ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码:

```asm
<sigchld_handler>:
  endbr64
  cmp    $0x84,%edi
  jg     L32
  cmp    $0x80,%edi
  jle    L22
  mov    D12(%rip),%rax
  mov    (%rax),%rdi
  jmp    1050 <exit@plt>
  nopl   0x0(%rax,%rax,1)
L32:
  cmp    $0x85,%edi
  je     L24
  cmp    $0x86,%edi
  jne    L38
  mov    $0x1,%edi
  jmp    1050 <exit@plt>
  nopw   0x0(%rax,%rax,1)
L22:
  jne    L38
  xor    %edi,%edi
  jmp    1050 <exit@plt>
  nopl   0x0(%rax)
L24:
  mov    $0x2,%edi
  jmp    1050 <exit@plt>
  nopw   0x0(%rax,%rax,1)
L38:
  mov    D18(%rip),%rax
  mov    (%rax),%rdi
  jmp    1050 <exit@plt>
```

assistant: ```c
void sigchld_handler(int s) {
  switch (s) {
  case SIGTSTP:
    exit((long)SIG_IGN);
    break;
  case SIGTTOU:
    exit((long)SIG_IGN);
    break;
  case SIGTTIN:
    exit((long)SIG_IGN);
    break;
  case SIGHUP:
    exit((long)SIG_IGN);
    break;
  case SIGALRM:
    exit(1);
    break;
  case SIGUSR1:
    exit(0);
    break;
  case SIGCHLD:
    exit(2);
    break;
  default:
    exit((long)SIG_DFL);
    break;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<solucionUnica>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x178,%rsp
  mov    %edi,%r12d
  mov    %esi,%ebp
  mov    %rdx,%r15
  mov    %rdx,0x10(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x168(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r14
  mov    %r14,%rdi
  call   <reiniciarSudoku@plt>
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <copyMatriz@plt>
  movslq %ebp,%rdx
  movslq %r12d,%rax
  lea    (%rax,%rax,8),%r13
  add    %rdx,%r13
  movl   $0x0,0x20(%rsp,%r13,4)
  mov    %r14,%rdi
  call   <resolverSudoku@plt>
  mov    %eax,%ebx
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <copyMatriz@plt>
  movl   $0x0,0x20(%rsp,%r13,4)
  test   %ebx,%ebx
  je     L55
  mov    $0x0,%r15d
  mov    $0x1,%ebx
  mov    %r14,%r13
  movslq %r12d,%rax
  lea    (%rax,%rax,8),%rax
  mov    %rax,0x18(%rsp)
  jmp    L65
L53:
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  je     L33
L65:
  mov    $0x0,%r8d
  mov    %ebx,%ecx
  mov    %ebp,%edx
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <verificarReglas@plt>
  test   %rax,%rax
  je     L53
  movslq %ebp,%r14
  add    0x18(%rsp),%r14
  mov    %ebx,0x20(%rsp,%r14,4)
  mov    %r13,%rdi
  call   <resolverSudoku@plt>
  mov    %eax,0xc(%rsp)
  mov    %r13,%rsi
  mov    0x10(%rsp),%rdi
  call   <copyMatriz@plt>
  movl   $0x0,0x20(%rsp,%r14,4)
  cmpl   $0x0,0xc(%rsp)
  je     L53
  add    $0x1,%r15d
  cmp    $0x1,%r15d
  jle    L53
  mov    $0x0,%ebx
  jmp    L55
L33:
  test   %r15d,%r15d
  setne  %bl
  movzbl %bl,%ebx
L55:
  mov    0x168(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L63
  mov    %ebx,%eax
  add    $0x178,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L63:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int solucionUnica(int iBorrar, int jBorrar, int sudokuAresolver[9][9]) {
  int sudokuSimulacion[9][9];
  reiniciarSudoku(sudokuSimulacion);
  copyMatriz(sudokuAresolver, sudokuSimulacion);
  sudokuSimulacion[iBorrar][jBorrar] = 0;
  int posibleCandidato;
  int sePuedeResolver;
  sePuedeResolver = resolverSudoku(sudokuSimulacion);
  copyMatriz(sudokuAresolver, sudokuSimulacion);
  sudokuSimulacion[iBorrar][jBorrar] = 0;
  if (!sePuedeResolver) {
    return 0;
  }
  int contadorSoluciones = 0;
  for (posibleCandidato = 1; posibleCandidato < 10; posibleCandidato++) {
    if (verificarReglas(sudokuSimulacion, iBorrar, jBorrar, posibleCandidato, 0)) {
