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
<array_print>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x5b,%edi
  push   %rbx
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  test   %eax,%eax
  jle    L6
  xor    %ebx,%ebx
  lea    D6(%rip),%r12
  nopl   (%rax)
L17:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,-0x4(%rbp)
  jg     L17
L6:
  pop    %rbx
  lea    D12(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <puts@plt>
```

tool_calls: name: read_data
arguments: label: D6
type_guessed: string
name: read_data
arguments: label: D12
type_guessed: string

tool: D6: %d 

tool: D12: ]

assistant: ```c
void array_print(int *arr) {
  int i;
  printf("[");
  for (i = 0; i < arr[-1]; i++)
    printf("%d ", arr[i]);
  printf("\b]\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<array_print>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x5b,%edi
  push   %rbx
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  test   %eax,%eax
  jle    L8
  xor    %ebx,%ebx
  lea    D15(%rip),%r12
  nopl   (%rax)
L3:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,-0x4(%rbp)
  jg     L3
L8:
  pop    %rbx
  lea    D9(%rip),%rdi
  pop    %rbp
  pop    %r12
  jmp    1080 <puts@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D15
datatype: string
name: extract_symbolic_data
arguments: label: D9
datatype: string

tool: D15: %d 

tool: D9: ]

assistant: ```c
void array_print(int *arr) {
  int i;
  printf("[");
  for (i = 0; i < arr[-1]; i++)
    printf("%d ", arr[i]);
  printf("\b]\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<put_zeros_int_1d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  addl   $0x1,-0x4(%rbp)
L17:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L10
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void put_zeros_int_1d(int *data, int len) {
  int i;
  for (i = 0; i < len; i++)
    data[i] = 0.;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<put_zeros_int_1d>:
  endbr64
  test   %esi,%esi
  jle    L6
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rdx
L14:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L14
L6:
  ret
```

assistant: ```c
void put_zeros_int_1d(int *data, int len) {
  int i;
  for (i = 0; i < len; i++)
    data[i] = 0.;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<put_zeros_int_1d>:
  endbr64
  test   %esi,%esi
  jle    L2
  movslq %esi,%rsi
  lea    0x0(,%rsi,4),%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  ret
```

assistant: ```c
void put_zeros_int_1d(int *data, int len) {
  int i;
  for (i = 0; i < len; i++)
    data[i] = 0.;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<put_zeros_int_1d>:
  endbr64
  test   %esi,%esi
  jle    L1
  movslq %esi,%rsi
  lea    0x0(,%rsi,4),%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L1:
  ret
```

assistant: ```c
void put_zeros_int_1d(int *data, int len) {
  int i;
  for (i = 0; i < len; i++)
    data[i] = 0.;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bytes_to_int64>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%rax
  movzwl %ax,%eax
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%rax
  and    $0xff0000,%eax
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x18,%rax
  mov    %eax,%eax
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x20,%rax
  mov    %rax,%rdx
  movabs $0xff00000000,%rax
  and    %rdx,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x5,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x28,%rax
  mov    %rax,%rdx
  movabs $0xff0000000000,%rax
  and    %rdx,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x6,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x30,%rax
  mov    %rax,%rdx
  movabs $0xff000000000000,%rax
  and    %rdx,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x7,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x38,%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  or     %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int bytes_to_int64(unsigned char *bytes) {
  long long num = bytes[0] & 0xff;
  num |= (((long long)bytes[1] << 8) & 0xff00);
  num |= (((long long)bytes[2] << 16) & 0xff0000);
  num |= (((long long)bytes[3] << 24) & 0xff000000);
  num |= (((long long)bytes[4] << 32) & 0xff00000000);
  num |= (((long long)bytes[5] << 40) & 0xff0000000000);
  num |= (((long long)bytes[6] << 48) & 0xff000000000000);
  num |= (((long long)bytes[7] << 56) & 0xff00000000000000);
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bytes_to_int64>:
  endbr64
  movzbl (%rdi),%ecx
  movzbl 0x1(%rdi),%edx
  shl    $0x8,%rdx
  movzbl 0x2(%rdi),%eax
  shl    $0x10,%rax
  or     %rax,%rdx
  or     %rcx,%rdx
  movzbl 0x3(%rdi),%eax
  shl    $0x18,%rax
  or     %edx,%eax
  ret
```

assistant: ```c
int bytes_to_int64(unsigned char *bytes) {
  long long num = bytes[0] & 0xff;
  num |= (((long long)bytes[1] << 8) & 0xff00);
  num |= (((long long)bytes[2] << 16) & 0xff0000);
  num |= (((long long)bytes[3] << 24) & 0xff000000);
  num |= (((long long)bytes[4] << 32) & 0xff00000000);
  num |= (((long long)bytes[5] << 40) & 0xff0000000000);
  num |= (((long long)bytes[6] << 48) & 0xff000000000000);
  num |= (((long long)bytes[7] << 56) & 0xff00000000000000);
  return num;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bytes_to_int64>:
  endbr64
  mov    (%rdi),%eax
  ret
```

