ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test_yajl_null>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D7
type_guessed: string

tool: D7: null


assistant: ```c
int test_yajl_null(void *ctx) {
  printf("null\n");
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bestMove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xf0,%rsp
  mov    %rdi,-0xd8(%rbp)
  mov    %edx,%eax
  mov    %rcx,-0xe8(%rbp)
  mov    %r8d,-0xec(%rbp)
  mov    %esi,%edx
  mov    %dl,-0xdc(%rbp)
  mov    %al,-0xe0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xffffd8f1,-0xbc(%rbp)
  movl   $0x0,-0xc0(%rbp)
  jmp    L26
L11:
  mov    -0xc0(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0xd8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x2e,%eax
  jne    L57
  movl   $0x0,-0xb4(%rbp)
  jmp    L69
L50:
  mov    -0xb4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0xd8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xb4(%rbp),%eax
  cltq
  mov    %edx,-0xb0(%rbp,%rax,4)
  addl   $0x1,-0xb4(%rbp)
L69:
  cmpl   $0x29,-0xb4(%rbp)
  jle    L50
  movsbl -0xdc(%rbp),%ecx
  mov    -0xc0(%rbp),%edx
  lea    -0xb0(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <dropCounter@plt>
  movsbl -0xe0(%rbp),%edx
  movsbl -0xdc(%rbp),%ecx
  lea    -0xb0(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <score@plt>
  mov    %eax,-0xb8(%rbp)
  cmpl   $0xfffffc19,-0xb8(%rbp)
  jl     L75
  cmpl   $0x3e7,-0xb8(%rbp)
  jg     L75
  cmpl   $0x0,-0xec(%rbp)
  jle    L75
  mov    -0xec(%rbp),%eax
  lea    -0x1(%rax),%edi
  movsbl -0xdc(%rbp),%edx
  movsbl -0xe0(%rbp),%esi
  lea    -0xc4(%rbp),%rcx
  lea    -0xb0(%rbp),%rax
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <bestMove>
  neg    %eax
  mov    %eax,-0xb8(%rbp)
L75:
  mov    -0xb8(%rbp),%eax
  cmp    -0xbc(%rbp),%eax
  jle    L57
  mov    -0xb8(%rbp),%eax
  mov    %eax,-0xbc(%rbp)
  mov    -0xe8(%rbp),%rax
  mov    -0xc0(%rbp),%edx
  mov    %edx,(%rax)
L57:
  addl   $0x1,-0xc0(%rbp)
L26:
  cmpl   $0x6,-0xc0(%rbp)
  jle    L11
  mov    -0xbc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L40
  call   <__stack_chk_fail@plt>
L40:
  leave
  ret
```

assistant: ```c
int bestMove(uint8_t board[42], char token, char oppToken, uint8_t *bestCol, uint8_t lookAhead) {
  uint8_t col;
  int bestScore = -9999;
  for (col = 0; col < 7; col++) {
    if (board[col] == '.') {
      int thisScore;
      uint8_t boardCopy[42], i;
      for (i = 0; i < 42; i++) {
        boardCopy[i] = board[i];
      }
      dropCounter(boardCopy, token, col);
      thisScore = score(boardCopy, token, oppToken);
      if (abs(thisScore) < 1000 && lookAhead > 0) {
        uint8_t oppBestCol;
        thisScore = -bestMove(boardCopy, oppToken, token, &oppBestCol, (lookAhead - 1));
      }
      if (thisScore > bestScore) {
        bestScore = thisScore;
        *bestCol = col;
      }
    }
  }
  return bestScore;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<bestMove>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xe8,%rsp
  mov    %rdi,%rbx
  mov    %rcx,0x18(%rsp)
  mov    %r8d,0x14(%rsp)
  mov    %dl,0x13(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebp
  movl   $0xffffd8f1,0xc(%rsp)
  movsbl %sil,%r13d
  jmp    L71
L43:
  cmp    0xc(%rsp),%eax
  jle    L51
  mov    0x18(%rsp),%rcx
  mov    %r12d,(%rcx)
  mov    %eax,0xc(%rsp)
L51:
  add    $0x1,%rbp
  cmp    $0x7,%rbp
  je     L64
L71:
  mov    %ebp,%r12d
  cmpl   $0x2e,(%rbx,%rbp,4)
  jne    L51
  mov    $0x0,%eax
L77:
