ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的？

```asm
<insertion_sort_r>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rcx,%r15
  mov    %r8,%rbp
  lea    0x1(%rdx),%r14
  cmp    %rcx,%r14
  jbe    L44
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L28:
  add    $0x1,%r14
  cmp    %r14,%r15
  jb     L3
L44:
  mov    %r14,%rdx
  test   %r14,%r14
  je     L28
L9:
  mov    -0x8(%rbp,%rdx,8),%rax
  cmp    %rax,0x0(%rbp,%rdx,8)
  jae    L28
  lea    -0x1(%rdx),%rbx
  mov    %rbp,%r8
  mov    %rbx,%rcx
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <swap@plt>
  test   %rbx,%rbx
  je     L28
  mov    %rbx,%rdx
  jmp    L9
```

assistant: ```c
void insertion_sort_r(char *base, size_t es, size_t lo, size_t hi, size_t *index) {
  for (++lo; lo <= hi; ++lo)
    for (size_t j = lo; j > 0 && index[j] < index[j - 1]; --j)
      swap(base, es, j, j - 1, index);
}
```
user: What is the c source code of the assembly code below:

```asm
<insertion_sort_r>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    0x1(%rdx),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    %rcx,%r13
  ja     L23
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rcx,%r14
  mov    %r8,%r15
  nopw   0x0(%rax,%rax,1)
L58:
  mov    %r13,%rdx
  test   %r13,%r13
  jne    L55
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L43:
  lea    -0x1(%rdx),%rbx
  mov    %r15,%r8
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    %rbx,%rcx
  call   <swap@plt>
  test   %rbx,%rbx
  je     L0
  mov    %rbx,%rdx
L55:
  mov    -0x8(%r15,%rdx,8),%rax
  cmp    %rax,(%r15,%rdx,8)
  jb     L43
L0:
  add    $0x1,%r13
  cmp    %r13,%r14
  jae    L58
L23:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void insertion_sort_r(char *base, size_t es, size_t lo, size_t hi, size_t *index) {
  for (++lo; lo <= hi; ++lo)
    for (size_t j = lo; j > 0 && index[j] < index[j - 1]; --j)
      swap(base, es, j, j - 1, index);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<insertion_sort_r>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  lea    0x1(%rdx),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    %rcx,%r13
  ja     L10
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    %rcx,%r14
  mov    %r8,%r15
  nopw   0x0(%rax,%rax,1)
L30:
  mov    %r13,%rdx
  test   %r13,%r13
  jne    L42
  jmp    L35
  nopw   0x0(%rax,%rax,1)
L5:
  lea    -0x1(%rdx),%rbx
  mov    %r15,%r8
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    %rbx,%rcx
  call   <swap@plt>
  test   %rbx,%rbx
  je     L35
  mov    %rbx,%rdx
L42:
  mov    -0x8(%r15,%rdx,8),%rax
  cmp    %rax,(%r15,%rdx,8)
  jb     L5
L35:
  add    $0x1,%r13
  cmp    %r13,%r14
  jae    L30
L10:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void insertion_sort_r(char *base, size_t es, size_t lo, size_t hi, size_t *index) {
  for (++lo; lo <= hi; ++lo)
    for (size_t j = lo; j > 0 && index[j] < index[j - 1]; --j)
      swap(base, es, j, j - 1, index);
}
```
user: Translate the below assembly code into C programming language:

```asm
<checkNum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L44
L14:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L35
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jle    L31
L35:
  mov    $0x0,%eax
  jmp    L20
L31:
  nop
  addl   $0x1,-0x8(%rbp)
L44:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L14
  mov    $0x1,%eax
L20:
  leave
  ret
```

assistant: ```c
int checkNum(char *hex) {
  int len = strlen(hex);
  for (int i = 0; i < len; i++) {
    if (hex[i] >= '0' && hex[i] <= '9')
      continue;
    return 0;
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<checkNum>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L44
  mov    %rbx,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rcx
L9:
  movzbl (%rdx),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L49
  mov    $0x0,%eax
L30:
  pop    %rbx
  ret
L49:
  add    $0x1,%rdx
  cmp    %rcx,%rdx
  jne    L9
  mov    $0x1,%eax
  jmp    L30
L44:
  mov    $0x1,%eax
  jmp    L30
```

