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

```asm
<get_resolvconf_addr>:
  endbr64
  push   %r13
  lea    D48(%rip),%rsi
  push   %r12
  lea    D39(%rip),%r12
  push   %rbp
  mov    %r12,%rdi
  push   %rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L63
L25:
  mov    %rsp,%rbx
  lea    D32(%rip),%r13
  lea    D9(%rip),%r12
  jmp    L68
  nopl   0x0(%rax,%rax,1)
L56:
  mov    %rbp,%rdx
  mov    $0x50,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <sscanf@plt>
  cmp    $0x1,%eax
  je     L47
L68:
  mov    %rbp,%rdi
  call   <feof@plt>
  test   %rax,%rax
  je     L56
  xor    %r12d,%r12d
L21:
  mov    %rbp,%rdi
  call   <fclose@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x68,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L47:
  lea    D32(%rip),%r12
  jmp    L21
  nopl   0x0(%rax)
L63:
  mov    %r12,%rsi
  mov    $0x1,%edi
  call   <err@plt>
  jmp    L25
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D48
datatype: string
name: parse_data
arguments: tag: D39
datatype: string
name: parse_data
arguments: tag: D9
datatype: string

tool: D48: r

tool: D39: /etc/resolv.conf

tool: D9: nameserver %15s

assistant: ```c
char *get_resolvconf_addr() {
  static char addr[16];
  char *rv;
  char buf[80];
  FILE *fp;
  rv = NULL;
  if ((fp = fopen("/etc/resolv.conf", "r")) == NULL)
    err(1, "/etc/resolv.conf");
  while (feof(fp) == 0) {
    fgets(buf, sizeof(buf), fp);
    if (sscanf(buf, "nameserver %15s", addr) == 1) {
      rv = addr;
      break;
    }
  }
  fclose(fp);
  return rv;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<store2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  mov    -0xc(%rbp),%edx
  sar    $0x8,%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void store2(uint8_t *buf, int16_t value) {
  buf[0] = value;
  buf[1] = value >> 8;
}
```
user: What is the c source code of the assembly code below:

```asm
<store2>:
  endbr64
  mov    %esi,(%rdi)
  sar    $0x8,%esi
  mov    %esi,0x4(%rdi)
  ret
```

assistant: ```c
void store2(uint8_t *buf, int16_t value) {
  buf[0] = value;
  buf[1] = value >> 8;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<store2>:
  endbr64
  mov    %esi,(%rdi)
  sar    $0x8,%esi
  mov    %esi,0x4(%rdi)
  ret
```

assistant: ```c
void store2(uint8_t *buf, int16_t value) {
  buf[0] = value;
  buf[1] = value >> 8;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<store2>:
  endbr64
  mov    %esi,%eax
  movd   %esi,%xmm0
  sar    $0x8,%eax
  movd   %eax,%xmm1
  punpckldq %xmm1,%xmm0
  movq   %xmm0,(%rdi)
  ret
```

