ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<motorHandle>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
L5:
  mov    $0x0,%eax
  call   <moveSpider@plt>
  jmp    L5
```

tool_calls: name: read_data
arguments: label: D0
type_infered: string

tool: D0: [INFO] Handle Motor Thread created. 


assistant: ```c
void *motorHandle(void *param) {
  printf("[INFO] Handle Motor Thread created. \n");
  unsigned int time = 1;
  while (1) {
    moveSpider();
    time++;
  }
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<motorHandle>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L7:
  xor    %eax,%eax
  call   <moveSpider@plt>
  jmp    L7
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
type_infered: string

tool: D3: [INFO] Handle Motor Thread created. 


assistant: ```c
void *motorHandle(void *param) {
  printf("[INFO] Handle Motor Thread created. \n");
  unsigned int time = 1;
  while (1) {
    moveSpider();
    time++;
  }
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<motorHandle>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  nopl   0x0(%rax)
L0:
  xor    %eax,%eax
  call   <moveSpider@plt>
  jmp    L0
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
data_type: string

tool: D5: [INFO] Handle Motor Thread created. 


assistant: ```c
void *motorHandle(void *param) {
  printf("[INFO] Handle Motor Thread created. \n");
  unsigned int time = 1;
  while (1) {
    moveSpider();
    time++;
  }
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<np>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L3
  mov    -0x8(%rbp),%rax
  jmp    L16
L3:
  lea    D9(%rip),%rax
L16:
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D9
data_type: string

tool: D9: [NULL]

assistant: ```c
const char *np(const char *str) {
  if (str)
    return str;
  else
    return "[NULL]";
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<np>:
  endbr64
  mov    %rdi,%rax
  test   %rdi,%rdi
  lea    D6(%rip),%rdx
  cmove  %rdx,%rax
  ret
```

tool_calls: name: read_data
arguments: data_tag: D6
datatype: string

tool: D6: [NULL]

assistant: ```c
const char *np(const char *str) {
  if (str)
    return str;
  else
    return "[NULL]";
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<np>:
  endbr64
  mov    %rdi,%rax
  test   %rdi,%rdi
  lea    D0(%rip),%rdx
  cmove  %rdx,%rax
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D0
type: string

tool: D0: [NULL]

assistant: ```c
const char *np(const char *str) {
  if (str)
    return str;
  else
    return "[NULL]";
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<np>:
  endbr64
  mov    %rdi,%rax
  test   %rdi,%rdi
  lea    D3(%rip),%rdx
  cmove  %rdx,%rax
  ret
```

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

tool: D3: [NULL]

assistant: ```c
const char *np(const char *str) {
  if (str)
    return str;
  else
    return "[NULL]";
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<lg2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <__builtin_constant_p@plt>
  test   %rax,%rax
  je     L8
  cmpl   $0x10000,-0x4(%rbp)
  je     L170
  cmpl   $0x10000,-0x4(%rbp)
  jg     L8
  cmpl   $0x8000,-0x4(%rbp)
  je     L280
  cmpl   $0x8000,-0x4(%rbp)
  jg     L8
  cmpl   $0x4000,-0x4(%rbp)
  je     L400
  cmpl   $0x4000,-0x4(%rbp)
  jg     L8
  cmpl   $0x2000,-0x4(%rbp)
  je     L98
  cmpl   $0x2000,-0x4(%rbp)
  jg     L8
  cmpl   $0x1000,-0x4(%rbp)
  je     L65
  cmpl   $0x1000,-0x4(%rbp)
  jg     L8
  cmpl   $0x800,-0x4(%rbp)
  je     L127
  cmpl   $0x800,-0x4(%rbp)
  jg     L8
  cmpl   $0x400,-0x4(%rbp)
  je     L116
  cmpl   $0x400,-0x4(%rbp)
  jg     L8
  cmpl   $0x200,-0x4(%rbp)
  je     L186
  cmpl   $0x200,-0x4(%rbp)
  jg     L8
  cmpl   $0x100,-0x4(%rbp)
  je     L90
  cmpl   $0x100,-0x4(%rbp)
  jg     L8
  cmpl   $0x80,-0x4(%rbp)
  je     L257
  cmpl   $0x80,-0x4(%rbp)
  jg     L8
  cmpl   $0x20,-0x4(%rbp)
  jg     L205
  cmpl   $0x0,-0x4(%rbp)
  jle    L8
  cmpl   $0x20,-0x4(%rbp)
  ja     L8
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D18(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D18(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
L205:
  cmpl   $0x40,-0x4(%rbp)
  je     L39
  jmp    L8
  mov    $0x0,%eax
  jmp    L142
  mov    $0x1,%eax
  jmp    L142
  mov    $0x2,%eax
  jmp    L142
  mov    $0x3,%eax
  jmp    L142
  mov    $0x4,%eax
  jmp    L142
  mov    $0x5,%eax
  jmp    L142
L39:
  mov    $0x6,%eax
  jmp    L142
L257:
  mov    $0x7,%eax
  jmp    L142
L90:
  mov    $0x8,%eax
  jmp    L142
L186:
  mov    $0x9,%eax
  jmp    L142
L116:
  mov    $0xa,%eax
  jmp    L142
L127:
  mov    $0xb,%eax
  jmp    L142
L65:
  mov    $0xc,%eax
  jmp    L142
L98:
  mov    $0xd,%eax
  jmp    L142
L400:
  mov    $0xe,%eax
  jmp    L142
L280:
  mov    $0xf,%eax
  jmp    L142
L170:
  mov    $0x10,%eax
  jmp    L142
L8:
  bsr    -0x4(%rbp),%edx
  xor    $0x1f,%edx
  mov    $0x1f,%eax
  sub    %edx,%eax
L142:
  leave
  ret
```

