ylfeng/ReF-Decompile-dataset huggingface dataset:
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
  mov    $0x4ee728,%edi
  call   <do_reverse_number@plt>
  mov    %eax,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0x63427,%edi
  call   <do_reverse_number@plt>
  mov    %eax,%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    $0x5a06e,%edi
  call   <do_reverse_number@plt>
  mov    %eax,%esi
  lea    D40(%rip),%rdi
  call   <printf@plt>
  mov    $0x2427e6,%edi
  call   <do_reverse_number@plt>
  mov    %eax,%esi
  lea    D32(%rip),%rdi
  call   <printf@plt>
  mov    $0x21507d,%edi
  call   <do_reverse_number@plt>
  mov    %eax,%esi
  lea    D36(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: label: D8
type_infered: string
name: read_data
arguments: label: D13
type_infered: string
name: read_data
arguments: label: D40
type_infered: string
name: read_data
arguments: label: D32
type_infered: string
name: read_data
arguments: label: D36
type_infered: string

tool: D8: 5170984 %d


tool: D13: 406567  %d


tool: D40: 368750  %d


tool: D32: 2369510 %d


tool: D36: 2183293 %d


assistant: ```c
int main() {
  printf("5170984 %d\n", do_reverse_number(5170984));
  printf("406567  %d\n", do_reverse_number(406567));
  printf("368750  %d\n", do_reverse_number(368750));
  printf("2369510 %d\n", do_reverse_number(2369510));
  printf("2183293 %d\n", do_reverse_number(2183293));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4ee728,%edi
  call   <do_reverse_number@plt>
  lea    D36(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x63427,%edi
  call   <do_reverse_number@plt>
  lea    D45(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x5a06e,%edi
  call   <do_reverse_number@plt>
  lea    D5(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x2427e6,%edi
  call   <do_reverse_number@plt>
  lea    D4(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x21507d,%edi
  call   <do_reverse_number@plt>
  lea    D31(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D36
type_infered: string
name: extract_symbolic_data
arguments: label: D45
type_infered: string
name: extract_symbolic_data
arguments: label: D5
type_infered: string
name: extract_symbolic_data
arguments: label: D4
type_infered: string
name: extract_symbolic_data
arguments: label: D31
type_infered: string

tool: D36: 5170984 %d


tool: D45: 406567  %d


tool: D5: 368750  %d


tool: D4: 2369510 %d


tool: D31: 2183293 %d


assistant: ```c
int main() {
  printf("5170984 %d\n", do_reverse_number(5170984));
  printf("406567  %d\n", do_reverse_number(406567));
  printf("368750  %d\n", do_reverse_number(368750));
  printf("2369510 %d\n", do_reverse_number(2369510));
  printf("2183293 %d\n", do_reverse_number(2183293));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x4ee728,%edi
  call   <do_reverse_number@plt>
  lea    D7(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x63427,%edi
  call   <do_reverse_number@plt>
  lea    D37(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x5a06e,%edi
  call   <do_reverse_number@plt>
  lea    D11(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x2427e6,%edi
  call   <do_reverse_number@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    $0x21507d,%edi
  call   <do_reverse_number@plt>
  lea    D48(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D7
datatype: string
name: unpack_data
arguments: tag: D37
datatype: string
name: unpack_data
arguments: tag: D11
datatype: string
name: unpack_data
arguments: tag: D6
datatype: string
name: unpack_data
arguments: tag: D48
datatype: string

tool: D7: 5170984 %d


tool: D37: 406567  %d


tool: D11: 368750  %d


tool: D6: 2369510 %d


tool: D48: 2183293 %d


assistant: ```c
int main() {
  printf("5170984 %d\n", do_reverse_number(5170984));
  printf("406567  %d\n", do_reverse_number(406567));
  printf("368750  %d\n", do_reverse_number(368750));
  printf("2369510 %d\n", do_reverse_number(2369510));
  printf("2183293 %d\n", do_reverse_number(2183293));
  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
<list_splice>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <list_empty@plt>
  test   %eax,%eax
  jne    L9
  mov    -0x10(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <__list_splice@plt>
L9:
  nop
  leave
  ret
```

assistant: ```c
void list_splice(struct list_head *list, struct list_head *head) {
  if (!list_empty(list)) {
    __list_splice(list, head);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<list_splice>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <list_empty@plt>
  test   %eax,%eax
  je     L6
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L6:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <__list_splice@plt>
  jmp    L4
```

