ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<randbuf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %edi,%rbx
  lea    0x1(%rbx),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebx,%ebx
  jle    L18
  mov    %rax,%r13
  lea    (%rax,%rbx,1),%rbp
  nopw   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
  add    $0x1,%r13
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x61,%eax
  mov    %al,-0x1(%r13)
  cmp    %rbp,%r13
  jne    L9
  add    %r12,%rbx
L25:
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L18:
  mov    %rax,%rbx
  jmp    L25
```

assistant: ```c
char *randbuf(int len) {
  char *buf;
  int i;
  buf = (char *)malloc(len + 1);
  for (i = 0; i < len; i++) {
    buf[i] = 'a' + (rand() % 26);
  }
  buf[i] = 0;
  return buf;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<randbuf>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %edi,%rbx
  lea    0x1(%rbx),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebx,%ebx
  jle    L8
  mov    %rax,%r13
  lea    (%rax,%rbx,1),%rbp
  nopw   0x0(%rax,%rax,1)
L11:
  xor    %eax,%eax
  add    $0x1,%r13
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x4ec4ec4f,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x23,%rdx
  sub    %ecx,%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  add    $0x61,%eax
  mov    %al,-0x1(%r13)
  cmp    %r13,%rbp
  jne    L11
  add    %r12,%rbx
L28:
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   (%rax)
L8:
  mov    %rax,%rbx
  jmp    L28
```

assistant: ```c
char *randbuf(int len) {
  char *buf;
  int i;
  buf = (char *)malloc(len + 1);
  for (i = 0; i < len; i++) {
    buf[i] = 'a' + (rand() % 26);
  }
  buf[i] = 0;
  return buf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<h2o_mem_swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x150,%rsp
  mov    %rdi,-0x138(%rbp)
  mov    %rsi,-0x140(%rbp)
  mov    %rdx,-0x148(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x138(%rbp),%rax
  mov    %rax,-0x128(%rbp)
  mov    -0x140(%rbp),%rax
  mov    %rax,-0x120(%rbp)
  jmp    L15
L26:
  mov    -0x148(%rbp),%rax
  mov    $0x100,%edx
  cmp    %rdx,%rax
  cmova  %rdx,%rax
  mov    %rax,-0x118(%rbp)
  mov    -0x118(%rbp),%rdx
  mov    -0x128(%rbp),%rcx
  lea    -0x110(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x118(%rbp),%rdx
  mov    -0x120(%rbp),%rcx
  mov    -0x128(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x118(%rbp),%rdx
  lea    -0x110(%rbp),%rcx
  mov    -0x120(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x118(%rbp),%rax
  sub    %rax,-0x148(%rbp)
L15:
  cmpq   $0x0,-0x148(%rbp)
  jne    L26
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

assistant: ```c
void h2o_mem_swap(void *_x, void *_y, size_t len) {
  char *x = _x, *y = _y;
  char buf[256];
  while (len != 0) {
    size_t blocksz = len < sizeof(buf) ? len : sizeof(buf);
    memcpy(buf, x, blocksz);
    memcpy(x, y, blocksz);
    memcpy(y, buf, blocksz);
    len -= blocksz;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<h2o_mem_swap>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L8
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%rbp
  mov    %rsp,%r14
L19:
  mov    $0x100,%ebx
  cmp    %rbx,%rbp
  cmovbe %rbp,%rbx
  mov    %rbx,%rdx
  mov    %r12,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  sub    %rbx,%rbp
  jne    L19
L8:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void h2o_mem_swap(void *_x, void *_y, size_t len) {
  char *x = _x, *y = _y;
  char buf[256];
  while (len != 0) {
    size_t blocksz = len < sizeof(buf) ? len : sizeof(buf);
    memcpy(buf, x, blocksz);
    memcpy(x, y, blocksz);
    memcpy(y, buf, blocksz);
    len -= blocksz;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<h2o_mem_swap>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L23
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%rbp
  mov    %rsp,%r14
  nopw   0x0(%rax,%rax,1)
L3:
  mov    $0x100,%ebx
  mov    %r12,%rsi
  mov    %r14,%rdi
  cmp    %rbx,%rbp
  cmovbe %rbp,%rbx
  mov    %rbx,%rdx
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  sub    %rbx,%rbp
  jne    L3
L23:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void h2o_mem_swap(void *_x, void *_y, size_t len) {
  char *x = _x, *y = _y;
  char buf[256];
  while (len != 0) {
    size_t blocksz = len < sizeof(buf) ? len : sizeof(buf);
    memcpy(buf, x, blocksz);
    memcpy(x, y, blocksz);
    memcpy(y, buf, blocksz);
    len -= blocksz;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<h2o_mem_swap>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x110,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  test   %rdx,%rdx
  je     L29
  mov    %rdi,%r12
  mov    %rsi,%r13
  mov    %rdx,%rbp
  mov    %rsp,%r14
  nopw   0x0(%rax,%rax,1)
L16:
  mov    $0x100,%ebx
  mov    %r12,%rsi
  mov    %r14,%rdi
  cmp    %rbx,%rbp
  cmovbe %rbp,%rbx
  mov    %rbx,%rdx
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    %rbx,%rdx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <memcpy@plt>
  sub    %rbx,%rbp
  jne    L16
L29:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x110,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void h2o_mem_swap(void *_x, void *_y, size_t len) {
  char *x = _x, *y = _y;
  char buf[256];
  while (len != 0) {
    size_t blocksz = len < sizeof(buf) ? len : sizeof(buf);
    memcpy(buf, x, blocksz);
    memcpy(x, y, blocksz);
    memcpy(y, buf, blocksz);
    len -= blocksz;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D8(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <_puts@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D8
type: string