assistant: ```c
int bytes_to_int64(unsigned char *bytes) {
  long long num = bytes[0] & 0xff;
  num |= (((long long)bytes[1] << 8) & 0xff00);
  num |= (((long long)bytes[2] << 16) & 0xff0000);
  num |= (((long long)bytes[3] << 24) & 0xff000000);
  num |= (((long long)bytes[4] << 32) & 0xff00000000);
  num |= (((long long)bytes[5] << 40) & 0xff0000000000);
  num |= (((long long)bytes[6] << 48) & 0xff000000000000);
  num |= (((long long)bytes[7] << 56) & 0xff00000000000000);
  return num;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<bytes_to_int64>:
  endbr64
  mov    (%rdi),%eax
  ret
```

assistant: ```c
int bytes_to_int64(unsigned char *bytes) {
  long long num = bytes[0] & 0xff;
  num |= (((long long)bytes[1] << 8) & 0xff00);
  num |= (((long long)bytes[2] << 16) & 0xff0000);
  num |= (((long long)bytes[3] << 24) & 0xff000000);
  num |= (((long long)bytes[4] << 32) & 0xff00000000);
  num |= (((long long)bytes[5] << 40) & 0xff0000000000);
  num |= (((long long)bytes[6] << 48) & 0xff000000000000);
  num |= (((long long)bytes[7] << 56) & 0xff00000000000000);
  return num;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <get_number_and_base@plt>
  mov    $0x0,%eax
  call   <convert_number@plt>
  mov    $0x0,%eax
  call   <display_converted_number@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  get_number_and_base();
  convert_number();
  display_converted_number();
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <get_number_and_base@plt>
  mov    $0x0,%eax
  call   <convert_number@plt>
  mov    $0x0,%eax
  call   <display_converted_number@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  get_number_and_base();
  convert_number();
  display_converted_number();
  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
  xor    %eax,%eax
  call   <get_number_and_base@plt>
  xor    %eax,%eax
  call   <convert_number@plt>
  xor    %eax,%eax
  call   <display_converted_number@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  get_number_and_base();
  convert_number();
  display_converted_number();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <get_number_and_base@plt>
  xor    %eax,%eax
  call   <convert_number@plt>
  xor    %eax,%eax
  call   <display_converted_number@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  get_number_and_base();
  convert_number();
  display_converted_number();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<vector_max>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x20(%rbp),%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  movss  %xmm0,(%rax)
  movl   $0x1,-0x4(%rbp)
  jmp    L14
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  movaps %xmm0,%xmm1
  movd   %eax,%xmm0
  call   <fmax@plt>
  movd   %xmm0,%eax
  mov    -0x18(%rbp),%rdx
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L10
  mov    $0x1,%eax
  leave
  ret
```

