ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    $0x5,%esi
  lea    D0(%rip),%rbx
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %rsp,%r12
  mov    %r12,%rsi
  lea    D24(%rip),%rbp
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x20(%rsp),%rdx
  mov    $0xc,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%esi
  lea    D4(%rip),%rbp
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x2c(%rsp),%rdx
  mov    $0xc,%r8d
  mov    $0x1,%ecx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D37(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x4,%esi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
type: string
name: unpack_data
arguments: data_label: D24
type: string
name: unpack_data
arguments: data_label: D4
type: string
name: unpack_data
arguments: data_label: D37
type: string

tool: D0: %d


tool: D24: %p


tool: D4: &matrix[%d]: %p sizeof(matrix[%d]): %d


tool: D37: -----------------

assistant: ```c
int main() {
  int vector[5];
  printf("%d\n", sizeof(vector) / sizeof(int));
  printf("%p\n", vector);
  printf("%p\n", &vector[0]);
  int matrix[2][3];
  for (int i = 0; i < 2; i++) {
    printf("&matrix[%d]: %p sizeof(matrix[%d]): %d\n", i, &matrix[i], i, sizeof(matrix[i]));
  }
  printf("-----------------\n");
  printf("%d\n", sizeof(char));
  printf("%d\n", sizeof('a'));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x5,%esi
  push   %r12
  lea    D28(%rip),%r12
  push   %rbp
  lea    D10(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <printf@plt>
  mov    %r12,%rdi
  mov    %r13,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  mov    %r13,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D16(%rip),%r12
  xor    %ecx,%ecx
  xor    %esi,%esi
  lea    0x20(%rsp),%rdx
  mov    $0xc,%r8d
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x2c(%rsp),%rdx
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    $0xc,%r8d
  mov    $0x1,%ecx
  mov    $0x1,%esi
  call   <printf@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D28
type_infered: string
name: unpack_data
arguments: tag: D10
type_infered: string
name: unpack_data
arguments: tag: D16
type_infered: string
name: unpack_data
arguments: tag: D38
type_infered: string

tool: D28: %p


tool: D10: %d


tool: D16: &matrix[%d]: %p sizeof(matrix[%d]): %d


tool: D38: -----------------

assistant: ```c
int main() {
  int vector[5];
  printf("%d\n", sizeof(vector) / sizeof(int));
  printf("%p\n", vector);
  printf("%p\n", &vector[0]);
  int matrix[2][3];
  for (int i = 0; i < 2; i++) {
    printf("&matrix[%d]: %p sizeof(matrix[%d]): %d\n", i, &matrix[i], i, sizeof(matrix[i]));
  }
  printf("-----------------\n");
  printf("%d\n", sizeof(char));
  printf("%d\n", sizeof('a'));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r13
  mov    $0x5,%esi
  push   %r12
  lea    D20(%rip),%r12
  push   %rbp
  lea    D39(%rip),%rbp
  mov    %rbp,%rdi
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r13
  call   <printf@plt>
  mov    %r12,%rdi
  mov    %r13,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rdi
  mov    %r13,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D3(%rip),%r12
  xor    %ecx,%ecx
  xor    %esi,%esi
  lea    0x20(%rsp),%rdx
  mov    $0xc,%r8d
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x2c(%rsp),%rdx
  mov    %r12,%rdi
  xor    %eax,%eax
  mov    $0xc,%r8d
  mov    $0x1,%ecx
  mov    $0x1,%esi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  call   <puts@plt>
  mov    $0x1,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    $0x4,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D20
type_infered: string
name: read_data
arguments: label: D39
type_infered: string
name: read_data
arguments: label: D3
type_infered: string
name: read_data
arguments: label: D5
type_infered: string

