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
<extractName>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %rsi,-0x90(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x70(%rbp)
  movq   $0x0,-0x68(%rbp)
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x88(%rbp),%rax
  mov    %rax,%rdi
  call   <indexOfNameBegin@plt>
  mov    %eax,-0x74(%rbp)
  mov    -0x74(%rbp),%eax
  mov    %eax,-0x7c(%rbp)
  movl   $0x0,-0x78(%rbp)
  jmp    L34
L8:
  mov    -0x7c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x88(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  je     L17
  mov    -0x7c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x88(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x78(%rbp),%eax
  cltq
  mov    %dl,-0x70(%rbp,%rax,1)
  addl   $0x1,-0x7c(%rbp)
  addl   $0x1,-0x78(%rbp)
L34:
  cmpl   $0x63,-0x7c(%rbp)
  jle    L8
  jmp    L16
L17:
  nop
L16:
  lea    -0x70(%rbp),%rdx
  mov    -0x90(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

assistant: ```c
void extractName(char *fullName, char *nameOnly) {
  char name[100] = {0};
  int i, j;
  int begin = indexOfNameBegin(fullName);
  for (i = begin, j = 0; i < 100; i++, j++) {
    if (fullName[i] == '.') {
      break;
    }
    name[j] = fullName[i];
  }
  strcpy(nameOnly, name);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<extractName>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movl   $0x0,0x60(%rsp)
  call   <indexOfNameBegin@plt>
  cmp    $0x63,%eax
  jg     L17
  cltq
  mov    %rax,%rdx
  mov    %rsp,%rcx
  sub    %rax,%rcx
L20:
  movzbl (%rbx,%rdx,1),%eax
  cmp    $0x2e,%al
  je     L17
  mov    %al,(%rcx,%rdx,1)
  add    $0x1,%rdx
  cmp    $0x63,%edx
  jle    L20
L17:
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void extractName(char *fullName, char *nameOnly) {
  char name[100] = {0};
  int i, j;
  int begin = indexOfNameBegin(fullName);
  for (i = begin, j = 0; i < 100; i++, j++) {
    if (fullName[i] == '.') {
      break;
    }
    name[j] = fullName[i];
  }
  strcpy(nameOnly, name);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<extractName>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x60(%rsp)
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  call   <indexOfNameBegin@plt>
  mov    %rsp,%rsi
  cmp    $0x63,%eax
  jg     L31
  mov    %rsp,%rsi
  cltq
  mov    %rsi,%rcx
  sub    %rax,%rcx
  jmp    L16
  nopl   0x0(%rax,%rax,1)
L43:
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    $0x63,%eax
  jg     L31
L16:
  movzbl (%rbx,%rax,1),%edx
  cmp    $0x2e,%dl
  jne    L43
L31:
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void extractName(char *fullName, char *nameOnly) {
  char name[100] = {0};
  int i, j;
  int begin = indexOfNameBegin(fullName);
  for (i = begin, j = 0; i < 100; i++, j++) {
    if (fullName[i] == '.') {
      break;
    }
    name[j] = fullName[i];
  }
  strcpy(nameOnly, name);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<extractName>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x60(%rsp)
  movaps %xmm0,(%rsp)
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  call   <indexOfNameBegin@plt>
  mov    %rsp,%rsi
  cmp    $0x63,%eax
  jg     L43
  mov    %rsp,%rsi
  cltq
  mov    %rsi,%rcx
  sub    %rax,%rcx
  jmp    L27
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %dl,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    $0x63,%eax
  jg     L43
L27:
  movzbl (%rbx,%rax,1),%edx
  cmp    $0x2e,%dl
  jne    L5
L43:
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  ret
L48:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void extractName(char *fullName, char *nameOnly) {
  char name[100] = {0};
  int i, j;
  int begin = indexOfNameBegin(fullName);
  for (i = begin, j = 0; i < 100; i++, j++) {
    if (fullName[i] == '.') {
      break;
    }
    name[j] = fullName[i];
  }
  strcpy(nameOnly, name);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<GPMCErrValStatusGet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x48,%eax
  mov    %eax,%eax
  mov    (%rax),%eax
  and    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int GPMCErrValStatusGet(unsigned int baseAddr) {
  unsigned int status;
  status = ((*((volatile unsigned int *)(baseAddr + (0x48)))) & (0x00000001u));
  return (status);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<GPMCErrValStatusGet>:
  endbr64
  mov    0x48(%edi),%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
unsigned int GPMCErrValStatusGet(unsigned int baseAddr) {
  unsigned int status;
  status = ((*((volatile unsigned int *)(baseAddr + (0x48)))) & (0x00000001u));
  return (status);
}
```
user: Translate the below assembly code into C programming language:

