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
<is_row_free>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L26
L14:
  mov    -0x1c(%rbp),%eax
  lea    0x0(,%rax,8),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L16
  mov    $0x0,%eax
  jmp    L20
L16:
  addl   $0x1,-0x4(%rbp)
L26:
  cmpl   $0x7,-0x4(%rbp)
  jle    L14
  mov    $0x1,%eax
L20:
  pop    %rbp
  ret
```

assistant: ```c
int is_row_free(int *board, int row) {
  for (int i = 0; i < 8; i++) {
    if (board[row * 8 + i]) {
      return 0;
    }
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<is_row_free>:
  endbr64
  lea    0x0(,%rsi,8),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  lea    (%rdi,%rdx,1),%rax
  lea    0x20(%rdi,%rdx,1),%rdx
L2:
  cmpl   $0x0,(%rax)
  jne    L4
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L2
  mov    $0x1,%eax
  ret
L4:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int is_row_free(int *board, int row) {
  for (int i = 0; i < 8; i++) {
    if (board[row * 8 + i]) {
      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
<is_row_free>:
  endbr64
  lea    0x0(,%rsi,8),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  lea    (%rdi,%rdx,1),%rax
  lea    0x20(%rdi,%rdx,1),%rdx
  nopl   0x0(%rax,%rax,1)
L3:
  mov    (%rax),%ecx
  test   %ecx,%ecx
  jne    L0
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L3
  mov    $0x1,%eax
  ret
  nopl   (%rax)
L0:
  xor    %eax,%eax
  ret
```

assistant: ```c
int is_row_free(int *board, int row) {
  for (int i = 0; i < 8; i++) {
    if (board[row * 8 + i]) {
      return 0;
    }
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_row_free>:
  endbr64
  shl    $0x3,%esi
  movslq %esi,%rax
  mov    (%rdi,%rax,4),%eax
  test   %eax,%eax
  jne    L19
  add    $0x1,%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r9d
  lea    0x0(,%rsi,4),%rdx
  test   %r9d,%r9d
  jne    L60
  mov    0x4(%rdi,%rdx,1),%eax
  test   %eax,%eax
  jne    L19
  mov    0x8(%rdi,%rdx,1),%r8d
  test   %r8d,%r8d
  jne    L60
  mov    0xc(%rdi,%rdx,1),%eax
  test   %eax,%eax
  jne    L19
  mov    0x10(%rdi,%rdx,1),%esi
  test   %esi,%esi
  jne    L60
  mov    0x14(%rdi,%rdx,1),%ecx
  test   %ecx,%ecx
  jne    L60
  mov    0x18(%rdi,%rdx,1),%edx
  xor    %eax,%eax
  test   %edx,%edx
  sete   %al
  ret
  xchg   %ax,%ax
L19:
  xor    %eax,%eax
L60:
  ret
```

assistant: ```c
int is_row_free(int *board, int row) {
  for (int i = 0; i < 8; i++) {
    if (board[row * 8 + i]) {
      return 0;
    }
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<bot>:
  endbr64
  add    $0xc,%rdi
  mov    $0x0,%r10d
  jmp    L84
L123:
  cmpl   $0xffffffff,(%rsi)
  je     L22
L2:
  add    $0x1,%ecx
  add    $0x1c,%r8
  add    $0x1c,%rsi
  add    $0x1c,%rax
  cmp    $0x6,%ecx
  je     L5
L71:
  test   %r9d,%r9d
  jne    L5
  mov    (%rax),%edx
  add    $0x1,%edx
  test   $0xfffffffd,%edx
  je     L77
  mov    -0x18(%rax),%edx
  cmp    $0xffffffff,%edx
  je     L123
  cmp    $0x1,%edx
  jne    L2
  cmpl   $0x1,(%rsi)
  jne    L2
  cmpl   $0x1,(%r8)
  jne    L2
  test   %ecx,%ecx
  je     L2
L77:
  mov    %r10d,%eax
  ret
L22:
  cmpl   $0xffffffff,(%r8)
  jne    L2
  jmp    L77
L5:
  add    $0x1,%r10d
  add    $0x4,%rdi
  cmp    $0x7,%r10d
  je     L77
L84:
  mov    -0xc(%rdi),%r9d
  lea    -0x4(%rdi),%rsi
  lea    0x10(%rdi),%rax
  mov    %rdi,%r8
  mov    $0x0,%ecx
  jmp    L71
```

