ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<util_utf8toutf16_alloc>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  call   <util_utf8toutf16_len@plt>
  cmp    $0xffffffff,%eax
  je     L3
  lea    0x2(%rax),%ebx
  mov    %ebx,%esi
  mov    $0x1,%edi
  call   <calloc@plt>
  mov    %rax,%r12
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %eax,%ecx
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <util_utf8toutf16@plt>
  test   %eax,%eax
  je     L23
L13:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  mov    %r12,%rdi
  call   <free@plt>
  mov    $0x0,%r12d
  jmp    L13
L3:
  mov    $0x0,%r12d
  jmp    L13
```

assistant: ```c
unsigned char *util_utf8toutf16_alloc(unsigned char *utf8) {
  unsigned char *out;
  int new_len;
  new_len = util_utf8toutf16_len(utf8);
  if (new_len == -1)
    return NULL;
  out = calloc(1, new_len + 2);
  if (!util_utf8toutf16(out, new_len + 2, utf8, (int)strlen((char *)utf8))) {
    free(out);
    return NULL;
  }
  return out;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<util_utf8toutf16_alloc>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  call   <util_utf8toutf16_len@plt>
  cmp    $0xffffffff,%eax
  je     L15
  lea    0x2(%rax),%ebp
  mov    $0x1,%edi
  mov    %ebp,%esi
  call   <calloc@plt>
  mov    %r12,%rdi
  mov    %rax,%r13
  call   <strlen@plt>
  mov    %r12,%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    %eax,%ecx
  call   <util_utf8toutf16@plt>
  test   %eax,%eax
  je     L9
L27:
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L15:
  xor    %r13d,%r13d
  pop    %rbp
  pop    %r12
  mov    %r13,%rax
  pop    %r13
  ret
  nopl   0x0(%rax)
L9:
  mov    %r13,%rdi
  xor    %r13d,%r13d
  call   <free@plt>
  jmp    L27
```

assistant: ```c
unsigned char *util_utf8toutf16_alloc(unsigned char *utf8) {
  unsigned char *out;
  int new_len;
  new_len = util_utf8toutf16_len(utf8);
  if (new_len == -1)
    return NULL;
  out = calloc(1, new_len + 2);
  if (!util_utf8toutf16(out, new_len + 2, utf8, (int)strlen((char *)utf8))) {
    free(out);
    return NULL;
  }
  return out;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<util_utf8toutf16_alloc>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  call   <util_utf8toutf16_len@plt>
  cmp    $0xffffffff,%eax
  je     L23
  lea    0x2(%rax),%ebp
  mov    $0x1,%edi
  mov    %ebp,%esi
  call   <calloc@plt>
  mov    %r12,%rdi
  mov    %rax,%r13
  call   <strlen@plt>
  mov    %r12,%rdx
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    %eax,%ecx
  call   <util_utf8toutf16@plt>
  test   %eax,%eax
  je     L10
L17:
  mov    %r13,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L23:
  xor    %r13d,%r13d
  pop    %rbp
  pop    %r12
  mov    %r13,%rax
  pop    %r13
  ret
  nopl   0x0(%rax)
L10:
  mov    %r13,%rdi
  xor    %r13d,%r13d
  call   <free@plt>
  jmp    L17
```

assistant: ```c
unsigned char *util_utf8toutf16_alloc(unsigned char *utf8) {
  unsigned char *out;
  int new_len;
  new_len = util_utf8toutf16_len(utf8);
  if (new_len == -1)
    return NULL;
  out = calloc(1, new_len + 2);
  if (!util_utf8toutf16(out, new_len + 2, utf8, (int)strlen((char *)utf8))) {
    free(out);
    return NULL;
  }
  return out;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<filepresent>:
  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   %eax,%eax
  jne    L9
  mov    $0x1,%eax
  jmp    L27
L9:
  mov    $0x0,%eax
L27:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int filepresent(char *run) {
  struct stat buf;
  if (!stat(run, &buf))
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<filepresent>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %eax,%eax
  sete   %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int filepresent(char *run) {
  struct stat buf;
  if (!stat(run, &buf))
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<filepresent>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %eax,%eax
  sete   %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int filepresent(char *run) {
  struct stat buf;
  if (!stat(run, &buf))
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<filepresent>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <stat@plt>
  test   %eax,%eax
  sete   %al
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  movzbl %al,%eax
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int filepresent(char *run) {
  struct stat buf;
  if (!stat(run, &buf))
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<numlength>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L12
L14:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L12:
  cmpl   $0x0,-0x14(%rbp)
  jne    L14
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int numlength(off_t num) {
  int x = 0;
  while (num) {
    num /= 10;
    ++x;
  }
  return x;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<numlength>:
  endbr64
  test   %edi,%edi
  je     L18
  mov    $0x0,%edx
L7:
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edi
  sub    %edi,%eax
  mov    %eax,%edi
  add    $0x1,%edx
  test   %eax,%eax
  jne    L7
L16:
  mov    %edx,%eax
  ret
L18:
  mov    %edi,%edx
  jmp    L16
```