assistant: ```c
int vector_max(float *r, float *a, int n) {
  *r = *a;
  for (int i = 1; i < n; i++)
    *r = fmax(*r, a[i]);
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<vector_max>:
  endbr64
  movss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  cmp    $0x1,%edx
  jle    L15
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  lea    0x4(%rsi),%rbx
  lea    -0x2(%rdx),%eax
  lea    0x8(%rsi,%rax,4),%r12
L12:
  movss  (%rbx),%xmm1
  movss  0x0(%rbp),%xmm0
  call   <fmax@plt>
  movss  %xmm0,0x0(%rbp)
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L12
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int vector_max(float *r, float *a, int n) {
  *r = *a;
  for (int i = 1; i < n; i++)
    *r = fmax(*r, a[i]);
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<vector_max>:
  endbr64
  movss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  cmp    $0x1,%edx
  jle    L0
  lea    -0x2(%rdx),%eax
  push   %r12
  lea    0x8(%rsi,%rax,4),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    0x4(%rsi),%rbx
  nopl   0x0(%rax)
L18:
  movss  (%rbx),%xmm1
  add    $0x4,%rbx
  call   <fmax@plt>
  movss  %xmm0,0x0(%rbp)
  cmp    %r12,%rbx
  jne    L18
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L0:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int vector_max(float *r, float *a, int n) {
  *r = *a;
  for (int i = 1; i < n; i++)
    *r = fmax(*r, a[i]);
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<vector_max>:
  endbr64
  movss  (%rsi),%xmm0
  movss  %xmm0,(%rdi)
  cmp    $0x1,%edx
  jle    L6
  lea    -0x2(%rdx),%eax
  push   %r12
  lea    0x8(%rsi,%rax,4),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  lea    0x4(%rsi),%rbx
  nopl   0x0(%rax)
L16:
  movss  (%rbx),%xmm1
  add    $0x4,%rbx
  call   <fmax@plt>
  movss  %xmm0,0x0(%rbp)
  cmp    %r12,%rbx
  jne    L16
  pop    %rbx
  mov    $0x1,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L6:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int vector_max(float *r, float *a, int n) {
  *r = *a;
  for (int i = 1; i < n; i++)
    *r = fmax(*r, a[i]);
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<SCE_String_ReplaceChar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L27
L28:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L16
  mov    -0x18(%rbp),%rax
  movzbl -0x20(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L16:
  addq   $0x1,-0x18(%rbp)
L27:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int SCE_String_ReplaceChar(char *str, char a, char b) {
  int rep = 0;
  while (*str) {
    if (*str == a) {
      *str = b;
      rep++;
    }
    str++;
  }
  return rep;
}
```
user: Translate the below assembly code into C programming language:

```asm
<SCE_String_ReplaceChar>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L39
  mov    $0x0,%ecx
  jmp    L8
L40:
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  je     L44
L8:
  cmp    %al,%sil
  jne    L40
  mov    %dl,(%rdi)
  add    $0x1,%ecx
  jmp    L40
L39:
  mov    $0x0,%ecx
L44:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int SCE_String_ReplaceChar(char *str, char a, char b) {
  int rep = 0;
  while (*str) {
    if (*str == a) {
      *str = b;
      rep++;
    }
    str++;
  }
  return rep;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<SCE_String_ReplaceChar>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L20
  xchg   %ax,%ax
L21:
  cmp    %al,%sil
  jne    L8
  mov    %dl,(%rdi)
  add    $0x1,%r8d
L8:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L21
L20:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int SCE_String_ReplaceChar(char *str, char a, char b) {
  int rep = 0;
  while (*str) {
    if (*str == a) {
      *str = b;
      rep++;
    }
    str++;
  }
  return rep;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<SCE_String_ReplaceChar>:
  endbr64
  movzbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L23
  xchg   %ax,%ax
L25:
  cmp    %al,%sil
  je     L33
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  jne    L25
L23:
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L33:
  mov    %dl,(%rdi)
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  add    $0x1,%r8d
  test   %al,%al
  jne    L25
  mov    %r8d,%eax
  ret
```

assistant: ```c
int SCE_String_ReplaceChar(char *str, char a, char b) {
  int rep = 0;
  while (*str) {
    if (*str == a) {
      *str = b;
      rep++;
    }
    str++;
  }
  return rep;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<more_call2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <more@plt>
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <more@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <more20@plt>
  mov    -0x8(%rbp),%rdx
  add    %rdx,%rax
  movzbl (%rax),%eax
  leave
  ret
```