      sudokuSimulacion[iBorrar][jBorrar] = posibleCandidato;
      sePuedeResolver = resolverSudoku(sudokuSimulacion);
      copyMatriz(sudokuAresolver, sudokuSimulacion);
      sudokuSimulacion[iBorrar][jBorrar] = 0;
      if (!sePuedeResolver) {
        continue;
      }
      contadorSoluciones++;
      if (contadorSoluciones > 1) {
        return 0;
      }
    }
  }
  if (contadorSoluciones == 0) {
    return 0;
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<solucionUnica>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x168,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x158(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdi
  call   <reiniciarSudoku@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <copyMatriz@plt>
  movslq %ebp,%rax
  movslq %r12d,%rdx
  mov    %rbx,%rdi
  lea    (%rax,%rax,8),%r13
  add    %rdx,%r13
  movl   $0x0,0x10(%rsp,%r13,4)
  call   <resolverSudoku@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  mov    %eax,%r15d
  call   <copyMatriz@plt>
  movl   $0x0,0x10(%rsp,%r13,4)
  test   %r15d,%r15d
  je     L71
  movl   $0x0,0x8(%rsp)
  mov    $0x1,%r15d
  jmp    L52
  cs nopw 0x0(%rax,%rax,1)
L28:
  add    $0x1,%r15d
  cmp    $0xa,%r15d
  je     L19
L52:
  xor    %r8d,%r8d
  mov    %r15d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <verificarReglas@plt>
  test   %rax,%rax
  je     L28
  mov    %rbx,%rdi
  mov    %r15d,0x10(%rsp,%r13,4)
  call   <resolverSudoku@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  mov    %eax,0xc(%rsp)
  call   <copyMatriz@plt>
  mov    0xc(%rsp),%edx
  movl   $0x0,0x10(%rsp,%r13,4)
  test   %edx,%edx
  je     L28
  cmpl   $0x1,0x8(%rsp)
  je     L71
  add    $0x1,%r15d
  movl   $0x1,0x8(%rsp)
  cmp    $0xa,%r15d
  jne    L52
  nopl   0x0(%rax,%rax,1)
L19:
  mov    0x158(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    0x8(%rsp),%eax
  add    $0x168,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L71:
  movl   $0x0,0x8(%rsp)
  jmp    L19
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int solucionUnica(int iBorrar, int jBorrar, int sudokuAresolver[9][9]) {
  int sudokuSimulacion[9][9];
  reiniciarSudoku(sudokuSimulacion);
  copyMatriz(sudokuAresolver, sudokuSimulacion);
  sudokuSimulacion[iBorrar][jBorrar] = 0;
  int posibleCandidato;
  int sePuedeResolver;
  sePuedeResolver = resolverSudoku(sudokuSimulacion);
  copyMatriz(sudokuAresolver, sudokuSimulacion);
  sudokuSimulacion[iBorrar][jBorrar] = 0;
  if (!sePuedeResolver) {
    return 0;
  }
  int contadorSoluciones = 0;
  for (posibleCandidato = 1; posibleCandidato < 10; posibleCandidato++) {
    if (verificarReglas(sudokuSimulacion, iBorrar, jBorrar, posibleCandidato, 0)) {
      sudokuSimulacion[iBorrar][jBorrar] = posibleCandidato;
      sePuedeResolver = resolverSudoku(sudokuSimulacion);
      copyMatriz(sudokuAresolver, sudokuSimulacion);
      sudokuSimulacion[iBorrar][jBorrar] = 0;
      if (!sePuedeResolver) {
        continue;
      }
      contadorSoluciones++;
      if (contadorSoluciones > 1) {
        return 0;
      }
    }
  }
  if (contadorSoluciones == 0) {
    return 0;
  }
  return 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<solucionUnica>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x168,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x158(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rdi
  call   <reiniciarSudoku@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  call   <copyMatriz@plt>