assistant: ```c
int bot(int grid[6][7], int move) {
  int i, j;
  for (j = 0; j < 7; j++) {
    for (i = 0; i < 6; i++) {
      if (grid[0][j] != 0)
        break;
      else {
        if ((grid[i + 1][j] == -1 && grid[i + 1][j] == -1 && grid[i + 1][j] == -1) || (grid[i + 1][j] == 1 && grid[i + 1][j] == 1 && grid[i + 1][j] == 1) || grid[i][j + 1] == -1 && grid[i][j + 2] == -1 && grid[i][j + 3] == -1 || grid[i][j + 1] == 1 && grid[i][j + 2] == 1 && grid[i][j + 3] == 1 && i != 0)
          return j;
      }
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<bot>:
  endbr64
  add    $0x8,%rdi
  mov    $0x1,%r9d
L72:
  mov    -0x8(%rdi),%edx
  mov    %rdi,%rsi
  add    $0x4,%rdi
  lea    -0x1(%r9),%r10d
  lea    0x14(%rsi),%rax
  mov    %rdi,%r8
  xor    %ecx,%ecx
  test   %edx,%edx
  jne    L94
L91:
  mov    (%rax),%edx
  add    $0x1,%edx
  and    $0xfffffffd,%edx
  je     L88
  mov    -0x18(%rax),%edx
  cmp    $0xffffffff,%edx
  je     L111
  cmp    $0x1,%edx
  jne    L32
  cmpl   $0x1,(%rsi)
  je     L44
L32:
  add    $0x1,%ecx
  add    $0x1c,%rsi
  add    $0x1c,%r8
  add    $0x1c,%rax
  cmp    $0x6,%ecx
  jne    L91
L94:
  add    $0x1,%r9d
  cmp    $0x8,%r9d
  jne    L72
L88:
  mov    %r10d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L111:
  cmpl   $0xffffffff,(%rsi)
  jne    L32
  cmpl   $0xffffffff,(%r8)
  jne    L32
  jmp    L88
  nopl   (%rax)
L44:
  cmpl   $0x1,(%r8)
  jne    L32
  test   %ecx,%ecx
  je     L32
  mov    %r10d,%eax
  ret
```

assistant: ```c
int bot(int grid[6][7], int move) {
  int i, j;
  for (j = 0; j < 7; j++) {
    for (i = 0; i < 6; i++) {
      if (grid[0][j] != 0)
        break;
      else {
        if ((grid[i + 1][j] == -1 && grid[i + 1][j] == -1 && grid[i + 1][j] == -1) || (grid[i + 1][j] == 1 && grid[i + 1][j] == 1 && grid[i + 1][j] == 1) || grid[i][j + 1] == -1 && grid[i][j + 2] == -1 && grid[i][j + 3] == -1 || grid[i][j + 1] == 1 && grid[i][j + 2] == 1 && grid[i][j + 3] == 1 && i != 0)
          return j;
      }
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<dstsub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %edx,-0x38(%rbp)
  mov    %rcx,-0x48(%rbp)
  mov    -0x34(%rbp),%eax
  sar    %eax
  mov    %eax,-0x24(%rbp)
  mov    -0x38(%rbp),%eax
  cltd
  idivl  -0x34(%rbp)
  mov    %eax,-0x20(%rbp)
  movl   $0x0,-0x28(%rbp)
  movl   $0x1,-0x2c(%rbp)
  jmp    L18
L0:
  mov    -0x34(%rbp),%eax
  sub    -0x2c(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x20(%rbp),%eax
  add    %eax,-0x28(%rbp)
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x38(%rbp),%eax
  sub    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x38(%rbp),%eax
  sub    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mulsd  -0x18(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x18(%rbp),%xmm1
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x2c(%rbp)
L18:
  mov    -0x2c(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L0
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x48(%rbp),%rax
  movsd  (%rax),%xmm0
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void dstsub(int n, double *a, int nc, double *c) {
  int j, k, kk, ks, m;
  double wkr, wki, xr;
  m = n >> 1;
  ks = nc / n;
  kk = 0;
  for (j = 1; j < m; j++) {
    k = n - j;
    kk += ks;
    wkr = c[kk] - c[nc - kk];
    wki = c[kk] + c[nc - kk];
    xr = wki * a[k] - wkr * a[j];
    a[k] = wkr * a[k] + wki * a[j];
    a[j] = xr;
  }
  a[m] *= c[0];
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<dstsub>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rsi,%r8
  mov    %edx,%ebp
  mov    %rcx,%r9
  mov    %edi,%r10d
  sar    %r10d
  mov    %edx,%eax
  cltd
  idiv   %edi
  cmp    $0x3,%edi
  jle    L18
  mov    %edi,%ebx
  movslq %eax,%rdx
  lea    0x0(,%rdx,8),%r11
  lea    (%rcx,%r11,1),%rsi
  neg    %eax
  movslq %eax,%rdi
  shl    $0x3,%rdi
  movslq %ebp,%rbp
  sub    %rdx,%rbp
  lea    (%rcx,%rbp,8),%rcx
  movslq %ebx,%rbx
  lea    (%r8,%rbx,8),%rdx
  lea    0x8(%r8),%rax
  lea    -0x2(%r10),%ebx
  lea    0x10(%r8,%rbx,8),%rbx
L14:
  movsd  (%rsi),%xmm1
  movsd  (%rcx),%xmm0
  movapd %xmm1,%xmm4
  subsd  %xmm0,%xmm4
  addsd  %xmm0,%xmm1
  movsd  -0x8(%rdx),%xmm0
  movsd  (%rax),%xmm2
  movapd %xmm0,%xmm3
  mulsd  %xmm4,%xmm3
  movapd %xmm2,%xmm5
  mulsd  %xmm1,%xmm5
  addsd  %xmm5,%xmm3
  movsd  %xmm3,-0x8(%rdx)
  mulsd  %xmm1,%xmm0
  mulsd  %xmm4,%xmm2
  subsd  %xmm2,%xmm0
  movsd  %xmm0,(%rax)
  add    %r11,%rsi
  add    %rdi,%rcx
  sub    $0x8,%rdx
  add    $0x8,%rax
  cmp    %rbx,%rax
  jne    L14
L18:
  movslq %r10d,%r10
  lea    (%r8,%r10,8),%rax
  movsd  (%rax),%xmm0
  mulsd  (%r9),%xmm0
  movsd  %xmm0,(%rax)
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void dstsub(int n, double *a, int nc, double *c) {
  int j, k, kk, ks, m;
  double wkr, wki, xr;
  m = n >> 1;
  ks = nc / n;
  kk = 0;
  for (j = 1; j < m; j++) {
    k = n - j;
    kk += ks;
    wkr = c[kk] - c[nc - kk];
    wki = c[kk] + c[nc - kk];
    xr = wki * a[k] - wkr * a[j];
    a[k] = wkr * a[k] + wki * a[j];
    a[j] = xr;
  }
  a[m] *= c[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
<dstsub>:
  endbr64
  movslq %edx,%r9
  movslq %edi,%r8
  push   %rbx
  mov    %rsi,%r10
  mov    %r9d,%eax
  mov    %r8d,%ebx
  mov    %rcx,%r11
  cltd
  sar    %ebx
  idiv   %r8d
  cmp    $0x3,%r8d
  jle    L0
  cltq
  lea    (%r10,%r8,8),%rdx
  lea    -0x2(%rbx),%r8d
  lea    0x0(,%rax,8),%rdi
  sub    %rax,%r9
  lea    0x10(%r10,%r8,8),%r8
  lea    (%rcx,%rdi,1),%rsi
  lea    0x8(%r10),%rax
  lea    (%rcx,%r9,8),%rcx
  nopl   (%rax)
L15:
  movsd  (%rsi),%xmm1
  movsd  (%rcx),%xmm0
  add    $0x8,%rax
  add    %rdi,%rsi
  movsd  -0x8(%rax),%xmm2
  sub    %rdi,%rcx
  sub    $0x8,%rdx
  movapd %xmm1,%xmm4
  addsd  %xmm0,%xmm1
  subsd  %xmm0,%xmm4
  movsd  (%rdx),%xmm0
  movapd %xmm2,%xmm5
  mulsd  %xmm1,%xmm5
  movapd %xmm0,%xmm3
  mulsd  %xmm4,%xmm3
  mulsd  %xmm1,%xmm0
  mulsd  %xmm4,%xmm2
  addsd  %xmm5,%xmm3
  subsd  %xmm2,%xmm0
  movsd  %xmm3,(%rdx)
  movsd  %xmm0,-0x8(%rax)
  cmp    %rax,%r8
  jne    L15
L0:
  movslq %ebx,%rbx
  lea    (%r10,%rbx,8),%rax
  movsd  (%rax),%xmm0
  mulsd  (%r11),%xmm0
  movsd  %xmm0,(%rax)
  pop    %rbx
  ret
```

