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

```asm
<main>:
  endbr64
  push   %r14
  lea    D36(%rip),%rdi
  push   %r13
  lea    D20(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L14
  mov    %rbx,%r12
  xor    %ebp,%ebp
  lea    D35(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L26:
  mov    %ebp,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x4,%r12
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %ebp,%eax
  jg     L26
  mov    (%rbx),%ebp
  cmp    $0x1,%eax
  jle    L67
  add    $0x4,%rbx
  mov    %ebp,%r12d
  mov    $0x1,%r13d
  xchg   %ax,%ax
L25:
  mov    (%rbx),%esi
  mov    %ebp,%edi
  add    $0x1,%r13d
  add    $0x4,%rbx
  call   <max@plt>
  mov    -0x4(%rbx),%esi
  mov    %r12d,%edi
  mov    %eax,%ebp
  call   <min@plt>
  mov    %eax,%r12d
  cmp    %r13d,0x4(%rsp)
  jg     L25
L31:
  mov    %r12d,%esi
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D49(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L59
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L14:
  mov    (%rbx),%ebp
  mov    %ebp,%r12d
  jmp    L31
L67:
  mov    %ebp,%r12d
  jmp    L31
L59:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D36
type_guessed: string
name: retrieve_data
arguments: label: D20
type_guessed: string
name: retrieve_data
arguments: label: D35
type_guessed: string
name: retrieve_data
arguments: label: D7
type_guessed: string
name: retrieve_data
arguments: label: D49
type_guessed: string

tool: D36: How many numbers? 

tool: D20: %d

tool: D35: N[%d] = 

tool: D7: The small number is %d


tool: D49: The big number is %d


assistant: ```c
int main() {
  int i, numbers_len;
  printf("How many numbers? ");
  scanf("%d", &numbers_len);
  int *numbers = (int *)malloc(numbers_len * sizeof(int));
  for (i = 0; i < numbers_len; i++) {
    printf("N[%d] = ", i);
    scanf("%d", numbers + i);
  }
  int big = *numbers;
  int small = numbers[0];
  for (i = 1; i < numbers_len; i++) {
    big = max(big, *(numbers + i));
    small = min(small, numbers[i]);
  }
  printf("The small number is %d\n", small);
  printf("The big number is %d\n", big);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r14
  lea    D7(%rip),%rdi
  push   %r13
  lea    D18(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L68
  mov    %rbx,%r12
  xor    %ebp,%ebp
  lea    D15(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %ebp,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%ebp
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  add    $0x4,%r12
  call   <scanf@plt>
  mov    0x4(%rsp),%eax
  cmp    %ebp,%eax
  jg     L12
  mov    (%rbx),%ebp
  cmp    $0x1,%eax
  jle    L21
  add    $0x4,%rbx
  mov    %ebp,%r12d
  mov    $0x1,%r13d
  xchg   %ax,%ax
L10:
  mov    (%rbx),%esi
  mov    %ebp,%edi
  add    $0x1,%r13d
  add    $0x4,%rbx
  call   <max@plt>
  mov    -0x4(%rbx),%esi
  mov    %r12d,%edi
  mov    %eax,%ebp
  call   <min@plt>
  mov    %eax,%r12d
  cmp    %r13d,0x4(%rsp)
  jg     L10
L14:
  mov    %r12d,%esi
  lea    D24(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D33(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L54
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L68:
  mov    (%rbx),%ebp
  mov    %ebp,%r12d
  jmp    L14
L21:
  mov    %ebp,%r12d
  jmp    L14
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D7
datatype: string
name: retrieve_data
arguments: data_label: D18
datatype: string
name: retrieve_data
arguments: data_label: D15
datatype: string
name: retrieve_data
arguments: data_label: D24
datatype: string
name: retrieve_data
arguments: data_label: D33
datatype: string

tool: D7: How many numbers? 

tool: D18: %d

tool: D15: N[%d] = 

tool: D24: The small number is %d


tool: D33: The big number is %d


assistant: ```c
int main() {
  int i, numbers_len;
  printf("How many numbers? ");
  scanf("%d", &numbers_len);
  int *numbers = (int *)malloc(numbers_len * sizeof(int));
  for (i = 0; i < numbers_len; i++) {
    printf("N[%d] = ", i);
    scanf("%d", numbers + i);
  }
  int big = *numbers;
  int small = numbers[0];
  for (i = 1; i < numbers_len; i++) {
    big = max(big, *(numbers + i));
    small = min(small, numbers[i]);
  }
  printf("The small number is %d\n", small);
  printf("The big number is %d\n", big);
}
```
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
  movl   $0x5,-0x4(%rbp)
  mov    $0x6,%edx
  lea    D5(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D5
data_type: string

