ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<doubleToString>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L33
  mov    %rsp,%rdi
  lea    D29(%rip),%rdx
  mov    $0x64,%esi
  call   <snprintf@plt>
L40:
  mov    %rsp,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L42
L11:
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    %rbx,%rax
  add    $0x70,%rsp
  pop    %rbx
  ret
L33:
  cvttsd2si %xmm0,%rax
  pxor   %xmm0,%xmm0
  cvtsi2sd %rax,%xmm0
  mov    %rsp,%rdi
  lea    D2(%rip),%rdx
  mov    $0x64,%esi
  call   <snprintf@plt>
  jmp    L40
L42:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L11
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D29
datatype: string
name: unpack_data
arguments: label: D2
datatype: string
name: unpack_data
arguments: label: D6
datatype: string

tool: D29: %f

tool: D2: %ld

tool: D6: Out of mem in numberToString

assistant: ```c
char *doubleToString(double number, int type) {
  char result_buffer[100];
  char *result_string;
  if (type == 2)
    snprintf(result_buffer, 100, "%ld", (long)number);
  else
    snprintf(result_buffer, 100, "%f", number);
  result_string = malloc(sizeof(char) * strlen(result_buffer) + 1);
  if (result_string == NULL) {
    printf("Out of mem in numberToString");
    exit(0);
  }
  strcpy(result_string, result_buffer);
  return result_string;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<doubleToString>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L15
  mov    %rsp,%rbp
  lea    D8(%rip),%rdx
  mov    $0x64,%esi
  mov    %rbp,%rdi
  call   <snprintf@plt>
L5:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L19
L37:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x78,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L15:
  cvttsd2si %xmm0,%rax
  pxor   %xmm0,%xmm0
  mov    %rsp,%rbp
  mov    $0x64,%esi
  lea    D2(%rip),%rdx
  mov    %rbp,%rdi
  cvtsi2sd %rax,%xmm0
  call   <snprintf@plt>
  jmp    L5
  nop
L19:
  lea    D5(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L37
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D8
type_guessed: string
name: read_data
arguments: data_tag: D2
type_guessed: string
name: read_data
arguments: data_tag: D5
type_guessed: string

tool: D8: %f

tool: D2: %ld

tool: D5: Out of mem in numberToString

assistant: ```c
char *doubleToString(double number, int type) {
  char result_buffer[100];
  char *result_string;
  if (type == 2)
    snprintf(result_buffer, 100, "%ld", (long)number);
  else
    snprintf(result_buffer, 100, "%f", number);
  result_string = malloc(sizeof(char) * strlen(result_buffer) + 1);
  if (result_string == NULL) {
    printf("Out of mem in numberToString");
    exit(0);
  }
  strcpy(result_string, result_buffer);
  return result_string;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<doubleToString>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  cmp    $0x2,%edi
  je     L6
  mov    %rsp,%rbp
  lea    D10(%rip),%rdx
  mov    $0x64,%esi
  mov    %rbp,%rdi
  call   <snprintf@plt>
L48:
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L45
L16:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <strcpy@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x78,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L6:
  cvttsd2si %xmm0,%rax
  pxor   %xmm0,%xmm0
  mov    %rsp,%rbp
  mov    $0x64,%esi
  lea    D0(%rip),%rdx
  mov    %rbp,%rdi
  cvtsi2sd %rax,%xmm0
  call   <snprintf@plt>
  jmp    L48
  nop
L45:
  lea    D23(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L16
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D10
data_type: string
name: retrieve_data
arguments: tag: D0
data_type: string
name: retrieve_data
arguments: tag: D23
data_type: string

tool: D10: %f

tool: D0: %ld

tool: D23: Out of mem in numberToString

assistant: ```c
char *doubleToString(double number, int type) {
  char result_buffer[100];
  char *result_string;
  if (type == 2)
    snprintf(result_buffer, 100, "%ld", (long)number);
  else
    snprintf(result_buffer, 100, "%f", number);
  result_string = malloc(sizeof(char) * strlen(result_buffer) + 1);
  if (result_string == NULL) {
    printf("Out of mem in numberToString");
    exit(0);
  }
  strcpy(result_string, result_buffer);
  return result_string;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dump_degs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L4
L1:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L1
  nop
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D3
data_type: string