assistant: ```c
int numlength(off_t num) {
  int x = 0;
  while (num) {
    num /= 10;
    ++x;
  }
  return x;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<numlength>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L0
  nopl   0x0(%rax,%rax,1)
L19:
  movslq %edi,%rax
  mov    %edi,%edx
  add    $0x1,%r8d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %eax,%edi
  jne    L19
L0:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int numlength(off_t num) {
  int x = 0;
  while (num) {
    num /= 10;
    ++x;
  }
  return x;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<numlength>:
  endbr64
  xor    %r8d,%r8d
  test   %edi,%edi
  je     L14
  nopl   0x0(%rax,%rax,1)
L17:
  movslq %edi,%rax
  mov    %edi,%edx
  add    $0x1,%r8d
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %eax,%edi
  jne    L17
L14:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int numlength(off_t num) {
  int x = 0;
  while (num) {
    num /= 10;
    ++x;
  }
  return x;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<print_thread>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <thrd_exit@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D9
data_type: string

tool: D9: %s


assistant: ```c
int print_thread(void *s) {
  printf("%s\n", (char *)s);
  thrd_exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<print_thread>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <thrd_exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D4
type_infered: string

tool: D4: %s


assistant: ```c
int print_thread(void *s) {
  printf("%s\n", (char *)s);
  thrd_exit(0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<print_thread>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <thrd_exit@plt>
```

tool_calls: name: read_data
arguments: data_label: D4
datatype: string

tool: D4: %s


assistant: ```c
int print_thread(void *s) {
  printf("%s\n", (char *)s);
  thrd_exit(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<print_thread>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <thrd_exit@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D5
datatype: string

tool: D5: %s


assistant: ```c
int print_thread(void *s) {
  printf("%s\n", (char *)s);
  thrd_exit(0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<dice_distance_bool>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L99
L92:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L11
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L11
  mov    $0x1,%eax
  jmp    L25
L11:
  mov    $0x0,%eax
L25:
  add    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L72
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L72
  mov    $0x1,%eax
  jmp    L86
L72:
  mov    $0x0,%eax
L86:
  add    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L100
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L100
  mov    $0x1,%eax
  jmp    L65
L100:
  mov    $0x0,%eax
L65:
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x10(%rbp)
L99:
  mov    -0x10(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L92
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  pxor   %xmm1,%xmm1
  cvtsi2sd %eax,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0xc(%rbp),%xmm0
  movapd %xmm0,%xmm2
  addsd  %xmm0,%xmm2
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x4(%rbp),%xmm0
  addsd  %xmm0,%xmm2
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x8(%rbp),%xmm0
  addsd  %xmm2,%xmm0
  divsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double dice_distance_bool(const char *u, const char *v, int n) {
  int i = 0;
  int ntt = 0, nft = 0, ntf = 0;
  for (i = 0; i < n; i++) {
    ntt += (u[i] && v[i]);
    ntf += (u[i] && !v[i]);
    nft += (!u[i] && v[i]);
  }
  return (double)(nft + ntf) / (double)(2.0 * ntt + ntf + nft);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dice_distance_bool>:
  endbr64
  test   %edx,%edx
  jle    L76
  mov    %edx,%edx
  mov    $0x0,%eax
  mov    $0x0,%r9d
  mov    $0x0,%ecx
  mov    $0x0,%r10d
  mov    $0x0,%r11d
  jmp    L34
L9:
  cmpb   $0x0,(%rsi,%rax,1)
  setne  %r8b
  movzbl %r8b,%r8d
L22:
  add    %r8d,%ecx
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L81
L34:
  movzbl (%rdi,%rax,1),%r8d
  test   %r8b,%r8b
  je     L9
  cmpb   $0x0,(%rsi,%rax,1)
  jne    L1
  jne    L74
  add    $0x1,%r9d
  mov    %r11d,%r8d
  jmp    L22
L76:
  mov    $0x0,%r9d
  mov    $0x0,%ecx
  mov    $0x0,%r10d
L81:
  lea    (%rcx,%r9,1),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2sd %r10d,%xmm1
  addsd  %xmm1,%xmm1
  pxor   %xmm2,%xmm2
  cvtsi2sd %r9d,%xmm2
  addsd  %xmm2,%xmm1
  pxor   %xmm2,%xmm2
  cvtsi2sd %ecx,%xmm2
  addsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  ret
L1:
  add    $0x1,%r10d
L74:
  test   %r8b,%r8b
  je     L9
  mov    %r11d,%r8d
  jmp    L22
```

assistant: ```c
double dice_distance_bool(const char *u, const char *v, int n) {
  int i = 0;
  int ntt = 0, nft = 0, ntf = 0;
  for (i = 0; i < n; i++) {
    ntt += (u[i] && v[i]);
    ntf += (u[i] && !v[i]);
    nft += (!u[i] && v[i]);
  }
  return (double)(nft + ntf) / (double)(2.0 * ntt + ntf + nft);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<dice_distance_bool>:
  endbr64
  test   %edx,%edx
  jle    L69
  movslq %edx,%rdx
  xor    %eax,%eax
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  xor    %r10d,%r10d
  jmp    L6
  nopl   0x0(%rax)
L13:
  test   %cl,%cl
  je     L57
  add    $0x1,%r10d
L31:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L41
L6:
  cmpb   $0x0,(%rdi,%rax,1)
  movzbl (%rsi,%rax,1),%ecx
  jne    L13
  cmp    $0x1,%cl
  sbb    $0xffffffff,%r8d
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L6
L41:
  pxor   %xmm1,%xmm1
  pxor   %xmm2,%xmm2
  lea    (%r9,%r8,1),%eax
  cvtsi2sd %r10d,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %r9d,%xmm2
  cvtsi2sd %eax,%xmm0
  addsd  %xmm1,%xmm1
  addsd  %xmm2,%xmm1
  pxor   %xmm2,%xmm2
  cvtsi2sd %r8d,%xmm2
  addsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L57:
  add    $0x1,%r9d
  jmp    L31
  xchg   %ax,%ax
L69:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  divsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double dice_distance_bool(const char *u, const char *v, int n) {
  int i = 0;
  int ntt = 0, nft = 0, ntf = 0;
  for (i = 0; i < n; i++) {
    ntt += (u[i] && v[i]);
    ntf += (u[i] && !v[i]);
    nft += (!u[i] && v[i]);
  }
  return (double)(nft + ntf) / (double)(2.0 * ntt + ntf + nft);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dice_distance_bool>:
  endbr64
  test   %edx,%edx
  jle    L32
  movslq %edx,%rdx
  xor    %eax,%eax
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  xor    %r10d,%r10d
  jmp    L8
  nopl   0x0(%rax)
L20:
  test   %cl,%cl
  je     L53
  add    $0x1,%r10d
L23:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L0
L8:
  cmpb   $0x0,(%rdi,%rax,1)
  movzbl (%rsi,%rax,1),%ecx
  jne    L20
  cmp    $0x1,%cl
  sbb    $0xffffffff,%r8d
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L8
L0:
  pxor   %xmm1,%xmm1
  pxor   %xmm2,%xmm2
  lea    (%r8,%r9,1),%eax
  cvtsi2sd %r10d,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sd %r9d,%xmm2
  cvtsi2sd %eax,%xmm0
  addsd  %xmm1,%xmm1
  addsd  %xmm2,%xmm1
  pxor   %xmm2,%xmm2
  cvtsi2sd %r8d,%xmm2
  addsd  %xmm2,%xmm1
  divsd  %xmm1,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L53:
  add    $0x1,%r9d
  jmp    L23
  xchg   %ax,%ax
L32:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  divsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double dice_distance_bool(const char *u, const char *v, int n) {
  int i = 0;
  int ntt = 0, nft = 0, ntf = 0;
  for (i = 0; i < n; i++) {
    ntt += (u[i] && v[i]);
    ntf += (u[i] && !v[i]);
    nft += (!u[i] && v[i]);
  }
  return (double)(nft + ntf) / (double)(2.0 * ntt + ntf + nft);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test_info_c90>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L4
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L4:
  mov    -0xc8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  sub    $0x8,%rsp
  mov    -0xb4(%rbp),%eax
  push   %rax
  push   -0xc8(%rbp)
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0xffffffff,%edx
  mov    $0x0,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <test_fail_message_va@plt>
  add    $0x20,%rsp
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
datatype: string

