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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L3
L20:
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L16
L38:
  addl   $0x1,-0x1c(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edx
  lea    D17(%rip),%rax
  mov    %rax,%rsi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0xc(%rbp),%eax
  movq   %rdx,%xmm0
  mov    %eax,%edi
  call   <print_results@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x14(%rbp)
L16:
  cmpl   $0x2,-0x14(%rbp)
  jle    L38
  addl   $0x1,-0x18(%rbp)
L3:
  cmpl   $0x6,-0x18(%rbp)
  jle    L20
  mov    -0x1c(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D17
datatype: string
name: disassemble_data
arguments: data_label: D26
datatype: string
name: disassemble_data
arguments: data_label: D23
datatype: string

tool: D17: dmaxabs

tool: D26: # c::%s:len=%d


tool: D23: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dmaxabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  mov    $0x3,%r12d
  mov    $0x1,%r14d
  lea    D19(%rip),%r15
  jmp    L36
L7:
  add    $0x1,%r14d
  add    $0x3,%r12d
  cmp    $0x15,%r12d
  je     L21
L36:
  lea    -0x3(%r12),%ebx
  mov    %r14d,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ebp
  lea    -0x1(%r14),%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%ecx
  mov    $0xf4240,%eax
  cltd
  idiv   %ecx
  mov    %eax,%r13d
L6:
  add    $0x1,%ebx
  mov    %ebp,%edx
  mov    %r15,%rsi
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%esi
  mov    %r13d,%edi
  call   <benchmark@plt>
  mov    %r13d,%edi
  call   <print_results@plt>
  mov    %ebx,%esi
  lea    D6(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    %r12d,%ebx
  jne    L6
  jmp    L7
L21:
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: label: D19
datatype: string
name: disassemble_data
arguments: label: D13
datatype: string
name: disassemble_data
arguments: label: D6
datatype: string

tool: D19: dmaxabs

tool: D13: # c::%s:len=%d


tool: D6: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dmaxabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D29(%rip),%r15
  push   %r14
  lea    D14(%rip),%r14
  push   %r13
  lea    D26(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L11:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L5:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L5
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L11
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: unpack_data
arguments: label: D29
datatype: string
name: unpack_data
arguments: label: D14
datatype: string
name: unpack_data
arguments: label: D26
datatype: string

tool: D29: dmaxabs

tool: D14: # c::%s:len=%d


tool: D26: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dmaxabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D16(%rip),%r15
  push   %r14
  lea    D11(%rip),%r14
  push   %r13
  lea    D9(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L7:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L14:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L14
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L7
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: tag: D16
type: string
name: parse_data
arguments: tag: D11
type: string
name: parse_data
arguments: tag: D9
type: string

tool: D16: dmaxabs

tool: D11: # c::%s:len=%d


tool: D9: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dmaxabs", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<suberase>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%ecx
  add    %ecx,%ecx
  add    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L37
  mov    $0x0,%eax
  jmp    L32
L30:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jle    L55
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0xc(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <moveable@plt>
  cmp    %eax,-0x4(%rbp)
  jne    L45
  movl   $0x0,-0x8(%rbp)
  jmp    L55
L45:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x8(%rbp)
L55:
  mov    -0xc(%rbp),%edx
  mov    -0x1c(%rbp),%eax
  add    %eax,%edx
  movslq %edx,%rax
  imul   $0x1948b0fd,%rax,%rax
  shr    $0x20,%rax
  sar    $0x3,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x3,%eax
  add    %ecx,%eax
  lea    0x0(,%rax,8),%ecx
  add    %ecx,%eax
  sub    %eax,%edx
  mov    %edx,-0xc(%rbp)
L37:
  cmpl   $0x54,-0x8(%rbp)
  jle    L30
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L33
L11:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jne    L40
  addl   $0x1,-0x8(%rbp)
L40:
  addl   $0x1,-0xc(%rbp)
L33:
  cmpl   $0x50,-0xc(%rbp)
  jle    L11
  mov    -0x8(%rbp),%eax
L32:
  leave
  ret
```

assistant: ```c
int suberase(sudoku, d, e)
int sudoku[];
int d;
int e;
{
  int a = e;
  int b;
  int c = 0;
  if (d % 3 == 0) {
    return (0);
  }
  while (c < 85) {
    if (sudoku[a] > 0) {
      b = sudoku[a];
      sudoku[a] = 0;
      if (moveable(sudoku, a) == b) {
        c = 0;
      } else {
        sudoku[a] = b;
        c++;
      }
    }
    a = (a + d) % 81;
  }
  c = 0;
  for (a = 0; a < 81; a++) {
    if (sudoku[a] == 0) {
      c++;
    }
  }
  return (c);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<suberase>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edx,%ebx
  movslq %esi,%rax
  imul   $0x55555556,%rax,%rax
  shr    $0x20,%rax
  mov    %esi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,2),%edx
  mov    %esi,%eax
  sub    %edx,%eax
  je     L71
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    $0x0,%r13d
  jmp    L39
L42:
  lea    (%r12,%rbx,1),%eax
  movslq %eax,%rbx
  imul   $0x1948b0fd,%rbx,%rbx
  sar    $0x23,%rbx
  cltd
  sub    %edx,%ebx
  lea    (%rbx,%rbx,8),%edx
  lea    (%rdx,%rdx,8),%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  cmp    $0x54,%r13d
  jg     L11
L39:
  movslq %ebx,%rax
  lea    0x0(%rbp,%rax,4),%r14
  mov    (%r14),%r15d
  test   %r15d,%r15d
  jle    L42
  movl   $0x0,(%r14)
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <moveable@plt>
  cmp    %eax,%r15d
  je     L21
  mov    %r15d,(%r14)
  add    $0x1,%r13d
  jmp    L42
L11:
  mov    %rbp,%rdx
  add    $0x144,%rbp
  mov    $0x0,%eax
L22:
  cmpl   $0x1,(%rdx)
  adc    $0x0,%eax
  add    $0x4,%rdx
  cmp    %rbp,%rdx
  jne    L22
L71:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L21:
  lea    (%r12,%rbx,1),%eax
  movslq %eax,%rbx
  imul   $0x1948b0fd,%rbx,%rbx
  sar    $0x23,%rbx
  cltd
  sub    %edx,%ebx
  lea    (%rbx,%rbx,8),%edx
  lea    (%rdx,%rdx,8),%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  mov    $0x0,%r13d
  jmp    L39
```