assistant: ```c
void store2(uint8_t *buf, int16_t value) {
  buf[0] = value;
  buf[1] = value >> 8;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<append>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L27
L7:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L99
  addl   $0x1,-0xc(%rbp)
L27:
  cmpl   $0x1f,-0xc(%rbp)
  jle    L7
  jmp    L72
L99:
  nop
L72:
  movl   $0x0,-0x8(%rbp)
  jmp    L90
L56:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L41
  addl   $0x1,-0x8(%rbp)
L90:
  cmpl   $0x1f,-0x8(%rbp)
  jle    L56
  jmp    L61
L41:
  nop
L61:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  cmp    $0x20,%eax
  jg     L1
  movl   $0x0,-0x4(%rbp)
  jmp    L96
L12:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%ecx
  mov    -0x4(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
L96:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L12
  mov    -0xc(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L1:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void append(char *o, char *p) {
  int i, j, k;
  for (i = 0; i < 32; i++)
    if (o[i] == '\0')
      break;
  for (j = 0; j < 32; j++)
    if (p[j] == '\0')
      break;
  if (i + j <= 32) {
    for (k = 0; k < j; k++)
      o[i + k] = p[k];
    o[i + k] = '\0';
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<append>:
  endbr64
  mov    %rdi,%r9
  mov    %rdi,%rdx
  mov    $0x0,%eax
L82:
  cmpb   $0x0,(%rdx)
  je     L67
  add    $0x1,%eax
  add    $0x1,%rdx
  cmp    $0x20,%eax
  jne    L82
L67:
  mov    %rsi,%rcx
  mov    $0x0,%edx
L34:
  cmpb   $0x0,(%rcx)
  je     L35
  add    $0x1,%edx
  add    $0x1,%rcx
  cmp    $0x20,%edx
  jne    L34
  test   %eax,%eax
  jle    L5
L51:
  ret
L35:
  lea    (%rax,%rdx,1),%ecx
  cmp    $0x20,%ecx
  jg     L51
  test   %edx,%edx
  jle    L42
L5:
  mov    %edx,%r10d
  mov    $0x0,%ecx
  movslq %eax,%r8
  add    %r9,%r8
L39:
  movzbl (%rsi,%rcx,1),%edi
  mov    %dil,(%r8,%rcx,1)
  add    $0x1,%rcx
  cmp    %r10,%rcx
  jne    L39
L53:
  add    %edx,%eax
  cltq
  movb   $0x0,(%r9,%rax,1)
  ret
L42:
  mov    $0x0,%edx
  jmp    L53
```

assistant: ```c
void append(char *o, char *p) {
  int i, j, k;
  for (i = 0; i < 32; i++)
    if (o[i] == '\0')
      break;
  for (j = 0; j < 32; j++)
    if (p[j] == '\0')
      break;
  if (i + j <= 32) {
    for (k = 0; k < j; k++)
      o[i + k] = p[k];
    o[i + k] = '\0';
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<append>:
  endbr64
  xor    %eax,%eax
  jmp    L19
  nopl   0x0(%rax,%rax,1)
L81:
  add    $0x1,%rax
  cmp    $0x20,%rax
  je     L49
L19:
  cmpb   $0x0,(%rdi,%rax,1)
  movslq %eax,%rdx
  jne    L81
L66:
  xor    %eax,%eax
  jmp    L76
  nopw   0x0(%rax,%rax,1)
L0:
  add    $0x1,%rax
  cmp    $0x20,%rax
  je     L43
L76:
  cmpb   $0x0,(%rsi,%rax,1)
  mov    %eax,%ecx
  jne    L0
  lea    (%rdx,%rax,1),%r9d
  cmp    $0x20,%r9d
  jg     L57
  test   %eax,%eax
  je     L22
L60:
  xor    %eax,%eax
  add    %rdi,%rdx
  nopl   (%rax)
L13:
  movzbl (%rsi,%rax,1),%r8d
  mov    %r8b,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%ecx
  jg     L13
  movslq %r9d,%rdx
L22:
  movb   $0x0,(%rdi,%rdx,1)
  ret
  nopl   0x0(%rax)
L43:
  test   %edx,%edx
  je     L73
L57:
  ret
  nopl   (%rax)
L49:
  mov    $0x20,%edx
  jmp    L66
  nopw   0x0(%rax,%rax,1)
L73:
  mov    $0x20,%ecx
  mov    $0x20,%r9d
  jmp    L60
```

assistant: ```c
void append(char *o, char *p) {
  int i, j, k;
  for (i = 0; i < 32; i++)
    if (o[i] == '\0')
      break;
  for (j = 0; j < 32; j++)
    if (p[j] == '\0')
      break;
  if (i + j <= 32) {
    for (k = 0; k < j; k++)
      o[i + k] = p[k];
    o[i + k] = '\0';
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <another@plt>
  cmp    $0x2,%eax
  je     L3
  mov    $0x0,%eax
  call   <abort@plt>
L3:
  mov    $0x0,%eax
  call   <null_is_ptr_type@plt>
  pop    %rbp
  ret
```

assistant: ```c
main() {
  if (another() != 2)
    abort();
  return null_is_ptr_type();
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <another@plt>
  cmp    $0x2,%eax
  jne    L16
L5:
  mov    $0x0,%eax
  call   <null_is_ptr_type@plt>
  add    $0x8,%rsp
  ret
L16:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L5
```