tool: D4: INFO

assistant: ```c
void test_info_c90(const char *desc, ...) {
  va_list ap;
  va_start(ap, desc);
  test_fail_message_va("INFO", NULL, -1, NULL, NULL, NULL, NULL, desc, ap);
  va_end(ap);
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_info_c90>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  sub    $0x8,%rsp
  push   %rbp
  push   %rbx
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0xffffffff,%edx
  mov    $0x0,%esi
  lea    D4(%rip),%rdi
  call   <test_fail_message_va@plt>
  add    $0x20,%rsp
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D4
type_infered: string

tool: D4: INFO

assistant: ```c
void test_info_c90(const char *desc, ...) {
  va_list ap;
  va_start(ap, desc);
  test_fail_message_va("INFO", NULL, -1, NULL, NULL, NULL, NULL, desc, ap);
  va_end(ap);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test_info_c90>:
  endbr64
  push   %rbp
  xor    %ebp,%ebp
  mov    %rdi,%rsi
  push   %rbx
  mov    %rdi,%rbx
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  sub    $0x8,%rsp
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  push   %rbp
  xor    %ecx,%ecx
  mov    $0xffffffff,%edx
  xor    %esi,%esi
  push   %rbx
  lea    D5(%rip),%rdi
  push   $0x0
  call   <test_fail_message_va@plt>
  add    $0x20,%rsp
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: label: D5
type_infered: string

tool: D5: INFO

assistant: ```c
void test_info_c90(const char *desc, ...) {
  va_list ap;
  va_start(ap, desc);
  test_fail_message_va("INFO", NULL, -1, NULL, NULL, NULL, NULL, desc, ap);
  va_end(ap);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test_info_c90>:
  endbr64
  push   %rbp
  xor    %ebp,%ebp
  mov    %rdi,%rsi
  push   %rbx
  mov    %rdi,%rbx
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  sub    $0x8,%rsp
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  push   %rbp
  xor    %ecx,%ecx
  mov    $0xffffffff,%edx
  xor    %esi,%esi
  push   %rbx
  lea    D6(%rip),%rdi
  push   $0x0
  call   <test_fail_message_va@plt>
  add    $0x20,%rsp
  mov    %ebp,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D6
type_infered: string

