ylfeng/ReF-Decompile-dataset huggingface dataset:
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   %r12
  mov    $0x8,%esi
  push   %rbp
  lea    D27(%rip),%rbp
  push   %rbx
  sub    $0x30,%rsp
  movdqa D5(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x20(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D28(%rip),%xmm0
  mov    %rbx,%rdi
  movaps %xmm0,0x10(%rsp)
  call   <radixSort@plt>
  nopl   0x0(%rax)
L7:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L7
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D27
type_infered: string
name: disassemble_data
arguments: data_tag: D5
type_infered: byte[16]
name: disassemble_data
arguments: data_tag: D28
type_infered: byte[16]

tool: D27: %d 

tool: {"D5": [0xec,0x2e,0x00,0x00,0x40,0x53,0x00,0x00,0x2b,0x00,0x00,0x00,0x21,0x02,0x00,0x00]}

tool: {"D28": [0x5a,0x00,0x00,0x00,0x07,0x00,0x00,0x00,0xf2,0x00,0x00,0x00,0xeb,0xa8,0x00,0x00]}

assistant: ```c
int main() {
  int A[] = {12012, 21312, 43, 545, 90, 7, 242, 43243};
  int n = sizeof(A) / sizeof(A[0]);
  radixSort(A, n);
  int i;
  for (i = 0; i < n; i++)
    printf("%d ", A[i]);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    $0x0,%eax
  call   <print_version@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L17:
  lea    D23(%rip),%rax
  mov    %rax,%rsi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  call   <print_results@plt>
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L0:
  cmpl   $0x2,-0xc(%rbp)
  jle    L17
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D23
data_type: string
name: retrieve_data
arguments: data_label: D1
data_type: string
name: retrieve_data
arguments: data_label: D13
data_type: string

tool: D23: inv

tool: D1: # c::native::%s


tool: D13: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "inv");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  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    $0x0,%ebx
  lea    D27(%rip),%r13
  lea    D1(%rip),%r12
  lea    D23(%rip),%rbp
L9:
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L9
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D27
type_guessed: string
name: disassemble_data
arguments: data_label: D1
type_guessed: string
name: disassemble_data
arguments: data_label: D23
type_guessed: string

tool: D27: inv

tool: D1: # c::native::%s


tool: D23: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "inv");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r13
  xor    %edi,%edi
  lea    D25(%rip),%r13
  push   %r12
  lea    D20(%rip),%r12
  push   %rbp
  lea    D9(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L4:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L4
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: data_tag: D25
data_type: string
name: read_data
arguments: data_tag: D20
data_type: string
name: read_data
arguments: data_tag: D9
data_type: string

tool: D25: inv

tool: D20: # c::native::%s


tool: D9: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "inv");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  xor    %edi,%edi
  lea    D2(%rip),%r13
  push   %r12
  lea    D20(%rip),%r12
  push   %rbp
  lea    D21(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
L1:
  mov    %r13,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <printf@plt>
  xor    %eax,%eax
  call   <benchmark@plt>
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  cmp    $0x3,%ebx
  jne    L1
  mov    $0x3,%esi
  mov    $0x3,%edi
  call   <print_summary@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D2
datatype: string
name: disassemble_data
arguments: data_tag: D20
datatype: string
name: disassemble_data
arguments: data_tag: D21
datatype: string