assistant: ```c
main() {
  if (another() != 2)
    abort();
  return null_is_ptr_type();
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <another@plt>
  cmp    $0x2,%eax
  je     L4
  xor    %eax,%eax
  call   <abort@plt>
L4:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <null_is_ptr_type@plt>
```

assistant: ```c
main() {
  if (another() != 2)
    abort();
  return null_is_ptr_type();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <another@plt>
  cmp    $0x2,%eax
  je     L4
  xor    %eax,%eax
  call   <abort@plt>
L4:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1090 <null_is_ptr_type@plt>
```

assistant: ```c
main() {
  if (another() != 2)
    abort();
  return null_is_ptr_type();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L17
L18:
  mov    -0xc(%rbp),%eax
  imul   $0x1003f,%eax,%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L17:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L18
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int hash(char *s0) {
  unsigned int h = 0;
  char *s;
  for (s = s0; *s; s++)
    h = h * 65599 + *s;
  return h;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hash>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L18
  mov    $0x0,%eax
L3:
  imul   $0x1003f,%eax,%eax
  movsbl %dl,%edx
  add    %edx,%eax
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L3
  ret
L18:
  mov    $0x0,%eax
  ret
```

assistant: ```c
unsigned int hash(char *s0) {
  unsigned int h = 0;
  char *s;
  for (s = s0; *s; s++)
    h = h * 65599 + *s;
  return h;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hash>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L19
  nopl   (%rax)
L12:
  imul   $0x1003f,%eax,%eax
  add    $0x1,%rdi
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L12
  ret
  nopl   0x0(%rax)
L19:
  ret
```

assistant: ```c
unsigned int hash(char *s0) {
  unsigned int h = 0;
  char *s;
  for (s = s0; *s; s++)
    h = h * 65599 + *s;
  return h;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<hash>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L1
  nopl   (%rax)
L7:
  imul   $0x1003f,%eax,%eax
  add    $0x1,%rdi
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L7
  ret
  nopl   0x0(%rax)
L1:
  ret
```

assistant: ```c
unsigned int hash(char *s0) {
  unsigned int h = 0;
  char *s;
  for (s = s0; *s; s++)
    h = h * 65599 + *s;
  return h;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stdinout_cleanup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <close@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <close@plt>
  test   %eax,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void stdinout_cleanup() {
  assert(!close(STDIN_FILENO));
  assert(!close(STDOUT_FILENO));
}
```
user: What is the c source code of the assembly code below?

```asm
<stdinout_cleanup>:
  endbr64
  sub    $0x8,%rsp
  mov    D16(%rip),%rax
  mov    (%rax),%edi
  call   <close@plt>
  test   %eax,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    D9(%rip),%rax
  mov    (%rax),%edi
  call   <close@plt>
  test   %eax,%eax
  sete   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void stdinout_cleanup() {
  assert(!close(STDIN_FILENO));
  assert(!close(STDOUT_FILENO));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<stdinout_cleanup>:
  endbr64
  sub    $0x8,%rsp
  mov    D17(%rip),%rax
  mov    (%rax),%edi
  call   <close@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  mov    D8(%rip),%rax
  mov    (%rax),%edi
  call   <close@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  add    $0x8,%rsp
  jmp    1070 <assert@plt>
```