  movslq %ebp,%rax
  movslq %r12d,%rdx
  mov    %rbx,%rdi
  lea    (%rax,%rax,8),%r13
  add    %rdx,%r13
  movl   $0x0,0x10(%rsp,%r13,4)
  call   <resolverSudoku@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  mov    %eax,%r15d
  call   <copyMatriz@plt>
  movl   $0x0,0x10(%rsp,%r13,4)
  test   %r15d,%r15d
  je     L38
  movl   $0x0,0x8(%rsp)
  mov    $0x1,%r15d
  jmp    L67
  cs nopw 0x0(%rax,%rax,1)
L77:
  add    $0x1,%r15d
  cmp    $0xa,%r15d
  je     L29
L67:
  xor    %r8d,%r8d
  mov    %r15d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <verificarReglas@plt>
  test   %rax,%rax
  je     L77
  mov    %rbx,%rdi
  mov    %r15d,0x10(%rsp,%r13,4)
  call   <resolverSudoku@plt>
  mov    %rbx,%rsi
  mov    %r14,%rdi
  mov    %eax,0xc(%rsp)
  call   <copyMatriz@plt>
  mov    0xc(%rsp),%edx
  movl   $0x0,0x10(%rsp,%r13,4)
  test   %edx,%edx
  je     L77
  cmpl   $0x1,0x8(%rsp)
  je     L38
  add    $0x1,%r15d
  movl   $0x1,0x8(%rsp)
  cmp    $0xa,%r15d
  jne    L67
  nopl   0x0(%rax,%rax,1)
L29:
  mov    0x158(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    0x8(%rsp),%eax
  add    $0x168,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L38:
  movl   $0x0,0x8(%rsp)
  jmp    L29
L13:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int solucionUnica(int iBorrar, int jBorrar, int sudokuAresolver[9][9]) {
  int sudokuSimulacion[9][9];
  reiniciarSudoku(sudokuSimulacion);
  copyMatriz(sudokuAresolver, sudokuSimulacion);
  sudokuSimulacion[iBorrar][jBorrar] = 0;
  int posibleCandidato;
  int sePuedeResolver;
  sePuedeResolver = resolverSudoku(sudokuSimulacion);
  copyMatriz(sudokuAresolver, sudokuSimulacion);
  sudokuSimulacion[iBorrar][jBorrar] = 0;
  if (!sePuedeResolver) {
    return 0;
  }
  int contadorSoluciones = 0;
  for (posibleCandidato = 1; posibleCandidato < 10; posibleCandidato++) {
    if (verificarReglas(sudokuSimulacion, iBorrar, jBorrar, posibleCandidato, 0)) {
      sudokuSimulacion[iBorrar][jBorrar] = posibleCandidato;
      sePuedeResolver = resolverSudoku(sudokuSimulacion);
      copyMatriz(sudokuAresolver, sudokuSimulacion);
      sudokuSimulacion[iBorrar][jBorrar] = 0;
      if (!sePuedeResolver) {
        continue;
      }
      contadorSoluciones++;
      if (contadorSoluciones > 1) {
        return 0;
      }
    }
  }
  if (contadorSoluciones == 0) {
    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
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  lea    -0x3c(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  jmp    L17
L48:
  mov    -0x3c(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,-0x34(%rbp)
  mov    -0x34(%rbp),%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  lea    0x1(%rax),%edx
  mov    -0x34(%rbp),%eax
  cltq
  mov    %edx,-0x30(%rbp,%rax,4)
  mov    -0x3c(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x3c(%rbp)
L17:
  mov    -0x3c(%rbp),%eax
  test   %eax,%eax
  jne    L48
  movl   $0x0,-0x38(%rbp)
  jmp    L41
L36:
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L49
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x38(%rbp),%eax
  mov    %eax,%esi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L49:
  addl   $0x1,-0x38(%rbp)
L41:
  cmpl   $0x9,-0x38(%rbp)
  jle    L36
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L58
  call   <__stack_chk_fail@plt>
L58:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D9
datatype: string
name: parse_data
arguments: tag: D8
datatype: string