assistant: ```c
void dstsub(int n, double *a, int nc, double *c) {
  int j, k, kk, ks, m;
  double wkr, wki, xr;
  m = n >> 1;
  ks = nc / n;
  kk = 0;
  for (j = 1; j < m; j++) {
    k = n - j;
    kk += ks;
    wkr = c[kk] - c[nc - kk];
    wki = c[kk] + c[nc - kk];
    xr = wki * a[k] - wkr * a[j];
    a[k] = wkr * a[k] + wki * a[j];
    a[j] = xr;
  }
  a[m] *= c[0];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
int main(void) {
  int a[] = {0};
  return a[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
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  int a[] = {0};
  return a[0];
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int a[] = {0};
  return a[0];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  int a[] = {0};
  return a[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
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D34(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%ecx
  movslq %ecx,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x24,%rax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  mov    %eax,%edx
  sub    %edi,%edx
  imul   $0x32,%edx,%esi
  mov    %ecx,%edx
  sub    %esi,%edx
  movslq %edx,%rbx
  imul   $0x66666667,%rbx,%rbx
  sar    $0x22,%rbx
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%ebx
  mov    %ebx,%ebp
  lea    (%rbx,%rbx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  mov    %edx,%ebx
  sub    %edi,%eax
  mov    %eax,%esi
  lea    D40(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movslq %ebx,%rbp
  imul   $0x66666667,%rbp,%rbp
  sar    $0x21,%rbp
  mov    %ebx,%eax
  sar    $0x1f,%eax
  sub    %eax,%ebp
  mov    %ebp,%esi
  lea    D7(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x0(%rbp,%rbp,4),%eax
  mov    %ebx,%esi
  sub    %eax,%esi
  lea    D52(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D17
type_infered: string
name: read_data
arguments: tag: D34
type_infered: string
name: read_data
arguments: tag: D40
type_infered: string
name: read_data
arguments: tag: D26
type_infered: string
name: read_data
arguments: tag: D7
type_infered: string
name: read_data
arguments: tag: D52
type_infered: string