assistant: ```c
int checkNum(char *hex) {
  int len = strlen(hex);
  for (int i = 0; i < len; i++) {
    if (hex[i] >= '0' && hex[i] <= '9')
      continue;
    return 0;
  }
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<checkNum>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L10
  sub    $0x1,%eax
  mov    %rbx,%rdi
  lea    0x1(%rbx,%rax,1),%rdx
L27:
  movzbl (%rdi),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L17
  xor    %eax,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  add    $0x1,%rdi
  cmp    %rdx,%rdi
  jne    L27
L10:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int checkNum(char *hex) {
  int len = strlen(hex);
  for (int i = 0; i < len; i++) {
    if (hex[i] >= '0' && hex[i] <= '9')
      continue;
    return 0;
  }
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<checkNum>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L8
  sub    $0x1,%eax
  mov    %rbx,%rdi
  lea    0x1(%rbx,%rax,1),%rdx
L0:
  movzbl (%rdi),%eax
  sub    $0x30,%eax
  cmp    $0x9,%al
  jbe    L29
  xor    %eax,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L29:
  add    $0x1,%rdi
  cmp    %rdx,%rdi
  jne    L0
L8:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int checkNum(char *hex) {
  int len = strlen(hex);
  for (int i = 0; i < len; i++) {
    if (hex[i] >= '0' && hex[i] <= '9')
      continue;
    return 0;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<test_example_main_no_args>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D13(%rip),%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <example_main@plt>
  mov    %eax,%edi
  mov    $0x17a,%ecx
  lea    D0(%rip),%rax
  mov    %rax,%rdx
  mov    $0x0,%esi
  call   <_assert_int_equal@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D13
type: string
name: parse_data
arguments: tag: D0
type: string

tool: D13: example

tool: D0: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/BellerophonMobile/xmp3/refs/heads/master/deps/cmockery/src/example/calculator_test.c

assistant: ```c
void test_example_main_no_args(void **state) {
  char *args[] = {
      "example",
  };
  _assert_int_equal(((unsigned long long)((unsigned)(example_main((sizeof(args) / sizeof((args)[0])), args)))), ((unsigned long long)((unsigned)(0))), "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/BellerophonMobile/xmp3/refs/heads/master/deps/cmockery/src/example/calculator_test.c", 378);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test_example_main_no_args>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D1(%rip),%rax
  mov    %rax,(%rsp)
  mov    %rsp,%rsi
  mov    $0x1,%edi
  call   <example_main@plt>
  mov    %eax,%edi
  mov    $0x17a,%ecx
  lea    D5(%rip),%rdx
  mov    $0x0,%esi
  call   <_assert_int_equal@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D1
datatype: string
name: unpack_data
arguments: label: D5
datatype: string

tool: D1: example

tool: D5: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/BellerophonMobile/xmp3/refs/heads/master/deps/cmockery/src/example/calculator_test.c

assistant: ```c
void test_example_main_no_args(void **state) {
  char *args[] = {
      "example",
  };
  _assert_int_equal(((unsigned long long)((unsigned)(example_main((sizeof(args) / sizeof((args)[0])), args)))), ((unsigned long long)((unsigned)(0))), "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/BellerophonMobile/xmp3/refs/heads/master/deps/cmockery/src/example/calculator_test.c", 378);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<test_example_main_no_args>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x1,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  lea    D19(%rip),%rax
  mov    %rsp,%rsi
  mov    %rax,(%rsp)
  call   <example_main@plt>
  xor    %esi,%esi
  mov    $0x17a,%ecx
  lea    D4(%rip),%rdx
  mov    %eax,%edi
  call   <_assert_int_equal@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D19
data_type: string
name: parse_data
arguments: data_tag: D4
data_type: string

tool: D19: example

tool: D4: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/BellerophonMobile/xmp3/refs/heads/master/deps/cmockery/src/example/calculator_test.c

assistant: ```c
void test_example_main_no_args(void **state) {
  char *args[] = {
      "example",
  };
  _assert_int_equal(((unsigned long long)((unsigned)(example_main((sizeof(args) / sizeof((args)[0])), args)))), ((unsigned long long)((unsigned)(0))), "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/BellerophonMobile/xmp3/refs/heads/master/deps/cmockery/src/example/calculator_test.c", 378);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test_example_main_no_args>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x1,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  lea    D8(%rip),%rax
  mov    %rsp,%rsi
  mov    %rax,(%rsp)
  call   <example_main@plt>
  xor    %esi,%esi
  mov    $0x17a,%ecx
  lea    D13(%rip),%rdx
  mov    %eax,%edi
  call   <_assert_int_equal@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D8
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D13
type_guessed: string

tool: D8: example

tool: D13: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/BellerophonMobile/xmp3/refs/heads/master/deps/cmockery/src/example/calculator_test.c