tool: D3: %g %g


assistant: ```c
void dump_degs(double *R1, double *R2, int N, int *pairing) {
  int i;
  for (i = 0; i < N; i++) {
    printf("%g %g\n", R1[i], R2[pairing[i]]);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<dump_degs>:
  endbr64
  test   %edx,%edx
  jle    L16
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rcx,%r14
  mov    %edx,%ebp
  mov    $0x0,%ebx
  lea    D2(%rip),%r15
L2:
  movslq (%r14,%rbx,4),%rax
  movsd  (%r12,%rbx,8),%xmm0
  movsd  0x0(%r13,%rax,8),%xmm1
  mov    %r15,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L2
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L16:
  ret
```

tool_calls: name: parse_data
arguments: tag: D2
type: string

tool: D2: %g %g


assistant: ```c
void dump_degs(double *R1, double *R2, int N, int *pairing) {
  int i;
  for (i = 0; i < N; i++) {
    printf("%g %g\n", R1[i], R2[pairing[i]]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dump_degs>:
  endbr64
  test   %edx,%edx
  jle    L18
  push   %r15
  mov    %rdi,%r15
  push   %r14
  lea    D4(%rip),%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edx,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L0:
  movslq 0x0(%r13,%rbx,4),%rax
  movsd  (%r15,%rbx,8),%xmm0
  mov    %r14,%rdi
  add    $0x1,%rbx
  movsd  (%r12,%rax,8),%xmm1
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L0
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L18:
  ret
```

tool_calls: name: unpack_data
arguments: tag: D4
data_type: string

tool: D4: %g %g


assistant: ```c
void dump_degs(double *R1, double *R2, int N, int *pairing) {
  int i;
  for (i = 0; i < N; i++) {
    printf("%g %g\n", R1[i], R2[pairing[i]]);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<dump_degs>:
  endbr64
  test   %edx,%edx
  jle    L11
  push   %r15
  mov    %rdi,%r15
  push   %r14
  lea    D1(%rip),%r14
  push   %r13
  mov    %rcx,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  movslq %edx,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L7:
  movslq 0x0(%r13,%rbx,4),%rax
  movsd  (%r15,%rbx,8),%xmm0
  mov    %r14,%rdi
  add    $0x1,%rbx
  movsd  (%r12,%rax,8),%xmm1
  call   <printf@plt>
  cmp    %rbx,%rbp
  jne    L7
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L11:
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D1
type_guessed: string

tool: D1: %g %g


assistant: ```c
void dump_degs(double *R1, double *R2, int N, int *pairing) {
  int i;
  for (i = 0; i < N; i++) {
    printf("%g %g\n", R1[i], R2[pairing[i]]);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<std_calloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L0:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type_guessed: string

tool: D9: calloc

assistant: ```c
void *std_calloc(size_t nmemb, size_t size) {
  void *p;
  if ((p = calloc(nmemb, size)) == NULL) {
    perror("calloc");
    exit(1);
  }
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<std_calloc>:
  endbr64
  push   %rbx
  call   <calloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L13
L9:
  mov    %rbx,%rax
  pop    %rbx
  ret
L13:
  lea    D9(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L9
```

tool_calls: name: unpack_data
arguments: data_label: D9
datatype: string

tool: D9: calloc

assistant: ```c
void *std_calloc(size_t nmemb, size_t size) {
  void *p;
  if ((p = calloc(nmemb, size)) == NULL) {
    perror("calloc");
    exit(1);
  }
  return p;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<std_calloc>:
  endbr64
  push   %r12
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L9
L1:
  mov    %r12,%rax
  pop    %r12
  ret
L9:
  lea    D6(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L1
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type_infered: string

tool: D6: calloc

assistant: ```c
void *std_calloc(size_t nmemb, size_t size) {
  void *p;
  if ((p = calloc(nmemb, size)) == NULL) {
    perror("calloc");
    exit(1);
  }
  return p;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<std_calloc>:
  endbr64
  push   %r12
  call   <calloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
L15:
  mov    %r12,%rax
  pop    %r12
  ret
L2:
  lea    D3(%rip),%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L15
```