assistant: ```c
int suberase(sudoku, d, e)
int sudoku[];
int d;
int e;
{
  int a = e;
  int b;
  int c = 0;
  if (d % 3 == 0) {
    return (0);
  }
  while (c < 85) {
    if (sudoku[a] > 0) {
      b = sudoku[a];
      sudoku[a] = 0;
      if (moveable(sudoku, a) == b) {
        c = 0;
      } else {
        sudoku[a] = b;
        c++;
      }
    }
    a = (a + d) % 81;
  }
  c = 0;
  for (a = 0; a < 81; a++) {
    if (sudoku[a] == 0) {
      c++;
    }
  }
  return (c);
}
```
user: Translate the below assembly code into C programming language:

```asm
<suberase>:
  endbr64
  movslq %esi,%rax
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rax,%r12
  imul   $0x55555556,%rax,%rax
  push   %rbp
  mov    %edx,%ebp
  mov    %r12d,%edx
  sar    $0x1f,%edx
  push   %rbx
  shr    $0x20,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,2),%edx
  mov    %r12d,%eax
  sub    $0x8,%rsp
  sub    %edx,%eax
  je     L44
  mov    %rdi,%rbx
  xor    %r15d,%r15d
  nopl   (%rax)
L20:
  movslq %ebp,%rax
  lea    (%rbx,%rax,4),%r14
  mov    (%r14),%r13d
  test   %r13d,%r13d
  jg     L2
L57:
  lea    (%r12,%rbp,1),%eax
  movslq %eax,%rbp
  cltd
  imul   $0x1948b0fd,%rbp,%rbp
  sar    $0x23,%rbp
  sub    %edx,%ebp
  lea    0x0(%rbp,%rbp,8),%edx
  lea    (%rdx,%rdx,8),%edx
  sub    %edx,%eax
  mov    %eax,%ebp
  cmp    $0x55,%r15d
  jne    L20
  mov    %rbx,%rdx
  lea    0x144(%rbx),%rcx
  xor    %eax,%eax
  nopl   (%rax)
L79:
  cmpl   $0x1,(%rdx)
  adc    $0x0,%eax
  add    $0x4,%rdx
  cmp    %rdx,%rcx
  jne    L79
L44:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L14:
  lea    (%r12,%rbp,1),%eax
  xor    %r15d,%r15d
  movslq %eax,%rbp
  cltd
  imul   $0x1948b0fd,%rbp,%rbp
  sar    $0x23,%rbp
  sub    %edx,%ebp
  lea    0x0(%rbp,%rbp,8),%edx
  lea    (%rdx,%rdx,8),%edx
  sub    %edx,%eax
  mov    %eax,%ebp
  cltq
  lea    (%rbx,%rax,4),%r14
  mov    (%r14),%r13d
  test   %r13d,%r13d
  jle    L46
  nopl   0x0(%rax)
L2:
  movl   $0x0,(%r14)
  xor    %eax,%eax
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <moveable@plt>
  cmp    %r13d,%eax
  je     L14
  mov    %r13d,(%r14)
  add    $0x1,%r15d
  jmp    L57
L46:
  lea    (%r12,%rbp,1),%eax
  movslq %eax,%rbp
  cltd
  imul   $0x1948b0fd,%rbp,%rbp
  sar    $0x23,%rbp
  sub    %edx,%ebp
  lea    0x0(%rbp,%rbp,8),%edx
  lea    (%rdx,%rdx,8),%edx
  sub    %edx,%eax
  mov    %eax,%ebp
  jmp    L20
```

assistant: ```c
int suberase(sudoku, d, e)
int sudoku[];
int d;
int e;
{
  int a = e;
  int b;
