ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  lea    D1(%rip),%r14
  push   %r13
  lea    -0x30(%rbp),%rsi
  mov    %r14,%rdi
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  movslq -0x30(%rbp),%rax
  mov    %rsp,%rsi
  mov    %rax,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L119
L97:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L97
L119:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L113
L75:
  mov    %rsp,%r12
  xor    %ebx,%ebx
  mov    %r12,%r13
  test   %ecx,%ecx
  jle    L35
  nop
L81:
  mov    %r13,%rsi
  mov    %r14,%rdi
  add    $0x1,%ebx
  add    $0x4,%r13
  call   <scanf@plt>
  cmp    %ebx,-0x30(%rbp)
  jg     L81
L35:
  lea    -0x2c(%rbp),%rsi
  mov    %r14,%rdi
  lea    D5(%rip),%r13
  xor    %ebx,%ebx
  call   <scanf@plt>
  mov    -0x30(%rbp),%eax
  test   %eax,%eax
  jle    L55
  nopl   0x0(%rax,%rax,1)
L43:
  mov    -0x2c(%rbp),%ecx
  add    $0x1,%ebx
  cmp    %ecx,(%r12)
  je     L77
  add    $0x4,%r12
  cmp    %ebx,%eax
  jg     L43
L55:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  lea    -0x20(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L77:
  mov    %ebx,%esi
  mov    %r13,%rdi
  add    $0x4,%r12
  call   <printf@plt>
  mov    -0x30(%rbp),%eax
  cmp    %ebx,%eax
  jg     L43
  jmp    L55
L113:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L75
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type_infered: string
name: disassemble_data
arguments: data_tag: D5
type_infered: string

tool: D1: %d

tool: D5: Key found at position %d 


assistant: ```c
int main() {
  int n, i, key;
  scanf("%d", &n);
  int a[n];
  for (i = 0; i < n; i++)
    scanf("%d", &a[i]);
  scanf("%d", &key);
  for (i = 0; i < n; i++)
    if (a[i] == key)
      printf("Key found at position %d \n", i + 1);
  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
<mkdir_exit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
data_type: string

tool: D6: %s

assistant: ```c
void mkdir_exit(char *message, int i) {
  printf("%s", message);
  exit(i);
}
```
user: What is the c source code of the assembly code below?

```asm
<mkdir_exit>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  mov    %rdi,%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%edi
  call   <exit@plt>
  pop    %rbx
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type: string

tool: D1: %s

assistant: ```c
void mkdir_exit(char *message, int i) {
  printf("%s", message);
  exit(i);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<mkdir_exit>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  mov    %rdi,%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <exit@plt>
```

tool_calls: name: retrieve_data
arguments: label: D1
type_infered: string

tool: D1: %s

assistant: ```c
void mkdir_exit(char *message, int i) {
  printf("%s", message);
  exit(i);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<mkdir_exit>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  mov    %rdi,%rsi
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%edi
  pop    %rbp
  jmp    1070 <exit@plt>
```

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

tool: D2: %s

assistant: ```c
void mkdir_exit(char *message, int i) {
  printf("%s", message);
  exit(i);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L28
L27:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L34
  mov    -0x8(%rbp),%rax
  jmp    L20
L34:
  addq   $0x1,-0x8(%rbp)
  subq   $0x1,-0x28(%rbp)
L28:
  cmpq   $0x0,-0x28(%rbp)
  jne    L27
  mov    $0x0,%eax
L20:
  pop    %rbp
  ret
```

assistant: ```c
void *memchr(const void *s, int c, size_t n) {
  const char *tmp = (const char *)s;
  while (n) {
    if (*tmp == c) {
      return (void *)tmp;
    }
    ++tmp;
    --n;
  }
  return NULL;
}
```
user: What is the c source code of the assembly code below?

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L19
  add    %rdi,%rdx
L8:
  movsbl (%rax),%ecx
  cmp    %esi,%ecx
  je     L17
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L8
  mov    $0x0,%eax
  ret
L19:
  mov    $0x0,%eax
L17:
  ret
```

assistant: ```c
void *memchr(const void *s, int c, size_t n) {
  const char *tmp = (const char *)s;
  while (n) {
    if (*tmp == c) {
      return (void *)tmp;
    }
    ++tmp;
    --n;
  }
  return NULL;
}
```
user: What is the c source code of the assembly code below:

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L27
  add    %rdi,%rdx
  jmp    L21
  nopl   0x0(%rax)
L28:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L27
L21:
  movsbl (%rax),%ecx
  cmp    %esi,%ecx
  jne    L28
  ret
  nopl   0x0(%rax)
L27:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memchr(const void *s, int c, size_t n) {
  const char *tmp = (const char *)s;
  while (n) {
    if (*tmp == c) {
      return (void *)tmp;
    }
    ++tmp;
    --n;
  }
  return NULL;
}
```
user: What is the c source code of the assembly code below?

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L32
  add    %rdi,%rdx
  jmp    L19
  nopl   0x0(%rax)
L33:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L32
L19:
  movsbl (%rax),%ecx
  cmp    %esi,%ecx
  jne    L33
  ret
  nopl   0x0(%rax)
L32:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memchr(const void *s, int c, size_t n) {
  const char *tmp = (const char *)s;
  while (n) {
    if (*tmp == c) {
      return (void *)tmp;
    }
    ++tmp;
    --n;
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<cmdCopyRight>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L35
L71:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    $0x0,%edi
  call   <cmdSet@plt>
  addl   $0x1,-0x4(%rbp)
L35:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L71
  movl   $0x0,-0x4(%rbp)
  jmp    L32
L34:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  addl   $0x1,-0x4(%rbp)
L32:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L34
  mov    $0x0,%eax
  call   <cmdLoopBegin@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L66
L9:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    $0x1,%edi
  call   <cmdAdd@plt>
  addl   $0x1,-0x4(%rbp)
L66:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L9
  movl   $0x0,-0x4(%rbp)
  jmp    L63
L18:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  addl   $0x1,-0x4(%rbp)
L63:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L18
  mov    $0x1,%edi
  call   <cmdSubtract@plt>
  mov    $0x0,%eax
  call   <cmdLoopEnd@plt>
  nop
  leave
  ret
```

assistant: ```c
void cmdCopyRight(int iCount) {
  int iPos;
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveRight(1);
    cmdSet(0);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveLeft(1);
  cmdLoopBegin();
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveRight(1);
    cmdAdd(1);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveLeft(1);
  cmdSubtract(1);
  cmdLoopEnd();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cmdCopyRight>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L59
  mov    %edi,%r12d
  mov    $0x0,%ebx
L37:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    $0x0,%edi
  call   <cmdSet@plt>
  mov    %ebx,%ebp
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jne    L37
  mov    $0x0,%ebx
L36:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L36
  mov    $0x0,%eax
  call   <cmdLoopBegin@plt>
  mov    $0x0,%ebx
L4:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    $0x1,%edi
  call   <cmdAdd@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L4
  mov    $0x0,%ebx
L10:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    %ebx,%eax
  add    $0x1,%ebx
  cmp    %eax,%ebp
  jne    L10
L53:
  mov    $0x1,%edi
  call   <cmdSubtract@plt>
  mov    $0x0,%eax
  call   <cmdLoopEnd@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L59:
  mov    $0x0,%eax
  call   <cmdLoopBegin@plt>
  jmp    L53
```

assistant: ```c
void cmdCopyRight(int iCount) {
  int iPos;
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveRight(1);
    cmdSet(0);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveLeft(1);
  cmdLoopBegin();
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveRight(1);
    cmdAdd(1);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveLeft(1);
  cmdSubtract(1);
  cmdLoopEnd();
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<cmdCopyRight>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L3
  mov    %edi,%r12d
  xor    %ebp,%ebp
  nopl   (%rax)
L1:
  mov    $0x1,%edi
  mov    %ebp,%ebx
  lea    0x1(%rbp),%ebp
  call   <cmdMoveRight@plt>
  xor    %edi,%edi
  call   <cmdSet@plt>
  cmp    %ebp,%r12d
  jne    L1
  xor    %ebp,%ebp
  nopl   (%rax)
L37:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L37
  xor    %eax,%eax
  xor    %ebp,%ebp
  call   <cmdLoopBegin@plt>
  nopl   0x0(%rax)
L10:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    $0x1,%edi
  call   <cmdAdd@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L10
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L56:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L56
L40:
  mov    $0x1,%edi
  call   <cmdSubtract@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    10e0 <cmdLoopEnd@plt>
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  call   <cmdLoopBegin@plt>
  jmp    L40
```

assistant: ```c
void cmdCopyRight(int iCount) {
  int iPos;
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveRight(1);
    cmdSet(0);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveLeft(1);
  cmdLoopBegin();
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveRight(1);
    cmdAdd(1);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveLeft(1);
  cmdSubtract(1);
  cmdLoopEnd();
}
```
user: Translate the below assembly code into C programming language:

```asm
<cmdCopyRight>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %edi,%edi
  jle    L15
  mov    %edi,%r12d
  xor    %ebp,%ebp
  nopl   (%rax)
L59:
  mov    $0x1,%edi
  mov    %ebp,%ebx
  lea    0x1(%rbp),%ebp
  call   <cmdMoveRight@plt>
  xor    %edi,%edi
  call   <cmdSet@plt>
  cmp    %ebp,%r12d
  jne    L59
  xor    %ebp,%ebp
  nopl   (%rax)
L11:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L11
  xor    %eax,%eax
  xor    %ebp,%ebp
  call   <cmdLoopBegin@plt>
  nopl   0x0(%rax)
L1:
  mov    $0x1,%edi
  call   <cmdMoveRight@plt>
  mov    $0x1,%edi
  call   <cmdAdd@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L1
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L42:
  mov    $0x1,%edi
  call   <cmdMoveLeft@plt>
  mov    %ebp,%eax
  add    $0x1,%ebp
  cmp    %eax,%ebx
  jne    L42
L3:
  mov    $0x1,%edi
  call   <cmdSubtract@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    10e0 <cmdLoopEnd@plt>
  nopl   0x0(%rax,%rax,1)
L15:
  xor    %eax,%eax
  call   <cmdLoopBegin@plt>
  jmp    L3
```

assistant: ```c
void cmdCopyRight(int iCount) {
  int iPos;
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveRight(1);
    cmdSet(0);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveLeft(1);
  cmdLoopBegin();
  for (iPos = 0; iPos < iCount; iPos++) {
    cmdMoveRight(1);
    cmdAdd(1);
  }
  for (iPos = 0; iPos < iCount; iPos++)
    cmdMoveLeft(1);
  cmdSubtract(1);
  cmdLoopEnd();
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <read@plt>
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <read@plt>
  mov    $0x0,%edi
  call   <read@plt>
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D20
datatype: string
name: extract_symbolic_data
arguments: tag: D12
datatype: string
name: extract_symbolic_data
arguments: tag: D14
datatype: string

tool: D20: Press any key to fault memory in


tool: D12: Done. Press any key to continue


tool: D14: Finished 1


assistant: ```c
int main() {
  printf("Press any key to fault memory in\n");
  read(0);
  printf("Done. Press any key to continue\n");
  read(0);
  read(0);
  printf("Finished 1\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <read@plt>
  lea    D18(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <read@plt>
  mov    $0x0,%edi
  call   <read@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D4
data_type: string
name: disassemble_data
arguments: data_label: D18
data_type: string
name: disassemble_data
arguments: data_label: D13
data_type: string

tool: D4: Press any key to fault memory in


tool: D18: Done. Press any key to continue


tool: D13: Finished 1


assistant: ```c
int main() {
  printf("Press any key to fault memory in\n");
  read(0);
  printf("Done. Press any key to continue\n");
  read(0);
  read(0);
  printf("Finished 1\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <read@plt>
  lea    D23(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <read@plt>
  xor    %edi,%edi
  call   <read@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D1
type_infered: string
name: parse_data
arguments: tag: D23
type_infered: string
name: parse_data
arguments: tag: D2
type_infered: string

tool: D1: Press any key to fault memory in


tool: D23: Done. Press any key to continue


tool: D2: Finished 1


assistant: ```c
int main() {
  printf("Press any key to fault memory in\n");
  read(0);
  printf("Done. Press any key to continue\n");
  read(0);
  read(0);
  printf("Finished 1\n");
  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    D15(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <read@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <read@plt>
  xor    %edi,%edi
  call   <read@plt>
  lea    D26(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D15
type: string
name: unpack_data
arguments: data_tag: D13
type: string
name: unpack_data
arguments: data_tag: D26
type: string

tool: D15: Press any key to fault memory in


tool: D13: Done. Press any key to continue


tool: D26: Finished 1


assistant: ```c
int main() {
  printf("Press any key to fault memory in\n");
  read(0);
  printf("Done. Press any key to continue\n");
  read(0);
  read(0);
  printf("Finished 1\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<valid_digit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L10
L13:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L15
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L15
  addq   $0x1,-0x18(%rbp)
  jmp    L10
L15:
  mov    $0x0,%eax
  jmp    L41
L10:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L13
  mov    $0x1,%eax
L41:
  pop    %rbp
  ret
```

assistant: ```c
int valid_digit(char *ip_str) {
  int err = 0;
  while (*ip_str) {
    if (*ip_str >= '0' && *ip_str <= '9')
      ++ip_str;
    else
      return 0;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<valid_digit>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L13
L2:
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L22
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L2
  mov    $0x1,%eax
  ret
L13:
  mov    $0x1,%eax
  ret
L22:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int valid_digit(char *ip_str) {
  int err = 0;
  while (*ip_str) {
    if (*ip_str >= '0' && *ip_str <= '9')
      ++ip_str;
    else
      return 0;
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<valid_digit>:
  endbr64
  jmp    L11
  cs nopw 0x0(%rax,%rax,1)
L25:
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L20
  add    $0x1,%rdi
L11:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L25
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  xor    %eax,%eax
  ret
```

assistant: ```c
int valid_digit(char *ip_str) {
  int err = 0;
  while (*ip_str) {
    if (*ip_str >= '0' && *ip_str <= '9')
      ++ip_str;
    else
      return 0;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<valid_digit>:
  endbr64
  jmp    L28
  cs nopw 0x0(%rax,%rax,1)
L11:
  sub    $0x30,%eax
  cmp    $0x9,%al
  ja     L2
  add    $0x1,%rdi
L28:
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L11
  mov    $0x1,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  xor    %eax,%eax
  ret
```