assistant: ```c
void stdinout_cleanup() {
  assert(!close(STDIN_FILENO));
  assert(!close(STDOUT_FILENO));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<stdinout_cleanup>:
  endbr64
  sub    $0x8,%rsp
  mov    D0(%rip),%rax
  mov    (%rax),%edi
  call   <close@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  call   <assert@plt>
  mov    D12(%rip),%rax
  mov    (%rax),%edi
  call   <close@plt>
  xor    %edi,%edi
  test   %eax,%eax
  sete   %dil
  add    $0x8,%rsp
  jmp    1070 <assert@plt>
```

assistant: ```c
void stdinout_cleanup() {
  assert(!close(STDIN_FILENO));
  assert(!close(STDOUT_FILENO));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0x8(%rbp)
  movl   $0x2,-0x10(%rbp)
L55:
  movl   $0x1,-0x4(%rbp)
  movl   $0x2,-0xc(%rbp)
  jmp    L51
L35:
  mov    -0x10(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L22
  movl   $0x0,-0x4(%rbp)
L22:
  addl   $0x1,-0xc(%rbp)
L51:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  cmp    %eax,-0xc(%rbp)
  jle    L35
  cmpl   $0x0,-0x4(%rbp)
  je     L29
  addl   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  movslq %edx,%rax
  imul   $0x14f8b589,%rax,%rax
  shr    $0x20,%rax
  sar    $0xd,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x186a0,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L29
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L29:
  addl   $0x1,-0x10(%rbp)
  jmp    L55
```

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

tool: D8: Found prime: %d


assistant: ```c
int main() {
  int a, b, d = 1;
  for (a = 2;; a++) {
    bool c = 1;
    for (b = 2; b < sqrt(a) + 1; b++) {
      if (a % b == 0) {
        c = 0;
      }
    }
    if (c) {
      d = d + 1;
      if (d % 100000 == 0) {
        printf("Found prime: %d\n", a);
      }
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x1,%r13d
  mov    $0x2,%ebp
  lea    D1(%rip),%r14
  jmp    L22
L48:
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  cmove  %edx,%r12d
  add    $0x1,%ebx
L33:
  mov    %ebp,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jge    L48
  test   %r12d,%r12d
  je     L10
  add    $0x1,%r13d
  movslq %r13d,%rax
  imul   $0x14f8b589,%rax,%rax
  sar    $0x2d,%rax
  mov    %r13d,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  imul   $0x186a0,%eax,%eax
  cmp    %eax,%r13d
  jne    L10
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <printf@plt>
L10:
  add    $0x1,%ebp
L22:
  mov    $0x1,%r12d
  mov    $0x2,%ebx
  jmp    L33
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
data_type: string