  mov    (%rbx,%rax,1),%edx
  mov    %edx,0x30(%rsp,%rax,1)
  add    $0x4,%rax
  cmp    $0xa8,%rax
  jne    L77
  lea    0x30(%rsp),%r14
  mov    %r12d,%edx
  mov    %r13d,%esi
  mov    %r14,%rdi
  call   <dropCounter@plt>
  movsbl 0x13(%rsp),%r15d
  mov    %r15d,%edx
  mov    %r13d,%esi
  mov    %r14,%rdi
  call   <score@plt>
  lea    0x3e7(%rax),%edx
  cmp    $0x7ce,%edx
  ja     L43
  cmpl   $0x0,0x14(%rsp)
  jle    L43
  lea    0x2c(%rsp),%rcx
  mov    %r14,%rdi
  mov    0x14(%rsp),%eax
  lea    -0x1(%rax),%r8d
  mov    %r13d,%edx
  mov    %r15d,%esi
  call   <bestMove>
  neg    %eax
  jmp    L43
L64:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  mov    0xc(%rsp),%eax
  add    $0xe8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L34:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bestMove(uint8_t board[42], char token, char oppToken, uint8_t *bestCol, uint8_t lookAhead) {
  uint8_t col;
  int bestScore = -9999;
  for (col = 0; col < 7; col++) {
    if (board[col] == '.') {
      int thisScore;
      uint8_t boardCopy[42], i;
      for (i = 0; i < 42; i++) {
        boardCopy[i] = board[i];
      }
      dropCounter(boardCopy, token, col);
      thisScore = score(boardCopy, token, oppToken);
      if (abs(thisScore) < 1000 && lookAhead > 0) {
        uint8_t oppBestCol;
        thisScore = -bestMove(boardCopy, oppToken, token, &oppBestCol, (lookAhead - 1));
      }
      if (thisScore > bestScore) {
        bestScore = thisScore;
        *bestCol = col;
      }
    }
  }
  return bestScore;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bestMove>:
  endbr64
  push   %r15
  mov    $0xffffd8f1,%r15d
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0xe8,%rsp
  mov    %rcx,0x18(%rsp)
  lea    0x30(%rsp),%r13
  mov    %r8d,0x10(%rsp)
  mov    %sil,0xe(%rsp)
  mov    %dl,0xf(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  jmp    L88
  nopl   (%rax)
L95:
  add    $0x1,%rbp
  cmp    $0x7,%rbp
  je     L38
L88:
  cmpl   $0x2e,(%rbx,%rbp,4)
  mov    %ebp,%r12d
  jne    L95
  xor    %eax,%eax
  nopl   0x0(%rax)
L86:
  mov    (%rbx,%rax,1),%edx
  mov    %edx,0x0(%r13,%rax,1)
  add    $0x4,%rax
  cmp    $0xa8,%rax
  jne    L86
  movsbl 0xe(%rsp),%r14d
  mov    %r12d,%edx
  mov    %r13,%rdi
  mov    %r14d,%esi
  call   <dropCounter@plt>
  movsbl 0xf(%rsp),%r9d
  mov    %r14d,%esi
  mov    %r13,%rdi
  mov    %r9d,%edx
  mov    %r9d,0x14(%rsp)
  call   <score@plt>
  lea    0x3e7(%rax),%edx
  cmp    $0x7ce,%edx
  ja     L20
  mov    0x10(%rsp),%edx
  mov    0x14(%rsp),%r9d
  test   %edx,%edx
  jg     L15
L20:
  cmp    %r15d,%eax
  jle    L95
  mov    0x18(%rsp),%rcx
  add    $0x1,%rbp
  mov    %eax,%r15d
  mov    %r12d,(%rcx)
  cmp    $0x7,%rbp
  jne    L88
  xchg   %ax,%ax
L38:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L60
  add    $0xe8,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  mov    0x10(%rsp),%eax
  lea    0x2c(%rsp),%rcx
  mov    %r14d,%edx
  mov    %r9d,%esi
  mov    %r13,%rdi
  lea    -0x1(%rax),%r8d
  call   <bestMove>
  neg    %eax
  jmp    L20
L60:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bestMove(uint8_t board[42], char token, char oppToken, uint8_t *bestCol, uint8_t lookAhead) {
  uint8_t col;
  int bestScore = -9999;
  for (col = 0; col < 7; col++) {
    if (board[col] == '.') {
      int thisScore;
      uint8_t boardCopy[42], i;
      for (i = 0; i < 42; i++) {
        boardCopy[i] = board[i];
      }
      dropCounter(boardCopy, token, col);
