ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L5
L31:
  mov    -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L5:
  cmpl   $0x7e,-0x8(%rbp)
  jle    L31
  movl   $0x0,-0x8(%rbp)
  jmp    L34
L0:
  mov    -0x8(%rbp),%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L34:
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  cmp    %eax,-0x8(%rbp)
  jl     L0
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
type: string

tool: D5: Sum is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int sum = 0;
  for (i = 0; i < 127; i++) {
    sum += i;
  }
  for (i = 0; i < argv[1][0]; i++) {
    sum += i;
  }
  printf("Sum is: %d\n", sum);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x7f,%eax
L1:
  sub    $0x1,%eax
  jne    L1
  mov    0x8(%rsi),%rax
  movsbl (%rax),%edx
  test   %edx,%edx
  jle    L20
  mov    $0x1f41,%esi
  mov    $0x0,%eax
L4:
  add    %eax,%esi
  add    $0x1,%eax
  cmp    %edx,%eax
  jne    L4
L17:
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L20:
  mov    $0x1f41,%esi
  jmp    L17
```

tool_calls: name: unpack_data
arguments: data_tag: D2
data_type: string

tool: D2: Sum is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int sum = 0;
  for (i = 0; i < 127; i++) {
    sum += i;
  }
  for (i = 0; i < argv[1][0]; i++) {
    sum += i;
  }
  printf("Sum is: %d\n", sum);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rax
  mov    $0x1f41,%esi
  movsbl (%rax),%edx
  test   %edx,%edx
  jle    L9
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L7:
  add    %eax,%esi
  add    $0x1,%eax
  cmp    %edx,%eax
  jne    L7
L9:
  lea    D3(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D3
type_guessed: string

tool: D3: Sum is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int sum = 0;
  for (i = 0; i < 127; i++) {
    sum += i;
  }
  for (i = 0; i < argv[1][0]; i++) {
    sum += i;
  }
  printf("Sum is: %d\n", sum);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    0x8(%rsi),%rax
  movsbl (%rax),%edx
  test   %edx,%edx
  jle    L96
  lea    -0x1(%rdx),%eax
  cmp    $0x9,%eax
  jbe    L63
  mov    %edx,%ecx
  movdqa D8(%rip),%xmm1
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movdqa D26(%rip),%xmm3
  shr    $0x2,%ecx
  xchg   %ax,%ax
L141:
  movdqa %xmm1,%xmm2
  add    $0x1,%eax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %eax,%ecx
  jne    L141
  movdqa %xmm0,%xmm1
  mov    %edx,%eax
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%eax
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%esi
  add    $0x1f41,%esi
  test   $0x3,%dl
  je     L133
L138:
  lea    0x1(%rax),%ecx
  add    %eax,%esi
  cmp    %edx,%ecx
  jge    L133
  add    %ecx,%esi
  lea    0x2(%rax),%ecx
  cmp    %ecx,%edx
  jle    L133
  add    %ecx,%esi
  lea    0x3(%rax),%ecx
  cmp    %ecx,%edx
  jle    L133
  add    %ecx,%esi
  lea    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L133
  add    %ecx,%esi
  lea    0x5(%rax),%ecx
  cmp    %ecx,%edx
  jle    L133
  add    %ecx,%esi
  lea    0x6(%rax),%ecx
  cmp    %ecx,%edx
  jle    L133
  add    %ecx,%esi
  lea    0x7(%rax),%ecx
  cmp    %ecx,%edx
  jle    L133
  add    %ecx,%esi
  lea    0x8(%rax),%ecx
  cmp    %ecx,%edx
  jle    L133
  add    $0x9,%eax
  add    %ecx,%esi
  cmp    %eax,%edx
  jle    L133
  add    %eax,%esi
L133:
  lea    D29(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L96:
  mov    $0x1f41,%esi
  jmp    L133
L63:
  mov    $0x1f41,%esi
  xor    %eax,%eax
  jmp    L138
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
type_infered: byte[16]
name: extract_symbolic_data
arguments: tag: D26
type_infered: byte[16]
name: extract_symbolic_data
arguments: tag: D29
type_infered: string

tool: {"D8": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D26": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: D29: Sum is: %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i;
  int sum = 0;
  for (i = 0; i < 127; i++) {
    sum += i;
  }
  for (i = 0; i < argv[1][0]; i++) {
    sum += i;
  }
