ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<quickSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L7
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <median@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x20(%rbp),%edx
  mov    -0x8(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <swap@plt>
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <partition@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quickSort>
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quickSort>
L7:
  nop
  leave
  ret
```

assistant: ```c
void quickSort(int *array, int start, int end) {
  if (start < end) {
    int medianIndex = median(array, start, end);
    swap(array, medianIndex, end);
    int mid = partition(array, start, end);
    quickSort(array, start, mid - 1);
    quickSort(array, mid + 1, end);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<quickSort>:
  endbr64
  cmp    %edx,%esi
  jl     L6
  ret
L6:
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %edx,%ebx
  call   <median@plt>
  mov    %eax,%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <swap@plt>
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <partition@plt>
  mov    %eax,%ebp
  lea    -0x1(%rax),%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <quickSort>
  lea    0x1(%rbp),%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <quickSort>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void quickSort(int *array, int start, int end) {
  if (start < end) {
    int medianIndex = median(array, start, end);
    swap(array, medianIndex, end);
    int mid = partition(array, start, end);
    quickSort(array, start, mid - 1);
    quickSort(array, mid + 1, end);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<quickSort>:
  endbr64
  cmp    %esi,%edx
  jle    L4
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
L9:
  mov    %ebp,%esi
  mov    %ebx,%edx
  mov    %r13,%rdi
  call   <median@plt>
  mov    %ebx,%edx
  mov    %r13,%rdi
  mov    %eax,%esi
  call   <swap@plt>
  mov    %ebp,%esi
  mov    %ebx,%edx
  mov    %r13,%rdi
  call   <partition@plt>
  mov    %ebp,%esi
  mov    %r13,%rdi
  mov    %eax,%r12d
  lea    -0x1(%rax),%edx
  call   <quickSort>
  lea    0x1(%r12),%ebp
  cmp    %ebx,%ebp
  jl     L9
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L4:
  ret
```

assistant: ```c
void quickSort(int *array, int start, int end) {
  if (start < end) {
    int medianIndex = median(array, start, end);
    swap(array, medianIndex, end);
    int mid = partition(array, start, end);
    quickSort(array, start, mid - 1);
    quickSort(array, mid + 1, end);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<strlcat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    -0x38(%rbp),%rbx
  mov    -0x40(%rbp),%r13
  mov    -0x48(%rbp),%r12
  jmp    L6
L21:
  add    $0x1,%rbx
L6:
  mov    %r12,%rax
  lea    -0x1(%rax),%r12
  test   %rax,%rax
  je     L11
  movzbl (%rbx),%eax
  test   %al,%al
  jne    L21
L11:
  mov    %rbx,%rax
  sub    -0x38(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x48(%rbp),%rax
  sub    -0x28(%rbp),%rax
  mov    %rax,%r12
  test   %r12,%r12
  jne    L35
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    -0x28(%rbp),%rdx
  add    %rdx,%rax
  jmp    L31
L36:
  cmp    $0x1,%r12
  je     L69
  mov    %rbx,%rax
  lea    0x1(%rax),%rbx
  movzbl 0x0(%r13),%edx
  mov    %dl,(%rax)
  sub    $0x1,%r12
L69:
  add    $0x1,%r13
