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
<check_funcptr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D7(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L19
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  jmp    L2
L19:
  mov    $0x0,%eax
L2:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D7
datatype: string

tool: D7: (*)

assistant: ```c
const char *check_funcptr(const char *type) {
  const char *str = strstr(type, "(*)");
  if (str) {
    return str + 2;
  }
  return NULL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<check_funcptr>:
  endbr64
  sub    $0x8,%rsp
  lea    D4(%rip),%rsi
  call   <strstr@plt>
  lea    0x2(%rax),%rdx
  test   %rax,%rax
  cmovne %rdx,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
type: string

tool: D4: (*)

assistant: ```c
const char *check_funcptr(const char *type) {
  const char *str = strstr(type, "(*)");
  if (str) {
    return str + 2;
  }
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<check_funcptr>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rsi
  call   <strstr@plt>
  test   %rax,%rax
  lea    0x2(%rax),%rdx
  cmovne %rdx,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D5
datatype: string

tool: D5: (*)

assistant: ```c
const char *check_funcptr(const char *type) {
  const char *str = strstr(type, "(*)");
  if (str) {
    return str + 2;
  }
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<check_funcptr>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rsi
  call   <strstr@plt>
  test   %rax,%rax
  lea    0x2(%rax),%rdx
  cmovne %rdx,%rax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
data_type: string

tool: D7: (*)