  printf("Sum is: %d\n", sum);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x100,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L2
L8:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x10(%rbp)
L2:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L8
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <acc_copyin@plt>
  mov    -0xc(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <acc_copyout@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  const int N = 256;
  int i;
  unsigned char *h;
  h = (unsigned char *)malloc(N);
  for (i = 0; i < N; i++) {
    h[i] = i;
  }
  (void)acc_copyin(h, N);
  acc_copyout(h + 1, N - 1);
  free(h);
  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   %rbx
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x0,%eax
L8:
  mov    %al,(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    $0x100,%rax
  jne    L8
  mov    $0x100,%esi
  mov    %rbx,%rdi
  call   <acc_copyin@plt>
  lea    0x1(%rbx),%rdi
  mov    $0xff,%esi
  call   <acc_copyout@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  const int N = 256;
  int i;
  unsigned char *h;
  h = (unsigned char *)malloc(N);
  for (i = 0; i < N; i++) {
    h[i] = i;
  }
  (void)acc_copyin(h, N);
  acc_copyout(h + 1, N - 1);
  free(h);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x100,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  xor    %eax,%eax
  nopl   0x0(%rax)
L2:
  mov    %al,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  cmp    $0x100,%rax
  jne    L2
  mov    %rbp,%rdi
  mov    $0x100,%esi
  call   <acc_copyin@plt>
  lea    0x1(%rbp),%rdi
  mov    $0xff,%esi
  call   <acc_copyout@plt>
  mov    %rbp,%rdi
  call   <free@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  const int N = 256;
  int i;
  unsigned char *h;
  h = (unsigned char *)malloc(N);
  for (i = 0; i < N; i++) {
    h[i] = i;
  }
  (void)acc_copyin(h, N);
  acc_copyout(h + 1, N - 1);
  free(h);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  movl   $0x17,-0x4(%rbp)
  jmp    L18
L8:
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L18:
  cmpl   $0x0,-0x8(%rbp)
  jns    L8
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x, y;
  x = __VERIFIER_nondet_int();
  y = 23;
  while (x >= 0) {
    x = x - y;
    y = y + 1;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  js     L12
  mov    $0x17,%edx
L5:
  sub    %edx,%eax
  add    $0x1,%edx
  test   %eax,%eax
  jns    L5
L12:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x, y;
  x = __VERIFIER_nondet_int();
  y = 23;
  while (x >= 0) {
    x = x - y;
    y = y + 1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x, y;
  x = __VERIFIER_nondet_int();
  y = 23;
  while (x >= 0) {
    x = x - y;
    y = y + 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <__VERIFIER_nondet_int@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int x, y;
  x = __VERIFIER_nondet_int();
  y = 23;
  while (x >= 0) {
    x = x - y;
    y = y + 1;
  }
  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   %rbp
  mov    %rsp,%rbp
  sub    $0x140,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x4645444342413938,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movl   $0x0,-0x130(%rbp)
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x128(%rbp),%rax
  mov    %rax,%rsi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x134(%rbp),%rax
  mov    %rax,%rsi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
L7:
  mov    -0x128(%rbp),%rax
  mov    -0x134(%rbp),%edx
  movslq %edx,%rcx
  cqto
  idiv   %rcx
  mov    %rdx,%rax
  mov    %eax,%edx
  mov    -0x130(%rbp),%eax
  cltq
  mov    %edx,-0x120(%rbp,%rax,4)
  addl   $0x1,-0x130(%rbp)
  mov    -0x128(%rbp),%rax
  mov    -0x134(%rbp),%edx
  movslq %edx,%rcx
  cqto
  idiv   %rcx
  mov    %rax,-0x128(%rbp)
  mov    -0x128(%rbp),%rax
  test   %rax,%rax
  jne    L7
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  subl   $0x1,-0x130(%rbp)
  jmp    L18
L4:
  mov    -0x130(%rbp),%eax
  cltq
  mov    -0x120(%rbp,%rax,4),%eax
  mov    %eax,-0x12c(%rbp)
  mov    -0x12c(%rbp),%eax
  cltq