L35:
  movzbl 0x0(%r13),%eax
  test   %al,%al
  jne    L36
  movb   $0x0,(%rbx)
  mov    %r13,%rax
  sub    -0x40(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
L31:
  add    $0x38,%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %rbp
  ret
```

assistant: ```c
size_t strlcat(char *dst, const char *src, size_t siz) {
  register char *d = dst;
  register const char *s = src;
  register size_t n = siz;
  size_t dlen;
  while ((n-- != 0) && (*d != '\0'))
    d++;
  dlen = d - dst;
  n = siz - dlen;
  if (!n)
    return dlen + strlen(s);
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return dlen + (s - src);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<strlcat>:
  endbr64
  push   %rbx
  test   %rdx,%rdx
  je     L58
  mov    %rdi,%r8
  lea    (%rdi,%rdx,1),%rcx
  mov    %rdi,%rax
L64:
  cmpb   $0x0,(%rax)
  je     L21
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L64
  mov    %rcx,%rax
L21:
  mov    %rax,%rbx
  sub    %r8,%rbx
  sub    %rbx,%rdx
  je     L88
  movzbl (%rsi),%r8d
  mov    %rsi,%rcx
  test   %r8b,%r8b
  jne    L80
L50:
  movb   $0x0,(%rax)
  sub    %rsi,%rcx
  lea    (%rcx,%rbx,1),%rax
L18:
  pop    %rbx
  ret
L58:
  mov    %rdx,%rbx
L88:
  mov    %rsi,%rdi
  call   <strlen@plt>
  add    %rbx,%rax
  jmp    L18
L73:
  add    $0x1,%rcx
  movzbl (%rcx),%r8d
  test   %r8b,%r8b
  je     L50
L80:
  cmp    $0x1,%rdx
  je     L73
  mov    %r8b,(%rax)
  sub    $0x1,%rdx
  lea    0x1(%rax),%rax
  jmp    L73
```

assistant: ```c
size_t strlcat(char *dst, const char *src, size_t siz) {
  register char *d = dst;
  register const char *s = src;
  register size_t n = siz;
  size_t dlen;
  while ((n-- != 0) && (*d != '\0'))
    d++;
  dlen = d - dst;
  n = siz - dlen;
  if (!n)
    return dlen + strlen(s);
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return dlen + (s - src);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strlcat>:
  endbr64
  push   %r12
  test   %rdx,%rdx
  je     L3
  mov    %rdi,%r8
  lea    (%rdi,%rdx,1),%rcx
  mov    %rdi,%rax
  jmp    L85
  nopl   0x0(%rax,%rax,1)
L29:
  add    $0x1,%rax
  cmp    %rcx,%rax
  je     L94
L85:
  cmpb   $0x0,(%rax)
  jne    L29
L94:
  mov    %rax,%r12
  sub    %r8,%r12
  sub    %r12,%rdx
  je     L77
  movzbl (%rsi),%ecx
  mov    %rsi,%r8
  test   %cl,%cl
  jne    L89
  jmp    L17
  nopl   (%rax)
L45:
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  mov    %r8,%rcx
  add    $0x1,%rax
L106:
  lea    0x1(%rcx),%r8
  movzbl 0x1(%rcx),%ecx
  test   %cl,%cl
  je     L92
L89:
  cmp    $0x1,%rdx
  jne    L45
  cmpb   $0x0,0x1(%r8)
  lea    0x1(%r8),%rcx
  jne    L106
  sub    %rsi,%rcx
  add    %rcx,%r12
L17:
  movb   $0x0,(%rax)
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L92:
  sub    %rsi,%r8
  add    %r8,%r12
  jmp    L17
L3:
  xor    %r12d,%r12d
L77:
  mov    %rsi,%rdi
  call   <strlen@plt>
  add    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
size_t strlcat(char *dst, const char *src, size_t siz) {
  register char *d = dst;
  register const char *s = src;
  register size_t n = siz;
  size_t dlen;
  while ((n-- != 0) && (*d != '\0'))
    d++;
  dlen = d - dst;
  n = siz - dlen;
  if (!n)
    return dlen + strlen(s);
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return dlen + (s - src);
}
```
user: What is the c source code of the assembly code below:

```asm
<strlcat>:
  endbr64
  push   %r12
  test   %rdx,%rdx
  je     L19