assistant: ```c
void test_example_main_no_args(void **state) {
  char *args[] = {
      "example",
  };
  _assert_int_equal(((unsigned long long)((unsigned)(example_main((sizeof(args) / sizeof((args)[0])), args)))), ((unsigned long long)((unsigned)(0))), "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/BellerophonMobile/xmp3/refs/heads/master/deps/cmockery/src/example/calculator_test.c", 378);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sig>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    $0x0,%eax
  call   <drop_all_storage@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void sig(int num) {
  drop_all_storage();
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sig>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <drop_all_storage@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void sig(int num) {
  drop_all_storage();
  exit(0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<sig>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <drop_all_storage@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void sig(int num) {
  drop_all_storage();
  exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<sig>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <drop_all_storage@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void sig(int num) {
  drop_all_storage();
  exit(0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<check_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x4(%rbp)
  jmp    L2
L18:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L21
  mov    $0x1,%eax
  jmp    L33
L21:
  addl   $0x1,-0x4(%rbp)
L2:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L18
  mov    $0x0,%eax
L33:
  pop    %rbp
  ret
```

assistant: ```c
int check_prime(int num) {
  int i = 2;
  while (i < num) {
    if (num % i == 0)
      return 1;
    i = i + 1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<check_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L18
  test   $0x1,%dil
  je     L10
  mov    $0x2,%ecx
L15:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  je     L25
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L15
  mov    $0x1,%eax
  ret
L25:
  mov    $0x0,%eax
  ret
L18:
  mov    $0x0,%eax
  ret
L10:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int check_prime(int num) {
  int i = 2;
  while (i < num) {
    if (num % i == 0)
      return 1;
    i = i + 1;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<check_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L16
  test   $0x1,%dil
  je     L31
  mov    $0x2,%ecx
  jmp    L27
  cs nopw 0x0(%rax,%rax,1)
L6:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L31
L27:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L6
L16:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L31:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int check_prime(int num) {
  int i = 2;
  while (i < num) {
    if (num % i == 0)
      return 1;
    i = i + 1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<check_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L12
  test   $0x1,%dil
  je     L40
  mov    $0x2,%ecx
  jmp    L36
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L40
L36:
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L1
L12:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L40:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int check_prime(int num) {
  int i = 2;
  while (i < num) {
    if (num % i == 0)
      return 1;
    i = i + 1;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_transition_matrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  mov    $0x4,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    -0x18(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x18(%rbp),%rax
  lea    0x8(%rax),%rbx
  mov    $0x4,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  lea    0x10(%rax),%rbx
  mov    $0x4,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,(%rbx)
  mov    -0x18(%rbp),%rax
  lea    0x18(%rax),%rbx
  mov    $0x4,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,(%rbx)
  movl   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  jmp    L29
L34:
  mov    -0x28(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0x24(%rbp)
  cmpl   $0xffffffff,-0x24(%rbp)
  je     L29
  cmpq   $0x0,-0x20(%rbp)
  je     L25
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
L25:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,-0x20(%rbp)
L29:
  mov    -0x3c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x3c(%rbp)
  test   %eax,%eax
  jg     L34
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int **get_transition_matrix(char *sample, int samplesize) {
  int **transitions = malloc(sizeof(int *) * 4);