  movzbl -0x20(%rbp,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
  subl   $0x1,-0x130(%rbp)
L18:
  cmpl   $0x0,-0x130(%rbp)
  jns    L4
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L38
  call   <__stack_chk_fail@plt>
L38:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D26
type_infered: string
name: disassemble_data
arguments: data_tag: D41
type_infered: string
name: disassemble_data
arguments: data_tag: D40
type_infered: string
name: disassemble_data
arguments: data_tag: D11
type_infered: string
name: disassemble_data
arguments: data_tag: D24
type_infered: string

tool: D26: Numero a convertir: 

tool: D41: %ld

tool: D40: Base: 

tool: D11: %i

tool: D24: Numero Convertido: 

assistant: ```c
int main(void) {
  const char digitos[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
  int numeroConvertido[64], siguienteDigito, base, indice = 0;
  long int numeroAConvertir;
  printf("Numero a convertir: ");
  scanf("%ld", &numeroAConvertir);
  printf("Base: ");
  scanf("%i", &base);
  do {
    numeroConvertido[indice] = numeroAConvertir % base;
    indice++;
    numeroAConvertir /= base;
  } while (numeroAConvertir != 0);
  printf("Numero Convertido: ");
  for (indice--; indice >= 0; indice--) {
    siguienteDigito = numeroConvertido[indice];
    printf("%c", digitos[siguienteDigito]);
  }
  printf("\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x130,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x128(%rsp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x4645444342413938,%rdx
  mov    %rax,0x110(%rsp)
  mov    %rdx,0x118(%rsp)
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D32(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%rdx
  movslq 0x4(%rsp),%rsi
  mov    $0x0,%ecx
L37:
  mov    %ecx,%ebx
  mov    %rdx,%rax
  cqto
  idiv   %rsi
  mov    %edx,0x10(%rsp,%rcx,4)
  mov    %rax,%rdx
  mov    %rcx,%rbp
  add    $0x1,%rcx
  test   %rax,%rax
  jne    L37
  movq   $0x0,0x8(%rsp)
  lea    D13(%rip),%rdi
  call   <printf@plt>
  test   %ebx,%ebx
  js     L35
  movslq %ebx,%rbx
  lea    0x0(,%rbx,4),%rax
  lea    0x10(%rsp,%rax,1),%rbx
  lea    0xc(%rsp,%rax,1),%r12
  mov    %ebp,%ebp
  shl    $0x2,%rbp
  sub    %rbp,%r12
L8:
  movslq (%rbx),%rax
  movsbl 0x110(%rsp,%rax,1),%edi
  call   <putchar@plt>
  sub    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L8
L35:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x128(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x130,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D3
type_infered: string
name: disassemble_data
arguments: data_label: D32
type_infered: string
name: disassemble_data
arguments: data_label: D4
type_infered: string
name: disassemble_data
arguments: data_label: D14
type_infered: string
name: disassemble_data
arguments: data_label: D13
type_infered: string

tool: D3: Numero a convertir: 

tool: D32: %ld

tool: D4: Base: 

tool: D14: %i

tool: D13: Numero Convertido: 

assistant: ```c
int main(void) {
  const char digitos[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
  int numeroConvertido[64], siguienteDigito, base, indice = 0;
  long int numeroAConvertir;
  printf("Numero a convertir: ");
  scanf("%ld", &numeroAConvertir);
  printf("Base: ");
  scanf("%i", &base);
  do {
    numeroConvertido[indice] = numeroAConvertir % base;
    indice++;
    numeroAConvertir /= base;
  } while (numeroAConvertir != 0);
  printf("Numero Convertido: ");
  for (indice--; indice >= 0; indice--) {
    siguienteDigito = numeroConvertido[indice];
    printf("%c", digitos[siguienteDigito]);
  }
  printf("\n");
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r12
  lea    D7(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x130,%rsp
  movdqa D11(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x128(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  movaps %xmm0,0x110(%rsp)
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D20(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D53(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  movslq 0x4(%rsp),%rsi