assistant: ```c
const char *check_funcptr(const char *type) {
  const char *str = strstr(type, "(*)");
  if (str) {
    return str + 2;
  }
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<knapSack>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r14
  mov    %rsi,0x18(%rsp)
  mov    %rdx,%r15
  mov    %rcx,0x10(%rsp)
  lea    0x8(,%rsi,8),%rax
  mov    %rax,0x20(%rsp)
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,0x28(%rsp)
  test   %r14,%r14
  js     L31
  lea    0x4(,%r12,4),%r13
  mov    %rax,0x8(%rsp)
  mov    %rax,%rbp
  mov    $0x0,%ebx
L66:
  mov    %r13,%rdi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  add    $0x1,%rbx
  add    $0x8,%rbp
  cmp    %rbx,%r14
  jge    L66
  mov    $0x0,%ebp
  jmp    L38
L29:
  mov    0x0(%r13),%rax
  movl   $0x0,(%rax,%rbx,4)
L2:
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jl     L83
L113:
  test   %rbp,%rbp
  je     L29
  test   %rbx,%rbx
  je     L29
  movslq -0x4(%r15,%rbp,4),%rax
  cmp    %rbx,%rax
  jle    L54
  mov    (%r14),%rax
  mov    (%rax,%rbx,4),%edx
  mov    0x0(%r13),%rax
  mov    %edx,(%rax,%rbx,4)
  jmp    L2
L54:
  mov    (%r14),%rdx
  mov    0x0(%r13),%rcx
  lea    (%rcx,%rbx,4),%rsi
  mov    %rsi,(%rsp)
  mov    (%rdx,%rbx,4),%esi
  mov    %rbx,%rcx
  sub    %rax,%rcx
  mov    0x10(%rsp),%rax
  mov    -0x4(%rax,%rbp,4),%edi
  add    (%rdx,%rcx,4),%edi
  call   <max@plt>
  mov    (%rsp),%rsi
  mov    %eax,(%rsi)
  jmp    L2
L83:
  add    $0x1,%rbp
  addq   $0x8,0x8(%rsp)
  cmp    %rbp,0x18(%rsp)
  jl     L31
L38:
  mov    0x8(%rsp),%r13
  lea    -0x8(%r13),%r14
  mov    $0x0,%ebx
  test   %r12,%r12
  jns    L113
  jmp    L83
L31:
  mov    0x20(%rsp),%rax
  mov    0x28(%rsp),%rdi
  mov    -0x8(%rax,%rdi,1),%rax
  mov    (%rax,%r12,4),%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int knapSack(long int W, long int N, int wt[], int val[]) {
  int **K;
  long int i, w;
  K = (int **)malloc((N + 1) * sizeof(int *));
  for (i = 0; i <= N; i++)
    K[i] = (int *)malloc((W + 1) * sizeof(int));
  for (i = 0; i <= N; i++) {
    for (w = 0; w <= W; w++) {
      if (i == 0 || w == 0)
        K[i][w] = 0;
      else if (wt[i - 1] <= w)
        K[i][w] = max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]);
      else
        K[i][w] = K[i - 1][w];
    }
  }
  return K[N][W];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<knapSack>:
  endbr64
  push   %r15
  lea    0x8(,%rsi,8),%rax
  mov    %rdi,%r15
  push   %r14
  mov    %rax,%rdi
  mov    %rcx,%r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x38,%rsp
  mov    %rsi,0x18(%rsp)
  mov    %rax,0x20(%rsp)
  call   <malloc@plt>
  mov    %rax,0x28(%rsp)
  test   %rbx,%rbx
  js     L38
  mov    %rax,(%rsp)
  mov    %rbx,%rsi
  xor    %r13d,%r13d
  mov    %rax,%rbp
  lea    0x4(,%r15,4),%rcx
  mov    %r12,0x8(%rsp)
  mov    %r13,%r12
  mov    %rsi,%r13
  mov    %rcx,%rbx
  nopl   0x0(%rax)
L96:
  mov    %rbx,%rdi
  add    $0x1,%r12
  add    $0x8,%rbp
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmp    %r12,%r13
  jge    L96
  mov    (%rsp),%rbp
  mov    0x8(%rsp),%r12
  xor    %r13d,%r13d
  mov    %r14,%r10
  nopl   0x0(%rax,%rax,1)
L66:
  test   %r15,%r15
  js     L116
  test   %r13,%r13
  mov    %r12,%r9
  lea    -0x8(%rbp),%r8
  sete   %r14b
  xor    %ebx,%ebx
  mov    %r14d,%r12d
  jmp    L12
  nopw   0x0(%rax,%rax,1)
L94:
  mov    (%r8),%rdx
  movslq -0x4(%r9,%r13,4),%rcx
  lea    (%rax,%rbx,4),%r14
  mov    (%rdx,%rbx,4),%esi
  cmp    %rbx,%rcx
  jle    L75
  mov    %esi,(%r14)
L48:
  add    $0x1,%rbx
  cmp    %rbx,%r15
  jl     L60
L12:
  mov    0x0(%rbp),%rax
  test   %rbx,%rbx
  je     L74
  test   %r12b,%r12b
  je     L94
L74:
  movl   $0x0,(%rax,%rbx,4)
  add    $0x1,%rbx
  cmp    %rbx,%r15
  jge    L12
L60:
  mov    %r9,%r12
L116:
  lea    0x1(%r13),%rax
  add    $0x8,%rbp
  cmp    %r13,0x18(%rsp)
  je     L38
  mov    %rax,%r13
  jmp    L66
  nopl   (%rax)
L75:
  mov    %rbx,%rax
  mov    -0x4(%r10,%r13,4),%edi
  mov    %r9,0x10(%rsp)
  sub    %rcx,%rax