tool: D8: I do not fear computers. I fear the lack of them - Isaac Asimov

assistant: ```c
int main(void) {
  char *str;
  str = "I do not fear computers. I fear the lack of them - Isaac Asimov";
  _puts(str);
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <_puts@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
data_type: string

tool: D6: I do not fear computers. I fear the lack of them - Isaac Asimov

assistant: ```c
int main(void) {
  char *str;
  str = "I do not fear computers. I fear the lack of them - Isaac Asimov";
  _puts(str);
  return (0);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D2(%rip),%rdi
  call   <_puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_infered: string

tool: D2: I do not fear computers. I fear the lack of them - Isaac Asimov

assistant: ```c
int main(void) {
  char *str;
  str = "I do not fear computers. I fear the lack of them - Isaac Asimov";
  _puts(str);
  return (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <_puts@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D0
type_guessed: string

tool: D0: I do not fear computers. I fear the lack of them - Isaac Asimov

assistant: ```c
int main(void) {
  char *str;
  str = "I do not fear computers. I fear the lack of them - Isaac Asimov";
  _puts(str);
  return (0);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ADCSequenceConfigure>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %rcx,-0x20(%rbp)
  shlq   $0x2,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    $0xf,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  not    %eax
  cltq
  mov    %rdx,%rsi
  and    %rax,%rsi
  mov    -0x18(%rbp),%rax
  and    $0xf,%eax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rcx
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  mov    %rax,%rdx
  mov    %rsi,%rax
  or     %rcx,%rax
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  add    $0x20,%rax
  mov    (%rax),%rdx
  mov    -0x10(%rbp),%rax
  mov    $0xf,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  not    %eax
  cltq
  and    %rax,%rdx
  mov    %rdx,%rsi
  mov    -0x20(%rbp),%rax
  and    $0x3,%eax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    %eax,%ecx
  shl    %cl,%rdx
  mov    %rdx,%rcx
  mov    -0x8(%rbp),%rax
  add    $0x20,%rax
  mov    %rax,%rdx
  mov    %rsi,%rax
  or     %rcx,%rax
  mov    %rax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ADCSequenceConfigure(unsigned long ulBase, unsigned long ulSequenceNum, unsigned long ulTrigger, unsigned long ulPriority) {
  ;
  ;