```asm
<GPMCErrValStatusGet>:
  endbr64
  mov    0x48(%edi),%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
unsigned int GPMCErrValStatusGet(unsigned int baseAddr) {
  unsigned int status;
  status = ((*((volatile unsigned int *)(baseAddr + (0x48)))) & (0x00000001u));
  return (status);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<GPMCErrValStatusGet>:
  endbr64
  mov    0x48(%edi),%eax
  and    $0x1,%eax
  ret
```

assistant: ```c
unsigned int GPMCErrValStatusGet(unsigned int baseAddr) {
  unsigned int status;
  status = ((*((volatile unsigned int *)(baseAddr + (0x48)))) & (0x00000001u));
  return (status);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<Quicksort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x30(%rbp),%rdi
  mov    -0x28(%rbp),%rcx
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%esi
  mov    -0x18(%rbp),%rax
  mov    %rdi,%r8
  mov    %rax,%rdi
  call   <quicksort@plt>
  nop
  leave
  ret
```

assistant: ```c
void Quicksort(void *array, int length, int Compare(void *, int, int), void Swap(void *, int, int)) {
  int low = 0;
  int high = length - 1;
  quicksort(array, low, high, Compare, Swap);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Quicksort>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdx,%rax
  mov    %rcx,%r8
  lea    -0x1(%rsi),%edx
  mov    %rax,%rcx
  mov    $0x0,%esi
  call   <quicksort@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void Quicksort(void *array, int length, int Compare(void *, int, int), void Swap(void *, int, int)) {
  int low = 0;
  int high = length - 1;
  quicksort(array, low, high, Compare, Swap);
}
```
user: What is the c source code of the assembly code below?

```asm
<Quicksort>:
  endbr64
  mov    %rdx,%r9
  mov    %rcx,%r8
  lea    -0x1(%rsi),%edx
  xor    %esi,%esi
  mov    %r9,%rcx
  jmp    1050 <quicksort@plt>
```

assistant: ```c
void Quicksort(void *array, int length, int Compare(void *, int, int), void Swap(void *, int, int)) {
  int low = 0;
  int high = length - 1;
  quicksort(array, low, high, Compare, Swap);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<Quicksort>:
  endbr64
  mov    %rdx,%r9
  mov    %rcx,%r8
  lea    -0x1(%rsi),%edx
  xor    %esi,%esi
  mov    %r9,%rcx
  jmp    1050 <quicksort@plt>
```

assistant: ```c
void Quicksort(void *array, int length, int Compare(void *, int, int), void Swap(void *, int, int)) {
  int low = 0;
  int high = length - 1;
  quicksort(array, low, high, Compare, Swap);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getstring1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movb   $0x61,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
char getstring1() {
  char c = 'a';
  return c;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<getstring1>:
  endbr64
  mov    $0x61,%eax
  ret
```

assistant: ```c
char getstring1() {
  char c = 'a';
  return c;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getstring1>:
  endbr64
  mov    $0x61,%eax
  ret
```

assistant: ```c
char getstring1() {
  char c = 'a';
  return c;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L14:
  addl   $0x1,-0x4(%rbp)
L17:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  test   %al,%al
  jne    L14
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int strlen(const char *s) {
  int len = 0;
  while (*s++)
    len++;
  return len;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strlen>:
  endbr64
  lea    0x1(%rdi),%rcx
  cmpb   $0x0,(%rdi)
  je     L0
  mov    %rcx,%rax
  mov    $0x1,%edx
  sub    %ecx,%edx
L23:
  lea    (%rdx,%rax,1),%ecx
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L23
L5:
  mov    %ecx,%eax
  ret
L0:
  mov    $0x0,%ecx
  jmp    L5
```

assistant: ```c
int strlen(const char *s) {
  int len = 0;
  while (*s++)
    len++;
  return len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  lea    0x1(%rdi),%rax
  je     L6
  mov    $0x1,%edx
  sub    %eax,%edx
  nopl   0x0(%rax)
L1:
  lea    (%rdx,%rax,1),%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L1
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int strlen(const char *s) {
  int len = 0;
  while (*s++)
    len++;
  return len;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  lea    0x1(%rdi),%rax
  je     L8
  mov    $0x1,%edx
  sub    %eax,%edx
  nopl   0x0(%rax)
L0:
  lea    (%rdx,%rax,1),%r8d
  add    $0x1,%rax
  cmpb   $0x0,-0x1(%rax)
  jne    L0
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int strlen(const char *s) {
  int len = 0;
  while (*s++)
    len++;
  return len;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  je     L6
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x1(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  movzbl -0x1(%rbp),%edx
  mov    %dl,(%rax)
L6:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(char *v1, char *v2) {
  if (v1 != v2) {
    char tmp = *v1;
    *v1 = *v2;
    *v2 = tmp;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<swap>:
  endbr64
  cmp    %rsi,%rdi
  je     L0
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  mov    %dl,(%rdi)
  mov    %al,(%rsi)
L0:
  ret
```

