ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<htalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x7,%rax
  and    $0xfffffffffffffff8,%rax
  add    $0x8,%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  imul   %edx,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L16
  mov    $0x0,%eax
  jmp    L8
L16:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc_usable_size@plt>
  cltq
  mov    -0x20(%rbp),%rdx
  add    $0x7,%rdx
  and    $0xfffffffffffffff8,%rdx
  lea    0x8(%rdx),%rcx
  mov    $0x0,%edx
  div    %rcx
  mov    %rax,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x7,%rax
  and    $0xfffffffffffffff8,%rax
  add    $0x8,%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  imul   %edx,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%rax
L8:
  leave
  ret
```

assistant: ```c
void *htalloc(size_t cap, size_t size) {
  void *table = malloc((sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))) * cap);
  if (!table)
    return NULL;
  cap = (malloc_usable_size(table) / (sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))));
  memset(table, 0, (sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))) * cap);
  return table;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<htalloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    0x7(%rsi),%rbx
  and    $0xfffffffffffffff8,%rbx
  add    $0x8,%rbx
  imul   %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L9
  mov    %rax,%rdi
  call   <malloc_usable_size@plt>
  cltq
  mov    $0x0,%edx
  div    %rbx
  mov    %ebx,%edx
  imul   %eax,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L9:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *htalloc(size_t cap, size_t size) {
  void *table = malloc((sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))) * cap);
  if (!table)
    return NULL;
  cap = (malloc_usable_size(table) / (sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))));
  memset(table, 0, (sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))) * cap);
  return table;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<htalloc>:
  endbr64
  push   %r12
  push   %rbx
  lea    0x7(%rsi),%rbx
  and    $0xfffffffffffffff8,%rbx
  add    $0x8,%rbx
  imul   %ebx,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L6
  mov    %rax,%rdi
  call   <malloc_usable_size@plt>
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %r12,%rdi
  cltq
  div    %rbx
  mov    %ebx,%edx
  imul   %eax,%edx
  call   <memset@plt>
L6:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
void *htalloc(size_t cap, size_t size) {
  void *table = malloc((sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))) * cap);
  if (!table)
    return NULL;
  cap = (malloc_usable_size(table) / (sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))));
  memset(table, 0, (sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))) * cap);
  return table;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<htalloc>:
  endbr64
  push   %r12
  push   %rbx
  lea    0x7(%rsi),%rbx
  and    $0xfffffffffffffff8,%rbx
  add    $0x8,%rbx
  imul   %ebx,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L8
  mov    %rax,%rdi
  call   <malloc_usable_size@plt>
  xor    %edx,%edx
  xor    %esi,%esi
  mov    %r12,%rdi
  cltq
  div    %rbx
  mov    %ebx,%edx
  imul   %eax,%edx
  call   <memset@plt>
L8:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
void *htalloc(size_t cap, size_t size) {
  void *table = malloc((sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))) * cap);
  if (!table)
    return NULL;
  cap = (malloc_usable_size(table) / (sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))));
  memset(table, 0, (sizeof(const char *) + (((size) + ((sizeof(void *)) - 1)) & ~((sizeof(void *)) - 1))) * cap);
  return table;
}
```