  ;
  ;
  ulSequenceNum *= 4;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) = (((*((volatile unsigned long *)(ulBase + 0x00000014))) & ~(0xf << ulSequenceNum)) | ((ulTrigger & 0xf) << ulSequenceNum));
  (*((volatile unsigned long *)(ulBase + 0x00000020))) = (((*((volatile unsigned long *)(ulBase + 0x00000020))) & ~(0xf << ulSequenceNum)) | ((ulPriority & 0x3) << ulSequenceNum));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ADCSequenceConfigure>:
  endbr64
  mov    %rdi,%r8
  mov    %rcx,%rdi
  mov    0x14(%r8),%r9
  lea    0x0(,%rsi,4),%rcx
  mov    $0xf,%eax
  shl    %cl,%eax
  not    %eax
  cltq
  and    $0xf,%edx
  shl    %cl,%rdx
  and    %rax,%r9
  or     %r9,%rdx
  mov    %rdx,0x14(%r8)
  mov    0x20(%r8),%rdx
  and    $0x3,%edi
  shl    %cl,%rdi
  and    %rdx,%rax
  or     %rax,%rdi
  mov    %rdi,0x20(%r8)
  ret
```

assistant: ```c
void ADCSequenceConfigure(unsigned long ulBase, unsigned long ulSequenceNum, unsigned long ulTrigger, unsigned long ulPriority) {
  ;
  ;

  ;
  ;
  ulSequenceNum *= 4;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) = (((*((volatile unsigned long *)(ulBase + 0x00000014))) & ~(0xf << ulSequenceNum)) | ((ulTrigger & 0xf) << ulSequenceNum));
  (*((volatile unsigned long *)(ulBase + 0x00000020))) = (((*((volatile unsigned long *)(ulBase + 0x00000020))) & ~(0xf << ulSequenceNum)) | ((ulPriority & 0x3) << ulSequenceNum));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ADCSequenceConfigure>:
  endbr64
  mov    %rcx,%r8
  mov    $0xf,%eax
  mov    0x14(%rdi),%r9
  and    $0xf,%edx
  lea    0x0(,%rsi,4),%rcx
  and    $0x3,%r8d
  shl    %cl,%eax
  shl    %cl,%rdx
  not    %eax
  shl    %cl,%r8
  cltq
  and    %rax,%r9
  or     %r9,%rdx
  mov    %rdx,0x14(%rdi)
  mov    0x20(%rdi),%rdx
  and    %rdx,%rax
  or     %rax,%r8
  mov    %r8,0x20(%rdi)
  ret
```

assistant: ```c
void ADCSequenceConfigure(unsigned long ulBase, unsigned long ulSequenceNum, unsigned long ulTrigger, unsigned long ulPriority) {
  ;
  ;

  ;
  ;
  ulSequenceNum *= 4;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) = (((*((volatile unsigned long *)(ulBase + 0x00000014))) & ~(0xf << ulSequenceNum)) | ((ulTrigger & 0xf) << ulSequenceNum));
  (*((volatile unsigned long *)(ulBase + 0x00000020))) = (((*((volatile unsigned long *)(ulBase + 0x00000020))) & ~(0xf << ulSequenceNum)) | ((ulPriority & 0x3) << ulSequenceNum));
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ADCSequenceConfigure>:
  endbr64
  mov    %rcx,%r8
  mov    $0xf,%eax
  mov    0x14(%rdi),%r9
  and    $0xf,%edx
  lea    0x0(,%rsi,4),%rcx
  and    $0x3,%r8d
  shl    %cl,%eax
  shl    %cl,%rdx
  not    %eax
  shl    %cl,%r8
  cltq
  and    %rax,%r9
  or     %r9,%rdx
  mov    %rdx,0x14(%rdi)
  mov    0x20(%rdi),%rdx
  and    %rdx,%rax
  or     %rax,%r8
  mov    %r8,0x20(%rdi)
  ret
```

assistant: ```c
void ADCSequenceConfigure(unsigned long ulBase, unsigned long ulSequenceNum, unsigned long ulTrigger, unsigned long ulPriority) {
  ;
  ;

  ;
  ;
  ulSequenceNum *= 4;
  (*((volatile unsigned long *)(ulBase + 0x00000014))) = (((*((volatile unsigned long *)(ulBase + 0x00000014))) & ~(0xf << ulSequenceNum)) | ((ulTrigger & 0xf) << ulSequenceNum));
  (*((volatile unsigned long *)(ulBase + 0x00000020))) = (((*((volatile unsigned long *)(ulBase + 0x00000020))) & ~(0xf << ulSequenceNum)) | ((ulPriority & 0x3) << ulSequenceNum));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cmemmove>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  jmp    L16
L1:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L16:
  mov    -0x18(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  test   %rax,%rax
  jne    L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void cmemmove(char *d, const char *s, long n) {
  while (n--)
    *d++ = *s++;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<cmemmove>:
  endbr64
  test   %rdx,%rdx
  je     L1
  mov    $0x0,%eax
L9:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L9
L1:
  ret
```