tool: D2: inv

tool: D20: # c::native::%s


tool: D21: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int i;
  srand(time(NULL));
  print_version();
  for (i = 0; i < 3; i++) {
    printf("# c::native::%s\n", "inv");
    elapsed = benchmark();
    print_results(elapsed);
    printf("ok %d benchmark finished\n", i + 1);
  }
  print_summary(3, 3);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<UniqueRowsDouble>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %rdx,-0x30(%rbp)
  mov    %rcx,-0x38(%rbp)
  mov    %r8,-0x40(%rbp)
  mov    %r9,-0x48(%rbp)
  mov    -0x28(%rbp),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <MatrixDouble@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdi
  mov    -0x10(%rbp),%rcx
  mov    -0x30(%rbp),%rdx
  mov    -0x28(%rbp),%esi
  mov    -0x24(%rbp),%eax
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <SortRowsDouble@plt>
  mov    -0x40(%rbp),%r8
  mov    -0x38(%rbp),%rdi
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%esi
  mov    -0x24(%rbp),%eax
  push   0x10(%rbp)
  push   -0x48(%rbp)
  mov    %r8,%r9
  mov    %rdi,%r8
  mov    %eax,%edi
  call   <CountSortedRowsDouble@plt>
  add    $0x10,%rsp
  mov    %eax,-0x14(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <FreeMatrixDouble@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x14(%rbp),%eax
  leave
  ret
```

assistant: ```c
int UniqueRowsDouble(int M, int N, double **list, double **uni_list, int *uni_i, int *uni_j, int *cnt) {
  double **sorted;
  int *sort_idx;
  int u;
  sorted = MatrixDouble(M, N);
  sort_idx = (int *)malloc(M * sizeof(int));
  SortRowsDouble(M, N, list, sorted, sort_idx);
  u = CountSortedRowsDouble(M, N, sorted, sort_idx, uni_list, uni_i, uni_j, cnt);
  FreeMatrixDouble(sorted);
  free(sort_idx);
  return u;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<UniqueRowsDouble>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebx
  mov    %esi,%r13d
  mov    %rdx,(%rsp)
  mov    %rcx,0x8(%rsp)
  mov    %r8,%r15
  mov    %r9,%r14
  call   <MatrixDouble@plt>
  mov    %rax,%r12
  lea    0x0(,%rbx,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rax,%r8
  mov    %r12,%rcx
  mov    (%rsp),%rdx
  mov    %r13d,%esi
  mov    %ebx,%edi
  call   <SortRowsDouble@plt>
  push   0x50(%rsp)
  push   %r14
  mov    %r15,%r9
  mov    0x18(%rsp),%r8
  mov    %rbp,%rcx
  mov    %r12,%rdx
  mov    %r13d,%esi
  mov    %ebx,%edi
  call   <CountSortedRowsDouble@plt>
  mov    %eax,%ebx
  mov    %r12,%rdi
  call   <FreeMatrixDouble@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  mov    %ebx,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int UniqueRowsDouble(int M, int N, double **list, double **uni_list, int *uni_i, int *uni_j, int *cnt) {
  double **sorted;
  int *sort_idx;
  int u;
  sorted = MatrixDouble(M, N);
  sort_idx = (int *)malloc(M * sizeof(int));
  SortRowsDouble(M, N, list, sorted, sort_idx);
  u = CountSortedRowsDouble(M, N, sorted, sort_idx, uni_list, uni_i, uni_j, cnt);
  FreeMatrixDouble(sorted);
  free(sort_idx);
  return u;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<UniqueRowsDouble>:
  endbr64
  push   %r15
  mov    %r8,%r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x18,%rsp
  mov    %r9,(%rsp)
  mov    %rdx,0x8(%rsp)
  call   <MatrixDouble@plt>
  lea    0x0(,%rbp,4),%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    0x8(%rsp),%rdx
  mov    %r13,%rcx
  mov    %r14d,%esi
  mov    %ebp,%edi
  mov    %rax,%r8
  mov    %rax,%r12
  call   <SortRowsDouble@plt>
  push   0x50(%rsp)
  mov    0x8(%rsp),%r9
  mov    %ebp,%edi
  mov    %rbx,%r8
  mov    %r12,%rcx
  mov    %r13,%rdx
  mov    %r14d,%esi
  push   %r9
  mov    %r15,%r9
  call   <CountSortedRowsDouble@plt>
  mov    %r13,%rdi
  mov    %eax,%r14d
  call   <FreeMatrixDouble@plt>
  mov    %r12,%rdi
  call   <free@plt>
  add    $0x28,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int UniqueRowsDouble(int M, int N, double **list, double **uni_list, int *uni_i, int *uni_j, int *cnt) {
  double **sorted;
  int *sort_idx;
  int u;
  sorted = MatrixDouble(M, N);
  sort_idx = (int *)malloc(M * sizeof(int));
  SortRowsDouble(M, N, list, sorted, sort_idx);
  u = CountSortedRowsDouble(M, N, sorted, sort_idx, uni_list, uni_i, uni_j, cnt);
  FreeMatrixDouble(sorted);
  free(sort_idx);
  return u;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<UniqueRowsDouble>:
  endbr64
  push   %r15
  mov    %r8,%r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rcx,%rbx
  sub    $0x18,%rsp
  mov    %r9,(%rsp)
  mov    %rdx,0x8(%rsp)
  call   <MatrixDouble@plt>
  lea    0x0(,%rbp,4),%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    0x8(%rsp),%rdx
  mov    %r13,%rcx
  mov    %r14d,%esi
  mov    %ebp,%edi
  mov    %rax,%r8
  mov    %rax,%r12
  call   <SortRowsDouble@plt>
  push   0x50(%rsp)
  mov    0x8(%rsp),%r9
  mov    %ebp,%edi
  mov    %rbx,%r8
  mov    %r12,%rcx
  mov    %r13,%rdx
  mov    %r14d,%esi
  push   %r9
  mov    %r15,%r9
  call   <CountSortedRowsDouble@plt>
  mov    %r13,%rdi
  mov    %eax,%r14d
  call   <FreeMatrixDouble@plt>
  mov    %r12,%rdi
  call   <free@plt>
  add    $0x28,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int UniqueRowsDouble(int M, int N, double **list, double **uni_list, int *uni_i, int *uni_j, int *cnt) {
  double **sorted;
  int *sort_idx;
  int u;
  sorted = MatrixDouble(M, N);
  sort_idx = (int *)malloc(M * sizeof(int));
  SortRowsDouble(M, N, list, sorted, sort_idx);
  u = CountSortedRowsDouble(M, N, sorted, sort_idx, uni_list, uni_i, uni_j, cnt);
  FreeMatrixDouble(sorted);
  free(sort_idx);
  return u;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<lint_to_binary>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L2
L12:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  cqto
  shr    $0x3f,%rdx
  add    %rdx,%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  mov    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x20(%rbp),%rax
  add    %rcx,%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L2:
  cmpl   $0x1f,-0xc(%rbp)
  jle    L12
  nop
  pop    %rbp
  ret
```