tool: D9: %d

tool: D8: %d repeat %d times 


assistant: ```c
int main() {
  int digits[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int num, digit, i;
  scanf("%d", &num);
  while (num != 0) {
    digit = num % 10;
    digits[digit]++;
    num /= 10;
  }
  for (i = 0; i < 10; i++) {
    if (digits[i] != 0) {
      printf("%d repeat %d times \n", i, digits[i]);
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  lea    0xc(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  je     L38
L34:
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  mov    %eax,%esi
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  cltq
  addl   $0x1,0x10(%rsp,%rax,4)
  mov    %esi,%edx
  test   %esi,%esi
  jne    L34
  movl   $0x0,0xc(%rsp)
L38:
  mov    $0x0,%ebx
  lea    D4(%rip),%rbp
  jmp    L69
L41:
  add    $0x1,%rbx
  cmp    $0xa,%rbx
  je     L36
L69:
  mov    %ebx,%esi
  mov    0x10(%rsp,%rbx,4),%edx
  test   %edx,%edx
  je     L41
  mov    %rbp,%rdi
  call   <printf@plt>
  jmp    L41
L36:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D11
data_type: string
name: read_data
arguments: data_label: D4
data_type: string

tool: D11: %d

tool: D4: %d repeat %d times 


assistant: ```c
int main() {
  int digits[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int num, digit, i;
  scanf("%d", &num);
  while (num != 0) {
    digit = num % 10;
    digits[digit]++;
    num /= 10;
  }
  for (i = 0; i < 10; i++) {
    if (digits[i] != 0) {
      printf("%d repeat %d times \n", i, digits[i]);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  pxor   %xmm0,%xmm0
  lea    D7(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  movaps %xmm0,0x10(%rsp)
  movq   $0x0,0x30(%rsp)
  movaps %xmm0,0x20(%rsp)
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  je     L62
  nopl   0x0(%rax)
L11:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%esi
  mov    %eax,%ecx
  mov    %edx,%eax
  add    %esi,%esi
  mov    %ecx,%edx
  sub    %esi,%eax
  cltq
  addl   $0x1,0x10(%rsp,%rax,4)
  test   %ecx,%ecx
  jne    L11
  movl   $0x0,0xc(%rsp)
L62:
  xor    %ebx,%ebx
  lea    0x10(%rsp),%rbp
  lea    D14(%rip),%r12
  jmp    L25
  nopl   (%rax)
L3:
  add    $0x1,%rbx
  cmp    $0xa,%rbx
  je     L36
L25:
  mov    0x0(%rbp,%rbx,4),%edx
  mov    %ebx,%esi
  test   %edx,%edx
  je     L3
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0xa,%rbx
  jne    L25
L36:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L51:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D7
data_type: string
name: parse_data
arguments: label: D14
data_type: string