tool: D20: %p


tool: D39: %d


tool: D3: &matrix[%d]: %p sizeof(matrix[%d]): %d


tool: D5: -----------------

assistant: ```c
int main() {
  int vector[5];
  printf("%d\n", sizeof(vector) / sizeof(int));
  printf("%p\n", vector);
  printf("%p\n", &vector[0]);
  int matrix[2][3];
  for (int i = 0; i < 2; i++) {
    printf("&matrix[%d]: %p sizeof(matrix[%d]): %d\n", i, &matrix[i], i, sizeof(matrix[i]));
  }
  printf("-----------------\n");
  printf("%d\n", sizeof(char));
  printf("%d\n", sizeof('a'));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<read_quaddobl_start_Laurent_system>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%r8d
  mov    %rax,%rsi
  mov    $0x2ff,%edi
  call   <_ada_use_c2phc4c@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  leave
  ret
```

assistant: ```c
int read_quaddobl_start_Laurent_system(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(767, a, b, c, 0);
  return fail;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_quaddobl_start_Laurent_system>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%r8d
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    $0x2ff,%edi
  call   <_ada_use_c2phc4c@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int read_quaddobl_start_Laurent_system(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(767, a, b, c, 0);
  return fail;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<read_quaddobl_start_Laurent_system>:
  endbr64
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x2ff,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int read_quaddobl_start_Laurent_system(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(767, a, b, c, 0);
  return fail;
}
```
user: What is the c source code of the assembly code below?

```asm
<read_quaddobl_start_Laurent_system>:
  endbr64
  xor    %r8d,%r8d
  xor    %ecx,%ecx
  xor    %edx,%edx
  xor    %esi,%esi
  mov    $0x2ff,%edi
  jmp    1050 <_ada_use_c2phc4c@plt>
```

assistant: ```c
int read_quaddobl_start_Laurent_system(void) {
  int *a, *b, fail;
  double *c;
  fail = _ada_use_c2phc4c(767, a, b, c, 0);
  return fail;
}
```
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    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <permute@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D3
datatype: string
name: retrieve_data
arguments: data_label: D10
datatype: string

tool: D3: %s

tool: D10: %s,Length =%d

assistant: ```c
int main() {
  char *a;
  scanf("%s", a);
  printf("%s,Length =%d", a, strlen(a));
  permute(a, 0, strlen(a) - 1);
  getchar();
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%ebx
  mov    %rbx,%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbx,%rsi
  lea    D10(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  lea    -0x1(%rax),%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <permute@plt>
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D9
type: string
name: disassemble_data
arguments: tag: D10
type: string

tool: D9: %s

tool: D10: %s,Length =%d

assistant: ```c
int main() {
  char *a;
  scanf("%s", a);
  printf("%s,Length =%d", a, strlen(a));
  permute(a, 0, strlen(a) - 1);
  getchar();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %ebp,%ebp
  lea    D9(%rip),%rdi
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  lea    D1(%rip),%rdi
  mov    %eax,%edx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  lea    -0x1(%rax),%edx
  call   <permute@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_guessed: string
name: unpack_data
arguments: data_tag: D1
type_guessed: string

tool: D9: %s

tool: D1: %s,Length =%d

assistant: ```c
int main() {
  char *a;
  scanf("%s", a);
  printf("%s,Length =%d", a, strlen(a));
  permute(a, 0, strlen(a) - 1);
  getchar();
  return 0;
}
```
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
  xor    %ebp,%ebp
  lea    D16(%rip),%rdi
  mov    %rbp,%rsi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  lea    D3(%rip),%rdi
  mov    %eax,%edx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rdi
  xor    %esi,%esi
  lea    -0x1(%rax),%edx
  call   <permute@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D16
type_guessed: string
name: extract_symbolic_data
arguments: tag: D3
type_guessed: string