user: Translate the below assembly code into C programming language:

```asm
<llistCursor_new>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x8,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L5
  mov    -0x8(%rbp),%rax
  movq   $0x0,(%rax)
L5:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
struct Node **llistCursor_new(void) {
  struct Node **cursor = malloc(sizeof(*cursor));
  if (cursor != NULL) {
    *cursor = NULL;
  }
  return cursor;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<llistCursor_new>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L1
  movq   $0x0,(%rax)
L1:
  add    $0x8,%rsp
  ret
```

assistant: ```c
struct Node **llistCursor_new(void) {
  struct Node **cursor = malloc(sizeof(*cursor));
  if (cursor != NULL) {
    *cursor = NULL;
  }
  return cursor;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<llistCursor_new>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L0
  movq   $0x0,(%rax)
L0:
  add    $0x8,%rsp
  ret
```

assistant: ```c
struct Node **llistCursor_new(void) {
  struct Node **cursor = malloc(sizeof(*cursor));
  if (cursor != NULL) {
    *cursor = NULL;
  }
  return cursor;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rand32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    D56(%rip),%eax
  imul   $0x41c64e6d,%eax,%eax
  add    $0x3039,%eax
  mov    %eax,D6(%rip)
  mov    D6(%rip),%eax
  sar    $0x10,%eax
  mov    %eax,-0x8(%rbp)
  mov    D6(%rip),%eax
  imul   $0x41c64e6d,%eax,%eax
  add    $0x3039,%eax
  mov    %eax,D6(%rip)
  mov    D6(%rip),%eax
  sar    $0x10,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  shl    $0x10,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t rand32() {
  static int seed = 0;
  uint32_t hi, lo;
  hi = (seed = seed * 1103515245 + 12345) >> 16;
  lo = (seed = seed * 1103515245 + 12345) >> 16;
  return (hi << 16) + lo;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rand32>:
  endbr64
  imul   $0x41c64e6d,0x2f1d(%rip),%eax        # 4024 <seed.0>
  add    $0x3039,%eax
  imul   $0x41c64e6d,%eax,%edx
  add    $0x3039,%edx
  mov    %edx,D2(%rip)
  mov    $0x0,%ax
  sar    $0x10,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint32_t rand32() {
  static int seed = 0;
  uint32_t hi, lo;
  hi = (seed = seed * 1103515245 + 12345) >> 16;
  lo = (seed = seed * 1103515245 + 12345) >> 16;
  return (hi << 16) + lo;
}
```
user: What is the c source code of the assembly code below?

```asm
<rand32>:
  endbr64
  imul   $0x41c64e6d,0x2f16(%rip),%eax        # 4024 <seed.0>
  add    $0x3039,%eax
  imul   $0x41c64e6d,%eax,%edx
  xor    %ax,%ax
  add    $0x3039,%edx
  mov    %edx,D3(%rip)
  sar    $0x10,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint32_t rand32() {
  static int seed = 0;
  uint32_t hi, lo;
  hi = (seed = seed * 1103515245 + 12345) >> 16;
  lo = (seed = seed * 1103515245 + 12345) >> 16;
  return (hi << 16) + lo;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rand32>:
  endbr64
  imul   $0x41c64e6d,0x2f16(%rip),%eax        # 4024 <seed.0>
  add    $0x3039,%eax
  imul   $0x41c64e6d,%eax,%edx
  xor    %ax,%ax
  add    $0x3039,%edx
  mov    %edx,D5(%rip)
  sar    $0x10,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint32_t rand32() {
  static int seed = 0;
  uint32_t hi, lo;
  hi = (seed = seed * 1103515245 + 12345) >> 16;
  lo = (seed = seed * 1103515245 + 12345) >> 16;
  return (hi << 16) + lo;
}
```
user: Translate the below assembly code into C programming language:

```asm
<alternative_bases>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,%eax
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x18(%rbp)
  mov    %al,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L11
L14:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x14(%rbp)
  je     L1
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    -0x4(%rbp),%edx
