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

```asm
<Fdmira>:
  endbr64
  movapd %xmm0,%xmm2
  movsd  D15(%rip),%xmm0
  subsd  %xmm1,%xmm0
  mulsd  D16(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D15
type: f64
name: disassemble_data
arguments: tag: D16
type: f64

tool: D15: 1

tool: D16: 4

assistant: ```c
double Fdmira(x, p)
double x, p;
{
  static double z;
  return ((4.0 * (1.0 - p) * x) / ((1.0 + z) * (1.0 + z)));
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Fdmira>:
  endbr64
  movapd %xmm0,%xmm2
  movsd  D0(%rip),%xmm0
  subsd  %xmm1,%xmm0
  mulsd  D11(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_infered: f64
name: unpack_data
arguments: data_label: D11
type_infered: f64

tool: D0: 1

tool: D11: 4

assistant: ```c
double Fdmira(x, p)
double x, p;
{
  static double z;
  return ((4.0 * (1.0 - p) * x) / ((1.0 + z) * (1.0 + z)));
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<Fdmira>:
  endbr64
  movapd %xmm0,%xmm2
  movsd  D10(%rip),%xmm0
  subsd  %xmm1,%xmm0
  mulsd  D5(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  ret
```

tool_calls: name: read_data
arguments: data_tag: D10
type: f64
name: read_data
arguments: data_tag: D5
type: f64

tool: D10: 1

tool: D5: 4

assistant: ```c
double Fdmira(x, p)
double x, p;
{
  static double z;
  return ((4.0 * (1.0 - p) * x) / ((1.0 + z) * (1.0 + z)));
}
```
user: Translate the below assembly code into C programming language:

```asm
<minl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jge    L18
  mov    -0x8(%rbp),%rax
  jmp    L19
L18:
  mov    -0x10(%rbp),%rax
L19:
  pop    %rbp
  ret
```

assistant: ```c
long minl(long int a, long int b) {
  if (a < b)
    return (a);
  else
    return (b);
}
```
user: What is the c source code of the assembly code below?

```asm
<minl>:
  endbr64
  cmp    %rsi,%rdi
  mov    %rsi,%rax
  cmovle %rdi,%rax
  ret
```

assistant: ```c
long minl(long int a, long int b) {
  if (a < b)
    return (a);
  else
    return (b);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<minl>:
  endbr64
  cmp    %rsi,%rdi
  mov    %rsi,%rax
  cmovle %rdi,%rax
  ret
```

assistant: ```c
long minl(long int a, long int b) {
  if (a < b)
    return (a);
  else
    return (b);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<minl>:
  endbr64
  cmp    %rsi,%rdi
  mov    %rsi,%rax
  cmovle %rdi,%rax
  ret
```