tool: D16: %s

tool: D3: %s,Length =%d

assistant: ```c
int main() {
  char *a;
  scanf("%s", a);
  printf("%s,Length =%d", a, strlen(a));
  permute(a, 0, strlen(a) - 1);
  getchar();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ranged_convert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    -0x48(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  mov    -0x38(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L21
  mov    -0x40(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L21
  mov    -0x40(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L57
L1:
  cmpq   $0x0,-0x18(%rbp)
  jns    L44
  mov    -0x10(%rbp),%rax
  sub    -0x18(%rbp),%rax
  sar    %rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  jmp    L7
L44:
  mov    -0x18(%rbp),%rax
  sub    -0x10(%rbp),%rax
  sar    %rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
L7:
  mov    -0x40(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x40(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x48(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  mov    -0x38(%rbp),%rcx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   *%rcx
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L70
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L57
L70:
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x18(%rbp)
L57:
  cmpq   $0x0,-0x18(%rbp)
  jns    L103
  mov    $0xffffffffffffffff,%rdx
  jmp    L9
L103:
  mov    $0x1,%edx
L9:
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x18(%rbp)
  jne    L1
  cmpq   $0x0,-0x20(%rbp)
  jne    L21
  cmpq   $0x0,-0x10(%rbp)
  je     L21
  mov    -0x40(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x48(%rbp),%rax
  mov    -0x24(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L21:
  mov    -0x20(%rbp),%rax
  leave
  ret
```

assistant: ```c
struct tm *ranged_convert(struct tm *(*convert)(const time_t *, struct tm *), time_t *t, struct tm *tp) {
  struct tm *r;
  if (!(r = (*convert)(t, tp)) && *t) {
    time_t bad = *t;
    time_t ok = 0;
    struct tm tm;
    while (bad != ok + (bad < 0 ? -1 : 1)) {
      time_t mid = *t = (bad < 0 ? bad + ((ok - bad) >> 1) : ok + ((bad - ok) >> 1));
      if ((r = (*convert)(t, tp))) {
        tm = *r;
        ok = mid;
      } else
        bad = mid;
    }
    if (!r && ok) {
      *t = ok;
      *tp = tm;
      r = tp;
    }
  }
  return r;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ranged_convert>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r14
  mov    %rsi,%rbp
  mov    %rdx,%r13
  mov    %rdx,%rsi
  mov    %rbp,%rdi
  call   *%r14
  test   %rax,%rax
  je     L63
L28:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L63:
  mov    0x0(%rbp),%r12
  test   %r12,%r12
  je     L28
  mov    $0x0,%r15d
  jmp    L39
L52:
  mov    %r12,%rbx
  sub    %r15,%rbx
  sar    %rbx
  add    %r15,%rbx
L46:
  mov    %rbx,0x0(%rbp)
  mov    %r13,%rsi
  mov    %rbp,%rdi
  call   *%r14
  test   %rax,%rax
  je     L24
  mov    (%rax),%ecx
  mov    %ecx,0xc(%rsp)
  mov    %rbx,%r15
L39:
  test   %r12,%r12
  js     L105
  lea    0x1(%r15),%rdx
  cmp    %rdx,%r12
  jne    L52
L41:
  test   %rax,%rax
  jne    L28
  test   %r15,%r15
  je     L28
  mov    %r15,0x0(%rbp)
  mov    0xc(%rsp),%eax
  mov    %eax,0x0(%r13)
  mov    %r13,%rax
  jmp    L28
L24:
  mov    %rbx,%r12
  jmp    L39
L105:
  lea    -0x1(%r15),%rdx
  cmp    %rdx,%r12
  je     L41
  mov    %r15,%rbx
  sub    %r12,%rbx
  sar    %rbx
  add    %r12,%rbx
  jmp    L46
```