assistant: ```c
void cmemmove(char *d, const char *s, long n) {
  while (n--)
    *d++ = *s++;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<cmemmove>:
  endbr64
  test   %rdx,%rdx
  je     L18
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L13:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L13
L18:
  ret
```

assistant: ```c
void cmemmove(char *d, const char *s, long n) {
  while (n--)
    *d++ = *s++;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<cmemmove>:
  endbr64
  mov    %rdi,%rcx
  lea    -0x1(%rdx),%rdi
  test   %rdx,%rdx
  je     L13
  lea    0x1(%rsi),%r8
  mov    %rcx,%rax
  sub    %r8,%rax
  cmp    $0xe,%rax
  jbe    L8
  cmp    $0x6,%rdi
  jbe    L8
  cmp    $0xe,%rdi
  jbe    L75
  mov    %rdx,%r8
  xor    %eax,%eax
  and    $0xfffffffffffffff0,%r8
  nopl   (%rax)
L88:
  movdqu (%rsi,%rax,1),%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %r8,%rax
  jne    L88
  mov    %rdx,%rax
  and    $0xfffffffffffffff0,%rax
  lea    (%rcx,%rax,1),%r8
  lea    (%rsi,%rax,1),%r9
  sub    %rax,%rdi
  cmp    %rax,%rdx
  je     L27
L20:
  sub    %rax,%rdx
  lea    -0x1(%rdx),%r10
  cmp    $0x6,%r10
  jbe    L7
  mov    (%rsi,%rax,1),%rsi
  mov    %rsi,(%rcx,%rax,1)
  mov    %rdx,%rax
  and    $0xfffffffffffffff8,%rax
  add    %rax,%r8
  add    %rax,%r9
  sub    %rax,%rdi
  cmp    %rax,%rdx
  je     L13
L7:
  movzbl (%r9),%eax
  mov    %al,(%r8)
  test   %rdi,%rdi
  je     L13
  movzbl 0x1(%r9),%eax
  mov    %al,0x1(%r8)
  cmp    $0x1,%rdi
  je     L13
  movzbl 0x2(%r9),%eax
  mov    %al,0x2(%r8)
  cmp    $0x2,%rdi
  je     L13
  movzbl 0x3(%r9),%eax
  mov    %al,0x3(%r8)
  cmp    $0x3,%rdi
  je     L13
  movzbl 0x4(%r9),%eax
  mov    %al,0x4(%r8)
  cmp    $0x4,%rdi
  je     L13
  movzbl 0x5(%r9),%eax
  mov    %al,0x5(%r8)
  cmp    $0x5,%rdi
  je     L13
  movzbl 0x6(%r9),%eax
  mov    %al,0x6(%r8)
  ret
L8:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L156:
  movzbl (%rsi,%rax,1),%edi
  mov    %dil,(%rcx,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L156
L13:
  ret
L75:
  mov    %rsi,%r9
  mov    %rcx,%r8
  xor    %eax,%eax
  jmp    L20
L27:
  ret
```