  mov    %r8,0x8(%rsp)
  add    (%rdx,%rax,4),%edi
  mov    %r10,(%rsp)
  call   <max@plt>
  mov    (%rsp),%r10
  mov    0x8(%rsp),%r8
  mov    %eax,(%r14)
  mov    0x10(%rsp),%r9
  jmp    L48
  nopl   0x0(%rax)
L38:
  mov    0x28(%rsp),%rax
  mov    0x20(%rsp),%rdx
  mov    -0x8(%rax,%rdx,1),%rax
  mov    (%rax,%r15,4),%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int knapSack(long int W, long int N, int wt[], int val[]) {
  int **K;
  long int i, w;
  K = (int **)malloc((N + 1) * sizeof(int *));
  for (i = 0; i <= N; i++)
    K[i] = (int *)malloc((W + 1) * sizeof(int));
  for (i = 0; i <= N; i++) {
    for (w = 0; w <= W; w++) {
      if (i == 0 || w == 0)
        K[i][w] = 0;
      else if (wt[i - 1] <= w)
        K[i][w] = max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]);
      else
        K[i][w] = K[i - 1][w];
    }
  }
  return K[N][W];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<knapSack>:
  endbr64
  push   %r15
  lea    0x8(,%rsi,8),%rax
  mov    %rdi,%r15
  push   %r14
  mov    %rax,%rdi
  mov    %rcx,%r14
  push   %r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x38,%rsp
  mov    %rsi,0x18(%rsp)
  mov    %rax,0x20(%rsp)
  call   <malloc@plt>
  mov    %rax,0x28(%rsp)
  test   %rbx,%rbx
  js     L76
  mov    %rax,(%rsp)
  xor    %r13d,%r13d
  lea    0x4(,%r15,4),%rcx
  mov    %r12,0x8(%rsp)
  mov    %r13,%rbp
  mov    %rax,%r12
  mov    %rbx,%r13
  mov    %rcx,%rbx
  cs nopw 0x0(%rax,%rax,1)
L93:
  mov    %rbx,%rdi
  add    $0x1,%rbp
  add    $0x8,%r12
  call   <malloc@plt>
  mov    %rax,-0x8(%r12)
  cmp    %rbp,%r13
  jge    L93
  mov    (%rsp),%rbp
  mov    0x8(%rsp),%r12
  xor    %r13d,%r13d
  mov    %r14,%r10
  test   %r15,%r15
  js     L76
  nopw   0x0(%rax,%rax,1)
L95:
  test   %r13,%r13
  mov    %r12,%r9
  lea    -0x8(%rbp),%r8
  sete   %r14b
  xor    %ebx,%ebx
  mov    %r14d,%r12d
  jmp    L40
  nopl   (%rax)
L92:
  mov    (%r8),%rdx
  movslq -0x4(%r9,%r13,4),%rcx
  lea    (%rax,%rbx,4),%r14
  mov    (%rdx,%rbx,4),%esi
  cmp    %rbx,%rcx
  jle    L44
  mov    %esi,(%r14)
L87:
  add    $0x1,%rbx
  cmp    %rbx,%r15
  jl     L36
L40:
  mov    0x0(%rbp),%rax
  test   %rbx,%rbx
  je     L66
  test   %r12b,%r12b
  je     L92
L66:
  movl   $0x0,(%rax,%rbx,4)
  add    $0x1,%rbx
  cmp    %rbx,%r15
  jge    L40
L36:
  mov    %r9,%r12
  lea    0x1(%r13),%rax
  add    $0x8,%rbp
  cmp    %r13,0x18(%rsp)
  je     L76
  mov    %rax,%r13
  jmp    L95
  nopl   (%rax)
L44:
  mov    %rbx,%rax
  mov    -0x4(%r10,%r13,4),%edi
  mov    %r9,0x10(%rsp)
  sub    %rcx,%rax
  mov    %r8,0x8(%rsp)
  add    (%rdx,%rax,4),%edi
  mov    %r10,(%rsp)
  call   <max@plt>
  mov    (%rsp),%r10
  mov    0x8(%rsp),%r8
  mov    %eax,(%r14)
  mov    0x10(%rsp),%r9
  jmp    L87
  nopl   0x0(%rax)
L76:
  mov    0x28(%rsp),%rax
  mov    0x20(%rsp),%rdx
  mov    -0x8(%rax,%rdx,1),%rax
  mov    (%rax,%r15,4),%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int knapSack(long int W, long int N, int wt[], int val[]) {
  int **K;
  long int i, w;
  K = (int **)malloc((N + 1) * sizeof(int *));
  for (i = 0; i <= N; i++)
    K[i] = (int *)malloc((W + 1) * sizeof(int));
  for (i = 0; i <= N; i++) {
    for (w = 0; w <= W; w++) {
      if (i == 0 || w == 0)
        K[i][w] = 0;
      else if (wt[i - 1] <= w)
        K[i][w] = max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]);
      else
        K[i][w] = K[i - 1][w];
    }
  }
  return K[N][W];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<SolveScalarWarning>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x400,%rsp
  mov    %rdi,-0x3f8(%rbp)
  movsd  %xmm0,-0x400(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x400(%rbp),%rcx
  mov    -0x3f8(%rbp),%rdx
  lea    -0x3f0(%rbp),%rax
  movq   %rcx,%xmm0
  lea    D8(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sprintf@plt>
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rdi
  call   <DymosimMessage@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D8
datatype: string