assistant: ```c
uint32_t lg2(uint32_t x) {
  if (__builtin_constant_p(x))
    switch (x) {
    case 1:
      return 0;
    case 2:
      return 1;
    case 4:
      return 2;
    case 8:
      return 3;
    case 16:
      return 4;
    case 32:
      return 5;
    case 64:
      return 6;
    case 128:
      return 7;
    case 256:
      return 8;
    case 512:
      return 9;
    case 1024:
      return 10;
    case 2048:
      return 11;
    case 4096:
      return 12;
    case 8192:
      return 13;
    case 16384:
      return 14;
    case 32768:
      return 15;
    case 65536:
      return 16;
    }
  return 31 - __builtin_clz(x);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<lg2>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <__builtin_constant_p@plt>
  test   %rax,%rax
  je     L104
  cmp    $0x100,%ebx
  je     L281
  jg     L206
  cmp    $0x20,%ebx
  jg     L40
  test   %ebx,%ebx
  jle    L104
  cmp    $0x20,%ebx
  ja     L104
  mov    %ebx,%eax
  lea    D2(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x1,%eax
  jmp    L173
L40:
  mov    $0x6,%eax
  cmp    $0x40,%ebx
  je     L173
  cmp    $0x80,%ebx
  jne    L104
  mov    $0x7,%eax
  jmp    L173
L206:
  cmp    $0x2000,%ebx
  je     L187
  jle    L57
  mov    $0xf,%eax
  cmp    $0x8000,%ebx
  je     L173
  cmp    $0x10000,%ebx
  jne    L125
  mov    $0x10,%eax
  jmp    L173
L57:
  mov    $0xb,%eax
  cmp    $0x800,%ebx
  je     L173
  jle    L102
  mov    $0xc,%eax
  cmp    $0x1000,%ebx
  je     L173
L104:
  bsr    %ebx,%ebx
  xor    $0x1f,%ebx
  mov    $0x1f,%eax
  sub    %ebx,%eax
L173:
  pop    %rbx
  ret
L102:
  mov    $0x9,%eax
  cmp    $0x200,%ebx
  je     L173
  cmp    $0x400,%ebx
  jne    L104
  mov    $0xa,%eax
  jmp    L173
L125:
  cmp    $0x4000,%ebx
  jne    L104
  mov    $0xe,%eax
  jmp    L173
  mov    $0x2,%eax
  jmp    L173
  mov    $0x3,%eax
  jmp    L173
  mov    $0x4,%eax
  jmp    L173
  mov    $0x5,%eax
  jmp    L173
L281:
  mov    $0x8,%eax
  jmp    L173
  mov    $0x0,%eax
  jmp    L173
L187:
  mov    $0xd,%eax
  jmp    L173
```

assistant: ```c
uint32_t lg2(uint32_t x) {
  if (__builtin_constant_p(x))
    switch (x) {
    case 1:
      return 0;
    case 2:
      return 1;
    case 4:
      return 2;
    case 8:
      return 3;
    case 16:
      return 4;
    case 32:
      return 5;
    case 64:
      return 6;
    case 128:
      return 7;
    case 256:
      return 8;
    case 512:
      return 9;
    case 1024:
      return 10;
    case 2048:
      return 11;
    case 4096:
      return 12;
    case 8192:
      return 13;
    case 16384:
      return 14;
    case 32768:
      return 15;
    case 65536:
      return 16;
    }
  return 31 - __builtin_clz(x);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<lg2>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <__builtin_constant_p@plt>
  test   %rax,%rax
  je     L87
  cmp    $0x100,%ebx
  je     L14
  jg     L12
  cmp    $0x20,%ebx
  jg     L55
  test   %ebx,%ebx
  jle    L87
  cmp    $0x20,%ebx
  ja     L87
  lea    D8(%rip),%rdx
  mov    %ebx,%eax
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
L12:
  cmp    $0x2000,%ebx
  je     L56
  jle    L128
  mov    $0xf,%eax
  cmp    $0x8000,%ebx
  je     L45
  cmp    $0x10000,%ebx
  jne    L70
  mov    $0x10,%eax
L45:
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L128:
  mov    $0xb,%eax
  cmp    $0x800,%ebx
  je     L45
  jle    L122
  mov    $0xc,%eax
  cmp    $0x1000,%ebx
  je     L45
L87:
  bsr    %ebx,%ebx
  mov    $0x1f,%eax
  xor    $0x1f,%ebx
  sub    %ebx,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L55:
  mov    $0x6,%eax
  cmp    $0x40,%ebx
  je     L45
  cmp    $0x80,%ebx
  jne    L87
  mov    $0x7,%eax
  pop    %rbx
  ret
L122:
  mov    $0x9,%eax
  cmp    $0x200,%ebx
  je     L45
  cmp    $0x400,%ebx
  jne    L87
  mov    $0xa,%eax
  pop    %rbx
  ret
L70:
  cmp    $0x4000,%ebx
  jne    L87
  mov    $0xe,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L56:
  mov    $0xd,%eax
  pop    %rbx
  ret
  nop
L14:
  mov    $0x8,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x2,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x5,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x3,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x4,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