tool: D7: %d

tool: D14: %d repeat %d times 


assistant: ```c
int main() {
  int digits[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int num, digit, i;
  scanf("%d", &num);
  while (num != 0) {
    digit = num % 10;
    digits[digit]++;
    num /= 10;
  }
  for (i = 0; i < 10; i++) {
    if (digits[i] != 0) {
      printf("%d repeat %d times \n", i, digits[i]);
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x48,%rsp
  pxor   %xmm0,%xmm0
  lea    D41(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  movaps %xmm0,0x10(%rsp)
  movq   $0x0,0x30(%rsp)
  movaps %xmm0,0x20(%rsp)
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  je     L22
  nopl   0x0(%rax,%rax,1)
L187:
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%esi
  mov    %eax,%ecx
  mov    %edx,%eax
  add    %esi,%esi
  mov    %ecx,%edx
  sub    %esi,%eax
  cltq
  addl   $0x1,0x10(%rsp,%rax,4)
  test   %ecx,%ecx
  jne    L187
  movl   $0x0,0xc(%rsp)
L22:
  mov    0x10(%rsp),%edx
  test   %edx,%edx
  jne    L15
L43:
  mov    0x14(%rsp),%edx
  test   %edx,%edx
  jne    L106
L80:
  mov    0x18(%rsp),%edx
  test   %edx,%edx
  jne    L225
L66:
  mov    0x1c(%rsp),%edx
  test   %edx,%edx
  jne    L165
L201:
  mov    0x20(%rsp),%edx
  test   %edx,%edx
  jne    L50
L36:
  mov    0x24(%rsp),%edx
  test   %edx,%edx
  jne    L146
L13:
  mov    0x28(%rsp),%edx
  test   %edx,%edx
  jne    L145
L177:
  mov    0x2c(%rsp),%edx
  test   %edx,%edx
  jne    L21
L93:
  mov    0x30(%rsp),%edx
  test   %edx,%edx
  jne    L193
L162:
  mov    0x34(%rsp),%edx
  test   %edx,%edx
  jne    L119
L195:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L104
  xor    %eax,%eax
  add    $0x48,%rsp
  ret
L15:
  xor    %esi,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L43
L119:
  mov    $0x9,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L195
L193:
  mov    $0x8,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L162
L21:
  mov    $0x7,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L93
L145:
  mov    $0x6,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L177
L146:
  mov    $0x5,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L13
L50:
  mov    $0x4,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L36
L165:
  mov    $0x3,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L201
L225:
  mov    $0x2,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L66
L106:
  mov    $0x1,%esi
  lea    D93(%rip),%rdi
  call   <printf@plt>
  jmp    L80
L104:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D41
datatype: string
name: parse_data
arguments: data_tag: D93
datatype: string