tool: D5: hello


assistant: ```c
int main() {
  int x = 5;
  write(1, "hello\n", 6);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%edx
  lea    D5(%rip),%rsi
  mov    $0x1,%edi
  call   <write@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D5
datatype: string

tool: D5: hello


assistant: ```c
int main() {
  int x = 5;
  write(1, "hello\n", 6);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%edx
  mov    $0x1,%edi
  lea    D5(%rip),%rsi
  call   <write@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D5
data_type: string

tool: D5: hello


assistant: ```c
int main() {
  int x = 5;
  write(1, "hello\n", 6);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%edx
  mov    $0x1,%edi
  lea    D3(%rip),%rsi
  call   <write@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D3
type_infered: string

tool: D3: hello


assistant: ```c
int main() {
  int x = 5;
  write(1, "hello\n", 6);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<xrealloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L11
  movq   $0x1,-0x20(%rbp)
L11:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L0
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <os_error@plt>
L0:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D1
datatype: string

tool: D1: Memory allocation failure in xrealloc

assistant: ```c
void *xrealloc(void *ptr, size_t size) {
  if (size == 0)
    size = 1;
  void *newp = realloc(ptr, size);
  if (!newp)
    os_error("Memory allocation failure in xrealloc");
  return newp;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<xrealloc>:
  endbr64
  push   %rbx
  test   %rsi,%rsi
  mov    $0x1,%eax
  cmove  %rax,%rsi
  call   <realloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L14
L3:
  mov    %rbx,%rax
  pop    %rbx
  ret
L14:
  lea    D3(%rip),%rdi
  call   <os_error@plt>
  jmp    L3
```

tool_calls: name: read_data
arguments: data_label: D3
data_type: string

tool: D3: Memory allocation failure in xrealloc

assistant: ```c
void *xrealloc(void *ptr, size_t size) {
  if (size == 0)
    size = 1;
  void *newp = realloc(ptr, size);
  if (!newp)
    os_error("Memory allocation failure in xrealloc");
  return newp;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<xrealloc>:
  endbr64
  push   %r12
  test   %rsi,%rsi
  mov    $0x1,%eax
  cmove  %rax,%rsi
  call   <realloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L16
L12:
  mov    %r12,%rax
  pop    %r12
  ret
L16:
  lea    D4(%rip),%rdi
  call   <os_error@plt>
  jmp    L12
```