assistant: ```c
struct tm *ranged_convert(struct tm *(*convert)(const time_t *, struct tm *), time_t *t, struct tm *tp) {
  struct tm *r;
  if (!(r = (*convert)(t, tp)) && *t) {
    time_t bad = *t;
    time_t ok = 0;
    struct tm tm;
    while (bad != ok + (bad < 0 ? -1 : 1)) {
      time_t mid = *t = (bad < 0 ? bad + ((ok - bad) >> 1) : ok + ((bad - ok) >> 1));
      if ((r = (*convert)(t, tp))) {
        tm = *r;
        ok = mid;
      } else
        bad = mid;
    }
    if (!r && ok) {
      *t = ok;
      *tp = tm;
      r = tp;
    }
  }
  return r;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ranged_convert>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  mov    %rdx,%rsi
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13,%rdi
  push   %rbx
  sub    $0x18,%rsp
  call   *%rbp
  test   %rax,%rax
  je     L2
L43:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mov    0x0(%r13),%r14
  xor    %r15d,%r15d
  test   %r14,%r14
  je     L43
L81:
  test   %r14,%r14
  js     L42
L34:
  lea    0x1(%r15),%rcx
  cmp    %rcx,%r14
  je     L45
  mov    %r14,%rbx
  sub    %r15,%rbx
  sar    %rbx
  add    %r15,%rbx
L30:
  mov    %rbx,0x0(%r13)
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   *%rbp
  test   %rax,%rax
  je     L19
  mov    (%rax),%edx
  mov    %rbx,%r15
  mov    %edx,0xc(%rsp)
  test   %r14,%r14
  jns    L34
L42:
  lea    -0x1(%r15),%rcx
  cmp    %rcx,%r14
  je     L45
  mov    %r15,%rbx
  sub    %r14,%rbx
  sar    %rbx
  add    %r14,%rbx
  jmp    L30
  nopl   0x0(%rax)
L19:
  mov    %rbx,%r14
  jmp    L81
  nopl   (%rax)
L45:
  test   %rax,%rax
  jne    L43
  test   %r15,%r15
  je     L43
  mov    0xc(%rsp),%eax
  mov    %r15,0x0(%r13)
  mov    %eax,(%r12)
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
struct tm *ranged_convert(struct tm *(*convert)(const time_t *, struct tm *), time_t *t, struct tm *tp) {
  struct tm *r;
  if (!(r = (*convert)(t, tp)) && *t) {
    time_t bad = *t;
    time_t ok = 0;
    struct tm tm;
    while (bad != ok + (bad < 0 ? -1 : 1)) {
      time_t mid = *t = (bad < 0 ? bad + ((ok - bad) >> 1) : ok + ((bad - ok) >> 1));
      if ((r = (*convert)(t, tp))) {
        tm = *r;
        ok = mid;
      } else
        bad = mid;
    }
    if (!r && ok) {
      *t = ok;
      *tp = tm;
      r = tp;
    }
  }
  return r;
}
```
user: What is the c source code of the assembly code below?