tool: D41: %d

tool: D93: %d repeat %d times 


assistant: ```c
int main() {
  int digits[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
  int num, digit, i;
  scanf("%d", &num);
  while (num != 0) {
    digit = num % 10;
    digits[digit]++;
    num /= 10;
  }
  for (i = 0; i < 10; i++) {
    if (digits[i] != 0) {
      printf("%d repeat %d times \n", i, digits[i]);
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ayuv2yuyv_ref>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edx,%r13d
  shr    $0x1f,%r13d
  add    %edx,%r13d
  sar    %r13d
  cmp    $0x1,%edx
  jle    L67
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    $0x0,%eax
L59:
  movswl 0x2(%r12,%rax,8),%edx
  mov    %dl,0x0(%rbp,%rax,4)
  mov    0x4(%r12,%rax,8),%edx
  movzbl %dh,%ecx
  mov    %cl,0x1(%rbp,%rax,4)
  movswl 0x2(%r12,%rax,8),%edx
  mov    %dl,0x2(%rbp,%rax,4)
  mov    (%r12,%rax,8),%edx
  mov    %dl,0x3(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r13d
  jg     L59
  mov    $0x0,%ebx
  jmp    L84
L58:
  mov    $0x0,%eax
  call   <abort@plt>
L37:
  add    $0x1,%rbx
  cmp    %ebx,%r13d
  jle    L67
L84:
  mov    (%r12,%rbx,8),%edx
  mov    %edx,%eax
  sar    $0x10,%eax
  movsbl 0x0(%rbp,%rbx,4),%ecx
  cmp    %eax,%ecx
  jne    L58
  movsbl 0x1(%rbp,%rbx,4),%esi
  mov    0x4(%r12,%rbx,8),%ecx
  sar    $0x8,%ecx
  cmp    %ecx,%esi
  jne    L58
  movsbl 0x2(%rbp,%rbx,4),%ecx
  cmp    %ecx,%eax
  jne    L58
  movsbl 0x3(%rbp,%rbx,4),%eax
  cmp    %eax,%edx
  jne    L58
  jmp    L37
L67:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
__attribute__((no)) void ayuv2yuyv_ref(int *d, int *src, int n) {
  char *dest = (char *)d;
  int i;
  for (i = 0; i < n / 2; i++) {
    dest[i * 4 + 0] = (src[i * 2 + 0]) >> 16;
    dest[i * 4 + 1] = (src[i * 2 + 1]) >> 8;
    dest[i * 4 + 2] = (src[i * 2 + 0]) >> 16;
    dest[i * 4 + 3] = (src[i * 2 + 0]) >> 0;
  }
  for (i = 0; i < n / 2; i++) {
    if (dest[i * 4 + 0] != (src[i * 2 + 0]) >> 16 || dest[i * 4 + 1] != (src[i * 2 + 1]) >> 8 || dest[i * 4 + 2] != (src[i * 2 + 0]) >> 16 || dest[i * 4 + 3] != (src[i * 2 + 0]) >> 0)
      abort();
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ayuv2yuyv_ref>:
  endbr64
  push   %r13
  mov    %edx,%r13d
  push   %r12
  shr    $0x1f,%r13d
  push   %rbp
  add    %edx,%r13d
  push   %rbx
  sar    %r13d
  sub    $0x8,%rsp
  cmp    $0x1,%edx
  jle    L2
  mov    %rdi,%rbp
  mov    %rsi,%r12
  xor    %eax,%eax
  nopl   0x0(%rax)
L60:
  movswl 0x2(%r12,%rax,8),%edx
  mov    %dl,0x0(%rbp,%rax,4)
  mov    0x4(%r12,%rax,8),%edx
  mov    %dh,0x1(%rbp,%rax,4)
  movswl 0x2(%r12,%rax,8),%edx
  mov    %dl,0x2(%rbp,%rax,4)
  mov    (%r12,%rax,8),%edx
  mov    %dl,0x3(%rbp,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r13d
  jg     L60
  xor    %ebx,%ebx
  jmp    L8
  nopw   0x0(%rax,%rax,1)
L58:
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <abort@plt>
  cmp    %ebx,%r13d
  jle    L2
L8:
  mov    (%r12,%rbx,8),%edx
  movsbl 0x0(%rbp,%rbx,4),%eax
  mov    %edx,%ecx
  sar    $0x10,%ecx
  cmp    %ecx,%eax
  jne    L58
  mov    0x4(%r12,%rbx,8),%ecx
  movsbl 0x1(%rbp,%rbx,4),%esi
  sar    $0x8,%ecx
  cmp    %ecx,%esi
  jne    L58
  movsbl 0x2(%rbp,%rbx,4),%ecx
  cmp    %ecx,%eax
  jne    L58
  movsbl 0x3(%rbp,%rbx,4),%eax
  cmp    %eax,%edx
  jne    L58
  add    $0x1,%rbx
  cmp    %ebx,%r13d
  jg     L8
L2:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
__attribute__((no)) void ayuv2yuyv_ref(int *d, int *src, int n) {
  char *dest = (char *)d;
  int i;
  for (i = 0; i < n / 2; i++) {
    dest[i * 4 + 0] = (src[i * 2 + 0]) >> 16;
    dest[i * 4 + 1] = (src[i * 2 + 1]) >> 8;
    dest[i * 4 + 2] = (src[i * 2 + 0]) >> 16;
    dest[i * 4 + 3] = (src[i * 2 + 0]) >> 0;
  }
  for (i = 0; i < n / 2; i++) {
    if (dest[i * 4 + 0] != (src[i * 2 + 0]) >> 16 || dest[i * 4 + 1] != (src[i * 2 + 1]) >> 8 || dest[i * 4 + 2] != (src[i * 2 + 0]) >> 16 || dest[i * 4 + 3] != (src[i * 2 + 0]) >> 0)
      abort();
  }
}
```
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
  mov    %rsp,%rbp
  call   <fork@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type: string

tool: D9: Novo processo criado!


assistant: ```c
int main(void) {
  fork();
  printf("Novo processo criado!\n");
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <fork@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D7
data_type: string

tool: D7: Novo processo criado!


assistant: ```c
int main(void) {
  fork();
  printf("Novo processo criado!\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  call   <fork@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D1
datatype: string

tool: D1: Novo processo criado!


assistant: ```c
int main(void) {
  fork();
  printf("Novo processo criado!\n");
}
```
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
  sub    $0x8,%rsp
  call   <fork@plt>
  lea    D7(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D7
data_type: string

tool: D7: Novo processo criado!


assistant: ```c
int main(void) {
  fork();
  printf("Novo processo criado!\n");
}
```