assistant: ```c
void cmemmove(char *d, const char *s, long n) {
  while (n--)
    *d++ = *s++;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<inicMatEntidades>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L32
L48:
  mov    -0x14(%rbp),%eax
  cltq
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rax,(%rdx)
  addl   $0x1,-0x10(%rbp)
L32:
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cmp    %eax,-0x10(%rbp)
  jl     L48
  movl   $0x0,-0x10(%rbp)
  jmp    L58
L12:
  movl   $0x0,-0xc(%rbp)
  jmp    L9
L15:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0xc(%rbp)
L9:
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L15
  addl   $0x1,-0x10(%rbp)
L58:
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cmp    %eax,-0x10(%rbp)
  jl     L12
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char **inicMatEntidades(int dim_modelo) {
  char **MatEntidades = (char **)malloc(3 * dim_modelo * sizeof(char *));
  int i, j;
  for (i = 0; i < 3 * dim_modelo; i++) {
    MatEntidades[i] = (char *)malloc(sizeof(char) * 3 * dim_modelo);
  }
  for (i = 0; i < 3 * dim_modelo; i++) {
    for (j = 0; j < 3 * dim_modelo; j++) {
      MatEntidades[i][j] = '\0';
    }
  }
  return MatEntidades;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inicMatEntidades>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    (%rdi,%rdi,2),%r13d
  lea    0x0(,%r13,8),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %r13d,%r13d
  jle    L8
  mov    %r13d,%r15d
  mov    %rax,%rbx
  mov    %r13d,%eax
  lea    (%r12,%rax,8),%rbp
  mov    %r12,%r14
L39:
  mov    %r15d,%edi
  call   <malloc@plt>
  mov    %rax,(%r14)
  add    $0x8,%r14
  cmp    %rbp,%r14
  jne    L39
  mov    %r13d,%esi
L35:
  mov    %rbx,%rcx
  mov    $0x0,%eax
L1:
  mov    (%rcx),%rdx
  movb   $0x0,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L1
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L35
L8:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char **inicMatEntidades(int dim_modelo) {
  char **MatEntidades = (char **)malloc(3 * dim_modelo * sizeof(char *));
  int i, j;
  for (i = 0; i < 3 * dim_modelo; i++) {
    MatEntidades[i] = (char *)malloc(sizeof(char) * 3 * dim_modelo);
  }
  for (i = 0; i < 3 * dim_modelo; i++) {
    for (j = 0; j < 3 * dim_modelo; j++) {
      MatEntidades[i][j] = '\0';
    }
  }
  return MatEntidades;
}
```
user: What is the c source code of the assembly code below?

```asm
<inicMatEntidades>:
  endbr64
  push   %r15
  lea    (%rdi,%rdi,2),%r15d
  push   %r14
  lea    0x0(,%r15,8),%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r15d,%r15d
  jle    L37
  mov    %r15d,%ebp
  mov    %rax,%rbx
  mov    %rax,%r14
  lea    (%rax,%rbp,8),%r12
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %r15d,%edi
  add    $0x8,%r14
  call   <malloc@plt>
  mov    %rax,-0x8(%r14)
  cmp    %r12,%r14
  jne    L9
  nopl   (%rax)
L3:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L17:
  mov    (%rbx),%rdx
  movb   $0x0,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbp
  jne    L17
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L3
L37:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char **inicMatEntidades(int dim_modelo) {
  char **MatEntidades = (char **)malloc(3 * dim_modelo * sizeof(char *));
  int i, j;
  for (i = 0; i < 3 * dim_modelo; i++) {
    MatEntidades[i] = (char *)malloc(sizeof(char) * 3 * dim_modelo);
  }
  for (i = 0; i < 3 * dim_modelo; i++) {
    for (j = 0; j < 3 * dim_modelo; j++) {
      MatEntidades[i][j] = '\0';
    }
  }
  return MatEntidades;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<inicMatEntidades>:
  endbr64
  push   %r15
  lea    (%rdi,%rdi,2),%r15d
  push   %r14
  lea    0x0(,%r15,8),%edi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r15d,%r15d
  jle    L33
  mov    %r15d,%ebp
  mov    %rax,%rbx
  mov    %rax,%r14
  lea    (%rax,%rbp,8),%r12
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %r15d,%edi
  add    $0x8,%r14
  call   <malloc@plt>
  mov    %rax,-0x8(%r14)
  cmp    %r12,%r14
  jne    L8
  nopl   (%rax)
L21:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L16:
  mov    (%rbx),%rdx
  movb   $0x0,(%rdx,%rax,1)
  add    $0x1,%rax
  cmp    %rax,%rbp
  jne    L16
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L21
L33:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char **inicMatEntidades(int dim_modelo) {
  char **MatEntidades = (char **)malloc(3 * dim_modelo * sizeof(char *));
  int i, j;
  for (i = 0; i < 3 * dim_modelo; i++) {
    MatEntidades[i] = (char *)malloc(sizeof(char) * 3 * dim_modelo);
  }
  for (i = 0; i < 3 * dim_modelo; i++) {
    for (j = 0; j < 3 * dim_modelo; j++) {
      MatEntidades[i][j] = '\0';
    }
  }
  return MatEntidades;
}
```