```asm
<ranged_convert>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  mov    %rdx,%rsi
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13,%rdi
  push   %rbx
  sub    $0x18,%rsp
  call   *%rbp
  test   %rax,%rax
  je     L87
L6:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L87:
  mov    0x0(%r13),%r14
  xor    %r15d,%r15d
  test   %r14,%r14
  je     L6
L35:
  test   %r14,%r14
  js     L104
L86:
  lea    0x1(%r15),%rcx
  cmp    %rcx,%r14
  je     L8
  mov    %r14,%rbx
  sub    %r15,%rbx
  sar    %rbx
  add    %r15,%rbx
L57:
  mov    %rbx,0x0(%r13)
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   *%rbp
  test   %rax,%rax
  je     L25
  mov    (%rax),%edx
  mov    %rbx,%r15
  mov    %edx,0xc(%rsp)
  test   %r14,%r14
  jns    L86
L104:
  lea    -0x1(%r15),%rcx
  cmp    %rcx,%r14
  je     L8
  mov    %r15,%rbx
  sub    %r14,%rbx
  sar    %rbx
  add    %r14,%rbx
  jmp    L57
  nopl   0x0(%rax)
L25:
  mov    %rbx,%r14
  jmp    L35
  nopl   (%rax)
L8:
  test   %rax,%rax
  jne    L6
  test   %r15,%r15
  je     L6
  mov    0xc(%rsp),%eax
  mov    %r15,0x0(%r13)
  mov    %eax,(%r12)
  add    $0x18,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
struct tm *ranged_convert(struct tm *(*convert)(const time_t *, struct tm *), time_t *t, struct tm *tp) {
  struct tm *r;
  if (!(r = (*convert)(t, tp)) && *t) {
    time_t bad = *t;
    time_t ok = 0;
    struct tm tm;
    while (bad != ok + (bad < 0 ? -1 : 1)) {
      time_t mid = *t = (bad < 0 ? bad + ((ok - bad) >> 1) : ok + ((bad - ok) >> 1));
      if ((r = (*convert)(t, tp))) {
        tm = *r;
        ok = mid;
      } else
        bad = mid;
    }
    if (!r && ok) {
      *t = ok;
      *tp = tm;
      r = tp;
    }
  }
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ReceiveData>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L13
L5:
  mov    -0x18(%rbp),%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rax,%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L35
  mov    $0xffffffff,%eax
  jmp    L2
L35:
  mov    -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
L13:
  mov    -0x8(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L5
  mov    -0x8(%rbp),%eax
L2:
  leave
  ret
```

assistant: ```c
int ReceiveData(int socket, void *data, int bytes) {
  int received = 0, b;
  while (received < bytes) {
    b = read(socket, (char *)data + received, bytes - received);
    if (b <= 0)
      return -1;
    received += b;
  }
  return received;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ReceiveData>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  jle    L43
  mov    %edi,%r12d
  mov    %rsi,%r13
  mov    %edx,%ebp
  mov    $0x0,%ebx
L28:
  mov    %ebp,%edx
  sub    %ebx,%edx
  movslq %ebx,%rsi
  add    %r13,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %eax,%eax
  jle    L4
  add    %eax,%ebx
  cmp    %ebx,%ebp
  jg     L28
  jmp    L11
L43:
  mov    $0x0,%ebx
  jmp    L11
L4:
  mov    $0xffffffff,%ebx
L11:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int ReceiveData(int socket, void *data, int bytes) {
  int received = 0, b;
  while (received < bytes) {
    b = read(socket, (char *)data + received, bytes - received);
    if (b <= 0)
      return -1;
    received += b;
  }
  return received;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ReceiveData>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  jle    L33
  mov    %edi,%ebp
  mov    %rsi,%r13
  mov    %edx,%ebx
  xor    %r12d,%r12d
  jmp    L25
  xchg   %ax,%ax
L38:
  add    %eax,%r12d
  cmp    %r12d,%ebx
  jle    L2
L25:
  mov    %ebx,%edx
  movslq %r12d,%rsi
  mov    %ebp,%edi
  sub    %r12d,%edx
  add    %r13,%rsi
  call   <read@plt>
  test   %eax,%eax
  jg     L38
  mov    $0xffffffff,%r12d
L2:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L33:
  xor    %r12d,%r12d
  jmp    L2
```

assistant: ```c
int ReceiveData(int socket, void *data, int bytes) {
  int received = 0, b;
  while (received < bytes) {
    b = read(socket, (char *)data + received, bytes - received);
    if (b <= 0)
      return -1;
    received += b;
  }
  return received;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ReceiveData>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  jle    L30
  mov    %edi,%ebp
  mov    %rsi,%r13
  mov    %edx,%ebx
  xor    %r12d,%r12d
  jmp    L41
  xchg   %ax,%ax
L49:
  add    %eax,%r12d
  cmp    %r12d,%ebx
  jle    L31
L41:
  mov    %ebx,%edx
  movslq %r12d,%rsi
  mov    %ebp,%edi
  sub    %r12d,%edx
  add    %r13,%rsi
  call   <read@plt>
  test   %eax,%eax
  jg     L49
  mov    $0xffffffff,%r12d
L31:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L30:
  xor    %r12d,%r12d
  jmp    L31
```

assistant: ```c
int ReceiveData(int socket, void *data, int bytes) {
  int received = 0, b;
  while (received < bytes) {
    b = read(socket, (char *)data + received, bytes - received);
    if (b <= 0)
      return -1;
    received += b;
  }
  return received;
}
```