tool: D8: Singular scalar system. Using minimum norm solution for %.400s = %g


assistant: ```c
void SolveScalarWarning(const char *xs, double x) {
  char s[1000];
  sprintf(s, "Singular scalar system. Using minimum norm solution for %.400s = %g\n", xs, x);
  DymosimMessage(s);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<SolveScalarWarning>:
  endbr64
  push   %rbx
  sub    $0x3f0,%rsp
  mov    %rdi,%rdx
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    D2(%rip),%rsi
  mov    %rbx,%rdi
  call   <sprintf@plt>
  mov    %rbx,%rdi
  call   <DymosimMessage@plt>
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x3f0,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D2
datatype: string

tool: D2: Singular scalar system. Using minimum norm solution for %.400s = %g


assistant: ```c
void SolveScalarWarning(const char *xs, double x) {
  char s[1000];
  sprintf(s, "Singular scalar system. Using minimum norm solution for %.400s = %g\n", xs, x);
  DymosimMessage(s);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<SolveScalarWarning>:
  endbr64
  push   %rbp
  mov    %rdi,%rdx
  lea    D7(%rip),%rsi
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <DymosimMessage@plt>
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x3f0,%rsp
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D7
data_type: string

tool: D7: Singular scalar system. Using minimum norm solution for %.400s = %g


assistant: ```c
void SolveScalarWarning(const char *xs, double x) {
  char s[1000];
  sprintf(s, "Singular scalar system. Using minimum norm solution for %.400s = %g\n", xs, x);
  DymosimMessage(s);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<SolveScalarWarning>:
  endbr64
  push   %rbp
  mov    %rdi,%rdx
  lea    D1(%rip),%rsi
  sub    $0x3f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <sprintf@plt>
  mov    %rbp,%rdi
  call   <DymosimMessage@plt>
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x3f0,%rsp
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D1
type: string

tool: D1: Singular scalar system. Using minimum norm solution for %.400s = %g


assistant: ```c
void SolveScalarWarning(const char *xs, double x) {
  char s[1000];
  sprintf(s, "Singular scalar system. Using minimum norm solution for %.400s = %g\n", xs, x);
  DymosimMessage(s);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<nmalloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L7
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <die@plt>
L7:
  cmpq   $0x0,-0x8(%rbp)
  jne    L19
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <die@plt>
L19:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D14
type: string
name: extract_symbolic_data
arguments: tag: D7
type: string

tool: D14: Allocating zero bytes.  Please report a bug.


tool: D7: Nano is out of memory!


assistant: ```c
void *nmalloc(size_t howmuch) {
  void *r = malloc(howmuch);
  if (howmuch == 0)
    die("Allocating zero bytes.  Please report a bug.\n");
  if (r == NULL)
    die(("Nano is out of memory!\n"));
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<nmalloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rbp,%rbp
  je     L1
L32:
  test   %rbx,%rbx
  je     L27
L16:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  lea    D12(%rip),%rdi
  call   <die@plt>
  jmp    L32
L27:
  lea    D9(%rip),%rdi
  call   <die@plt>
  jmp    L16
```

tool_calls: name: read_data
arguments: tag: D12
type_guessed: string
name: read_data
arguments: tag: D9
type_guessed: string

tool: D12: Allocating zero bytes.  Please report a bug.


tool: D9: Nano is out of memory!


assistant: ```c
void *nmalloc(size_t howmuch) {
  void *r = malloc(howmuch);
  if (howmuch == 0)
    die("Allocating zero bytes.  Please report a bug.\n");
  if (r == NULL)
    die(("Nano is out of memory!\n"));
  return r;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<nmalloc>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rbx,%rbx
  je     L10
L4:
  test   %r12,%r12
  je     L1
L25:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  lea    D4(%rip),%rdi
  call   <die@plt>
  jmp    L4
L1:
  lea    D17(%rip),%rdi
  call   <die@plt>
  jmp    L25
```

tool_calls: name: parse_data
arguments: data_label: D4
type_infered: string
name: parse_data
arguments: data_label: D17
type_infered: string

tool: D4: Allocating zero bytes.  Please report a bug.


tool: D17: Nano is out of memory!