  mov    -0x28(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <check_if_char_in_string@plt>
  test   %rax,%rax
  jne    L1
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2c,(%rax)
  addl   $0x1,-0x4(%rbp)
L1:
  addl   $0x1,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L14
  cmpl   $0x0,-0x4(%rbp)
  jle    L19
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2c,%al
  jne    L19
  mov    -0x4(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L24
L19:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
L24:
  nop
  leave
  ret
```

assistant: ```c
void alternative_bases(char reference_base, char *bases_for_snp, char alt_bases[], int number_of_samples) {
  int i;
  int num_alt_bases = 0;
  for (i = 0; i < number_of_samples; i++) {
    if ((bases_for_snp[i] != reference_base)) {
      if (check_if_char_in_string(alt_bases, bases_for_snp[i], num_alt_bases) == 0) {
        alt_bases[num_alt_bases] = bases_for_snp[i];
        num_alt_bases++;
        alt_bases[num_alt_bases] = ',';
        num_alt_bases++;
      }
    }
  }
  if (num_alt_bases > 0 && alt_bases[num_alt_bases - 1] == ',') {
    alt_bases[num_alt_bases - 1] = '\0';
  } else {
    alt_bases[num_alt_bases] = '\0';
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<alternative_bases>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%r12
  test   %ecx,%ecx
  jle    L93
  mov    %edi,%r13d
  mov    %rsi,%rbx
  lea    -0x1(%rcx),%eax
  lea    0x1(%rsi,%rax,1),%r15
  mov    $0x0,%ebp
  jmp    L83
L36:
  add    $0x1,%rbx
  cmp    %r15,%rbx
  je     L26
L83:
  movzbl (%rbx),%esi
  cmp    %r13b,%sil
  je     L36
  movsbl %sil,%esi
  mov    %ebp,%edx
  mov    %r12,%rdi
  call   <check_if_char_in_string@plt>
  test   %rax,%rax
  jne    L36
  movzbl (%rbx),%edx
  movslq %ebp,%rax
  mov    %dl,(%r12,%rax,1)
  lea    0x1(%rbp),%eax
  cltq
  movb   $0x2c,(%r12,%rax,1)
  add    $0x2,%ebp
  jmp    L36
L26:
  test   %ebp,%ebp
  jle    L97
  movslq %ebp,%rax
  lea    -0x1(%r12,%rax,1),%rax
  cmpb   $0x2c,(%rax)
  je     L76
L97:
  movslq %ebp,%rbp
  movb   $0x0,(%r12,%rbp,1)
L40:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L76:
  movb   $0x0,(%rax)
  jmp    L40
L93:
  mov    $0x0,%ebp
  jmp    L97
```

assistant: ```c
void alternative_bases(char reference_base, char *bases_for_snp, char alt_bases[], int number_of_samples) {
  int i;
  int num_alt_bases = 0;
  for (i = 0; i < number_of_samples; i++) {
    if ((bases_for_snp[i] != reference_base)) {
      if (check_if_char_in_string(alt_bases, bases_for_snp[i], num_alt_bases) == 0) {
        alt_bases[num_alt_bases] = bases_for_snp[i];
        num_alt_bases++;
        alt_bases[num_alt_bases] = ',';
        num_alt_bases++;
      }
    }
  }
  if (num_alt_bases > 0 && alt_bases[num_alt_bases - 1] == ',') {
    alt_bases[num_alt_bases - 1] = '\0';
  } else {
    alt_bases[num_alt_bases] = '\0';
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<alternative_bases>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %ecx,%ecx
  jle    L27
  lea    -0x1(%rcx),%eax
  mov    %edi,%r12d
  mov    %rsi,%rbx
  xor    %ebp,%ebp
  lea    0x1(%rsi,%rax,1),%r13
  nopl   0x0(%rax,%rax,1)
L35:
  movsbl (%rbx),%esi
  cmp    %r12b,%sil
  je     L1
  mov    %ebp,%edx
  mov    %r14,%rdi
  call   <check_if_char_in_string@plt>
  test   %rax,%rax
  jne    L1
  movzbl (%rbx),%edx
  movslq %ebp,%rax
  mov    %dl,(%r14,%rax,1)
  lea    0x1(%rbp),%eax
  add    $0x2,%ebp
  cltq
  movb   $0x2c,(%r14,%rax,1)
L1:
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L35
  movslq %ebp,%rax
  test   %ebp,%ebp
  je     L27
  lea    -0x1(%r14,%rax,1),%rdx
  add    %rax,%r14
  cmpb   $0x2c,(%rdx)
  je     L23
L27:
  pop    %rbx
  pop    %rbp
  movb   $0x0,(%r14)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L23:
  movb   $0x0,(%rdx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void alternative_bases(char reference_base, char *bases_for_snp, char alt_bases[], int number_of_samples) {
  int i;
  int num_alt_bases = 0;
  for (i = 0; i < number_of_samples; i++) {
    if ((bases_for_snp[i] != reference_base)) {
      if (check_if_char_in_string(alt_bases, bases_for_snp[i], num_alt_bases) == 0) {
        alt_bases[num_alt_bases] = bases_for_snp[i];
        num_alt_bases++;
        alt_bases[num_alt_bases] = ',';
        num_alt_bases++;
      }
    }
  }
  if (num_alt_bases > 0 && alt_bases[num_alt_bases - 1] == ',') {
    alt_bases[num_alt_bases - 1] = '\0';
  } else {
    alt_bases[num_alt_bases] = '\0';
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<alternative_bases>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  test   %ecx,%ecx
  jle    L43
  lea    -0x1(%rcx),%eax
  mov    %edi,%r12d
  mov    %rsi,%rbx
  xor    %ebp,%ebp
  lea    0x1(%rsi,%rax,1),%r13
  nopl   0x0(%rax,%rax,1)
L22:
  movsbl (%rbx),%esi
  cmp    %r12b,%sil
  je     L12
  mov    %ebp,%edx
  mov    %r14,%rdi
  call   <check_if_char_in_string@plt>
  test   %rax,%rax
  jne    L12
  movzbl (%rbx),%edx
  movslq %ebp,%rax
  mov    %dl,(%r14,%rax,1)
  lea    0x1(%rbp),%eax
  add    $0x2,%ebp
  cltq
  movb   $0x2c,(%r14,%rax,1)
L12:
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L22
  movslq %ebp,%rax
  test   %ebp,%ebp
  je     L43
  lea    -0x1(%r14,%rax,1),%rdx
  add    %rax,%r14
  cmpb   $0x2c,(%rdx)
  je     L57
L43:
  pop    %rbx
  pop    %rbp
  movb   $0x0,(%r14)
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L57:
  movb   $0x0,(%rdx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void alternative_bases(char reference_base, char *bases_for_snp, char alt_bases[], int number_of_samples) {
  int i;
  int num_alt_bases = 0;
  for (i = 0; i < number_of_samples; i++) {
    if ((bases_for_snp[i] != reference_base)) {
      if (check_if_char_in_string(alt_bases, bases_for_snp[i], num_alt_bases) == 0) {
        alt_bases[num_alt_bases] = bases_for_snp[i];
        num_alt_bases++;
        alt_bases[num_alt_bases] = ',';
        num_alt_bases++;
      }
    }
  }
  if (num_alt_bases > 0 && alt_bases[num_alt_bases - 1] == ',') {
    alt_bases[num_alt_bases - 1] = '\0';
  } else {
    alt_bases[num_alt_bases] = '\0';
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x24(%rbp)
  movl   $0xa,-0x20(%rbp)
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <print@plt>
  mov    -0x20(%rbp),%eax
  mov    %eax,%edi
  call   <printid@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <print@plt>
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <printid@plt>
  movl   $0x1,-0x1c(%rbp)
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <print@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <printid@plt>
  mov    $0x21,%eax
  sub    -0x20(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  movl   $0x1,-0x14(%rbp)
  movl   $0x2,-0x10(%rbp)
  movl   $0x3,-0xc(%rbp)
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <print@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <printid@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  lea    D48(%rip),%rax
  mov    %rax,%rdi
  call   <print@plt>
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <printid@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D13
type_guessed: string
name: read_data
arguments: data_label: D28
type_guessed: string
name: read_data
arguments: data_label: D18
type_guessed: string
name: read_data
arguments: data_label: D27
type_guessed: string
name: read_data
arguments: data_label: D48
type_guessed: string

tool: D13: j  10

tool: D28: i  0

tool: D18: k  1

tool: D27: l  23

tool: D48: i  3

assistant: ```c
int main() {
  int i = 0, j = 10;
  int k;