      thisScore = score(boardCopy, token, oppToken);
      if (abs(thisScore) < 1000 && lookAhead > 0) {
        uint8_t oppBestCol;
        thisScore = -bestMove(boardCopy, oppToken, token, &oppBestCol, (lookAhead - 1));
      }
      if (thisScore > bestScore) {
        bestScore = thisScore;
        *bestCol = col;
      }
    }
  }
  return bestScore;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bestMove>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movsbl %sil,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0xe8,%rsp
  mov    %rcx,0x18(%rsp)
  mov    %r8d,0x14(%rsp)
  mov    %dl,0x13(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0xd8(%rsp)
  xor    %eax,%eax
  movl   $0xffffd8f1,0xc(%rsp)
  jmp    L3
  nopl   0x0(%rax)
L16:
  add    $0x1,%rbx
  cmp    $0x7,%rbx
  je     L59
L3:
  cmpl   $0x2e,(%r15,%rbx,4)
  mov    %ebx,%r13d
  jne    L16
  movdqu (%r15),%xmm0
  movdqu 0x10(%r15),%xmm1
  mov    %ebx,%edx
  mov    %ebp,%esi
  movdqu 0x20(%r15),%xmm2
  movdqu 0x30(%r15),%xmm3
  lea    0x30(%rsp),%r12
  movdqu 0x40(%r15),%xmm4
  movdqu 0x50(%r15),%xmm5
  mov    %r12,%rdi
  movaps %xmm0,0x30(%rsp)
  movdqu 0x60(%r15),%xmm6
  movdqu 0x70(%r15),%xmm7
  movaps %xmm1,0x40(%rsp)
  movdqu 0x80(%r15),%xmm0
  mov    0xa0(%r15),%rax
  movaps %xmm2,0x50(%rsp)
  movdqu 0x90(%r15),%xmm1
  movaps %xmm3,0x60(%rsp)
  movaps %xmm4,0x70(%rsp)
  movaps %xmm5,0x80(%rsp)
  movaps %xmm6,0x90(%rsp)
  movaps %xmm7,0xa0(%rsp)
  movaps %xmm0,0xb0(%rsp)
  movaps %xmm1,0xc0(%rsp)
  mov    %rax,0xd0(%rsp)
  call   <dropCounter@plt>
  movsbl 0x13(%rsp),%r14d
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    %r14d,%edx
  call   <score@plt>
  lea    0x3e7(%rax),%edx
  cmp    $0x7ce,%edx
  ja     L48
  mov    0x14(%rsp),%edx
  test   %edx,%edx
  jg     L36
L48:
  cmp    0xc(%rsp),%eax
  jle    L16
  mov    0x18(%rsp),%rsi
  add    $0x1,%rbx
  mov    %eax,0xc(%rsp)
  mov    %r13d,(%rsi)
  cmp    $0x7,%rbx
  jne    L3
  nopw   0x0(%rax,%rax,1)
L59:
  mov    0xd8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L66
  mov    0xc(%rsp),%eax
  add    $0xe8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L36:
  mov    0x14(%rsp),%eax
  lea    0x2c(%rsp),%rcx
  mov    %ebp,%edx
  mov    %r14d,%esi
  mov    %r12,%rdi
  lea    -0x1(%rax),%r8d
  call   <bestMove>
  neg    %eax
  jmp    L48
L66:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int bestMove(uint8_t board[42], char token, char oppToken, uint8_t *bestCol, uint8_t lookAhead) {
  uint8_t col;
  int bestScore = -9999;
  for (col = 0; col < 7; col++) {
    if (board[col] == '.') {
      int thisScore;
      uint8_t boardCopy[42], i;
      for (i = 0; i < 42; i++) {
        boardCopy[i] = board[i];
      }
      dropCounter(boardCopy, token, col);
      thisScore = score(boardCopy, token, oppToken);
      if (abs(thisScore) < 1000 && lookAhead > 0) {
        uint8_t oppBestCol;
        thisScore = -bestMove(boardCopy, oppToken, token, &oppBestCol, (lookAhead - 1));
      }
      if (thisScore > bestScore) {
        bestScore = thisScore;
        *bestCol = col;
      }
    }
  }
  return bestScore;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ehnlc_strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L29
  mov    $0x0,%eax
  jmp    L8
L29:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L7
L21:
  addq   $0x1,-0x8(%rbp)
L7:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L21
  mov    -0x8(%rbp),%rax
  sub    -0x18(%rbp),%rax
L8:
  pop    %rbp
  ret
```