assistant: ```c
void lint_to_binary(long number, int *binary) {
  int i;
  long quotient;
  quotient = number;
  for (i = 0; i < 32; i++) {
    binary[i] = 0;
    binary[i] = quotient % 2;
    quotient = quotient / 2;
  }
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<lint_to_binary>:
  endbr64
  lea    0x80(%rsi),%rcx
L2:
  mov    %rdi,%rdx
  shr    $0x3f,%rdx
  lea    (%rdi,%rdx,1),%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  mov    %eax,(%rsi)
  add    %rdx,%rdi
  sar    %rdi
  add    $0x4,%rsi
  cmp    %rcx,%rsi
  jne    L2
  ret
```

assistant: ```c
void lint_to_binary(long number, int *binary) {
  int i;
  long quotient;
  quotient = number;
  for (i = 0; i < 32; i++) {
    binary[i] = 0;
    binary[i] = quotient % 2;
    quotient = quotient / 2;
  }
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<lint_to_binary>:
  endbr64
  lea    0x80(%rsi),%rcx
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %rdi,%rdx
  add    $0x4,%rsi
  shr    $0x3f,%rdx
  lea    (%rdi,%rdx,1),%rax
  add    %rdx,%rdi
  and    $0x1,%eax
  sar    %rdi
  sub    %rdx,%rax
  mov    %eax,-0x4(%rsi)
  cmp    %rcx,%rsi
  jne    L9
  ret
```