  xor    %ecx,%ecx
  nopl   0x0(%rax,%rax,1)
L35:
  cqto
  mov    %ecx,%ebx
  idiv   %rsi
  mov    %edx,0x0(%rbp,%rcx,4)
  movslq %edx,%r12
  add    $0x1,%rcx
  test   %rax,%rax
  jne    L35
  lea    D50(%rip),%rdi
  sub    $0x1,%ebx
  movq   $0x0,0x8(%rsp)
  call   <printf@plt>
  movslq %ebx,%rbx
  jmp    L4
  nopl   0x0(%rax)
L18:
  movslq 0x0(%rbp,%rbx,4),%r12
  sub    $0x1,%rbx
L4:
  movsbl 0x110(%rsp,%r12,1),%edi
  call   <putchar@plt>
  cmp    $0xffffffff,%ebx
  jne    L18
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x128(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x130,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D7
type_guessed: string
name: parse_data
arguments: label: D11
type_guessed: byte[16]
name: parse_data
arguments: label: D26
type_guessed: string
name: parse_data
arguments: label: D20
type_guessed: string
name: parse_data
arguments: label: D53
type_guessed: string
name: parse_data
arguments: label: D50
type_guessed: string

tool: D7: Numero a convertir: 

tool: {"D11": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x41,0x42,0x43,0x44,0x45,0x46]}

tool: D26: %ld

tool: D20: Base: 

tool: D53: %i

tool: D50: Numero Convertido: 

assistant: ```c
int main(void) {
  const char digitos[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
  int numeroConvertido[64], siguienteDigito, base, indice = 0;
  long int numeroAConvertir;
  printf("Numero a convertir: ");
  scanf("%ld", &numeroAConvertir);
  printf("Base: ");
  scanf("%i", &base);
  do {
    numeroConvertido[indice] = numeroAConvertir % base;
    indice++;
    numeroAConvertir /= base;
  } while (numeroAConvertir != 0);
  printf("Numero Convertido: ");
  for (indice--; indice >= 0; indice--) {
    siguienteDigito = numeroConvertido[indice];
    printf("%c", digitos[siguienteDigito]);
  }
  printf("\n");
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  lea    D11(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x130,%rsp
  movdqa D27(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x128(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  movaps %xmm0,0x110(%rsp)
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D37(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D47(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x8(%rsp),%rax
  movslq 0x4(%rsp),%rsi
  xor    %ecx,%ecx
  nopl   0x0(%rax,%rax,1)
L4:
  cqto
  movslq %ecx,%rbp
  mov    %rcx,%r12
  idiv   %rsi
  mov    %edx,(%rbx,%rcx,4)
  add    $0x1,%rcx
  test   %rax,%rax
  jne    L4
  shl    $0x2,%rbp
  lea    D21(%rip),%rdi
  shl    $0x2,%r12
  movq   $0x0,0x8(%rsp)
  call   <printf@plt>
  add    %rbp,%rbx
  lea    0xc(%rsp,%rbp,1),%rbp
  sub    %r12,%rbp
  nopw   0x0(%rax,%rax,1)
L29:
  movslq (%rbx),%rax
  sub    $0x4,%rbx
  movsbl 0x110(%rsp,%rax,1),%edi
  call   <putchar@plt>
  cmp    %rbp,%rbx
  jne    L29
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x128(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x130,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D11
type_infered: string
name: extract_symbolic_data
arguments: tag: D27
type_infered: byte[16]
name: extract_symbolic_data
arguments: tag: D41
type_infered: string
name: extract_symbolic_data
arguments: tag: D37
type_infered: string
name: extract_symbolic_data
arguments: tag: D47
type_infered: string
name: extract_symbolic_data
arguments: tag: D21
type_infered: string

tool: D11: Numero a convertir: 

tool: {"D27": [0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x41,0x42,0x43,0x44,0x45,0x46]}

tool: D41: %ld

tool: D37: Base: 

tool: D47: %i

tool: D21: Numero Convertido: 

assistant: ```c
int main(void) {
  const char digitos[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
  int numeroConvertido[64], siguienteDigito, base, indice = 0;
  long int numeroAConvertir;
  printf("Numero a convertir: ");
  scanf("%ld", &numeroAConvertir);
  printf("Base: ");
  scanf("%i", &base);
  do {
    numeroConvertido[indice] = numeroAConvertir % base;
    indice++;
    numeroAConvertir /= base;
  } while (numeroAConvertir != 0);