assistant: ```c
size_t ehnlc_strlen(char const *str) {
  char const *p;
  if (!str) {
    return 0;
  }
  for (p = str; *p; ++p)
    ;
  return p - str;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<ehnlc_strlen>:
  endbr64
  test   %rdi,%rdi
  je     L27
  cmpb   $0x0,(%rdi)
  je     L35
  mov    %rdi,%rax
L10:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L10
L29:
  sub    %rdi,%rax
  ret
L35:
  mov    %rdi,%rax
  jmp    L29
L27:
  mov    $0x0,%eax
  ret
```

assistant: ```c
size_t ehnlc_strlen(char const *str) {
  char const *p;
  if (!str) {
    return 0;
  }
  for (p = str; *p; ++p)
    ;
  return p - str;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ehnlc_strlen>:
  endbr64
  test   %rdi,%rdi
  je     L12
  cmpb   $0x0,(%rdi)
  je     L12
  mov    %rdi,%rax
  nopl   0x0(%rax)
L9:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L9
  sub    %rdi,%rax
  ret
  nopl   (%rax)
L12:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t ehnlc_strlen(char const *str) {
  char const *p;
  if (!str) {
    return 0;
  }
  for (p = str; *p; ++p)
    ;
  return p - str;
}
```
user: What is the c source code of the assembly code below?

```asm
<ehnlc_strlen>:
  endbr64
  test   %rdi,%rdi
  je     L27
  cmpb   $0x0,(%rdi)
  je     L27
  mov    %rdi,%rax
  nopl   0x0(%rax)
L28:
  add    $0x1,%rax
  cmpb   $0x0,(%rax)
  jne    L28
  sub    %rdi,%rax
  ret
  nopl   (%rax)
L27:
  xor    %eax,%eax
  ret
```