assistant: ```c
void list_splice(struct list_head *list, struct list_head *head) {
  if (!list_empty(list)) {
    __list_splice(list, head);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<list_splice>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <list_empty@plt>
  test   %eax,%eax
  je     L9
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <__list_splice@plt>
```

assistant: ```c
void list_splice(struct list_head *list, struct list_head *head) {
  if (!list_empty(list)) {
    __list_splice(list, head);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<list_splice>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <list_empty@plt>
  test   %eax,%eax
  je     L4
  add    $0x8,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L4:
  add    $0x8,%rsp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <__list_splice@plt>
```

assistant: ```c
void list_splice(struct list_head *list, struct list_head *head) {
  if (!list_empty(list)) {
    __list_splice(list, head);
  }
}
```
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    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L15
L11:
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  add    %eax,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L15:
  cmpl   $0xa,-0x14(%rbp)
  jle    L11
  mov    -0x10(%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,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D1
type_infered: string
name: parse_data
arguments: label: D24
type_infered: string
name: parse_data
arguments: label: D13
type_infered: string

tool: D1: Enter grade: 

tool: D24: %d

tool: D13: Class average is %d


assistant: ```c
int main() {
  int counter, grade, total, average;
  total = 0;
  counter = 0;
  while (counter <= 10) {
    printf("Enter grade: ");
    scanf("%d", &grade);
    total = total + grade;
    counter = counter + 1;
  }
  average = total / 10;
  printf("Class average is %d\n", average);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0xb,%ebx
  mov    $0x0,%ebp
  lea    D18(%rip),%r14
  lea    0x4(%rsp),%r13
  lea    D24(%rip),%r12
L17:
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    0x4(%rsp),%ebp
  sub    $0x1,%ebx
  jne    L17
  movslq %ebp,%rsi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x22,%rsi
  sar    $0x1f,%ebp
  sub    %ebp,%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D18
data_type: string
name: read_data
arguments: data_tag: D24
data_type: string
name: read_data
arguments: data_tag: D16
data_type: string

tool: D18: Enter grade: 

tool: D24: %d

tool: D16: Class average is %d


assistant: ```c
int main() {
  int counter, grade, total, average;
  total = 0;
  counter = 0;
  while (counter <= 10) {
    printf("Enter grade: ");
    scanf("%d", &grade);
    total = total + grade;
    counter = counter + 1;
  }
  average = total / 10;
  printf("Class average is %d\n", average);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D18(%rip),%r14
  push   %r13
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0xb,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  nopw   0x0(%rax,%rax,1)
L16:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    0x4(%rsp),%ebp
  sub    $0x1,%ebx
  jne    L16
  movslq %ebp,%rsi
  sar    $0x1f,%ebp
  xor    %eax,%eax
  lea    D1(%rip),%rdi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x22,%rsi
  sub    %ebp,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
type_guessed: string
name: extract_symbolic_data
arguments: tag: D4
type_guessed: string
name: extract_symbolic_data
arguments: tag: D1
type_guessed: string

tool: D18: Enter grade: 

tool: D4: %d

tool: D1: Class average is %d


assistant: ```c
int main() {
  int counter, grade, total, average;
  total = 0;
  counter = 0;
  while (counter <= 10) {
    printf("Enter grade: ");
    scanf("%d", &grade);
    total = total + grade;
    counter = counter + 1;
  }
  average = total / 10;
  printf("Class average is %d\n", average);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D23(%rip),%r14
  push   %r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    $0xb,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r13
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    0x4(%rsp),%ebp
  sub    $0x1,%ebx
  jne    L6
  movslq %ebp,%rsi
  sar    $0x1f,%ebp
  xor    %eax,%eax
  lea    D28(%rip),%rdi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x22,%rsi
  sub    %ebp,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D23
type_infered: string
name: unpack_data
arguments: tag: D0
type_infered: string
name: unpack_data
arguments: tag: D28
type_infered: string

tool: D23: Enter grade: 

tool: D0: %d

tool: D28: Class average is %d


assistant: ```c
int main() {
  int counter, grade, total, average;
  total = 0;
  counter = 0;
  while (counter <= 10) {
    printf("Enter grade: ");
    scanf("%d", &grade);
    total = total + grade;
    counter = counter + 1;
  }
  average = total / 10;
  printf("Class average is %d\n", average);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<getLargestMultipleForPosition>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int getLargestMultipleForPosition(int **grid, int x, int y, int range) {
  int largest = 0;
  return largest;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<getLargestMultipleForPosition>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int getLargestMultipleForPosition(int **grid, int x, int y, int range) {
  int largest = 0;
  return largest;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getLargestMultipleForPosition>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int getLargestMultipleForPosition(int **grid, int x, int y, int range) {
  int largest = 0;
  return largest;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getLargestMultipleForPosition>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int getLargestMultipleForPosition(int **grid, int x, int y, int range) {
  int largest = 0;
  return largest;
}
```
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
  mov    $0x0,%eax
  call   <library@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  library();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <library@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  library();
  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
  xor    %eax,%eax
  call   <library@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  library();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int main(void) {
  library();
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<new_mem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L9
  mov    $0x0,%eax
  jmp    L6
L9:
  mov    -0x8(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <calloc@plt>
L6:
  leave
  ret
```

assistant: ```c
void *new_mem(size_t size) {
  if (size == 0)
    return NULL;
  return calloc(size, 1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<new_mem>:
  endbr64
  mov    $0x0,%eax
  test   %rdi,%rdi
  je     L0
  sub    $0x8,%rsp
  mov    $0x1,%esi
  call   <calloc@plt>
  add    $0x8,%rsp
  ret
L0:
  ret
```