tool_calls: name: read_data
arguments: tag: D4
datatype: string

tool: D4: Memory allocation failure in xrealloc

assistant: ```c
void *xrealloc(void *ptr, size_t size) {
  if (size == 0)
    size = 1;
  void *newp = realloc(ptr, size);
  if (!newp)
    os_error("Memory allocation failure in xrealloc");
  return newp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<xrealloc>:
  endbr64
  push   %r12
  test   %rsi,%rsi
  mov    $0x1,%eax
  cmove  %rax,%rsi
  call   <realloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L0
L14:
  mov    %r12,%rax
  pop    %r12
  ret
L0:
  lea    D7(%rip),%rdi
  call   <os_error@plt>
  jmp    L14
```

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

tool: D7: Memory allocation failure in xrealloc

assistant: ```c
void *xrealloc(void *ptr, size_t size) {
  if (size == 0)
    size = 1;
  void *newp = realloc(ptr, size);
  if (!newp)
    os_error("Memory allocation failure in xrealloc");
  return newp;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x410,%rsp
  mov    %edi,-0x404(%rbp)
  mov    %rsi,-0x410(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x1,-0x404(%rbp)
  jle    L6
  mov    -0x410(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0x1a4,%esi
  mov    %rax,%rdi
  call   <creat@plt>
  mov    %eax,-0x3f8(%rbp)
  mov    $0x1,%edi
  call   <close@plt>
  mov    -0x3f8(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <dup2@plt>
  jmp    L6
L35:
  mov    -0x3f4(%rbp),%edx
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L6:
  lea    -0x3f0(%rbp),%rax
  mov    $0x3e8,%edx
  mov    %rax,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    %eax,-0x3f4(%rbp)
  cmpl   $0x0,-0x3f4(%rbp)
  jne    L35
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L32
  call   <__stack_chk_fail@plt>
L32:
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc > 1) {
    int fd = creat(argv[1], 0644);
    close(1);
    dup2(fd, 1);
  }
  char buf[1000];
  int len;
  while ((len = read(0, buf, 1000)))
    write(1, buf, len);
  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   %rbx
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jg     L38
L11:
  mov    %rsp,%rbx
  jmp    L42
L38:
  mov    0x8(%rsi),%rdi
  mov    $0x1a4,%esi
  call   <creat@plt>
  mov    %eax,%ebx
  mov    $0x1,%edi
  call   <close@plt>
  mov    $0x1,%esi
  mov    %ebx,%edi
  call   <dup2@plt>
  jmp    L11
L33:
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L42:
  mov    $0x3e8,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <read@plt>
  mov    %eax,%edx
  test   %eax,%eax
  jne    L33
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  mov    $0x0,%eax
  add    $0x3f0,%rsp
  pop    %rbx
  ret
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc > 1) {
    int fd = creat(argv[1], 0644);
    close(1);
    dup2(fd, 1);
  }
  char buf[1000];
  int len;
  while ((len = read(0, buf, 1000)))
    write(1, buf, len);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L5
  mov    0x8(%rsi),%rdi
  mov    $0x1a4,%esi
  call   <creat@plt>
  mov    $0x1,%edi
  mov    %eax,%ebp
  call   <close@plt>
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <dup2@plt>
L5:
  mov    %rsp,%rbx
  jmp    L22
L1:
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L22:
  mov    $0x3e8,%edx
  mov    %rbx,%rsi
  xor    %edi,%edi
  call   <read@plt>
  mov    %eax,%edx
  test   %eax,%eax
  jne    L1
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x3f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc > 1) {
    int fd = creat(argv[1], 0644);
    close(1);
    dup2(fd, 1);
  }
  char buf[1000];
  int len;
  while ((len = read(0, buf, 1000)))
    write(1, buf, len);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x3f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  cmp    $0x1,%edi
  jle    L28
  mov    0x8(%rsi),%rdi
  mov    $0x1a4,%esi
  call   <creat@plt>
  mov    $0x1,%edi
  mov    %eax,%ebp
  call   <close@plt>
  mov    $0x1,%esi
  mov    %ebp,%edi
  call   <dup2@plt>
L28:
  mov    %rsp,%rbx
  jmp    L3
L14:
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
L3:
  mov    $0x3e8,%edx
  mov    %rbx,%rsi
  xor    %edi,%edi
  call   <read@plt>
  mov    %eax,%edx
  test   %eax,%eax
  jne    L14
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x3f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  if (argc > 1) {
    int fd = creat(argv[1], 0644);
    close(1);
    dup2(fd, 1);
  }
  char buf[1000];
  int len;
  while ((len = read(0, buf, 1000)))
    write(1, buf, len);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<rearrange>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L5
L39:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x10(%rbp)
L5:
  mov    -0x10(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L39
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L7:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L15:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L7
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void rearrange(int a[], int an, int r[], int rn) {
  int *u = malloc(sizeof(int) * an);
  for (int i = 0; i < an; ++i)
    u[r[i]] = a[i];
  for (int i = 0; i < an; ++i)
    a[i] = u[i];
  free(u);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rearrange>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %rdx,%r12
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  test   %ebx,%ebx
  jle    L7
  mov    %ebx,%ecx
  mov    $0x0,%edx
L22:
  mov    0x0(%rbp,%rdx,4),%esi
  movslq (%r12,%rdx,4),%rax
  mov    %esi,(%rdi,%rax,4)
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L22
  mov    $0x0,%edx
L10:
  mov    (%rdi,%rdx,4),%eax
  mov    %eax,0x0(%rbp,%rdx,4)
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L10
L7:
  call   <free@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void rearrange(int a[], int an, int r[], int rn) {
  int *u = malloc(sizeof(int) * an);
  for (int i = 0; i < an; ++i)
    u[r[i]] = a[i];
  for (int i = 0; i < an; ++i)
    a[i] = u[i];
  free(u);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<rearrange>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  movslq %esi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rbp,4),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%rdi
  test   %ebp,%ebp
  jle    L11
  xor    %ecx,%ecx
  nopl   0x0(%rax)
L26:
  mov    (%rbx,%rcx,4),%esi
  movslq (%r12,%rcx,4),%rax
  add    $0x1,%rcx
  mov    %esi,(%rdi,%rax,4)
  cmp    %rcx,%rbp
  jne    L26
  xor    %edx,%edx
  nopl   (%rax)
L27:
  mov    (%rdi,%rdx,4),%eax
  mov    %eax,(%rbx,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%rbp
  jne    L27
L11:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <free@plt>
```