assistant: ```c
long minl(long int a, long int b) {
  if (a < b)
    return (a);
  else
    return (b);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dcopy___>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jle    L25
  cmpl   $0x1,-0x24(%rbp)
  jne    L108
  cmpl   $0x1,-0x28(%rbp)
  jne    L108
  mov    -0xc(%rbp),%eax
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x30(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  jmp    L30
L108:
  cmpl   $0x0,-0x24(%rbp)
  jne    L117
  cmpl   $0x1,-0x28(%rbp)
  jne    L117
  mov    -0x20(%rbp),%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L8
L80:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x10(%rbp)
L8:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L80
  jmp    L30
L117:
  movl   $0x0,-0x10(%rbp)
  jmp    L107
L101:
  mov    -0x10(%rbp),%eax
  imul   -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  imul   -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x10(%rbp)
L107:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L101
  jmp    L30
L25:
  nop
L30:
  leave
  ret
```

assistant: ```c
void dcopy___(int *n_, const double *dx, int incx, double *dy, int incy) {
  int i, n = *n_;
  if (n <= 0)
    return;
  if (incx == 1 && incy == 1)
    memcpy(dy, dx, sizeof(double) * ((unsigned)n));
  else if (incx == 0 && incy == 1) {
    double x = dx[0];
    for (i = 0; i < n; ++i)
      dy[i] = x;
  } else {
    for (i = 0; i < n; ++i)
      dy[i * incy] = dx[i * incx];
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dcopy___>:
  endbr64
  mov    (%rdi),%r9d
  test   %r9d,%r9d
  jle    L22
  mov    %rcx,%rdi
  cmp    $0x1,%r8d
  sete   %al
  cmp    $0x1,%edx
  jne    L64
  test   %al,%al
  jne    L31
L64:
  test   %edx,%edx
  jne    L13
  test   %al,%al
  je     L13
  movsd  (%rsi),%xmm0
  mov    %rdi,%rax
  lea    -0x1(%r9),%edx
  lea    0x8(%rdi,%rdx,8),%rdx
L67:
  movsd  %xmm0,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L67
  ret
L31:
  sub    $0x8,%rsp
  lea    0x0(,%r9,8),%edx
  call   <memcpy@plt>
  add    $0x8,%rsp
  ret
L13:
  movslq %edx,%rdx
  shl    $0x3,%rdx
  mov    %rsi,%rcx
  movslq %r8d,%r8
  lea    0x0(,%r8,8),%rsi
  mov    %rdi,%rax
  mov    $0x0,%edi
L58:
  movsd  (%rcx),%xmm0
  movsd  %xmm0,(%rax)
  add    $0x1,%edi
  add    %rdx,%rcx
  add    %rsi,%rax
  cmp    %edi,%r9d
  jne    L58
  ret
L22:
  ret
```