assistant: ```c
void *new_mem(size_t size) {
  if (size == 0)
    return NULL;
  return calloc(size, 1);
}
```
user: What is the c source code of the assembly code below:

```asm
<new_mem>:
  endbr64
  test   %rdi,%rdi
  je     L7
  mov    $0x1,%esi
  jmp    1050 <calloc@plt>
  nopl   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *new_mem(size_t size) {
  if (size == 0)
    return NULL;
  return calloc(size, 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<new_mem>:
  endbr64
  test   %rdi,%rdi
  je     L4
  mov    $0x1,%esi
  jmp    1050 <calloc@plt>
  nopl   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *new_mem(size_t size) {
  if (size == 0)
    return NULL;
  return calloc(size, 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<scsi_fixstring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L45
L6:
  addq   $0x1,-0x18(%rbp)
L45:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  je     L104
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L6
  jmp    L104
L108:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L48
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  je     L104
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L104
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L104
L48:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  movzbl -0x1(%rdx),%edx
  mov    %dl,(%rax)
L104:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  je     L4
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L108
  jmp    L4
L74:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x0,(%rax)
L4:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jne    L74
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void scsi_fixstring(unsigned char *s, int bytecount) {
  unsigned char *p;
  unsigned char *end;
  p = s;
  end = s + bytecount;
  while (s != end && *s == ' ')
    ++s;
  while (s != end && *s) {
    if (*s++ != ' ' || (s != end && *s && *s != ' '))
      *p++ = *(s - 1);
  }
  while (p != end)
    *p++ = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<scsi_fixstring>:
  endbr64
  movslq %esi,%rsi
  add    %rdi,%rsi
  mov    %rdi,%rax
  cmp    %rsi,%rdi
  je     L108
L45:
  cmpb   $0x20,(%rax)
  jne    L72
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L45
  jmp    L98
L108:
  ret
L72:
  cmp    %rax,%rsi
  jne    L18
L7:
  cmp    %rdi,%rsi
  je     L78
L98:
  add    $0x1,%rdi
  movb   $0x0,-0x1(%rdi)
  cmp    %rdi,%rsi
  jne    L98
  ret
L58:
  lea    0x1(%rdi),%rcx
  mov    %dl,(%rdi)
  cmp    %rax,%rsi
  je     L90
L100:
  mov    %rcx,%rdi
L18:
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L7
  add    $0x1,%rax
  cmp    $0x20,%dl
  jne    L58
  cmp    %rax,%rsi
  je     L7
  mov    %rdi,%rcx
  testb  $0xdf,(%rax)
  je     L100
  lea    0x1(%rdi),%rcx
  mov    %dl,(%rdi)
  jmp    L100
L90:
  mov    %rcx,%rdi
  jmp    L7
L78:
  ret
```

assistant: ```c
void scsi_fixstring(unsigned char *s, int bytecount) {
  unsigned char *p;
  unsigned char *end;
  p = s;
  end = s + bytecount;
  while (s != end && *s == ' ')
    ++s;
  while (s != end && *s) {
    if (*s++ != ' ' || (s != end && *s && *s != ' '))
      *p++ = *(s - 1);
  }
  while (p != end)
    *p++ = '\0';
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<scsi_fixstring>:
  endbr64
  movslq %esi,%rdx
  mov    %rdi,%rax
  add    %rdi,%rdx
  cmp    %rdx,%rdi
  jne    L117
  jmp    L107
  nopl   0x0(%rax)
L17:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L8
L117:
  movzbl (%rax),%ecx
  cmp    $0x20,%cl
  je     L17
  cmp    %rax,%rdx
  jne    L32
  jmp    L8
L29:
  mov    %cl,(%rdi)
  lea    0x1(%rdi),%rsi
  cmp    %rax,%rdx
  je     L81
  movzbl (%rax),%ecx
L109:
  mov    %rsi,%rdi
L32:
  test   %cl,%cl
  je     L87
  add    $0x1,%rax
  cmp    $0x20,%cl
  jne    L29
  cmp    %rax,%rdx
  je     L87
  movzbl (%rax),%ecx
  mov    %rdi,%rsi
  test   $0xdf,%cl
  je     L109
  movb   $0x20,(%rdi)
  lea    0x1(%rdi),%rsi
  movzbl (%rax),%ecx
  jmp    L109
L81:
  mov    %rsi,%rdi
  nopl   (%rax)
L87:
  cmp    %rdi,%rdx
  je     L3
L8:
  sub    %rdi,%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
L3:
  ret
L107:
  ret
```