assistant: ```c
size_t ehnlc_strlen(char const *str) {
  char const *p;
  if (!str) {
    return 0;
  }
  for (p = str; *p; ++p)
    ;
  return p - str;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<fmatprt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    %r8,-0x30(%rbp)
  movl   $0x0,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L39
L9:
  movl   $0x0,-0xc(%rbp)
  jmp    L33
L17:
  mov    -0x8(%rbp),%rax
  lea    0x8(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    (%rax),%rcx
  mov    -0x30(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  movq   %rcx,%xmm0
  mov    %rdx,%rsi
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  addl   $0x1,-0xc(%rbp)
L33:
  mov    -0xc(%rbp),%eax
  cmp    -0x28(%rbp),%eax
  jl     L17
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  addl   $0x1,-0x10(%rbp)
L39:
  mov    -0x10(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L9
  nop
  nop
  leave
  ret
```

assistant: ```c
void fmatprt(FILE *fp, double *a, int n, int m, char *fmt) {
  int i, j;
  double *p;
  for (i = 0, p = a; i < n; ++i) {
    for (j = 0; j < m; ++j)
      fprintf(fp, fmt, *p++);
    fprintf(fp, "\n");
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fmatprt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0x4(%rsp)
  test   %edx,%edx
  jle    L39
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %ecx,%r15d
  mov    %r8,%r13
  lea    -0x1(%rcx),%eax
  lea    0x8(,%rax,8),%rax
  mov    %rax,0x8(%rsp)
  mov    $0x0,%r14d
  jmp    L10
L23:
  add    $0x8,%rbx
  movsd  -0x8(%rbx),%xmm0
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  cmp    %rbp,%rbx
  jne    L23
  mov    %rbp,%rbx
L47:
  mov    %r12,%rsi
  mov    $0xa,%edi
  call   <fputc@plt>
  add    $0x1,%r14d
  cmp    %r14d,0x4(%rsp)
  je     L39
L10:
  mov    0x8(%rsp),%rax
  lea    (%rbx,%rax,1),%rbp
  test   %r15d,%r15d
  jg     L23
  jmp    L47
L39:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void fmatprt(FILE *fp, double *a, int n, int m, char *fmt) {
  int i, j;
  double *p;
  for (i = 0, p = a; i < n; ++i) {
    for (j = 0; j < m; ++j)
      fprintf(fp, fmt, *p++);
    fprintf(fp, "\n");
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fmatprt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0x4(%rsp)
  test   %edx,%edx
  jle    L23
  lea    -0x1(%rcx),%eax
  mov    %rdi,%rbx
  mov    %rsi,%r15
  mov    %ecx,%r14d
  lea    0x8(,%rax,8),%rax
  mov    %r8,%r12
  xor    %r13d,%r13d
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax)
L19:
  mov    0x8(%rsp),%rax
  lea    (%r15,%rax,1),%rbp
  test   %r14d,%r14d
  jle    L5
  xchg   %ax,%ax
L17:
  movsd  (%r15),%xmm0
  add    $0x8,%r15
  mov    %r12,%rsi
  mov    %rbx,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  cmp    %rbp,%r15
  jne    L17
L5:
  mov    %rbx,%rsi
  mov    $0xa,%edi
  add    $0x1,%r13d
  call   <fputc@plt>
  cmp    %r13d,0x4(%rsp)
  jne    L19
L23:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void fmatprt(FILE *fp, double *a, int n, int m, char *fmt) {
  int i, j;
  double *p;
  for (i = 0, p = a; i < n; ++i) {
    for (j = 0; j < m; ++j)
      fprintf(fp, fmt, *p++);
    fprintf(fp, "\n");
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<fmatprt>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0x4(%rsp)
  test   %edx,%edx
  jle    L37
  lea    -0x1(%rcx),%eax
  mov    %rdi,%rbx
  mov    %rsi,%r15
  mov    %ecx,%r14d
  lea    0x8(,%rax,8),%rax
  mov    %r8,%r12
  xor    %r13d,%r13d
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax)
L9:
  mov    0x8(%rsp),%rax
  lea    (%r15,%rax,1),%rbp
  test   %r14d,%r14d
  jle    L13
  xchg   %ax,%ax
L31:
  movsd  (%r15),%xmm0
  add    $0x8,%r15
  mov    %r12,%rsi
  mov    %rbx,%rdi
  mov    $0x1,%eax
  call   <fprintf@plt>
  cmp    %rbp,%r15
  jne    L31
L13:
  mov    %rbx,%rsi
  mov    $0xa,%edi
  add    $0x1,%r13d
  call   <fputc@plt>
  cmp    %r13d,0x4(%rsp)
  jne    L9
L37:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void fmatprt(FILE *fp, double *a, int n, int m, char *fmt) {
  int i, j;
  double *p;
  for (i = 0, p = a; i < n; ++i) {
    for (j = 0; j < m; ++j)
      fprintf(fp, fmt, *p++);
    fprintf(fp, "\n");
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    %edx,%ecx
  mov    $0x1,%edx
  mov    %eax,%edi
  call   <_jc_invoke@plt>
  test   %eax,%eax
  je     L21
  test   %eax,%eax
  js     L20
  sub    $0x1,%eax
  cmp    $0x1,%eax
  ja     L20
  movl   $0x1,-0x4(%rbp)
  jmp    L20
L21:
  nop
L20:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int main(int ac, char **av) {
  int status = 0;
  switch (_jc_invoke(ac, (const char **)av, 1, vfprintf)) {
  case 0:
    break;
  case 2:
  case 1:
    status = 1;
    break;
  }
  return status;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    D0(%rip),%rax
  mov    (%rax),%ecx
  mov    $0x1,%edx
  call   <_jc_invoke@plt>
  sub    $0x1,%eax
  cmp    $0x1,%eax
  setbe  %al
  movzbl %al,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int ac, char **av) {
  int status = 0;
  switch (_jc_invoke(ac, (const char **)av, 1, vfprintf)) {
  case 0:
    break;
  case 2:
  case 1:
    status = 1;
    break;
  }
  return status;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    D8(%rip),%rax
  mov    $0x1,%edx
  mov    (%rax),%ecx
  call   <_jc_invoke@plt>
  sub    $0x1,%eax
  cmp    $0x1,%eax
  setbe  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int main(int ac, char **av) {
  int status = 0;
  switch (_jc_invoke(ac, (const char **)av, 1, vfprintf)) {
  case 0:
    break;
  case 2:
  case 1:
    status = 1;
    break;
  }
  return status;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    D6(%rip),%rax
  mov    $0x1,%edx
  mov    (%rax),%ecx
  call   <_jc_invoke@plt>
  sub    $0x1,%eax
  cmp    $0x1,%eax
  setbe  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  ret
```