  print("j  10");
  printid(j);
  print("i  0");
  printid(i);
  k = (10 >= 9);
  print("k  1");
  printid(k);
  {
    int l = 11 * 3 - j;
    int m[3] = {1, 2, 3};
    print("l  23");
    printid(l);
    i = m[2];
    print("i  3");
    printid(i);
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D43(%rip),%rdi
  call   <print@plt>
  mov    $0xa,%edi
  call   <printid@plt>
  lea    D13(%rip),%rdi
  call   <print@plt>
  mov    $0x0,%edi
  call   <printid@plt>
  lea    D46(%rip),%rdi
  call   <print@plt>
  mov    $0x1,%edi
  call   <printid@plt>
  lea    D48(%rip),%rdi
  call   <print@plt>
  mov    $0x17,%edi
  call   <printid@plt>
  lea    D27(%rip),%rdi
  call   <print@plt>
  mov    $0x3,%edi
  call   <printid@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D43
type_infered: string
name: read_data
arguments: data_tag: D13
type_infered: string
name: read_data
arguments: data_tag: D46
type_infered: string
name: read_data
arguments: data_tag: D48
type_infered: string
name: read_data
arguments: data_tag: D27
type_infered: string

tool: D43: j  10

tool: D13: i  0

tool: D46: k  1

tool: D48: l  23

tool: D27: i  3

assistant: ```c
int main() {
  int i = 0, j = 10;
  int k;
  print("j  10");
  printid(j);
  print("i  0");
  printid(i);
  k = (10 >= 9);
  print("k  1");
  printid(k);
  {
    int l = 11 * 3 - j;
    int m[3] = {1, 2, 3};
    print("l  23");
    printid(l);
    i = m[2];
    print("i  3");
    printid(i);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <print@plt>
  mov    $0xa,%edi
  call   <printid@plt>
  lea    D41(%rip),%rdi
  call   <print@plt>
  xor    %edi,%edi
  call   <printid@plt>
  lea    D12(%rip),%rdi
  call   <print@plt>
  mov    $0x1,%edi
  call   <printid@plt>
  lea    D26(%rip),%rdi
  call   <print@plt>
  mov    $0x17,%edi
  call   <printid@plt>
  lea    D14(%rip),%rdi
  call   <print@plt>
  mov    $0x3,%edi
  call   <printid@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D5
data_type: string
name: unpack_data
arguments: data_tag: D41
data_type: string
name: unpack_data
arguments: data_tag: D12
data_type: string
name: unpack_data
arguments: data_tag: D26
data_type: string
name: unpack_data
arguments: data_tag: D14
data_type: string

tool: D5: j  10

tool: D41: i  0

tool: D12: k  1

tool: D26: l  23

tool: D14: i  3

assistant: ```c
int main() {
  int i = 0, j = 10;
  int k;
  print("j  10");
  printid(j);
  print("i  0");
  printid(i);
  k = (10 >= 9);
  print("k  1");
  printid(k);
  {
    int l = 11 * 3 - j;
    int m[3] = {1, 2, 3};
    print("l  23");
    printid(l);
    i = m[2];
    print("i  3");
    printid(i);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <print@plt>
  mov    $0xa,%edi
  call   <printid@plt>
  lea    D16(%rip),%rdi
  call   <print@plt>
  xor    %edi,%edi
  call   <printid@plt>
  lea    D48(%rip),%rdi
  call   <print@plt>
  mov    $0x1,%edi
  call   <printid@plt>
  lea    D18(%rip),%rdi
  call   <print@plt>
  mov    $0x17,%edi
  call   <printid@plt>
  lea    D31(%rip),%rdi
  call   <print@plt>
  mov    $0x3,%edi
  call   <printid@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D6
datatype: string
name: unpack_data
arguments: data_label: D16
datatype: string
name: unpack_data
arguments: data_label: D48
datatype: string
name: unpack_data
arguments: data_label: D18
datatype: string
name: unpack_data
arguments: data_label: D31
datatype: string

tool: D6: j  10

tool: D16: i  0

tool: D48: k  1

tool: D18: l  23

tool: D31: i  3

assistant: ```c
int main() {
  int i = 0, j = 10;
  int k;
  print("j  10");
  printid(j);
  print("i  0");
  printid(i);
  k = (10 >= 9);
  print("k  1");
  printid(k);
  {
    int l = 11 * 3 - j;
    int m[3] = {1, 2, 3};
    print("l  23");
    printid(l);
    i = m[2];
    print("i  3");
    printid(i);
  }
  return 0;
}
```