tool: D17: Informe o valor: 

tool: D34: %d

tool: D40: %d nota(s) de 50 reais


tool: D26: %d nota(s) de 10 reais


tool: D7: %d nota(s) de 5 reais


tool: D52: %d nota(s) de 1 real


assistant: ```c
int main(void) {
  int val, aux, n50, n10, n5, n1;
  printf("Informe o valor: ");
  scanf("%d", &val);
  aux = val;
  n50 = aux / 50;
  aux = aux % 50;
  n10 = aux / 10;
  aux = aux % 10;
  n5 = aux / 5;
  n1 = aux % 5;
  printf("%d nota(s) de 50 reais\n", n50);
  printf("%d nota(s) de 10 reais\n", n10);
  printf("%d nota(s) de 5 reais\n", n5);
  printf("%d nota(s) de 1 real\n", n1);
  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
  push   %rbp
  lea    D8(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D23(%rip),%rdi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rax
  mov    %rax,%rcx
  imul   $0x51eb851f,%rax,%rax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sar    $0x24,%rax
  mov    %eax,%edx
  sub    %edi,%eax
  sub    %edi,%edx
  lea    D33(%rip),%rdi
  imul   $0x32,%edx,%esi
  mov    %ecx,%edx
  sub    %esi,%edx
  mov    %eax,%esi
  xor    %eax,%eax
  movslq %edx,%rbx
  mov    %edx,%ecx
  imul   $0x66666667,%rbx,%rbx
  sar    $0x1f,%ecx
  sar    $0x22,%rbx
  sub    %ecx,%ebx
  lea    (%rbx,%rbx,4),%ecx
  mov    %ebx,%ebp
  add    %ecx,%ecx
  sub    %ecx,%edx
  mov    %edx,%ebx
  call   <printf@plt>
  mov    %ebp,%esi
  movslq %ebx,%rbp
  xor    %eax,%eax
  imul   $0x66666667,%rbp,%rbp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%eax
  lea    D29(%rip),%rdi
  sar    $0x1f,%eax
  sar    $0x21,%rbp
  sub    %eax,%ebp
  xor    %eax,%eax
  mov    %ebp,%esi
  call   <printf@plt>
  lea    0x0(%rbp,%rbp,4),%eax
  lea    D47(%rip),%rdi
  sub    %eax,%ebx
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D8
type_guessed: string
name: unpack_data
arguments: data_label: D23
type_guessed: string
name: unpack_data
arguments: data_label: D33
type_guessed: string
name: unpack_data
arguments: data_label: D3
type_guessed: string
name: unpack_data
arguments: data_label: D29
type_guessed: string
name: unpack_data
arguments: data_label: D47
type_guessed: string