assistant: ```c
void dcopy___(int *n_, const double *dx, int incx, double *dy, int incy) {
  int i, n = *n_;
  if (n <= 0)
    return;
  if (incx == 1 && incy == 1)
    memcpy(dy, dx, sizeof(double) * ((unsigned)n));
  else if (incx == 0 && incy == 1) {
    double x = dx[0];
    for (i = 0; i < n; ++i)
      dy[i] = x;
  } else {
    for (i = 0; i < n; ++i)
      dy[i * incy] = dx[i * incx];
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dcopy___>:
  endbr64
  mov    (%rdi),%r9d
  mov    %rcx,%rax
  test   %r9d,%r9d
  jle    L36
  cmp    $0x1,%r8d
  sete   %cl
  cmp    $0x1,%edx
  jne    L50
  test   %cl,%cl
  je     L50
  lea    0x0(,%r9,8),%edx
  mov    %rax,%rdi
  jmp    1050 <memcpy@plt>
  nop
L47:
  movslq %edx,%rdx
  movslq %r8d,%r8
  mov    %rsi,%rcx
  xor    %edi,%edi
  shl    $0x3,%rdx
  lea    0x0(,%r8,8),%rsi
  nopw   0x0(%rax,%rax,1)
L51:
  movsd  (%rcx),%xmm0
  add    $0x1,%edi
  add    %rdx,%rcx
  movsd  %xmm0,(%rax)
  add    %rsi,%rax
  cmp    %edi,%r9d
  jne    L51
L36:
  ret
  nopw   0x0(%rax,%rax,1)
L50:
  movsd  (%rsi),%xmm0
  test   %edx,%edx
  jne    L47
  test   %cl,%cl
  je     L47
  lea    -0x1(%r9),%edx
  lea    0x8(%rax,%rdx,8),%rdx
  nopl   (%rax)
L41:
  movsd  %xmm0,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L41
  ret
```

assistant: ```c
void dcopy___(int *n_, const double *dx, int incx, double *dy, int incy) {
  int i, n = *n_;
  if (n <= 0)
    return;
  if (incx == 1 && incy == 1)
    memcpy(dy, dx, sizeof(double) * ((unsigned)n));
  else if (incx == 0 && incy == 1) {
    double x = dx[0];
    for (i = 0; i < n; ++i)
      dy[i] = x;
  } else {
    for (i = 0; i < n; ++i)
      dy[i * incy] = dx[i * incx];
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<dcopy___>:
  endbr64
  mov    %rdi,%r9
  mov    %rcx,%rdi
  mov    (%r9),%ecx
  test   %ecx,%ecx
  jle    L140
  cmp    $0x1,%r8d
  sete   %al
  cmp    $0x1,%edx
  jne    L16
  test   %al,%al
  je     L16
  lea    0x0(,%rcx,8),%edx
  jmp    1050 <memcpy@plt>
  nopl   (%rax)
L68:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L58:
  movsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rdi,%rax,8)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L58
L140:
  ret
  nopl   0x0(%rax)
L16:
  movsd  (%rsi),%xmm0
  test   %edx,%edx
  jne    L48
  test   %al,%al
  je     L48
  cmp    $0x1,%ecx
  je     L87
  mov    %ecx,%edx
  movapd %xmm0,%xmm1
  mov    %rdi,%rax
  shr    %edx
  unpcklpd %xmm1,%xmm1
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
L148:
  movups %xmm1,(%rax)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L148
