    $0x1,%rax
  cmpq   $0x0,-0x10(%rbp,%rax,8)
  jne    L23
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    0x0(%rbp),%rdi
  mov    %rax,%r12
  test   %rdi,%rdi
  je     L6
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L8:
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,(%r12,%rbx,1)
  mov    0x0(%rbp,%rbx,1),%rsi
  mov    %rax,%rdi
  add    $0x8,%rbx
  call   <strcpy@plt>
  mov    0x0(%rbp,%rbx,1),%rdi
  test   %rdi,%rdi
  jne    L8
  add    %r12,%rbx
  mov    %r12,%rax
  movq   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  xchg   %ax,%ax
L6:
  mov    %rax,%rbx
  mov    %r12,%rax
  movq   $0x0,(%rbx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char **strcpyArray(char **strings) {
  assert(strings);
  int count = 0;
  while (strings[count++])
    ;
  char **newArray = malloc(sizeof(char *) * count);
  count = 0;
  while (strings[count]) {
    newArray[count] = malloc(sizeof(char) * (1 + strlen(strings[count])));
    strcpy(newArray[count], strings[count]);
    ++count;
  }
  newArray[count] = 0;
  return newArray;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<func1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <crash@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D2
data_type: string

tool: D2: func1


assistant: ```c
__attribute__((visibility("default"))) __attribute__((no)) void func1() {
  printf("func1\n");
  crash();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<func1>:
  endbr64
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <crash@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D8
type_infered: string

tool: D8: func1


assistant: ```c
__attribute__((visibility("default"))) __attribute__((no)) void func1() {
  printf("func1\n");
  crash();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<func1>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <crash@plt>
```

tool_calls: name: read_data
arguments: label: D7
data_type: string

tool: D7: func1


assistant: ```c
__attribute__((visibility("default"))) __attribute__((no)) void func1() {
  printf("func1\n");
  crash();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<func1>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <crash@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
data_type: string

tool: D2: func1


assistant: ```c
__attribute__((visibility("default"))) __attribute__((no)) void func1() {
  printf("func1\n");
  crash();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<simple>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L14
L5:
  mov    -0x4(%rbp),%eax
  add    %eax,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jle    L5
  mov    -0x14(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int simple(int a, int n) {
  for (int i = 1; i <= n; i++) {
    a = a + i;
  }
  return a;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<simple>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  jle    L16
  add    $0x1,%esi
  mov    $0x1,%edx
L5:
  add    %edx,%eax
  add    $0x1,%edx
  cmp    %esi,%edx
  jne    L5
L16:
  ret
```

assistant: ```c
int simple(int a, int n) {
  for (int i = 1; i <= n; i++) {
    a = a + i;
  }
  return a;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<simple>:
  endbr64
  mov    %edi,%eax
  test   %esi,%esi
  jle    L8
  add    $0x1,%esi
  mov    $0x1,%edx
  nopw   0x0(%rax,%rax,1)
L4:
  add    %edx,%eax
  add    $0x1,%edx
  cmp    %esi,%edx
  jne    L4
L8:
  ret
```

assistant: ```c
int simple(int a, int n) {
  for (int i = 1; i <= n; i++) {
    a = a + i;
  }
  return a;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<simple>:
  endbr64
  mov    %esi,%ecx
  test   %esi,%esi
  jle    L0
  lea    -0x1(%rsi),%eax
  cmp    $0x9,%eax
  jbe    L93
  mov    %esi,%eax
  movdqa D1(%rip),%xmm1
  xor    %edx,%edx
  pxor   %xmm0,%xmm0
  movdqa D9(%rip),%xmm3
  shr    $0x2,%eax
  nopl   (%rax)
L30:
  movdqa %xmm1,%xmm2
  add    $0x1,%edx
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %edx,%eax
  jne    L30
  movdqa %xmm0,%xmm1
  mov    %ecx,%esi
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%esi
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%edx
  lea    (%rdx,%rdi,1),%eax
  lea    0x1(%rsi),%edx
  cmp    %esi,%ecx
  je     L60
L28:
  lea    0x1(%rdx),%esi
  add    %edx,%eax
  cmp    %esi,%ecx
  jl     L65
  add    %esi,%eax
  lea    0x2(%rdx),%esi
  cmp    %esi,%ecx
  jl     L65
  add    %esi,%eax
  lea    0x3(%rdx),%esi
  cmp    %esi,%ecx
  jl     L65
  add    %esi,%eax
  lea    0x4(%rdx),%esi
  cmp    %esi,%ecx
  jl     L65
  add    %esi,%eax
  lea    0x5(%rdx),%esi
  cmp    %esi,%ecx
  jl     L65
  add    %esi,%eax
  lea    0x6(%rdx),%esi
  cmp    %esi,%ecx
  jl     L65
  add    %esi,%eax
  lea    0x7(%rdx),%esi
  cmp    %esi,%ecx
  jl     L65
  add    %esi,%eax
  lea    0x8(%rdx),%esi
  cmp    %esi,%ecx
  jl     L65
  add    %esi,%eax
  add    $0x9,%edx
  lea    (%rax,%rdx,1),%esi
  cmp    %edx,%ecx
  cmovge %esi,%eax
  ret
  xchg   %ax,%ax
L0:
  mov    %edi,%eax
L65:
  ret
  nopl   0x0(%rax,%rax,1)
L60:
  ret
L93:
  mov    %edi,%eax
  mov    $0x1,%edx
  jmp    L28
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_guessed: byte[16]
name: unpack_data
arguments: data_label: D9
type_guessed: byte[16]

tool: {"D1": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D9": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
int simple(int a, int n) {
  for (int i = 1; i <= n; i++) {
    a = a + i;
  }
  return a;
}
```