tool: D8: Informe o valor: 

tool: D23: %d

tool: D33: %d nota(s) de 50 reais


tool: D3: %d nota(s) de 10 reais


tool: D29: %d nota(s) de 5 reais


tool: D47: %d nota(s) de 1 real


assistant: ```c
int main(void) {
  int val, aux, n50, n10, n5, n1;
  printf("Informe o valor: ");
  scanf("%d", &val);
  aux = val;
  n50 = aux / 50;
  aux = aux % 50;
  n10 = aux / 10;
  aux = aux % 10;
  n5 = aux / 5;
  n1 = aux % 5;
  printf("%d nota(s) de 50 reais\n", n50);
  printf("%d nota(s) de 10 reais\n", n10);
  printf("%d nota(s) de 5 reais\n", n5);
  printf("%d nota(s) de 1 real\n", n1);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D26(%rip),%rdi
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  movslq 0x4(%rsp),%rax
  mov    %rax,%rcx
  imul   $0x51eb851f,%rax,%rax
  mov    %ecx,%edi
  sar    $0x1f,%edi
  sar    $0x24,%rax
  mov    %eax,%edx
  sub    %edi,%eax
  sub    %edi,%edx
  lea    D47(%rip),%rdi
  imul   $0x32,%edx,%esi
  mov    %ecx,%edx
  sub    %esi,%edx
  mov    %eax,%esi
  xor    %eax,%eax
  movslq %edx,%rbx
  mov    %edx,%ecx
  imul   $0x66666667,%rbx,%rbx
  sar    $0x1f,%ecx
  sar    $0x22,%rbx
  sub    %ecx,%ebx
  lea    (%rbx,%rbx,4),%ecx
  mov    %ebx,%ebp
  add    %ecx,%ecx
  sub    %ecx,%edx
  mov    %edx,%ebx
  call   <printf@plt>
  mov    %ebp,%esi
  movslq %ebx,%rbp
  xor    %eax,%eax
  imul   $0x66666667,%rbp,%rbp
  lea    D54(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%eax
  lea    D33(%rip),%rdi
  sar    $0x1f,%eax
  sar    $0x21,%rbp
  sub    %eax,%ebp
  xor    %eax,%eax
  mov    %ebp,%esi
  call   <printf@plt>
  lea    0x0(%rbp,%rbp,4),%eax
  lea    D40(%rip),%rdi
  sub    %eax,%ebx
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D26
type_infered: string
name: read_data
arguments: tag: D14
type_infered: string
name: read_data
arguments: tag: D47
type_infered: string
name: read_data
arguments: tag: D54
type_infered: string
name: read_data
arguments: tag: D33
type_infered: string
name: read_data
arguments: tag: D40
type_infered: string

tool: D26: Informe o valor: 