assistant: ```c
int valid_digit(char *ip_str) {
  int err = 0;
  while (*ip_str) {
    if (*ip_str >= '0' && *ip_str <= '9')
      ++ip_str;
    else
      return 0;
  }
  return 1;
}
```
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   <print_current_date@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

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

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

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

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

assistant: ```c
int main() {
  print_current_date();
  return (0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_zip_memmem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x20(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jb     L17
  cmpq   $0x0,-0x30(%rbp)
  jne    L22
L17:
  mov    $0x0,%eax
  jmp    L27
L22:
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L9
L31:
  mov    -0x30(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L9
  mov    -0x8(%rbp),%rax
  jmp    L27
L9:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  sub    %rdx,%rax
  mov    %rax,%rdx
  mov    -0x20(%rbp),%rax
  sub    -0x30(%rbp),%rax
  add    %rdx,%rax
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    -0x8(%rbp),%rcx
  add    $0x1,%rcx
  mov    %eax,%esi
  mov    %rcx,%rdi
  call   <memchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L31
  mov    $0x0,%eax
L27:
  leave
  ret
```

assistant: ```c
unsigned char *_zip_memmem(const unsigned char *big, size_t biglen, const unsigned char *little, size_t littlelen) {
  const unsigned char *p;
  if ((biglen < littlelen) || (littlelen == 0))
    return NULL;
  p = big - 1;
  while ((p = (const unsigned char *)memchr(p + 1, little[0], (size_t)(big - (p + 1)) + (size_t)(biglen - littlelen) + 1)) != NULL) {
    if (memcmp(p + 1, little + 1, littlelen - 1) == 0)
      return (unsigned char *)p;
  }
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<_zip_memmem>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rcx,0x8(%rsp)
  lea    -0x1(%rcx),%r14
  cmp    %r14,%rsi
  jbe    L37
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rdx,%r12
  lea    -0x1(%rdi),%rbx
  lea    0x1(%rdx),%r15
L21:
  lea    0x1(%rbx),%rdi
  mov    %r13,%rax
  sub    %rdi,%rax
  lea    0x1(%rbp,%rax,1),%rdx
  sub    0x8(%rsp),%rdx
  movzbl (%r12),%esi
  call   <memchr@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L26
  lea    0x1(%rbx),%rdi
  mov    %r14,%rdx
  mov    %r15,%rsi
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L21
L26:
  mov    %rbx,%rax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  mov    $0x0,%ebx
  jmp    L26
```

assistant: ```c
unsigned char *_zip_memmem(const unsigned char *big, size_t biglen, const unsigned char *little, size_t littlelen) {
  const unsigned char *p;
  if ((biglen < littlelen) || (littlelen == 0))
    return NULL;
  p = big - 1;
  while ((p = (const unsigned char *)memchr(p + 1, little[0], (size_t)(big - (p + 1)) + (size_t)(biglen - littlelen) + 1)) != NULL) {
    if (memcmp(p + 1, little + 1, littlelen - 1) == 0)
      return (unsigned char *)p;
  }
  return NULL;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_zip_memmem>:
  endbr64
  push   %r15
  push   %r14
  lea    -0x1(%rcx),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,(%rsp)
  mov    %rdx,0x8(%rsp)
  cmp    %r14,%rsi
  jbe    L28
  mov    %rsi,%rbx
  mov    %rcx,%r13
  lea    0x1(%rdx),%r15
  mov    %rdi,%rbp
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L4:
  lea    0x1(%r12),%rbp
  mov    %r14,%rdx
  mov    %r15,%rsi
  mov    %rbp,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  je     L7
L33:
  mov    (%rsp),%rax
  mov    %rbp,%rdi
  sub    %rbp,%rax
  lea    0x1(%rbx,%rax,1),%rdx
  mov    0x8(%rsp),%rax
  sub    %r13,%rdx
  movzbl (%rax),%esi
  call   <memchr@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L4
L28:
  xor    %r12d,%r12d
L7:
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned char *_zip_memmem(const unsigned char *big, size_t biglen, const unsigned char *little, size_t littlelen) {
  const unsigned char *p;
  if ((biglen < littlelen) || (littlelen == 0))
    return NULL;
  p = big - 1;
  while ((p = (const unsigned char *)memchr(p + 1, little[0], (size_t)(big - (p + 1)) + (size_t)(biglen - littlelen) + 1)) != NULL) {
    if (memcmp(p + 1, little + 1, littlelen - 1) == 0)
      return (unsigned char *)p;
  }
  return NULL;
}
```