  printf("Numero Convertido: ");
  for (indice--; indice >= 0; indice--) {
    siguienteDigito = numeroConvertido[indice];
    printf("%c", digitos[siguienteDigito]);
  }
  printf("\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<div>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L3
L27:
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
L15:
  shll   -0x4(%rbp)
  shll   -0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jg     L31
  cmpl   $0x1fff,-0x4(%rbp)
  jle    L15
L31:
  sarl   -0x4(%rbp)
  sarl   -0x8(%rbp)
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x14(%rbp)
  mov    -0x8(%rbp),%eax
  add    %eax,-0xc(%rbp)
L3:
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jge    L27
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int div(int a, int b) {
  int quot;
  int q;
  int c;
  quot = 0;
  while (a >= b) {
    c = b;
    q = 1;
    do {
      c <<= 1;
      q <<= 1;
    } while (c <= a && c < 8192);
    c >>= 1;
    q >>= 1;
    a -= c;
    quot += q;
  }
  return quot;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<div>:
  endbr64
  cmp    %esi,%edi
  jl     L14
  mov    $0x0,%r8d
L22:
  mov    %esi,%eax
  mov    $0x1,%edx
  mov    $0x1fff,%ecx
  cmp    %ecx,%edi
  cmovle %edi,%ecx
L33:
  add    %eax,%eax
  add    %edx,%edx
  cmp    %eax,%ecx
  jge    L33
  sar    %eax
  sub    %eax,%edi
  sar    %edx
  add    %edx,%r8d
  cmp    %edi,%esi
  jle    L22
L39:
  mov    %r8d,%eax
  ret
L14:
  mov    $0x0,%r8d
  jmp    L39
```

assistant: ```c
int div(int a, int b) {
  int quot;
  int q;
  int c;
  quot = 0;
  while (a >= b) {
    c = b;
    q = 1;
    do {
      c <<= 1;
      q <<= 1;
    } while (c <= a && c < 8192);
    c >>= 1;
    q >>= 1;
    a -= c;
    quot += q;
  }
  return quot;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<div>:
  endbr64
  xor    %r8d,%r8d
  cmp    %esi,%edi
  jl     L5
  nopl   0x0(%rax,%rax,1)
L14:
  mov    $0x1fff,%ecx
  mov    %esi,%eax
  mov    $0x1,%edx
  cmp    %ecx,%edi
  cmovle %edi,%ecx
  nopl   0x0(%rax)
L12:
  add    %eax,%eax
  add    %edx,%edx
  cmp    %ecx,%eax
  jle    L12
  sar    %eax
  sar    %edx
  sub    %eax,%edi
  add    %edx,%r8d
  cmp    %edi,%esi
  jle    L14
L5:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int div(int a, int b) {
  int quot;
  int q;
  int c;
  quot = 0;
  while (a >= b) {
    c = b;
    q = 1;
    do {
      c <<= 1;
      q <<= 1;
    } while (c <= a && c < 8192);
    c >>= 1;
    q >>= 1;
    a -= c;
    quot += q;
  }
  return quot;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<div>:
  endbr64
  xor    %r8d,%r8d
  cmp    %esi,%edi
  jl     L17
  nopl   0x0(%rax,%rax,1)
L20:
  mov    $0x1fff,%ecx
  mov    %esi,%eax
  mov    $0x1,%edx
  cmp    %ecx,%edi
  cmovle %edi,%ecx
  nopl   0x0(%rax)
L9:
  add    %eax,%eax
  add    %edx,%edx
  cmp    %ecx,%eax
  jle    L9
  sar    %eax
  sar    %edx
  sub    %eax,%edi
  add    %edx,%r8d
  cmp    %edi,%esi
  jle    L20
L17:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int div(int a, int b) {
  int quot;
  int q;
  int c;
  quot = 0;
  while (a >= b) {
    c = b;
    q = 1;
    do {
      c <<= 1;
      q <<= 1;
    } while (c <= a && c < 8192);
    c >>= 1;
    q >>= 1;
    a -= c;
    quot += q;
  }
  return quot;
}
```
user: What is the c source code of the assembly code below:

```asm
<myHashCodeInt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  pop    %rbp
  ret
```

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

```asm
<myHashCodeInt>:
  endbr64
  mov    (%rdi),%eax
  ret
```

assistant: ```c
int myHashCodeInt(void *a) {
  int *aa = (int *)a;
  return *aa;
}
```
user: Translate the below assembly code into C programming language:

```asm
<myHashCodeInt>:
  endbr64
  mov    (%rdi),%eax
  ret
```

assistant: ```c
int myHashCodeInt(void *a) {
  int *aa = (int *)a;
  return *aa;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<myHashCodeInt>:
  endbr64
  mov    (%rdi),%eax
  ret
```

assistant: ```c
int myHashCodeInt(void *a) {
  int *aa = (int *)a;
  return *aa;
}
```