tool: D14: %d

tool: D47: %d nota(s) de 50 reais


tool: D54: %d nota(s) de 10 reais


tool: D33: %d nota(s) de 5 reais


tool: D40: %d nota(s) de 1 real


assistant: ```c
int main(void) {
  int val, aux, n50, n10, n5, n1;
  printf("Informe o valor: ");
  scanf("%d", &val);
  aux = val;
  n50 = aux / 50;
  aux = aux % 50;
  n10 = aux / 10;
  aux = aux % 10;
  n5 = aux / 5;
  n1 = aux % 5;
  printf("%d nota(s) de 50 reais\n", n50);
  printf("%d nota(s) de 10 reais\n", n10);
  printf("%d nota(s) de 5 reais\n", n5);
  printf("%d nota(s) de 1 real\n", n1);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Ungetc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  cmpl   $0xa,-0x4(%rbp)
  jne    L5
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    -0x1(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
L5:
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <ungetc@plt>
  nop
  leave
  ret
```

assistant: ```c
void Ungetc(int c, FILE *f, int *linep) {
  if (c == '\n')
    --(*linep);
  ungetc(c, f);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<Ungetc>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0xa,%edi
  je     L14
L15:
  call   <ungetc@plt>
  add    $0x8,%rsp
  ret
L14:
  subl   $0x1,(%rdx)
  jmp    L15
```

assistant: ```c
void Ungetc(int c, FILE *f, int *linep) {
  if (c == '\n')
    --(*linep);
  ungetc(c, f);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Ungetc>:
  endbr64
  cmp    $0xa,%edi
  jne    L3
  subl   $0x1,(%rdx)
L3:
  jmp    1050 <ungetc@plt>
```

assistant: ```c
void Ungetc(int c, FILE *f, int *linep) {
  if (c == '\n')
    --(*linep);
  ungetc(c, f);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Ungetc>:
  endbr64
  cmp    $0xa,%edi
  jne    L4
  subl   $0x1,(%rdx)
L4:
  jmp    1050 <ungetc@plt>
```

assistant: ```c
void Ungetc(int c, FILE *f, int *linep) {
  if (c == '\n')
    --(*linep);
  ungetc(c, f);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ft_strncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L11
L14:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L26
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x20(%rbp)
  jmp    L23
L26:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
L23:
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L14
  mov    -0x8(%rbp),%rax
  neg    %rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  pop    %rbp
  ret
```

assistant: ```c
char *ft_strncpy(char *dst, const char *src, size_t len) {
  size_t i;
  i = 0;
  while (i < len) {
    if (*src) {
      *dst = *src;
      src++;
    } else
      *dst = '\0';
    dst++;
    i++;
  }
  return (dst - i);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ft_strncpy>:
  endbr64
  test   %rdx,%rdx
  je     L0
  lea    (%rdi,%rdx,1),%rax
L13:
  movzbl (%rsi),%ecx
  cmp    $0x1,%cl
  sbb    $0xffffffffffffffff,%rsi
  mov    %cl,(%rdi)
  add    $0x1,%rdi
  cmp    %rdi,%rax
  jne    L13
L11:
  sub    %rdx,%rax
  ret
L0:
  mov    %rdi,%rax
  jmp    L11
```

assistant: ```c
char *ft_strncpy(char *dst, const char *src, size_t len) {
  size_t i;
  i = 0;
  while (i < len) {
    if (*src) {
      *dst = *src;
      src++;
    } else
      *dst = '\0';
    dst++;
    i++;
  }
  return (dst - i);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ft_strncpy>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L4
  add    %rdi,%rdx
  mov    %rdi,%rcx
  nopw   0x0(%rax,%rax,1)
L14:
  movzbl (%rsi),%r8d
  cmp    $0x1,%r8b
  mov    %r8b,(%rcx)
  sbb    $0xffffffffffffffff,%rsi
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L14
L4:
  ret
```

assistant: ```c
char *ft_strncpy(char *dst, const char *src, size_t len) {
  size_t i;
  i = 0;
  while (i < len) {
    if (*src) {
      *dst = *src;
      src++;
    } else
      *dst = '\0';
    dst++;
    i++;
  }
  return (dst - i);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ft_strncpy>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L14
  add    %rdi,%rdx
  mov    %rdi,%rcx
  nopw   0x0(%rax,%rax,1)
L3:
  movzbl (%rsi),%r8d
  cmp    $0x1,%r8b
  mov    %r8b,(%rcx)
  sbb    $0xffffffffffffffff,%rsi
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L3
L14:
  ret
```

assistant: ```c
char *ft_strncpy(char *dst, const char *src, size_t len) {
  size_t i;
  i = 0;
  while (i < len) {
    if (*src) {
      *dst = *src;
      src++;
    } else
      *dst = '\0';
    dst++;
    i++;
  }
  return (dst - i);
}
```