uint32_t lg2(uint32_t x) {
  if (__builtin_constant_p(x))
    switch (x) {
    case 1:
      return 0;
    case 2:
      return 1;
    case 4:
      return 2;
    case 8:
      return 3;
    case 16:
      return 4;
    case 32:
      return 5;
    case 64:
      return 6;
    case 128:
      return 7;
    case 256:
      return 8;
    case 512:
      return 9;
    case 1024:
      return 10;
    case 2048:
      return 11;
    case 4096:
      return 12;
    case 8192:
      return 13;
    case 16384:
      return 14;
    case 32768:
      return 15;
    case 65536:
      return 16;
    }
  return 31 - __builtin_clz(x);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<lg2>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <__builtin_constant_p@plt>
  test   %rax,%rax
  je     L112
  cmp    $0x100,%ebx
  je     L65
  jg     L16
  cmp    $0x20,%ebx
  jg     L20
  test   %ebx,%ebx
  jle    L112
  cmp    $0x20,%ebx
  ja     L112
  lea    D1(%rip),%rdx
  mov    %ebx,%eax
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
L16:
  cmp    $0x2000,%ebx
  je     L121
  jle    L103
  mov    $0xf,%eax
  cmp    $0x8000,%ebx
  je     L91
  cmp    $0x10000,%ebx
  jne    L8
  mov    $0x10,%eax
L91:
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
L103:
  mov    $0xb,%eax
  cmp    $0x800,%ebx
  je     L91
  jle    L143
  mov    $0xc,%eax
  cmp    $0x1000,%ebx
  je     L91
L112:
  bsr    %ebx,%ebx
  mov    $0x1f,%eax
  xor    $0x1f,%ebx
  sub    %ebx,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  mov    $0x6,%eax
  cmp    $0x40,%ebx
  je     L91
  cmp    $0x80,%ebx
  jne    L112
  mov    $0x7,%eax
  pop    %rbx
  ret
L143:
  mov    $0x9,%eax
  cmp    $0x200,%ebx
  je     L91
  cmp    $0x400,%ebx
  jne    L112
  mov    $0xa,%eax
  pop    %rbx
  ret
L8:
  cmp    $0x4000,%ebx
  jne    L112
  mov    $0xe,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L121:
  mov    $0xd,%eax
  pop    %rbx
  ret
  nop
L65:
  mov    $0x8,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x2,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x5,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x3,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  mov    $0x4,%eax
  pop    %rbx
  ret
  nopw   0x0(%rax,%rax,1)
  xor    %eax,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
uint32_t lg2(uint32_t x) {
  if (__builtin_constant_p(x))
    switch (x) {
    case 1:
      return 0;
    case 2:
      return 1;
    case 4:
      return 2;
    case 8:
      return 3;
    case 16:
      return 4;
    case 32:
      return 5;
    case 64:
      return 6;
    case 128:
      return 7;
    case 256:
      return 8;
    case 512:
      return 9;
    case 1024:
      return 10;
    case 2048:
      return 11;
    case 4096:
      return 12;
    case 8192:
      return 13;
    case 16384:
      return 14;
    case 32768:
      return 15;
    case 65536:
      return 16;
    }
  return 31 - __builtin_clz(x);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<__cxa_atexit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int __cxa_atexit(void func(void *), void *arg, void *dso_handle) {
  (void)func;
  (void)arg;
  (void)dso_handle;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int __cxa_atexit(void func(void *), void *arg, void *dso_handle) {
  (void)func;
  (void)arg;
  (void)dso_handle;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int __cxa_atexit(void func(void *), void *arg, void *dso_handle) {
  (void)func;
  (void)arg;
  (void)dso_handle;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int __cxa_atexit(void func(void *), void *arg, void *dso_handle) {
  (void)func;
  (void)arg;
  (void)dso_handle;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<free_dcuboid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void free_dcuboid(double ***m) {
  free(m[0][0]);
  free(m[0]);
  free(m);
}
```
user: Translate the below assembly code into C programming language:

```asm
<free_dcuboid>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%rax
  mov    (%rax),%rdi
  call   <free@plt>
  mov    (%rbx),%rdi
  call   <free@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  pop    %rbx
  ret
```