assistant: ```c
void swap(char *v1, char *v2) {
  if (v1 != v2) {
    char tmp = *v1;
    *v1 = *v2;
    *v2 = tmp;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<swap>:
  endbr64
  cmp    %rsi,%rdi
  je     L4
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  mov    %dl,(%rdi)
  mov    %al,(%rsi)
L4:
  ret
```

assistant: ```c
void swap(char *v1, char *v2) {
  if (v1 != v2) {
    char tmp = *v1;
    *v1 = *v2;
    *v2 = tmp;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<swap>:
  endbr64
  cmp    %rsi,%rdi
  je     L5
  movzbl (%rdi),%eax
  movzbl (%rsi),%edx
  mov    %dl,(%rdi)
  mov    %al,(%rsi)
L5:
  ret
```

assistant: ```c
void swap(char *v1, char *v2) {
  if (v1 != v2) {
    char tmp = *v1;
    *v1 = *v2;
    *v2 = tmp;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<destroy_share_memory>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <shmdt@plt>
  cmp    $0xffffffff,%eax
  jne    L22
  mov    $0xffffffff,%eax
  jmp    L3
L22:
  mov    D0(%rip),%rax
  mov    (%rax),%ecx
  mov    -0xc(%rbp),%eax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <shmctl@plt>
  cmp    $0xffffffff,%eax
  jne    L37
  mov    $0xffffffff,%eax
  jmp    L3
L37:
  mov    $0x0,%eax
L3:
  leave
  ret
```

assistant: ```c
int destroy_share_memory(void *shm, int shmid) {
  if (shmdt(shm) == -1) {
    return -1;
  }
  if (shmctl(shmid, IPC_RMID, 0) == -1) {
    return -1;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<destroy_share_memory>:
  endbr64
  push   %rbx
  mov    %esi,%ebx
  call   <shmdt@plt>
  cmp    $0xffffffff,%eax
  je     L3
  mov    $0x0,%edx
  mov    D9(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <shmctl@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L3:
  pop    %rbx
  ret
```

assistant: ```c
int destroy_share_memory(void *shm, int shmid) {
  if (shmdt(shm) == -1) {
    return -1;
  }
  if (shmctl(shmid, IPC_RMID, 0) == -1) {
    return -1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<destroy_share_memory>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  call   <shmdt@plt>
  cmp    $0xffffffff,%eax
  je     L5
  mov    D8(%rip),%rax
  xor    %edx,%edx
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <shmctl@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int destroy_share_memory(void *shm, int shmid) {
  if (shmdt(shm) == -1) {
    return -1;
  }
  if (shmctl(shmid, IPC_RMID, 0) == -1) {
    return -1;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<destroy_share_memory>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  call   <shmdt@plt>
  cmp    $0xffffffff,%eax
  je     L7
  mov    D9(%rip),%rax
  xor    %edx,%edx
  mov    %ebp,%edi
  mov    (%rax),%esi
  call   <shmctl@plt>
  cmp    $0xffffffff,%eax
  sete   %al
  movzbl %al,%eax
  neg    %eax
L7:
  pop    %rbp
  ret
```

assistant: ```c
int destroy_share_memory(void *shm, int shmid) {
  if (shmdt(shm) == -1) {
    return -1;
  }
  if (shmctl(shmid, IPC_RMID, 0) == -1) {
    return -1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0x24(%rbp)
  mov    $0x0,%eax
  call   <test_FILE@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  char array[20];
  int n = 10;
  test_FILE();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

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

assistant: ```c
int main(int argc, char *argv[]) {
  char array[20];
  int n = 10;
  test_FILE();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

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

assistant: ```c
int main(int argc, char *argv[]) {
  char array[20];
  int n = 10;
  test_FILE();
  return 0;
}
```
user: What is the c source code of the assembly code below?