assistant: ```c
void scsi_fixstring(unsigned char *s, int bytecount) {
  unsigned char *p;
  unsigned char *end;
  p = s;
  end = s + bytecount;
  while (s != end && *s == ' ')
    ++s;
  while (s != end && *s) {
    if (*s++ != ' ' || (s != end && *s && *s != ' '))
      *p++ = *(s - 1);
  }
  while (p != end)
    *p++ = '\0';
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<scsi_fixstring>:
  endbr64
  movslq %esi,%rdx
  mov    %rdi,%rax
  add    %rdi,%rdx
  cmp    %rdx,%rdi
  jne    L50
  jmp    L11
  nopl   0x0(%rax)
L70:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L26
L50:
  cmpb   $0x20,(%rax)
  je     L70
  cmp    %rax,%rdx
  jne    L110
  jmp    L26
  nopl   (%rax)
L100:
  mov    %cl,(%rdi)
  lea    0x1(%rdi),%rsi
  cmp    %rax,%rdx
  je     L74
L86:
  mov    %rsi,%rdi
L110:
  movzbl (%rax),%ecx
  test   %cl,%cl
  je     L107
  add    $0x1,%rax
  cmp    $0x20,%cl
  jne    L100
  cmp    %rax,%rdx
  je     L107
  mov    %rdi,%rsi
  testb  $0xdf,(%rax)
  je     L86
  movb   $0x20,(%rdi)
  lea    0x1(%rdi),%rsi
  jmp    L86
L74:
  mov    %rsi,%rdi
  nopw   0x0(%rax,%rax,1)
L107:
  cmp    %rdi,%rdx
  je     L19
L26:
  sub    %rdi,%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
L19:
  ret
L11:
  ret
```

assistant: ```c
void scsi_fixstring(unsigned char *s, int bytecount) {
  unsigned char *p;
  unsigned char *end;
  p = s;
  end = s + bytecount;
  while (s != end && *s == ' ')
    ++s;
  while (s != end && *s) {
    if (*s++ != ' ' || (s != end && *s && *s != ' '))
      *p++ = *(s - 1);
  }
  while (p != end)
    *p++ = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<syscon_load_quaddobl_Laurential>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  lea    -0x28(%rbp),%rdx
  mov    -0x34(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <syscon_quaddobl_Laurent_size_limit@plt>
  mov    %eax,-0x1c(%rbp)
  mov    -0x28(%rbp),%eax
  mov    $0x4,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rdx
  lea    -0x24(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x239,%edi
  call   <_ada_use_c2phc@plt>
  mov    %eax,-0x1c(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L11
L14:
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x20(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x48(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x20(%rbp)
L11:
  mov    -0x24(%rbp),%eax
  cmp    %eax,-0x20(%rbp)
  jl     L14
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x1c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
int syscon_load_quaddobl_Laurential(int k, int *nc, char *p) {
  int fail, i, szl;
  int *buffer;
  int size = k;
  double *c;
  fail = syscon_quaddobl_Laurent_size_limit(k, &szl);
  buffer = (int *)calloc(szl, sizeof(int));
  fail = _ada_use_c2phc(569, &size, buffer, c);
  for (i = 0; i < size; i++)
    p[i] = (char)buffer[i];
  p[size] = '\0';
  free(buffer);
  return fail;
}
```