tool: D1: Found prime: %d


assistant: ```c
int main() {
  int a, b, d = 1;
  for (a = 2;; a++) {
    bool c = 1;
    for (b = 2; b < sqrt(a) + 1; b++) {
      if (a % b == 0) {
        c = 0;
      }
    }
    if (c) {
      d = d + 1;
      if (d % 100000 == 0) {
        printf("Found prime: %d\n", a);
      }
    }
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  lea    D8(%rip),%r15
  push   %r14
  mov    $0x1,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0x2,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L46:
  mov    $0x1,%ebp
  mov    $0x2,%ebx
  jmp    L18
  nopl   0x0(%rax)
L11:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  cmove  %r13d,%ebp
  add    $0x1,%ebx
L18:
  mov    %r12d,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jge    L11
  test   %ebp,%ebp
  je     L4
  add    $0x1,%r14d
  imul   $0xbcbe61d,%r14d,%eax
  ror    $0x5,%eax
  cmp    $0xa7c5,%eax
  ja     L4
  mov    %r12d,%esi
  mov    %r15,%rdi
  call   <printf@plt>
L4:
  add    $0x1,%r12d
  jmp    L46
```

tool_calls: name: disassemble_data
arguments: tag: D8
type_guessed: string

tool: D8: Found prime: %d


assistant: ```c
int main() {
  int a, b, d = 1;
  for (a = 2;; a++) {
    bool c = 1;
    for (b = 2; b < sqrt(a) + 1; b++) {
      if (a % b == 0) {
        c = 0;
      }
    }
    if (c) {
      d = d + 1;
      if (d % 100000 == 0) {
        printf("Found prime: %d\n", a);
      }
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D6(%rip),%r15
  push   %r14
  mov    $0x1,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  mov    $0x2,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L33:
  mov    $0x1,%ebp
  mov    $0x2,%ebx
  jmp    L9
  nopl   0x0(%rax)
L10:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  test   %edx,%edx
  cmove  %r13d,%ebp
  add    $0x1,%ebx
L9:
  mov    %r12d,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jge    L10
  test   %ebp,%ebp
  je     L34
  add    $0x1,%r14d
  imul   $0xbcbe61d,%r14d,%eax
  ror    $0x5,%eax
  cmp    $0xa7c5,%eax
  ja     L34
  mov    %r12d,%esi
  mov    %r15,%rdi
  call   <printf@plt>
L34:
  add    $0x1,%r12d
  jmp    L33
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
type_guessed: string

tool: D6: Found prime: %d


assistant: ```c
int main() {
  int a, b, d = 1;
  for (a = 2;; a++) {
    bool c = 1;
    for (b = 2; b < sqrt(a) + 1; b++) {
      if (a % b == 0) {
        c = 0;
      }
    }
    if (c) {
      d = d + 1;
      if (d % 100000 == 0) {
        printf("Found prime: %d\n", a);
      }
    }
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<test13>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x24(%rbp)
  lea    -0x12(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <ObjectSize0@plt>
  addl   $0x1,-0x24(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <ObjectSize0@plt>
  mov    -0x24(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <ObjectSize0@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void test13() {
  char c[10];
  unsigned i = 0;
  char *p = c;
  ObjectSize0(p);
  ObjectSize0(p + ++i);
  ObjectSize0(p + i++);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test13>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rdi
  call   <ObjectSize0@plt>
  lea    0xf(%rsp),%rbx
  mov    %rbx,%rdi
  call   <ObjectSize0@plt>
  mov    %rbx,%rdi
  call   <ObjectSize0@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x20,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test13() {
  char c[10];
  unsigned i = 0;
  char *p = c;
  ObjectSize0(p);
  ObjectSize0(p + ++i);
  ObjectSize0(p + i++);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test13>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rdi
  lea    0xf(%rsp),%rbp
  call   <ObjectSize0@plt>
  mov    %rbp,%rdi
  call   <ObjectSize0@plt>
  mov    %rbp,%rdi
  call   <ObjectSize0@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x20,%rsp
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test13() {
  char c[10];
  unsigned i = 0;
  char *p = c;
  ObjectSize0(p);
  ObjectSize0(p + ++i);
  ObjectSize0(p + i++);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test13>:
  endbr64
  push   %rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rdi
  lea    0xf(%rsp),%rbp
  call   <ObjectSize0@plt>
  mov    %rbp,%rdi
  call   <ObjectSize0@plt>
  mov    %rbp,%rdi
  call   <ObjectSize0@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x20,%rsp
  pop    %rbp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test13() {
  char c[10];
  unsigned i = 0;
  char *p = c;
  ObjectSize0(p);
  ObjectSize0(p + ++i);
  ObjectSize0(p + i++);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  sub    $0xa4d,%ax
  mov    %ax,-0xa(%rbp)
  movw   $0x0,-0xc(%rbp)
  nop
  movzwl -0xa(%rbp),%eax
  lea    0x1c96(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  char copy12;
  char copy14;
  {
    state[0UL] = (input[0UL] + 914778474UL) - (unsigned short)29623;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      if (state[0UL] < local1) {
        if (state[0UL] != local1) {
          copy11 = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = copy11;
          copy11 = *((char *)(&state[local1]) + 1);
          *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
          *((char *)(&state[local1]) + 0) = copy11;
          copy12 = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = copy12;
          copy12 = *((char *)(&state[0UL]) + 1);
          *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
          *((char *)(&state[0UL]) + 0) = copy12;
        } else {
          state[local1] += state[local1];
        }
      } else {
        copy14 = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = copy14;
        state[local1] += state[local1];
      }
      local1++;
    }
    output[0UL] = (state[0UL] + 681723154UL) - (unsigned short)10364;
  }
}
```