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

assistant: ```c
int main(int argc, char *argv[]) {
  char array[20];
  int n = 10;
  test_FILE();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x1,-0x20(%rbp)
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L13
L15:
  mov    -0x30(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  imul   -0x20(%rbp),%rax
  add    %rax,-0x28(%rbp)
  shlq   -0x20(%rbp)
  mov    -0x30(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x30(%rbp)
L13:
  mov    -0x30(%rbp),%rax
  test   %rax,%rax
  jne    L15
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D36
type_guessed: string
name: retrieve_data
arguments: label: D14
type_guessed: string
name: retrieve_data
arguments: label: D38
type_guessed: string
name: retrieve_data
arguments: label: D32
type_guessed: string

tool: D36: Enter binary value: 

tool: D14: %ld

tool: D38: Decimal: %d


tool: D32: Octal: %lo

assistant: ```c
int main(void) {
  long int binum, decnum = 0, octnum = 0, j = 1, rem;
  printf("Enter binary value: ");
  scanf("%ld", &binum);
  while (binum != 0) {
    rem = binum % 10;
    decnum += rem * j;
    j *= 2;
    binum /= 10;
  }
  octnum = decnum;
  printf("Decimal: %d\n", decnum);
  printf("Octal: %lo", octnum);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L12
  mov    $0x1,%esi
  mov    $0x0,%ebx
  movabs $0x6666666666666667,%rdi
L25:
  mov    %rcx,%rax
  imul   %rdi
  sar    $0x2,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  imul   %rsi,%rcx
  add    %rcx,%rbx
  add    %rsi,%rsi
  mov    %rdx,%rcx
  test   %rdx,%rdx
  jne    L25
  movq   $0x0,(%rsp)
L9:
  mov    %rbx,%rsi
  lea    D32(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  ret
L12:
  mov    %rcx,%rbx
  jmp    L9
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D28
type: string
name: unpack_data
arguments: tag: D22
type: string
name: unpack_data
arguments: tag: D32
type: string
name: unpack_data
arguments: tag: D2
type: string

tool: D28: Enter binary value: 

tool: D22: %ld

tool: D32: Decimal: %d


tool: D2: Octal: %lo

assistant: ```c
int main(void) {
  long int binum, decnum = 0, octnum = 0, j = 1, rem;
  printf("Enter binary value: ");
  scanf("%ld", &binum);
  while (binum != 0) {
    rem = binum % 10;
    decnum += rem * j;
    j *= 2;
    binum /= 10;
  }
  octnum = decnum;
  printf("Decimal: %d\n", decnum);
  printf("Octal: %lo", octnum);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D38(%rip),%rdi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D4(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rcx
  test   %rcx,%rcx
  je     L8
  mov    $0x1,%esi
  xor    %ebp,%ebp
  movabs $0x6666666666666667,%rdi
  nopw   0x0(%rax,%rax,1)
L25:
  mov    %rcx,%rax
  imul   %rdi
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sar    $0x2,%rdx
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rcx
  imul   %rsi,%rcx
  add    %rsi,%rsi
  add    %rcx,%rbp
  mov    %rdx,%rcx
  test   %rdx,%rdx
  jne    L25
  movq   $0x0,(%rsp)
L11:
  mov    %rbp,%rsi
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  xor    %ebp,%ebp
  jmp    L11
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D38
type_guessed: string
name: read_data
arguments: tag: D4
type_guessed: string
name: read_data
arguments: tag: D17
type_guessed: string
name: read_data
arguments: tag: D28
type_guessed: string

tool: D38: Enter binary value: 

tool: D4: %ld

tool: D17: Decimal: %d


tool: D28: Octal: %lo

assistant: ```c
int main(void) {
  long int binum, decnum = 0, octnum = 0, j = 1, rem;
  printf("Enter binary value: ");
  scanf("%ld", &binum);
  while (binum != 0) {
    rem = binum % 10;
    decnum += rem * j;
    j *= 2;
    binum /= 10;
  }
  octnum = decnum;
  printf("Decimal: %d\n", decnum);
  printf("Octal: %lo", octnum);
  return 0;
}
```