assistant: ```c
void free_dcuboid(double ***m) {
  free(m[0][0]);
  free(m[0]);
  free(m);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<free_dcuboid>:
  endbr64
  push   %rbp
  mov    (%rdi),%rax
  mov    %rdi,%rbp
  mov    (%rax),%rdi
  call   <free@plt>
  mov    0x0(%rbp),%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1050 <free@plt>
```

assistant: ```c
void free_dcuboid(double ***m) {
  free(m[0][0]);
  free(m[0]);
  free(m);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<free_dcuboid>:
  endbr64
  push   %rbp
  mov    (%rdi),%rax
  mov    %rdi,%rbp
  mov    (%rax),%rdi
  call   <free@plt>
  mov    0x0(%rbp),%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1050 <free@plt>
```

assistant: ```c
void free_dcuboid(double ***m) {
  free(m[0][0]);
  free(m[0]);
  free(m);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<display_smatrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    %r8,-0x38(%rbp)
  mov    %r9,-0x40(%rbp)
  cmpq   $0x0,0x10(%rbp)
  je     L32
  mov    0x10(%rbp),%rdi
  call   <puts@plt>
L32:
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L15
L11:
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L49
L14:
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  movswl %ax,%edx
  mov    -0x40(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <printf@plt>
  addq   $0x1,-0x8(%rbp)
L49:
  mov    -0x8(%rbp),%rax
  cmp    -0x38(%rbp),%rax
  jle    L14
  mov    $0xa,%edi
  call   <putchar@plt>
  addq   $0x1,-0x10(%rbp)
L15:
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jle    L11
  nop
  nop
  leave
  ret
```

assistant: ```c
void display_smatrix(short **m, long nrl, long nrh, long ncl, long nch, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = nrl; i <= nrh; i++) {
    for (j = ncl; j <= nch; j++) {
      printf(format, m[i][j]);
    }
    putchar('\n');
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<display_smatrix>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r14
  mov    %rdx,0x8(%rsp)
  mov    %rcx,%r15
  mov    %r8,%r12
  mov    %r9,%r13
  mov    0x50(%rsp),%rdi
  test   %rdi,%rdi
  je     L33
  call   <puts@plt>
L33:
  cmp    0x8(%rsp),%r14
  jg     L37
  lea    (%rbx,%r14,8),%rbp
  jmp    L42
L61:
  mov    0x0(%rbp),%rax
  movswl (%rax,%rbx,2),%esi
  mov    %r13,%rdi
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jge    L61
L17:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r14
  add    $0x8,%rbp
  cmp    %r14,0x8(%rsp)
  jl     L37
L42:
  mov    %r15,%rbx
  cmp    %r12,%r15
  jle    L61
  jmp    L17
L37:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void display_smatrix(short **m, long nrl, long nrh, long ncl, long nch, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = nrl; i <= nrh; i++) {
    for (j = ncl; j <= nch; j++) {
      printf(format, m[i][j]);
    }
    putchar('\n');
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<display_smatrix>:
  endbr64
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %r9,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%rdi
  mov    %rdx,0x8(%rsp)
  test   %rdi,%rdi
  je     L42
  call   <puts@plt>
L42:
  cmp    0x8(%rsp),%r13
  jg     L40
  lea    (%rbx,%r13,8),%rbx
L46:
  mov    %r14,%r15
  cmp    %rbp,%r14
  jg     L25
  nopl   0x0(%rax,%rax,1)
L41:
  mov    (%rbx),%rax
  mov    %r12,%rdi
  movswl (%rax,%r15,2),%esi
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jge    L41
L25:
  mov    $0xa,%edi
  add    $0x1,%r13
  add    $0x8,%rbx
  call   <putchar@plt>
  cmp    %r13,0x8(%rsp)
  jge    L46
L40:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void display_smatrix(short **m, long nrl, long nrh, long ncl, long nch, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = nrl; i <= nrh; i++) {
    for (j = ncl; j <= nch; j++) {
      printf(format, m[i][j]);
    }
    putchar('\n');
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<display_smatrix>:
  endbr64
  push   %r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %r9,%r12
  push   %rbp
  mov    %r8,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    0x50(%rsp),%rdi
  mov    %rdx,0x8(%rsp)
  test   %rdi,%rdi
  je     L32
  call   <puts@plt>
L32:
  cmp    0x8(%rsp),%r13
  jg     L10
  lea    (%rbx,%r13,8),%rbx
L33:
  mov    %r14,%r15
  cmp    %rbp,%r14
  jg     L18
  nopl   0x0(%rax,%rax,1)
L42:
  mov    (%rbx),%rax
  mov    %r12,%rdi
  movswl (%rax,%r15,2),%esi
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %r15,%rbp
  jge    L42
L18:
  mov    $0xa,%edi
  add    $0x1,%r13
  add    $0x8,%rbx
  call   <putchar@plt>
  cmp    %r13,0x8(%rsp)
  jge    L33
L10:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void display_smatrix(short **m, long nrl, long nrh, long ncl, long nch, char *format, char *name) {
  long i, j;
  if (name != NULL)
    puts(name);
  for (i = nrl; i <= nrh; i++) {
    for (j = ncl; j <= nch; j++) {
      printf(format, m[i][j]);
    }
    putchar('\n');
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L18
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
L18:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <nsize@plt>
  mov    -0x8(%rbp),%edx
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L5:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%edi
  call   <abs@plt>
  lea    0x30(%rax),%ecx
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x14(%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,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L5
  nop
  nop
  leave
  ret
```