  int c = 0;
  if (d % 3 == 0) {
    return (0);
  }
  while (c < 85) {
    if (sudoku[a] > 0) {
      b = sudoku[a];
      sudoku[a] = 0;
      if (moveable(sudoku, a) == b) {
        c = 0;
      } else {
        sudoku[a] = b;
        c++;
      }
    }
    a = (a + d) % 81;
  }
  c = 0;
  for (a = 0; a < 81; a++) {
    if (sudoku[a] == 0) {
      c++;
    }
  }
  return (c);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<suberase>:
  endbr64
  movslq %esi,%rax
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rax,%r12
  imul   $0x55555556,%rax,%rax
  push   %rbp
  push   %rbx
  mov    %edx,%ebx
  mov    %r12d,%edx
  sar    $0x1f,%edx
  shr    $0x20,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,2),%edx
  mov    %r12d,%eax
  sub    $0x8,%rsp
  sub    %edx,%eax
  je     L76
  mov    %rdi,%rbp
  xor    %r15d,%r15d
  nopl   0x0(%rax)
L41:
  movslq %ebx,%rax
  lea    0x0(%rbp,%rax,4),%r14
  mov    (%r14),%r13d
  test   %r13d,%r13d
  jg     L47
L45:
  lea    (%r12,%rbx,1),%eax
  movslq %eax,%rbx
  cltd
  imul   $0x1948b0fd,%rbx,%rbx
  sar    $0x23,%rbx
  sub    %edx,%ebx
  lea    (%rbx,%rbx,8),%edx
  lea    (%rdx,%rdx,8),%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  cmp    $0x55,%r15d
  jne    L41
  pxor   %xmm0,%xmm0
  mov    %rbp,%rax
  lea    0x140(%rbp),%rdx
  movdqa %xmm0,%xmm2
  nopl   0x0(%rax,%rax,1)
L28:
  movdqu (%rax),%xmm1
  add    $0x10,%rax
  pcmpeqd %xmm2,%xmm1
  psubd  %xmm1,%xmm0
  cmp    %rax,%rdx
  jne    L28
  movdqa %xmm0,%xmm1
  cmpl   $0x1,0x140(%rbp)
  psrldq $0x8,%xmm1
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  adc    $0x0,%eax
L76:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L67:
  lea    (%r12,%rbx,1),%eax
  xor    %r15d,%r15d
  movslq %eax,%rbx
  cltd
  imul   $0x1948b0fd,%rbx,%rbx
  sar    $0x23,%rbx
  sub    %edx,%ebx
  lea    (%rbx,%rbx,8),%edx
  lea    (%rdx,%rdx,8),%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  cltq
  lea    0x0(%rbp,%rax,4),%r14
  mov    (%r14),%r13d
  test   %r13d,%r13d
  jle    L31
  nopl   0x0(%rax)
L47:
  movl   $0x0,(%r14)
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <moveable@plt>
  cmp    %r13d,%eax
  je     L67
  mov    %r13d,(%r14)
  add    $0x1,%r15d
  jmp    L45
L31:
  lea    (%r12,%rbx,1),%eax
  movslq %eax,%rbx
  cltd
  imul   $0x1948b0fd,%rbx,%rbx
  sar    $0x23,%rbx
  sub    %edx,%ebx
  lea    (%rbx,%rbx,8),%edx
  lea    (%rdx,%rdx,8),%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  jmp    L41
```