  mov    %ecx,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%ecx
  je     L142
L162:
  cltq
  movsd  %xmm0,(%rdi,%rax,8)
  ret
  xchg   %ax,%ax
L48:
  cmp    $0x1,%r8d
  jne    L29
  cmp    $0x1,%edx
  jne    L29
  lea    -0x1(%rcx),%eax
  cmp    $0x2,%eax
  jbe    L68
  lea    0x8(%rsi),%rax
  cmp    %rax,%rdi
  je     L68
  mov    %ecx,%edx
  xor    %eax,%eax
  shr    %edx
  shl    $0x4,%rdx
  xchg   %ax,%ax
L135:
  movupd (%rsi,%rax,1),%xmm2
  movups %xmm2,(%rdi,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rdx
  jne    L135
  mov    %ecx,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%ecx
  je     L140
  movsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rdi,%rax,8)
  ret
  nopl   0x0(%rax,%rax,1)
L142:
  ret
  nopl   0x0(%rax)
L29:
  movslq %edx,%rdx
  movslq %r8d,%r8
  xor    %eax,%eax
  shl    $0x3,%rdx
  shl    $0x3,%r8
  nopl   0x0(%rax,%rax,1)
L114:
  movsd  (%rsi),%xmm0
  add    $0x1,%eax
  add    %rdx,%rsi
  movsd  %xmm0,(%rdi)
  add    %r8,%rdi
  cmp    %eax,%ecx
  jne    L114
  ret
  cs nopw 0x0(%rax,%rax,1)
L87:
  xor    %eax,%eax
  jmp    L162
```

assistant: ```c
void dcopy___(int *n_, const double *dx, int incx, double *dy, int incy) {
  int i, n = *n_;
  if (n <= 0)
    return;
  if (incx == 1 && incy == 1)
    memcpy(dy, dx, sizeof(double) * ((unsigned)n));
  else if (incx == 0 && incy == 1) {
    double x = dx[0];
    for (i = 0; i < n; ++i)
      dy[i] = x;
  } else {
    for (i = 0; i < n; ++i)
      dy[i * incy] = dx[i * incx];
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<solve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x18(%rbp),%rdx
  lea    -0x1c(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <next_empty@plt>
  test   %eax,%eax
  je     L79
  movl   $0x1,-0xc(%rbp)
  jmp    L10
L25:
  mov    -0x18(%rbp),%edx
  mov    -0x1c(%rbp),%esi
  mov    -0xc(%rbp),%ecx
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <can_put@plt>
  test   %rax,%rax
  je     L55
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%eax
  cltq
  mov    -0xc(%rbp),%edx
  mov    %edx,(%rcx,%rax,4)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <solve>
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <next_empty@plt>
  test   %eax,%eax
  je     L22
L55:
  addl   $0x1,-0xc(%rbp)
L10:
  cmpl   $0x9,-0xc(%rbp)
  jle    L25
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  shl    $0x2,%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%eax
  cltq
  movl   $0x0,(%rdx,%rax,4)
  jmp    L32
L79:
  nop
  jmp    L32
L22:
  nop
L32:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