assistant: ```c
char more_call2(char *p, unsigned int x) {
  more(x);
  more(x + 1);
  return p[more20(x)];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<more_call2>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %esi,%edi
  call   <more@plt>
  lea    0x1(%rbx),%edi
  call   <more@plt>
  mov    %ebx,%edi
  call   <more20@plt>
  movzbl 0x0(%rbp,%rax,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char more_call2(char *p, unsigned int x) {
  more(x);
  more(x + 1);
  return p[more20(x)];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<more_call2>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  sub    $0x8,%rsp
  call   <more@plt>
  lea    0x1(%rbp),%edi
  call   <more@plt>
  mov    %ebp,%edi
  call   <more20@plt>
  movzbl (%rbx,%rax,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char more_call2(char *p, unsigned int x) {
  more(x);
  more(x + 1);
  return p[more20(x)];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<more_call2>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%edi
  sub    $0x8,%rsp
  call   <more@plt>
  lea    0x1(%rbp),%edi
  call   <more@plt>
  mov    %ebp,%edi
  call   <more20@plt>
  movzbl (%rbx,%rax,1),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char more_call2(char *p, unsigned int x) {
  more(x);
  more(x + 1);
  return p[more20(x)];
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<UI_dup_info_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L2
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <BUF_strdup@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L2
  mov    D17(%rip),%rax
  mov    (%rax),%edx
  mov    D29(%rip),%rax
  mov    (%rax),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <UIerr@plt>
  mov    $0xffffffff,%eax
  jmp    L15
L2:
  mov    D27(%rip),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  sub    $0x8,%rsp
  push   $0x0
  push   $0x0
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    %edx,%ecx
  mov    $0x1,%edx
  mov    %rax,%rdi
  call   <general_allocate_string@plt>
  add    $0x20,%rsp
L15:
  leave
  ret
```

assistant: ```c
int UI_dup_info_string(UI *ui, const char *text) {
  char *text_copy = NULL;
  if (text) {
    text_copy = BUF_strdup(text);
    if (text_copy == NULL) {
      UIerr(UI_F_UI_DUP_INFO_STRING, ERR_R_MALLOC_FAILURE);
      return -1;
    }
  }
  return general_allocate_string(ui, text_copy, 1, UIT_INFO, 0, NULL, 0, 0, NULL);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<UI_dup_info_string>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  test   %rsi,%rsi
  je     L14
  mov    %rsi,%rdi
  call   <BUF_strdup@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  je     L17
L14:
  sub    $0x8,%rsp
  push   $0x0
  push   $0x0
  push   $0x0
  mov    $0x0,%r9d
  mov    $0x0,%r8d
  mov    D18(%rip),%rax
  mov    (%rax),%ecx
  mov    $0x1,%edx
  mov    %rbx,%rdi
  call   <general_allocate_string@plt>
  add    $0x20,%rsp
L4:
  pop    %rbx
  ret
L17:
  mov    D10(%rip),%rax
  mov    (%rax),%esi
  mov    D29(%rip),%rax
  mov    (%rax),%edi
  call   <UIerr@plt>
  mov    $0xffffffff,%eax
  jmp    L4
```

assistant: ```c
int UI_dup_info_string(UI *ui, const char *text) {
  char *text_copy = NULL;
  if (text) {
    text_copy = BUF_strdup(text);
    if (text_copy == NULL) {
      UIerr(UI_F_UI_DUP_INFO_STRING, ERR_R_MALLOC_FAILURE);
      return -1;
    }
  }
  return general_allocate_string(ui, text_copy, 1, UIT_INFO, 0, NULL, 0, 0, NULL);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<UI_dup_info_string>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  test   %rsi,%rsi
  je     L6
  mov    %rsi,%rdi
  call   <BUF_strdup@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  je     L0
L6:
  sub    $0x8,%rsp
  mov    %rbp,%rdi
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    D6(%rip),%rax
  push   $0x0
  mov    $0x1,%edx
  push   $0x0
  push   $0x0
  mov    (%rax),%ecx
  call   <general_allocate_string@plt>
  add    $0x20,%rsp
  pop    %rbp
  ret
L0:
  mov    D15(%rip),%rax
  mov    (%rax),%esi
  mov    D23(%rip),%rax
  mov    (%rax),%edi
  call   <UIerr@plt>
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int UI_dup_info_string(UI *ui, const char *text) {
  char *text_copy = NULL;
  if (text) {
    text_copy = BUF_strdup(text);
    if (text_copy == NULL) {
      UIerr(UI_F_UI_DUP_INFO_STRING, ERR_R_MALLOC_FAILURE);
      return -1;
    }
  }
  return general_allocate_string(ui, text_copy, 1, UIT_INFO, 0, NULL, 0, 0, NULL);
}
```
user: Translate the below assembly code into C programming language:

```asm
<UI_dup_info_string>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  test   %rsi,%rsi
  je     L8
  mov    %rsi,%rdi
  call   <BUF_strdup@plt>
  mov    %rax,%rsi
  test   %rax,%rax
  je     L2
L8:
  sub    $0x8,%rsp
  mov    %rbp,%rdi
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  mov    D29(%rip),%rax
  push   $0x0
  mov    $0x1,%edx
  push   $0x0
  push   $0x0
  mov    (%rax),%ecx
  call   <general_allocate_string@plt>
  add    $0x20,%rsp
  pop    %rbp
  ret
L2:
  mov    D11(%rip),%rax
  mov    (%rax),%esi
  mov    D14(%rip),%rax
  mov    (%rax),%edi
  call   <UIerr@plt>
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
int UI_dup_info_string(UI *ui, const char *text) {
  char *text_copy = NULL;
  if (text) {
    text_copy = BUF_strdup(text);
    if (text_copy == NULL) {
      UIerr(UI_F_UI_DUP_INFO_STRING, ERR_R_MALLOC_FAILURE);
      return -1;
    }
  }
  return general_allocate_string(ui, text_copy, 1, UIT_INFO, 0, NULL, 0, 0, NULL);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<aesc_encrypt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %rdx,-0x78(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x78(%rbp),%rdx
  mov    -0x68(%rbp),%rcx
  lea    -0x50(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <aes_first_addroundkey@plt>
  movl   $0x1,-0x54(%rbp)
  jmp    L22
L6:
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <aes_subbyte_shiftrows_mixcols@plt>
  mov    -0x54(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x78(%rbp),%rax
  add    %rax,%rdx
  lea    -0x50(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <aes_addroundkey@plt>
  addl   $0x1,-0x54(%rbp)
L22:
  cmpl   $0x9,-0x54(%rbp)
  jbe    L6
  lea    -0x50(%rbp),%rax
  mov    %rax,%rdi
  call   <aes_subbytes_shiftrows@plt>
  mov    -0x78(%rbp),%rax
  lea    0x280(%rax),%rdx
  lea    -0x50(%rbp),%rcx
  mov    -0x70(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <aes_last_addroundkey@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

assistant: ```c
void aesc_encrypt(const uint8_t plain_text[], uint8_t cipher_text[], const uint8_t key_sched[]) {
  uint8_t state[16];
  unsigned int i;
  aes_first_addroundkey(state, plain_text, key_sched);
  for (i = 1; i < 10; ++i) {
    aes_subbyte_shiftrows_mixcols(state);
    aes_addroundkey(state, key_sched + i * 16);
  }
  aes_subbytes_shiftrows(state);
  aes_last_addroundkey(cipher_text, state, key_sched + 10 * 16);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<aesc_encrypt>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x50,%rsp
  mov    %rsi,%r14
  mov    %rdx,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rdi,%rsi
  mov    %rax,%rdi
  call   <aes_first_addroundkey@plt>
  lea    0x40(%r13),%rbx
  lea    0x280(%r13),%r12
  mov    %rsp,%rbp
L12:
  mov    %rbp,%rdi
  call   <aes_subbyte_shiftrows_mixcols@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <aes_addroundkey@plt>
  add    $0x40,%rbx
  cmp    %r12,%rbx
  jne    L12
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <aes_subbytes_shiftrows@plt>
  lea    0x280(%r13),%rdx
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <aes_last_addroundkey@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x50,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void aesc_encrypt(const uint8_t plain_text[], uint8_t cipher_text[], const uint8_t key_sched[]) {
  uint8_t state[16];
  unsigned int i;
  aes_first_addroundkey(state, plain_text, key_sched);
  for (i = 1; i < 10; ++i) {
    aes_subbyte_shiftrows_mixcols(state);
    aes_addroundkey(state, key_sched + i * 16);
  }
  aes_subbytes_shiftrows(state);
  aes_last_addroundkey(cipher_text, state, key_sched + 10 * 16);
}
```