assistant: ```c
void lint_to_binary(long number, int *binary) {
  int i;
  long quotient;
  quotient = number;
  for (i = 0; i < 32; i++) {
    binary[i] = 0;
    binary[i] = quotient % 2;
    quotient = quotient / 2;
  }
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<lint_to_binary>:
  endbr64
  lea    0x80(%rsi),%rcx
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %rdi,%rdx
  add    $0x4,%rsi
  shr    $0x3f,%rdx
  lea    (%rdi,%rdx,1),%rax
  add    %rdx,%rdi
  and    $0x1,%eax
  sar    %rdi
  sub    %rdx,%rax
  mov    %eax,-0x4(%rsi)
  cmp    %rsi,%rcx
  jne    L3
  ret
```

assistant: ```c
void lint_to_binary(long number, int *binary) {
  int i;
  long quotient;
  quotient = number;
  for (i = 0; i < 32; i++) {
    binary[i] = 0;
    binary[i] = quotient % 2;
    quotient = quotient / 2;
  }
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ac_get_yweek>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0xffffffff,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L33
  mov    $0xffffffff,%eax
  jmp    L29
L33:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    0x7(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%eax
  test   %eax,%eax
  je     L27
  mov    -0x18(%rbp),%rax
  mov    0x4(%rax),%eax
  lea    -0x1(%rax),%ecx
  jmp    L39
L27:
  mov    $0x6,%ecx
L39:
  mov    %edx,%eax
  sub    %ecx,%eax
  movslq %eax,%rdx
  imul   $0xffffffff92492493,%rdx,%rdx
  shr    $0x20,%rdx
  add    %eax,%edx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
L29:
  pop    %rbp
  ret
```

assistant: ```c
int ac_get_yweek(struct tm *_tm) {
  int week = -1;
  if (!_tm)
    return -1;
  week = (int)(((_tm)->tm_yday + 7 - ((_tm)->tm_wday ? (_tm)->tm_wday - 1 : 6)) / 7);
  return week;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ac_get_yweek>:
  endbr64
  test   %rdi,%rdi
  je     L6
  mov    (%rdi),%eax
  lea    0x7(%rax),%edx
  mov    0x4(%rdi),%ecx
  lea    -0x1(%rcx),%eax
  test   %ecx,%ecx
  mov    $0x6,%ecx
  cmove  %ecx,%eax
  sub    %eax,%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x2,%eax
  sar    $0x1f,%edx
  sub    %edx,%eax
  ret
L6:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int ac_get_yweek(struct tm *_tm) {
  int week = -1;
  if (!_tm)
    return -1;
  week = (int)(((_tm)->tm_yday + 7 - ((_tm)->tm_wday ? (_tm)->tm_wday - 1 : 6)) / 7);
  return week;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ac_get_yweek>:
  endbr64
  test   %rdi,%rdi
  je     L9
  mov    0x4(%rdi),%ecx
  mov    (%rdi),%eax
  test   %ecx,%ecx
  lea    0x7(%rax),%edx
  lea    -0x1(%rcx),%eax
  mov    $0x6,%ecx
  cmove  %ecx,%eax
  sub    %eax,%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x1f,%edx
  sar    $0x2,%eax
  sub    %edx,%eax
  ret
L9:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int ac_get_yweek(struct tm *_tm) {
  int week = -1;
  if (!_tm)
    return -1;
  week = (int)(((_tm)->tm_yday + 7 - ((_tm)->tm_wday ? (_tm)->tm_wday - 1 : 6)) / 7);
  return week;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ac_get_yweek>:
  endbr64
  test   %rdi,%rdi
  je     L9
  mov    0x4(%rdi),%ecx
  mov    (%rdi),%eax
  test   %ecx,%ecx
  lea    0x7(%rax),%edx
  lea    -0x1(%rcx),%eax
  mov    $0x6,%ecx
  cmove  %ecx,%eax
  sub    %eax,%edx
  movslq %edx,%rax
  imul   $0xffffffff92492493,%rax,%rax
  shr    $0x20,%rax
  add    %edx,%eax
  sar    $0x1f,%edx
  sar    $0x2,%eax
  sub    %edx,%eax
  ret
L9:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int ac_get_yweek(struct tm *_tm) {
  int week = -1;
  if (!_tm)
    return -1;
  week = (int)(((_tm)->tm_yday + 7 - ((_tm)->tm_wday ? (_tm)->tm_wday - 1 : 6)) / 7);
  return week;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %edi,-0xd4(%rbp)
  mov    %rsi,-0xe0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0xa8(%rbp)
  movq   $0x0,-0xa0(%rbp)
  movb   $0x1,-0xc1(%rbp)
  lea    -0xb8(%rbp),%rcx
  lea    -0xc0(%rbp),%rdx
  lea    -0x80(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc0(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0xc0(%rbp)
  mov    -0xb8(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0xb8(%rbp)
  lea    -0x80(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x98(%rbp)
  mov    -0xc0(%rbp),%rax
  cqto
  idivq  -0x98(%rbp)
  mov    %rdx,-0x90(%rbp)
  mov    -0xb8(%rbp),%rax
  add    $0x1,%rax
  cqto
  idivq  -0x98(%rbp)
  mov    %rdx,-0x88(%rbp)
  mov    -0x90(%rbp),%rax
  mov    %rax,-0xb0(%rbp)
  jmp    L52
L6:
  movb   $0x0,-0xc1(%rbp)
  mov    -0xb0(%rbp),%rax
  cmp    -0x88(%rbp),%rax
  jne    L29
  mov    -0xa8(%rbp),%rax
  mov    %rax,-0xa0(%rbp)
L29:
  lea    -0x80(%rbp),%rdx
  mov    -0xb0(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x42,%al
  jne    L27
  addq   $0x1,-0xa8(%rbp)
L27:
  mov    -0xb0(%rbp),%rax
  add    $0x1,%rax
  cqto
  idivq  -0x98(%rbp)
  mov    %rdx,-0xb0(%rbp)
L52:
  cmpb   $0x0,-0xc1(%rbp)
  jne    L6
  mov    -0xb0(%rbp),%rax
  cmp    -0x90(%rbp),%rax
  jne    L6
  mov    -0xb8(%rbp),%rax
  mov    -0xc0(%rbp),%rdx
  sub    %rdx,%rax
  add    $0x1,%rax
  cqto
  idivq  -0x98(%rbp)
  imul   -0xa8(%rbp),%rax
  mov    %rax,%rdx
  mov    -0xa0(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D18
type: string
name: parse_data
arguments: data_label: D19
type: string