assistant: ```c
int suberase(sudoku, d, e)
int sudoku[];
int d;
int e;
{
  int a = e;
  int b;
  int c = 0;
  if (d % 3 == 0) {
    return (0);
  }
  while (c < 85) {
    if (sudoku[a] > 0) {
      b = sudoku[a];
      sudoku[a] = 0;
      if (moveable(sudoku, a) == b) {
        c = 0;
      } else {
        sudoku[a] = b;
        c++;
      }
    }
    a = (a + d) % 81;
  }
  c = 0;
  for (a = 0; a < 81; a++) {
    if (sudoku[a] == 0) {
      c++;
    }
  }
  return (c);
}
```
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
  movl   $0x1,-0x4(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void main() {
  int32_t a;
  a = 2 - 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main() {
  int32_t a;
  a = 2 - 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main() {
  int32_t a;
  a = 2 - 1;
}
```
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
  ret
```

assistant: ```c
void main() {
  int32_t a;
  a = 2 - 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<gpi_exif_get_lilend>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  subl   $0x1,-0x1c(%rbp)
  jmp    L12
L3:
  shlq   $0x8,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  add    %rax,-0x8(%rbp)
  subl   $0x1,-0x1c(%rbp)
L12:
  cmpl   $0x0,-0x1c(%rbp)
  jns    L3
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
long gpi_exif_get_lilend(unsigned char *data, int size) {
  long total;
  total = 0;
  for (--size; size >= 0; size--) {
    total <<= 8;
    total += (unsigned char)(data[size]);
  }
  return (total);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<gpi_exif_get_lilend>:
  endbr64
  mov    %esi,%eax
  mov    %esi,%edx
  sub    $0x1,%edx
  js     L18
  movslq %edx,%rdx
  add    %rdi,%rdx
  movslq %esi,%rcx
  lea    -0x2(%rdi,%rcx,1),%rsi
  lea    -0x1(%rax),%eax
  sub    %rax,%rsi
  mov    $0x0,%eax
L3:
  shl    $0x8,%rax
  movzbl (%rdx),%ecx
  add    %rcx,%rax
  sub    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L3
  ret
L18:
  mov    $0x0,%eax
  ret
```

assistant: ```c
long gpi_exif_get_lilend(unsigned char *data, int size) {
  long total;
  total = 0;
  for (--size; size >= 0; size--) {
    total <<= 8;
    total += (unsigned char)(data[size]);
  }
  return (total);
}
```
user: Translate the below assembly code into C programming language:

```asm
<gpi_exif_get_lilend>:
  endbr64
  mov    %esi,%eax
  sub    $0x1,%eax
  js     L7
  movslq %esi,%rsi
  movslq %eax,%rdx
  mov    %eax,%eax
  lea    -0x2(%rdi,%rsi,1),%rsi
  add    %rdi,%rdx
  sub    %rax,%rsi
  xor    %eax,%eax
L16:
  movzbl (%rdx),%ecx
  shl    $0x8,%rax
  sub    $0x1,%rdx
  add    %rcx,%rax
  cmp    %rsi,%rdx
  jne    L16
  ret
  nopl   0x0(%rax)
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
long gpi_exif_get_lilend(unsigned char *data, int size) {
  long total;
  total = 0;
  for (--size; size >= 0; size--) {
    total <<= 8;
    total += (unsigned char)(data[size]);
  }
  return (total);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<gpi_exif_get_lilend>:
  endbr64
  mov    %esi,%eax
  sub    $0x1,%eax
  js     L13
  movslq %esi,%rsi
  movslq %eax,%rdx
  mov    %eax,%eax
  lea    -0x2(%rdi,%rsi,1),%rsi
  add    %rdi,%rdx
  sub    %rax,%rsi
  xor    %eax,%eax
L5:
  movzbl (%rdx),%ecx
  shl    $0x8,%rax
  sub    $0x1,%rdx
  add    %rcx,%rax
  cmp    %rdx,%rsi
  jne    L5
  ret
  nopl   0x0(%rax)
L13:
  xor    %eax,%eax
  ret
```

assistant: ```c
long gpi_exif_get_lilend(unsigned char *data, int size) {
  long total;
  total = 0;
  for (--size; size >= 0; size--) {
    total <<= 8;
    total += (unsigned char)(data[size]);
  }
  return (total);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fun1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L19
  mov    $0x1,%eax
  jmp    L1
L19:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <fun1>
  imul   -0x4(%rbp),%eax
L1:
  leave
  ret
```

assistant: ```c
int fun1(int n) {
  if (n == 1)
    return 1;
  return fun1(n - 1) * n;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fun1>:
  endbr64
  mov    %edi,%eax
  cmp    $0x1,%edi
  je     L7
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <fun1>
  imul   %ebx,%eax
  pop    %rbx
  ret
L7:
  ret
```

assistant: ```c
int fun1(int n) {
  if (n == 1)
    return 1;
  return fun1(n - 1) * n;
}
```
user: What is the c source code of the assembly code below:

```asm
<fun1>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  je     L3
  xchg   %ax,%ax
L4:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L4
L3:
  ret
```

assistant: ```c
int fun1(int n) {
  if (n == 1)
    return 1;
  return fun1(n - 1) * n;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<fun1>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  je     L2
  xchg   %ax,%ax
L9:
  mov    %edi,%edx
  sub    $0x1,%edi
  imul   %edx,%eax
  cmp    $0x1,%edi
  jne    L9
L2:
  ret
```

assistant: ```c
int fun1(int n) {
  if (n == 1)
    return 1;
  return fun1(n - 1) * n;
}
```