assistant: ```c
void *nmalloc(size_t howmuch) {
  void *r = malloc(howmuch);
  if (howmuch == 0)
    die("Allocating zero bytes.  Please report a bug.\n");
  if (r == NULL)
    die(("Nano is out of memory!\n"));
  return r;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<nmalloc>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rbx,%rbx
  je     L26
L39:
  test   %r12,%r12
  je     L24
L10:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L26:
  lea    D13(%rip),%rdi
  call   <die@plt>
  jmp    L39
L24:
  lea    D12(%rip),%rdi
  call   <die@plt>
  jmp    L10
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
data_type: string
name: retrieve_data
arguments: data_tag: D12
data_type: string

tool: D13: Allocating zero bytes.  Please report a bug.


tool: D12: Nano is out of memory!


assistant: ```c
void *nmalloc(size_t howmuch) {
  void *r = malloc(howmuch);
  if (howmuch == 0)
    die("Allocating zero bytes.  Please report a bug.\n");
  if (r == NULL)
    die(("Nano is out of memory!\n"));
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<b_subtractDigits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x3c(%rbp)
  mov    %r8d,-0x40(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L62
L77:
  mov    -0x40(%rbp),%eax
  sub    $0x1,%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x3c(%rbp),%eax
  sub    $0x1,%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L72
  addl   $0xa,-0x10(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L6
L72:
  movl   $0x0,-0x18(%rbp)
L6:
  mov    -0x8(%rbp),%eax
  sub    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L62:
  mov    -0xc(%rbp),%eax
  cmp    -0x40(%rbp),%eax
  jl     L77
  subl   $0x1,-0x14(%rbp)
  jmp    L49
L40:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jge    L41
  addl   $0xa,-0x10(%rbp)
  movl   $0x1,-0x18(%rbp)
  jmp    L23
L41:
  movl   $0x0,-0x18(%rbp)
L23:
  mov    -0x8(%rbp),%eax
  sub    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x38(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  mov    %eax,(%rdx)
L49:
  cmpl   $0x0,-0x14(%rbp)
  jns    L40
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void b_subtractDigits(unsigned *lrg, unsigned *sml, unsigned *buf, int lrg_size, int sml_size) {
  int carry = 0, j, a, b;
  for (int i = 0; i < sml_size; i++) {
    int k = sml_size - 1 - i;
    j = lrg_size - 1 - i;
    a = lrg[j];
    b = sml[k] + carry;
    if (a < b) {
      a += 10;
      carry = 1;
    } else
      carry = 0;
    a -= b;
    buf[j] = a;
  }
  j--;
  while (j >= 0) {
    a = lrg[j];
    b = carry;
    if (a < b) {
      a += 10;
      carry = 1;
    } else
      carry = 0;
    a -= b;
    buf[j--] = a;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<b_subtractDigits>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%r9
  test   %r8d,%r8d
  jle    L79
  mov    %ecx,%r12d
  movslq %ecx,%rcx
  lea    -0x4(,%rcx,4),%rax
  lea    -0x1(%r8),%r10d
  mov    %rcx,%rdx
  sub    %r10,%rdx
  lea    -0x8(,%rdx,4),%rbx
  movslq %r8d,%rdx
  sub    %rcx,%rdx
  lea    (%rsi,%rdx,4),%r11
  mov    $0x0,%ecx
  mov    $0x0,%r10d
  mov    $0x1,%ebp
  jmp    L68
L91:
  sub    %esi,%edx
  mov    %edx,(%r9,%rax,1)
  sub    $0x4,%rax
  cmp    %rbx,%rax
  je     L11
L68:
  mov    (%rdi,%rax,1),%edx