tool: D18: %s %lli %lli

tool: D19: %lli

assistant: ```c
int main(int argc, const char *argv[]) {
  long long i, n, l, r, ans = 0, ln, rn, j = 0;
  char a[105], started = 1;
  scanf("%s %lli %lli", a, &l, &r);
  --l, --r;
  n = strlen(a);
  ln = l % n;
  rn = (r + 1) % n;
  for (i = ln; started || i != ln; i = (i + 1) % n) {
    started = 0;
    if (i == rn)
      j = ans;
    if (a[i] == 'B')
      ++ans;
  }
  printf("%lli", (r - l + 1) / n * ans + j);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x98,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x88(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rcx
  mov    %rsp,%rdx
  lea    0x10(%rsp),%r13
  mov    %r13,%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rbp
  lea    -0x1(%rbp),%r12
  mov    %r12,(%rsp)
  mov    0x8(%rsp),%rbx
  lea    -0x1(%rbx),%rax
  mov    %rax,0x8(%rsp)
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %rax,%rcx
  mov    %r12,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%r8
  mov    %rbx,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%r9
  mov    %r8,%rax
  mov    $0x0,%edi
  mov    $0x0,%esi
  jmp    L19
L10:
  add    $0x1,%rax
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  cmp    %rdx,%r8
  je     L21
L19:
  cmp    %rax,%r9
  cmove  %rsi,%rdi
  cmpb   $0x42,0x10(%rsp,%rax,1)
  jne    L10
  add    $0x1,%rsi
  jmp    L10
L21:
  sub    %rbp,%rbx
  lea    0x1(%rbx),%rax
  cqto
  idiv   %rcx
  imul   %rsi,%rax
  lea    (%rax,%rdi,1),%rsi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  mov    0x88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  mov    $0x0,%eax
  add    $0x98,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D6
type: string
name: parse_data
arguments: data_tag: D12
type: string

tool: D6: %s %lli %lli

tool: D12: %lli

assistant: ```c
int main(int argc, const char *argv[]) {
  long long i, n, l, r, ans = 0, ln, rn, j = 0;
  char a[105], started = 1;
  scanf("%s %lli %lli", a, &l, &r);
  --l, --r;
  n = strlen(a);
  ln = l % n;
  rn = (r + 1) % n;
  for (i = ln; started || i != ln; i = (i + 1) % n) {
    started = 0;
    if (i == rn)
      j = ans;
    if (a[i] == 'B')
      ++ans;
  }
  printf("%lli", (r - l + 1) / n * ans + j);
  return 0;
}
```