assistant: ```c
void solve(int sudoku[9][9]) {
  int x, y, i;
  if (!next_empty(sudoku, &x, &y)) {
    return;
  }
  for (i = 1; i < 10; i++) {
    if (can_put(sudoku, x, y, i)) {
      sudoku[x][y] = i;
      solve(sudoku);
      int a, b;
      if (!next_empty(sudoku, &a, &b)) {
        return;
      }
    }
  }
  sudoku[x][y] = 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<solve>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <next_empty@plt>
  test   %eax,%eax
  je     L60
  mov    $0x1,%ebx
  lea    0x14(%rsp),%r12
  jmp    L51
L21:
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  je     L31
L51:
  mov    %ebx,%ecx
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  mov    %rbp,%rdi
  call   <can_put@plt>
  test   %rax,%rax
  je     L21
  movslq 0x8(%rsp),%rax
  lea    (%rax,%rax,8),%rax
  lea    0x0(%rbp,%rax,4),%rax
  movslq 0xc(%rsp),%rdx
  mov    %ebx,(%rax,%rdx,4)
  mov    %rbp,%rdi
  call   <solve>
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <next_empty@plt>
  test   %eax,%eax
  jne    L21
  jmp    L60
L31:
  movslq 0x8(%rsp),%rax
  lea    (%rax,%rax,8),%rax
  lea    0x0(%rbp,%rax,4),%rax
  movslq 0xc(%rsp),%rdx
  movl   $0x0,(%rax,%rdx,4)
L60:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void solve(int sudoku[9][9]) {
  int x, y, i;
  if (!next_empty(sudoku, &x, &y)) {
    return;
  }
  for (i = 1; i < 10; i++) {
    if (can_put(sudoku, x, y, i)) {
      sudoku[x][y] = i;
      solve(sudoku);
      int a, b;
      if (!next_empty(sudoku, &a, &b)) {
        return;
      }
    }
  }
  sudoku[x][y] = 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
<solve>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <next_empty@plt>
  test   %eax,%eax
  je     L56
  mov    $0x1,%ebx
  lea    0x14(%rsp),%r12
  jmp    L22
  xchg   %ax,%ax
L24:
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  je     L48
L22:
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  mov    %ebx,%ecx
  mov    %rbp,%rdi
  call   <can_put@plt>
  test   %rax,%rax
  je     L24
  movslq 0x8(%rsp),%rax
  movslq 0xc(%rsp),%rdx
  mov    %rbp,%rdi
  lea    (%rax,%rax,8),%rax
  lea    0x0(%rbp,%rax,4),%rax
  mov    %ebx,(%rax,%rdx,4)
  call   <solve>
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <next_empty@plt>
  test   %eax,%eax
  jne    L24
  nopl   0x0(%rax)
L56:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L48:
  movslq 0x8(%rsp),%rax
  movslq 0xc(%rsp),%rdx
  lea    (%rax,%rax,8),%rax
  lea    0x0(%rbp,%rax,4),%rax
  movl   $0x0,(%rax,%rdx,4)
  jmp    L56
L25:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void solve(int sudoku[9][9]) {
  int x, y, i;
  if (!next_empty(sudoku, &x, &y)) {
    return;
  }
  for (i = 1; i < 10; i++) {
    if (can_put(sudoku, x, y, i)) {
      sudoku[x][y] = i;
      solve(sudoku);
      int a, b;
      if (!next_empty(sudoku, &a, &b)) {
        return;
      }
    }
  }
  sudoku[x][y] = 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<solve>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  call   <next_empty@plt>
  test   %eax,%eax
  je     L27
  mov    $0x1,%ebx
  lea    0x14(%rsp),%r12
  jmp    L23
  xchg   %ax,%ax
L50:
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  je     L10
L23:
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%esi
  mov    %ebx,%ecx
  mov    %rbp,%rdi
  call   <can_put@plt>
  test   %rax,%rax
  je     L50
  movslq 0x8(%rsp),%rax
  movslq 0xc(%rsp),%rdx
  mov    %rbp,%rdi
  lea    (%rax,%rax,8),%rax
  lea    0x0(%rbp,%rax,4),%rax
  mov    %ebx,(%rax,%rdx,4)
  call   <solve>
  lea    0x10(%rsp),%rsi
  mov    %r12,%rdx
  mov    %rbp,%rdi
  call   <next_empty@plt>
  test   %eax,%eax
  jne    L50
  nopl   0x0(%rax)
L27:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L10:
  movslq 0x8(%rsp),%rax
  movslq 0xc(%rsp),%rdx
  lea    (%rax,%rax,8),%rax
  lea    0x0(%rbp,%rax,4),%rax
  movl   $0x0,(%rax,%rdx,4)
  jmp    L27
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void solve(int sudoku[9][9]) {
  int x, y, i;
  if (!next_empty(sudoku, &x, &y)) {
    return;
  }
  for (i = 1; i < 10; i++) {
    if (can_put(sudoku, x, y, i)) {
      sudoku[x][y] = i;
      solve(sudoku);
      int a, b;
      if (!next_empty(sudoku, &a, &b)) {
        return;
      }
    }
  }
  sudoku[x][y] = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xL>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  and    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
uint64_t xL(uint64_t x) {
  uint32_t maxVal = 0xffffffff;
  return x & (uint64_t)maxVal;
}
```
user: What is the c source code of the assembly code below:

```asm
<xL>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
uint64_t xL(uint64_t x) {
  uint32_t maxVal = 0xffffffff;
  return x & (uint64_t)maxVal;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<xL>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
uint64_t xL(uint64_t x) {
  uint32_t maxVal = 0xffffffff;
  return x & (uint64_t)maxVal;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<xL>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
uint64_t xL(uint64_t x) {
  uint32_t maxVal = 0xffffffff;
  return x & (uint64_t)maxVal;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<checkit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L69
L43:
  mov    -0x18(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x14(%rbp)
  jmp    L27
L29:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    0x4(%rax),%edx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    0x4(%rax),%eax
  cmp    %eax,%edx
  jne    L40
  mov    $0x0,%eax
  jmp    L5
L40:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jne    L76
  mov    $0x0,%eax
  jmp    L5
L76:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%edx
  sub    %edx,%eax
  mov    %eax,%edi
  call   <mod@plt>
  mov    %rax,%rbx
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    0x4(%rax),%eax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    0x4(%rdx),%edx
  sub    %edx,%eax
  mov    %eax,%edi
  call   <mod@plt>
  cmp    %rax,%rbx
  jne    L21
  mov    $0x0,%eax
  jmp    L5
L21:
  addl   $0x1,-0x14(%rbp)
L27:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L29
  addl   $0x1,-0x18(%rbp)
L69:
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x18(%rbp)
  jl     L43
  mov    $0x1,%eax
L5:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int checkit(int mat[][2], int n) {
  int i = 0;
  while (i < n - 1) {
    int j = i + 1;
    while (j < n) {
      if (mat[i][1] == mat[j][1])
        return 0;
      if (mat[i][0] == mat[j][0])
        return 0;
      if (mod(mat[i][0] - mat[j][0]) == mod(mat[i][1] - mat[j][1]))
        return 0;
      j++;
    }
    i++;
  }
  return 1;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<checkit>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,0x10(%rsp)
  cmp    $0x1,%esi
  jle    L33
  lea    -0x2(%rsi),%eax
  add    $0x2,%rax
  mov    %rax,0x18(%rsp)
  mov    %rdi,%r12
  movq   $0x1,(%rsp)
  lea    -0x1(%rsi),%eax
  mov    %eax,0x14(%rsp)
L12:
  mov    (%rsp),%rdx
  cmp    %edx,0x10(%rsp)
  jle    L19
  mov    0x14(%rsp),%eax
  sub    %edx,%eax
  lea    -0x1(%rdx,%rax,1),%rax
  mov    0x8(%rsp),%rsi
  lea    0x8(%rsi,%rax,8),%r15
  mov    %r12,%rbx
L43:
  mov    0xc(%rbx),%eax
  cmp    %eax,0x4(%r12)
  je     L4
  mov    (%r12),%edi
  mov    0x8(%rbx),%eax
  cmp    %eax,%edi
  je     L70
  sub    %eax,%edi
  call   <mod@plt>
  mov    %rax,%rbp
  mov    0x4(%r12),%edi
  sub    0xc(%rbx),%edi
  call   <mod@plt>
  cmp    %rax,%rbp
  je     L9
  add    $0x8,%rbx
  cmp    %r15,%rbx
  jne    L43
L19:
  addq   $0x1,(%rsp)
  mov    (%rsp),%rax
  add    $0x8,%r12
  cmp    0x18(%rsp),%rax
  jne    L12
  mov    $0x1,%eax
  jmp    L40
L33:
  mov    $0x1,%eax
  jmp    L40
L4:
  mov    $0x0,%eax
L40:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L70:
  mov    $0x0,%eax
  jmp    L40
L9:
  mov    $0x0,%eax
  jmp    L40
```

assistant: ```c
int checkit(int mat[][2], int n) {
  int i = 0;
  while (i < n - 1) {
    int j = i + 1;
    while (j < n) {
      if (mat[i][1] == mat[j][1])
        return 0;
      if (mat[i][0] == mat[j][0])
        return 0;
      if (mod(mat[i][0] - mat[j][0]) == mod(mat[i][1] - mat[j][1]))
        return 0;
      j++;
    }
    i++;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<checkit>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  cmp    $0x1,%esi
  jle    L13
  lea    -0x2(%rsi),%eax
  mov    %esi,%r12d
  mov    %rdi,%r15
  mov    $0x1,%r13d
  add    $0x2,%rax
  mov    %rax,0x18(%rsp)
  lea    -0x1(%rsi),%eax
  mov    %eax,0x14(%rsp)
L17:
  cmp    %r13d,%r12d
  jle    L38
  mov    0x14(%rsp),%eax
  mov    0x8(%rsp),%rdx
  mov    %r15,%r14
  sub    %r13d,%eax
  lea    -0x1(%r13,%rax,1),%rax
  lea    0x8(%rdx,%rax,8),%rbp
  jmp    L37
  nop
L51:
  mov    (%r15),%edi
  mov    0x8(%r14),%eax
  cmp    %eax,%edi
  je     L9
  sub    %eax,%edi
  call   <mod@plt>
  mov    0x4(%r15),%edi
  sub    0xc(%r14),%edi
  mov    %rax,%rbx
  call   <mod@plt>
  cmp    %rax,%rbx
  je     L9
  add    $0x8,%r14
  cmp    %rbp,%r14
  je     L38
L37:
  mov    0xc(%r14),%eax
  cmp    %eax,0x4(%r15)
  jne    L51
L9:
  xor    %eax,%eax
L73:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  add    $0x1,%r13
  add    $0x8,%r15
  cmp    0x18(%rsp),%r13
  jne    L17
L13:
  mov    $0x1,%eax
  jmp    L73
```

assistant: ```c
int checkit(int mat[][2], int n) {
  int i = 0;
  while (i < n - 1) {
    int j = i + 1;
    while (j < n) {
      if (mat[i][1] == mat[j][1])
        return 0;
      if (mat[i][0] == mat[j][0])
        return 0;
      if (mod(mat[i][0] - mat[j][0]) == mod(mat[i][1] - mat[j][1]))
        return 0;
      j++;
    }
    i++;
  }
  return 1;
}
```