assistant: ```c
void itoa(int n, char s[]) {
  int i, k;
  i = 0;
  if (n < 0) {
    s[i++] = '-';
  }
  k = nsize(n) + i;
  s[k--] = '\0';
  do {
    s[k--] = abs(n % 10) + '0';
  } while (n /= 10);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<itoa>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    $0x0,%ebx
  test   %edi,%edi
  js     L22
L20:
  mov    %ebp,%edi
  call   <nsize@plt>
  add    %ebx,%eax
  movslq %eax,%rdx
  movb   $0x0,(%r12,%rdx,1)
  sub    $0x1,%eax
  cltq
  add    %rax,%r12
L15:
  movslq %ebp,%rbx
  imul   $0x66666667,%rbx,%rbx
  sar    $0x22,%rbx
  mov    %ebp,%eax
  sar    $0x1f,%eax
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  add    %eax,%eax
  mov    %ebp,%edi
  sub    %eax,%edi
  call   <abs@plt>
  add    $0x30,%eax
  mov    %al,(%r12)
  mov    %ebx,%ebp
  sub    $0x1,%r12
  test   %ebx,%ebx
  jne    L15
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L22:
  movb   $0x2d,(%rsi)
  mov    $0x1,%ebx
  jmp    L20
```

assistant: ```c
void itoa(int n, char s[]) {
  int i, k;
  i = 0;
  if (n < 0) {
    s[i++] = '-';
  }
  k = nsize(n) + i;
  s[k--] = '\0';
  do {
    s[k--] = abs(n % 10) + '0';
  } while (n /= 10);
}
```