  transitions[0] = calloc(sizeof(int), 4);
  transitions[1] = calloc(sizeof(int), 4);
  transitions[2] = calloc(sizeof(int), 4);
  transitions[3] = calloc(sizeof(int), 4);
  int ctr = 0;
  int *last = 0;
  while (samplesize-- > 0) {
    int indx = sample[ctr++];
    if (indx != -1) {
      if (last != 0) {
        last[indx]++;
      }
      last = transitions[indx];
    }
  }
  return transitions;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<get_transition_matrix>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebp
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    $0x4,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,(%rbx)
  mov    $0x4,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,0x8(%rbx)
  mov    $0x4,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,0x10(%rbx)
  mov    $0x4,%esi
  mov    $0x4,%edi
  call   <calloc@plt>
  mov    %rax,0x18(%rbx)
  test   %ebp,%ebp
  jle    L1
  mov    %r12,%rdx
  lea    -0x1(%rbp),%eax
  lea    0x1(%r12,%rax,1),%rdi
  mov    $0x0,%ecx
  jmp    L7
L23:
  movsbq %al,%rax
  mov    (%rbx,%rax,8),%rcx
L37:
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  je     L1
L7:
  movzbl (%rdx),%eax
  cmp    $0xff,%al
  je     L37
  test   %rcx,%rcx
  je     L23
  movsbq %al,%rsi
  addl   $0x1,(%rcx,%rsi,4)
  jmp    L23
L1:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int **get_transition_matrix(char *sample, int samplesize) {
  int **transitions = malloc(sizeof(int *) * 4);
  transitions[0] = calloc(sizeof(int), 4);
  transitions[1] = calloc(sizeof(int), 4);
  transitions[2] = calloc(sizeof(int), 4);
  transitions[3] = calloc(sizeof(int), 4);
  int ctr = 0;
  int *last = 0;
  while (samplesize-- > 0) {
    int indx = sample[ctr++];
    if (indx != -1) {
      if (last != 0) {
        last[indx]++;
      }
      last = transitions[indx];
    }
  }
  return transitions;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<get_transition_matrix>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    $0x4,%esi
  mov    $0x4,%edi
  mov    %rax,%r12
  call   <calloc@plt>
  mov    $0x4,%esi
  mov    $0x4,%edi
  mov    %rax,(%r12)
  call   <calloc@plt>
  mov    $0x4,%esi
  mov    $0x4,%edi
  mov    %rax,0x8(%r12)
  call   <calloc@plt>
  mov    $0x4,%esi
  mov    $0x4,%edi
  mov    %rax,0x10(%r12)
  call   <calloc@plt>
  mov    %rax,0x18(%r12)
  test   %ebp,%ebp
  jle    L26
  lea    -0x1(%rbp),%eax
  mov    %rbx,%rdx
  xor    %ecx,%ecx
  lea    0x1(%rbx,%rax,1),%rsi
  nopw   0x0(%rax,%rax,1)
L33:
  movsbq (%rdx),%rax
  cmp    $0xff,%al
  je     L9
  test   %rcx,%rcx
  je     L18
  addl   $0x1,(%rcx,%rax,4)
L18:
  mov    (%r12,%rax,8),%rcx
L9:
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L33
L26:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int **get_transition_matrix(char *sample, int samplesize) {
  int **transitions = malloc(sizeof(int *) * 4);
  transitions[0] = calloc(sizeof(int), 4);
  transitions[1] = calloc(sizeof(int), 4);
  transitions[2] = calloc(sizeof(int), 4);
  transitions[3] = calloc(sizeof(int), 4);
  int ctr = 0;
  int *last = 0;
  while (samplesize-- > 0) {
    int indx = sample[ctr++];
    if (indx != -1) {
      if (last != 0) {
        last[indx]++;
      }
      last = transitions[indx];
    }
  }
  return transitions;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_transition_matrix>:
  endbr64
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    $0x4,%esi
  mov    $0x4,%edi
  mov    %rax,%r12
  call   <calloc@plt>
  mov    $0x4,%esi
  mov    $0x4,%edi
  mov    %rax,(%r12)
  call   <calloc@plt>
  mov    $0x4,%esi
  mov    $0x4,%edi
  mov    %rax,0x8(%r12)
  call   <calloc@plt>
  mov    $0x4,%esi
  mov    $0x4,%edi
  mov    %rax,0x10(%r12)
  call   <calloc@plt>
  mov    %rax,0x18(%r12)
  test   %ebp,%ebp
  jle    L28
  lea    -0x1(%rbp),%eax
  mov    %rbx,%rdx
  xor    %ecx,%ecx
  lea    0x1(%rbx,%rax,1),%rsi
  nopw   0x0(%rax,%rax,1)
L14:
  movsbq (%rdx),%rax
  cmp    $0xff,%al
  je     L31
  test   %rcx,%rcx
  je     L33
  addl   $0x1,(%rcx,%rax,4)
L33:
  mov    (%r12,%rax,8),%rcx
L31:
  add    $0x1,%rdx
  cmp    %rdx,%rsi
  jne    L14
L28:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int **get_transition_matrix(char *sample, int samplesize) {
  int **transitions = malloc(sizeof(int *) * 4);
  transitions[0] = calloc(sizeof(int), 4);
  transitions[1] = calloc(sizeof(int), 4);
  transitions[2] = calloc(sizeof(int), 4);
  transitions[3] = calloc(sizeof(int), 4);
  int ctr = 0;
  int *last = 0;
  while (samplesize-- > 0) {
    int indx = sample[ctr++];
    if (indx != -1) {
      if (last != 0) {
        last[indx]++;
      }
      last = transitions[indx];
    }
  }
  return transitions;
}
```