assistant: ```c
void rearrange(int a[], int an, int r[], int rn) {
  int *u = malloc(sizeof(int) * an);
  for (int i = 0; i < an; ++i)
    u[r[i]] = a[i];
  for (int i = 0; i < an; ++i)
    a[i] = u[i];
  free(u);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<rearrange>:
  endbr64
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x0(,%rsi,4),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  test   %r12d,%r12d
  jle    L21
  lea    -0x1(%r12),%r8d
  mov    %r12d,%edx
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L2:
  mov    (%rbx,%rax,4),%esi
  movslq 0x0(%rbp,%rax,4),%rcx
  add    $0x1,%rax
  mov    %esi,(%rdi,%rcx,4)
  cmp    %rdx,%rax
  jne    L2
  lea    0x4(%rdi),%rcx
  mov    %rbx,%rax
  sub    %rcx,%rax
  cmp    $0x8,%rax
  jbe    L40
  cmp    $0x2,%r8d
  jbe    L40
  mov    %r12d,%edx
  xor    %eax,%eax
  shr    $0x2,%edx
  shl    $0x4,%rdx
  xchg   %ax,%ax
L15:
  movdqu (%rdi,%rax,1),%xmm0
  movups %xmm0,(%rbx,%rax,1)
  add    $0x10,%rax
  cmp    %rdx,%rax
  jne    L15
  mov    %r12d,%eax
  and    $0xfffffffc,%eax
  test   $0x3,%r12b
  je     L21
  mov    %eax,%ecx
  mov    (%rdi,%rcx,4),%esi
  lea    0x0(,%rcx,4),%rdx
  mov    %esi,(%rbx,%rcx,4)
  lea    0x1(%rax),%ecx
  cmp    %ecx,%r12d
  jle    L21
  mov    0x4(%rdi,%rdx,1),%ecx
  add    $0x2,%eax
  mov    %ecx,0x4(%rbx,%rdx,1)
  cmp    %eax,%r12d
  jle    L21
  mov    0x8(%rdi,%rdx,1),%eax
  mov    %eax,0x8(%rbx,%rdx,1)
L21:
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1060 <free@plt>
  nopw   0x0(%rax,%rax,1)
L40:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L20:
  mov    (%rdi,%rax,4),%ecx
  mov    %ecx,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L20
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    1060 <free@plt>
```

assistant: ```c
void rearrange(int a[], int an, int r[], int rn) {
  int *u = malloc(sizeof(int) * an);
  for (int i = 0; i < an; ++i)
    u[r[i]] = a[i];
  for (int i = 0; i < an; ++i)
    a[i] = u[i];
  free(u);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<getcomponent>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L27
  mov    $0x0,%eax
  jmp    L54
L27:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L43
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  movb   $0x2f,(%rax)
  jmp    L68
L43:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L68
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L43
L68:
  mov    -0x10(%rbp),%rax
  movb   $0x0,(%rax)
  jmp    L8
L62:
  addq   $0x1,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L62
  mov    -0x8(%rbp),%rax
L54:
  pop    %rbp
  ret
```