assistant: ```c
int main(int ac, char **av) {
  int status = 0;
  switch (_jc_invoke(ac, (const char **)av, 1, vfprintf)) {
  case 0:
    break;
  case 2:
  case 1:
    status = 1;
    break;
  }
  return status;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xe,-0xc(%rbp)
  movl   $0x3,-0x8(%rbp)
  movl   $0x4c,-0x4(%rbp)
  mov    -0x4(%rbp),%ecx
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D6
datatype: string

tool: D6: n1=%d ; n2=%d ; n3=%d ;


assistant: ```c
int main() {
  int n1, n2, n3;
  n1 = 5 + 3 * 8 / 2 - 3;
  n2 = 2 % 2 + 2 * 2 - 2 / 2;
  n3 = 2 * 4 * (5 + 9.0 / 2);
  printf("n1=%d ; n2=%d ; n3=%d ;\n", n1, n2, n3);
  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
  mov    $0x4c,%ecx
  mov    $0x3,%edx
  mov    $0xe,%esi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: label: D0
type_guessed: string

tool: D0: n1=%d ; n2=%d ; n3=%d ;


assistant: ```c
int main() {
  int n1, n2, n3;
  n1 = 5 + 3 * 8 / 2 - 3;
  n2 = 2 % 2 + 2 * 2 - 2 / 2;
  n3 = 2 * 4 * (5 + 9.0 / 2);
  printf("n1=%d ; n2=%d ; n3=%d ;\n", n1, n2, n3);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4c,%ecx
  mov    $0x3,%edx
  mov    $0xe,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D4
data_type: string

tool: D4: n1=%d ; n2=%d ; n3=%d ;


assistant: ```c
int main() {
  int n1, n2, n3;
  n1 = 5 + 3 * 8 / 2 - 3;
  n2 = 2 % 2 + 2 * 2 - 2 / 2;
  n3 = 2 * 4 * (5 + 9.0 / 2);
  printf("n1=%d ; n2=%d ; n3=%d ;\n", n1, n2, n3);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4c,%ecx
  mov    $0x3,%edx
  mov    $0xe,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D6
type_infered: string

tool: D6: n1=%d ; n2=%d ; n3=%d ;


assistant: ```c
int main() {
  int n1, n2, n3;
  n1 = 5 + 3 * 8 / 2 - 3;
  n2 = 2 % 2 + 2 * 2 - 2 / 2;
  n3 = 2 * 4 * (5 + 9.0 / 2);
  printf("n1=%d ; n2=%d ; n3=%d ;\n", n1, n2, n3);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<USBDevDisconnect>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x40050000,-0x4(%rbp)
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <ASSERT@plt>
  mov    -0x4(%rbp),%eax
  add    $0x1,%eax
  cltq
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  add    $0x1,%edx
  movslq %edx,%rdx
  and    $0xffffffbf,%eax
  mov    %eax,(%rdx)
  nop
  leave
  ret
```

assistant: ```c
void USBDevDisconnect(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x40050000);
  (*((volatile uint8_t *)(ui32Base + 0x00000001))) &= (~0x00000040);
}
```
user: What is the c source code of the assembly code below:

```asm
<USBDevDisconnect>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  cmp    $0x40050000,%edi
  sete   %dil
  movzbl %dil,%edi
  call   <ASSERT@plt>
  add    $0x1,%ebx
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  and    $0xffffffbf,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void USBDevDisconnect(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x40050000);
  (*((volatile uint8_t *)(ui32Base + 0x00000001))) &= (~0x00000040);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<USBDevDisconnect>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0x40050000,%ebx
  sete   %dil
  add    $0x1,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  and    $0xffffffbf,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void USBDevDisconnect(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x40050000);
  (*((volatile uint8_t *)(ui32Base + 0x00000001))) &= (~0x00000040);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<USBDevDisconnect>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  cmp    $0x40050000,%ebx
  sete   %dil
  add    $0x1,%ebx
  call   <ASSERT@plt>
  movslq %ebx,%rbx
  mov    (%rbx),%eax
  and    $0xffffffbf,%eax
  mov    %eax,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
void USBDevDisconnect(uint32_t ui32Base) {
  ASSERT(ui32Base == 0x40050000);
  (*((volatile uint8_t *)(ui32Base + 0x00000001))) &= (~0x00000040);
}
```