tool: D6: INFO

assistant: ```c
void test_info_c90(const char *desc, ...) {
  va_list ap;
  va_start(ap, desc);
  test_fail_message_va("INFO", NULL, -1, NULL, NULL, NULL, NULL, desc, ap);
  va_end(ap);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<make_hex_string_learning>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x6665646362613938,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movb   $0x0,-0x10(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <num_hex_digits@plt>
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  and    $0x1,%eax
  test   %rax,%rax
  je     L2
  mov    -0x40(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  movb   $0x30,(%rax)
L2:
  mov    -0x28(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  subq   $0x1,-0x28(%rbp)
  jmp    L26
L27:
  mov    -0x38(%rbp),%rax
  and    $0xf,%eax
  mov    -0x28(%rbp),%rcx
  mov    -0x40(%rbp),%rdx
  add    %rcx,%rdx
  movzbl -0x20(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
  sarq   $0x4,-0x38(%rbp)
  subq   $0x1,-0x28(%rbp)
L26:
  cmpq   $0x0,-0x28(%rbp)
  jns    L27
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
void make_hex_string_learning(long n, char *s) {
  const char hex_lookup[] = "0123456789abcdef";
  long len = num_hex_digits(n);
  if (len & 1) {
    *s++ = '0';
  }
  s[len] = '\0';
  for (--len; len >= 0; n >>= 4, --len) {
    s[len] = hex_lookup[n & 0xf];
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<make_hex_string_learning>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x6665646362613938,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movb   $0x0,0x10(%rsp)
  call   <num_hex_digits@plt>
  test   $0x1,%al
  je     L8
  movb   $0x30,0x0(%rbp)
  lea    0x1(%rbp),%rbp
L8:
  movb   $0x0,0x0(%rbp,%rax,1)
  sub    $0x1,%rax
  js     L2
L32:
  mov    %rbx,%rdx
  and    $0xf,%edx
  movzbl (%rsp,%rdx,1),%edx
  mov    %dl,0x0(%rbp,%rax,1)
  sar    $0x4,%rbx
  sub    $0x1,%rax
  cmp    $0xffffffffffffffff,%rax
  jne    L32
L2:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void make_hex_string_learning(long n, char *s) {
  const char hex_lookup[] = "0123456789abcdef";
  long len = num_hex_digits(n);
  if (len & 1) {
    *s++ = '0';
  }
  s[len] = '\0';
  for (--len; len >= 0; n >>= 4, --len) {
    s[len] = hex_lookup[n & 0xf];
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<make_hex_string_learning>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  movdqa D0(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movb   $0x0,0x10(%rsp)
  movaps %xmm0,(%rsp)
  call   <num_hex_digits@plt>
  test   $0x1,%al
  je     L10
  movb   $0x30,0x0(%rbp)
  add    $0x1,%rbp
L10:
  movb   $0x0,0x0(%rbp,%rax,1)
  sub    $0x1,%rax
  js     L35
  nopl   (%rax)
L25:
  mov    %rbx,%rdx
  sar    $0x4,%rbx
  and    $0xf,%edx
  movzbl (%rsp,%rdx,1),%edx
  mov    %dl,0x0(%rbp,%rax,1)
  sub    $0x1,%rax
  jae    L25
L35:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D0
type_infered: byte[16]

tool: {"D0": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x61,0x62,0x63,0x64,0x65,0x66]}

assistant: ```c
void make_hex_string_learning(long n, char *s) {
  const char hex_lookup[] = "0123456789abcdef";
  long len = num_hex_digits(n);
  if (len & 1) {
    *s++ = '0';
  }
  s[len] = '\0';
  for (--len; len >= 0; n >>= 4, --len) {
    s[len] = hex_lookup[n & 0xf];
  }
}
```