  mov    %ecx,%esi
  add    (%r11,%rax,1),%esi
  mov    %r10d,%ecx
  cmp    %esi,%edx
  jge    L91
  add    $0xa,%edx
  mov    %ebp,%ecx
  jmp    L91
L11:
  mov    %r12d,%eax
  sub    %r8d,%eax
L102:
  sub    $0x1,%eax
  js     L100
  movslq %eax,%rdx
  mov    $0x0,%r8d
  mov    $0x1,%r10d
  jmp    L78
L79:
  mov    $0x0,%ecx
  jmp    L102
L66:
  sub    %ecx,%eax
  mov    %eax,(%r9,%rdx,4)
  sub    $0x1,%rdx
  test   %edx,%edx
  js     L100
  mov    %esi,%ecx
L78:
  mov    (%rdi,%rdx,4),%eax
  mov    %r8d,%esi
  cmp    %ecx,%eax
  jge    L66
  add    $0xa,%eax
  mov    %r10d,%esi
  jmp    L66
L100:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void b_subtractDigits(unsigned *lrg, unsigned *sml, unsigned *buf, int lrg_size, int sml_size) {
  int carry = 0, j, a, b;
  for (int i = 0; i < sml_size; i++) {
    int k = sml_size - 1 - i;
    j = lrg_size - 1 - i;
    a = lrg[j];
    b = sml[k] + carry;
    if (a < b) {
      a += 10;
      carry = 1;
    } else
      carry = 0;
    a -= b;
    buf[j] = a;
  }
  j--;
  while (j >= 0) {
    a = lrg[j];
    b = carry;
    if (a < b) {
      a += 10;
      carry = 1;
    } else
      carry = 0;
    a -= b;
    buf[j--] = a;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<b_subtractDigits>:
  endbr64
  mov    %rdx,%r9
  test   %r8d,%r8d
  jle    L52
  movslq %ecx,%rdx
  lea    -0x1(%rcx),%eax
  lea    -0x1(%r8),%r10d
  push   %rbx
  mov    %ecx,%ebx
  mov    %rdx,%rcx
  cltq
  sub    %r10,%rcx
  shl    $0x2,%rax
  lea    -0x8(,%rcx,4),%r11
  movslq %r8d,%rcx
  sub    %rdx,%rcx
  lea    (%rsi,%rcx,4),%r10
  xor    %ecx,%ecx
  nopl   (%rax)
L10:
  mov    (%r10,%rax,1),%esi
  mov    (%rdi,%rax,1),%edx
  add    %ecx,%esi
  xor    %ecx,%ecx
  cmp    %esi,%edx
  jge    L20
  add    $0xa,%edx
  mov    $0x1,%ecx
L20:
  sub    %esi,%edx
  mov    %edx,(%r9,%rax,1)
  sub    $0x4,%rax
  cmp    %rax,%r11
  jne    L10
  mov    %ebx,%eax
  sub    %r8d,%eax
  sub    $0x1,%eax
  movslq %eax,%rdx
  jns    L38
  jmp    L88
  nopl   (%rax)
L108:
  mov    %esi,%ecx
L38:
  mov    (%rdi,%rdx,4),%eax
  xor    %esi,%esi
  cmp    %ecx,%eax
  jge    L62
  add    $0xa,%eax
  mov    $0x1,%esi
L62:
  sub    %ecx,%eax
  mov    %eax,(%r9,%rdx,4)
  sub    $0x1,%rdx
  test   %edx,%edx
  jns    L108
L88:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L52:
  xor    %ecx,%ecx
  sub    $0x1,%eax
  movslq %eax,%rdx
  jns    L87
  jmp    L79
  nopl   0x0(%rax)
L97:
  mov    %esi,%ecx
L87:
  mov    (%rdi,%rdx,4),%eax
  xor    %esi,%esi
  cmp    %ecx,%eax
  jge    L24
  add    $0xa,%eax
  mov    $0x1,%esi
L24:
  sub    %ecx,%eax
  mov    %eax,(%r9,%rdx,4)
  sub    $0x1,%rdx
  test   %edx,%edx
  jns    L97
  ret
L79:
  ret
```

assistant: ```c
void b_subtractDigits(unsigned *lrg, unsigned *sml, unsigned *buf, int lrg_size, int sml_size) {
  int carry = 0, j, a, b;
  for (int i = 0; i < sml_size; i++) {
    int k = sml_size - 1 - i;
    j = lrg_size - 1 - i;
    a = lrg[j];
    b = sml[k] + carry;
    if (a < b) {
      a += 10;
      carry = 1;
    } else
      carry = 0;
    a -= b;
    buf[j] = a;
  }
  j--;
  while (j >= 0) {
    a = lrg[j];
    b = carry;
    if (a < b) {
      a += 10;
      carry = 1;
    } else
      carry = 0;
    a -= b;
    buf[j--] = a;
  }
}
```