assistant: ```c
char *getcomponent(char *path, char *name) {
  if (*path == '\0')
    return 0;
  if (*path == '/') {
    *name++ = '/';
  } else {
    do {
      *name++ = *path++;
    } while (*path != '/' && *path != '\0');
  }
  *name = '\0';
  while (*path == '/')
    path++;
  return path;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getcomponent>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L52
  cmp    $0x2f,%dl
  je     L46
L69:
  add    $0x1,%rax
  add    $0x1,%rsi
  movzbl -0x1(%rax),%edx
  mov    %dl,-0x1(%rsi)
  movzbl (%rax),%edx
  cmp    $0x2f,%dl
  je     L37
  test   %dl,%dl
  jne    L69
  jmp    L37
L46:
  movb   $0x2f,(%rsi)
  lea    0x1(%rsi),%rsi
L37:
  movb   $0x0,(%rsi)
  cmpb   $0x2f,(%rax)
  jne    L48
L1:
  add    $0x1,%rax
  cmpb   $0x2f,(%rax)
  je     L1
  ret
L52:
  mov    $0x0,%eax
L48:
  ret
```

assistant: ```c
char *getcomponent(char *path, char *name) {
  if (*path == '\0')
    return 0;
  if (*path == '/') {
    *name++ = '/';
  } else {
    do {
      *name++ = *path++;
    } while (*path != '/' && *path != '\0');
  }
  *name = '\0';
  while (*path == '/')
    path++;
  return path;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<getcomponent>:
  endbr64
  movzbl (%rdi),%edx
  mov    %rdi,%rax
  test   %dl,%dl
  je     L66
  cmp    $0x2f,%dl
  je     L12
  nopl   0x0(%rax,%rax,1)
L34:
  add    $0x1,%rax
  mov    %dl,(%rsi)
  add    $0x1,%rsi
  movzbl (%rax),%edx
  cmp    $0x2f,%dl
  je     L33
  test   %dl,%dl
  jne    L34
L33:
  movb   $0x0,(%rsi)
  cmpb   $0x2f,(%rax)
  jne    L22
  cs nopw 0x0(%rax,%rax,1)
L43:
  add    $0x1,%rax
  cmpb   $0x2f,(%rax)
  je     L43
  ret
  nopw   0x0(%rax,%rax,1)
L66:
  xor    %eax,%eax
L22:
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  movb   $0x2f,(%rsi)
  add    $0x1,%rsi
  jmp    L33
```

assistant: ```c
char *getcomponent(char *path, char *name) {
  if (*path == '\0')
    return 0;
  if (*path == '/') {
    *name++ = '/';
  } else {
    do {
      *name++ = *path++;
    } while (*path != '/' && *path != '\0');
  }
  *name = '\0';
  while (*path == '/')
    path++;
  return path;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getcomponent>:
  endbr64
  mov    %rdi,%rax
  mov    %rsi,%rdi
  movzbl (%rax),%edx
  test   %dl,%dl
  je     L43
  cmp    $0x2f,%dl
  je     L33
  cs nopw 0x0(%rax,%rax,1)
L23:
  mov    %rax,%rsi
  movsb  %ds:(%rsi),%es:(%rdi)
  movzbl (%rsi),%edx
  mov    %rsi,%rax
  cmp    $0x2f,%dl
  je     L5
  test   %dl,%dl
  jne    L23
L5:
  movb   $0x0,(%rdi)
  cmpb   $0x2f,(%rax)
  jne    L47
  nopl   0x0(%rax,%rax,1)
L6:
  add    $0x1,%rax
  cmpb   $0x2f,(%rax)
  je     L6
  ret
  nopw   0x0(%rax,%rax,1)
L43:
  xor    %eax,%eax
L47:
  ret
  nopl   0x0(%rax,%rax,1)
L33:
  movb   $0x2f,(%rsi)
  add    $0x1,%rdi
  jmp    L5
```

assistant: ```c
char *getcomponent(char *path, char *name) {
  if (*path == '\0')
    return 0;
  if (*path == '/') {
    *name++ = '/';
  } else {
    do {
      *name++ = *path++;
    } while (*path != '/' && *path != '\0');
  }
  *name = '\0';
  while (*path == '/')
    path++;
  return path;
}
```