tool_calls: name: parse_data
arguments: data_tag: D3
type: string

tool: D3: calloc

assistant: ```c
void *std_calloc(size_t nmemb, size_t size) {
  void *p;
  if ((p = calloc(nmemb, size)) == NULL) {
    perror("calloc");
    exit(1);
  }
  return p;
}
```
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    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    %eax,-0xc(%rbp)
  jmp    L25
L35:
  mov    -0x10(%rbp),%eax
  movd   %eax,%xmm0
  call   <InverseSquareRoot@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    %eax,-0xc(%rbp)
L25:
  movss  -0x10(%rbp),%xmm0
  pxor   %xmm1,%xmm1
  ucomiss %xmm1,%xmm0
  jp     L35
  pxor   %xmm1,%xmm1
  ucomiss %xmm1,%xmm0
  jne    L35
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D16
datatype: string
name: parse_data
arguments: tag: D0
datatype: string

tool: D16: Hello, llibrae ~

tool: D0: %f

assistant: ```c
int main(void) {
  float num;
  int ret;
  printf("Hello, llibrae ~\n");
  ret = scanf("%f", &num);
  while (num != 0) {
    printf("%f", InverseSquareRoot(num));
    ret = scanf("%f", &num);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D37(%rip),%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  ucomiss D14(%rip),%xmm0
  jp     L22
  je     L10
L22:
  lea    D5(%rip),%rbx
  lea    0x4(%rsp),%rbp
L33:
  call   <InverseSquareRoot@plt>
  mov    %rbx,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  pxor   %xmm1,%xmm1
  ucomiss %xmm1,%xmm0
  jp     L33
  jne    L33
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D37
datatype: string
name: retrieve_data
arguments: tag: D5
datatype: string
name: retrieve_data
arguments: tag: D14
datatype: byte[4]

tool: D37: Hello, llibrae ~

tool: D5: %f

tool: {"D14": [0x00,0x00,0x00,0x00]}

assistant: ```c
int main(void) {
  float num;
  int ret;
  printf("Hello, llibrae ~\n");
  ret = scanf("%f", &num);
  while (num != 0) {
    printf("%f", InverseSquareRoot(num));
    ret = scanf("%f", &num);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rdi
  push   %rbx
  lea    D10(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  pxor   %xmm2,%xmm2
  ucomiss %xmm2,%xmm0
  jp     L27
  je     L34
  xchg   %ax,%ax
L27:
  call   <InverseSquareRoot@plt>
  mov    %rbx,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  pxor   %xmm1,%xmm1
  ucomiss %xmm1,%xmm0
  jp     L27
  jne    L27
L34:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D9
type_guessed: string
name: retrieve_data
arguments: label: D10
type_guessed: string

tool: D9: Hello, llibrae ~

tool: D10: %f

assistant: ```c
int main(void) {
  float num;
  int ret;
  printf("Hello, llibrae ~\n");
  ret = scanf("%f", &num);
  while (num != 0) {
    printf("%f", InverseSquareRoot(num));
    ret = scanf("%f", &num);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D1(%rip),%rdi
  push   %rbx
  lea    D14(%rip),%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rbp
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  pxor   %xmm2,%xmm2
  ucomiss %xmm2,%xmm0
  jp     L42
  je     L0
  xchg   %ax,%ax
L42:
  call   <InverseSquareRoot@plt>
  mov    %rbx,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  pxor   %xmm1,%xmm1
  ucomiss %xmm1,%xmm0
  jp     L42
  jne    L42
L0:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D1
data_type: string
name: unpack_data
arguments: data_tag: D14
data_type: string

tool: D1: Hello, llibrae ~

tool: D14: %f

assistant: ```c
int main(void) {
  float num;
  int ret;
  printf("Hello, llibrae ~\n");
  ret = scanf("%f", &num);
  while (num != 0) {
    printf("%f", InverseSquareRoot(num));
    ret = scanf("%f", &num);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<hash_rest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x40,-0x14(%rbp)
  jmp    L12
L0:
  mov    -0x10(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sha1_proc@plt>
  mov    -0x14(%rbp),%eax
  cltq
  add    %rax,-0x10(%rbp)
L12:
  mov    -0x8(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,%rdx
  jge    L0
  mov    -0x3c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%rax
  sub    -0x10(%rbp),%rax
  mov    %eax,%edi
  mov    -0x10(%rbp),%rsi
  mov    -0x28(%rbp),%rax
  mov    %rdx,%rcx
  mov    %edi,%edx
  mov    %rax,%rdi
  call   <sha1_last@plt>
  nop
  leave
  ret
```

assistant: ```c
void hash_rest(struct sha1 *sh, char *input, long inlen, int prev) {
  char *ptr = input;
  char *end = input + inlen;
  int block = 64;
  while (end - ptr >= block) {
    sha1_proc(sh, ptr);
    ptr += block;
  }
  sha1_last(sh, ptr, end - ptr, inlen + prev);
}
```
user: Translate the below assembly code into C programming language:

```asm
<hash_rest>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %rdx,%r14
  mov    %ecx,%r13d
  lea    (%rsi,%rdx,1),%r12
  cmp    $0x3f,%rdx
  jle    L1
L9:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <sha1_proc@plt>
  add    $0x40,%rbx
  mov    %r12,%rdx
  sub    %rbx,%rdx
  cmp    $0x3f,%rdx
  jg     L9
L1:
  movslq %r13d,%rcx
  add    %r14,%rcx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <sha1_last@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void hash_rest(struct sha1 *sh, char *input, long inlen, int prev) {
  char *ptr = input;
  char *end = input + inlen;
  int block = 64;
  while (end - ptr >= block) {
    sha1_proc(sh, ptr);
    ptr += block;
  }
  sha1_last(sh, ptr, end - ptr, inlen + prev);
}
```
user: Translate the below assembly code into C programming language:

```asm
<hash_rest>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  lea    (%rsi,%rdx,1),%rbx
  cmp    $0x3f,%rdx
  jle    L17
  nopw   0x0(%rax,%rax,1)
L0:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x40,%rbp
  call   <sha1_proc@plt>
  mov    %rbx,%rdx
  sub    %rbp,%rdx
  cmp    $0x3f,%rdx
  jg     L0
L17:
  movslq %r13d,%rcx
  pop    %rbx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  add    %r14,%rcx
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <sha1_last@plt>
```

assistant: ```c
void hash_rest(struct sha1 *sh, char *input, long inlen, int prev) {
  char *ptr = input;
  char *end = input + inlen;
  int block = 64;
  while (end - ptr >= block) {
    sha1_proc(sh, ptr);
    ptr += block;
  }
  sha1_last(sh, ptr, end - ptr, inlen + prev);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<hash_rest>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  lea    (%rsi,%rdx,1),%rbx
  cmp    $0x3f,%rdx
  jle    L6
  nopw   0x0(%rax,%rax,1)
L5:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x40,%rbp
  call   <sha1_proc@plt>
  mov    %rbx,%rdx
  sub    %rbp,%rdx
  cmp    $0x3f,%rdx
  jg     L5
L6:
  movslq %r13d,%rcx
  pop    %rbx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  add    %r14,%rcx
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <sha1_last@plt>
```

assistant: ```c
void hash_rest(struct sha1 *sh, char *input, long inlen, int prev) {
  char *ptr = input;
  char *end = input + inlen;
  int block = 64;
  while (end - ptr >= block) {
    sha1_proc(sh, ptr);
    ptr += block;
  }
  sha1_last(sh, ptr, end - ptr, inlen + prev);
}
```
user: Translate the below assembly code into C programming language:

```asm
<rootprocess>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  jmp    L140
L98:
  mov    -0x1c(%rbp),%eax
  cmp    $0x80,%eax
  jne    L126
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <is_allowed_port@plt>
  test   %eax,%eax
  jne    L134
  mov    D67(%rip),%rax
  mov    (%rax),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <send_err@plt>
  test   %eax,%eax
  jne    L52
  jmp    L140
L134:
  movl   $0x4,-0x14(%rbp)
  mov    D73(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  nop
  mov    D58(%rip),%rax
  mov    (%rax),%edx
  mov    D48(%rip),%rax
  mov    (%rax),%ecx
  mov    -0x10(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <socket@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0xffffffff,-0xc(%rbp)
  je     L112
  jmp    L103
L126:
  mov    D72(%rip),%rax
  mov    (%rax),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <send_err@plt>
  test   %eax,%eax
  jne    L72
  jmp    L140
L103:
  movl   $0x1,-0x18(%rbp)
  mov    D55(%rip),%rax
  mov    (%rax),%edx
  mov    D79(%rip),%rax
  mov    (%rax),%esi
  lea    -0x18(%rbp),%rcx
  mov    -0xc(%rbp),%eax
  mov    $0x4,%r8d
  mov    %eax,%edi
  call   <setsockopt@plt>
  mov    -0x14(%rbp),%edx
  lea    -0x1c(%rbp),%rcx
  mov    -0xc(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <bind@plt>
  test   %rax,%rax
  jne    L112
  mov    -0xc(%rbp),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <send_fd@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  jmp    L140
L112:
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    -0x24(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <send_err@plt>
L140:
  lea    -0x1c(%rbp),%rsi
  mov    -0x24(%rbp),%eax
  mov    $0x0,%ecx
  mov    $0x4,%edx
  mov    %eax,%edi
  call   <recv@plt>
  cmp    $0x4,%eax
  je     L98
  jmp    L50
L52:
  nop
  jmp    L50
L72:
  nop
L50:
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L61
  call   <__stack_chk_fail@plt>
L61:
  leave
  ret
```

assistant: ```c
void rootprocess(int fd) {
  union {
    struct sockaddr sa;
    struct sockaddr_storage ss;
    struct sockaddr_in sin;
  } addr;
  while (recv(fd, &addr.ss, sizeof(addr.ss), 0) == sizeof(addr.ss)) {
    unsigned len;
    int sock;
    int family;
    switch (addr.sa.sa_family) {
    case AF_INET:
      if (!is_allowed_port(addr.sin.sin_port)) {
        if (send_err(fd, EACCES))
          return;
        continue;
      }
      len = sizeof(struct sockaddr_in);
      family = PF_INET;
      break;
    default:
      if (send_err(fd, EAFNOSUPPORT))
        return;
      continue;
    }
    sock = socket(family, SOCK_STREAM, IPPROTO_TCP);
    if (sock != -1) {
      const int val = 1;
      setsockopt(sock, SOL_SOCKET, SO_REUSEADDR, &val, sizeof(val));
      if (bind(sock, &addr.sa, len) == 0) {
        send_fd(fd, sock);
        close(sock);
        continue;
      }
    }
    send_err(fd, errno);
  }
}
```