  mov    %rdi,%r8
  lea    (%rdi,%rdx,1),%rcx
  mov    %rdi,%rax
  jmp    L75
  nopl   0x0(%rax,%rax,1)
L3:
  add    $0x1,%rax
  cmp    %rcx,%rax
  je     L83
L75:
  cmpb   $0x0,(%rax)
  jne    L3
L83:
  mov    %rax,%r12
  sub    %r8,%r12
  sub    %r12,%rdx
  je     L44
  movzbl (%rsi),%ecx
  mov    %rsi,%r8
  test   %cl,%cl
  jne    L67
  jmp    L80
  nopl   (%rax)
L92:
  mov    %cl,(%rax)
  sub    $0x1,%rdx
  mov    %r8,%rcx
  add    $0x1,%rax
L58:
  lea    0x1(%rcx),%r8
  movzbl 0x1(%rcx),%ecx
  test   %cl,%cl
  je     L107
L67:
  cmp    $0x1,%rdx
  jne    L92
  cmpb   $0x0,0x1(%r8)
  lea    0x1(%r8),%rcx
  jne    L58
  sub    %rsi,%rcx
  add    %rcx,%r12
L80:
  movb   $0x0,(%rax)
  mov    %r12,%rax
  pop    %r12
  ret
  nopl   0x0(%rax)
L107:
  sub    %rsi,%r8
  add    %r8,%r12
  jmp    L80
L19:
  xor    %r12d,%r12d
L44:
  mov    %rsi,%rdi
  call   <strlen@plt>
  add    %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
size_t strlcat(char *dst, const char *src, size_t siz) {
  register char *d = dst;
  register const char *s = src;
  register size_t n = siz;
  size_t dlen;
  while ((n-- != 0) && (*d != '\0'))
    d++;
  dlen = d - dst;
  n = siz - dlen;
  if (!n)
    return dlen + strlen(s);
  while (*s != '\0') {
    if (n != 1) {
      *d++ = *s;
      n--;
    }
    s++;
  }
  *d = '\0';
  return dlen + (s - src);
}
```
user: What is the c source code of the assembly code below?

```asm
<circle_display>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  movl   $0x14,-0x20(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movl   $0x3,-0x18(%rbp)
  mov    D11(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <glClear@plt>
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <glBegin@plt>
  movl   $0x0,-0x24(%rbp)
  jmp    L16
L8:
  mov    -0x18(%rbp),%eax
  add    %eax,%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  imul   %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <sin@plt>
  imul   -0x1c(%rbp),%eax
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <cos@plt>
  imul   -0x1c(%rbp),%eax
  mov    %ebx,%esi
  mov    %eax,%edi
  call   <glVertex2f@plt>
  addl   $0x1,-0x24(%rbp)
L16:
  mov    -0x24(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L8
  mov    $0x0,%eax
  call   <glEnd@plt>
  mov    $0x0,%eax
  call   <glFlush@plt>
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void circle_display(void) {
  const int n = 20;
  const GLfloat R = 0.5f;
  const GLfloat Pi = 3.1415926536f;
  int i;
  glClear(GL_COLOR_BUFFER_BIT);
  glBegin(GL_TRIANGLES);
  for (i = 0; i < n; i++) {
    GLfloat x = 2 * Pi / n * i;
    glVertex2f(R * cos(x), R * sin(x));
  }
  glEnd();
  glFlush();
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<circle_display>:
  endbr64
  push   %rbx
  mov    D16(%rip),%rax
  mov    (%rax),%edi
  call   <glClear@plt>
  mov    D17(%rip),%rax
  mov    (%rax),%edi
  call   <glBegin@plt>
  mov    $0x14,%ebx
L7:
  mov    $0x0,%edi
  call   <sin@plt>
  mov    $0x0,%edi
  call   <cos@plt>
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <glVertex2f@plt>
  sub    $0x1,%ebx
  jne    L7
  mov    $0x0,%eax
  call   <glEnd@plt>
  mov    $0x0,%eax
  call   <glFlush@plt>
  pop    %rbx
  ret
```

assistant: ```c
void circle_display(void) {
  const int n = 20;
  const GLfloat R = 0.5f;
  const GLfloat Pi = 3.1415926536f;
  int i;
  glClear(GL_COLOR_BUFFER_BIT);
  glBegin(GL_TRIANGLES);
  for (i = 0; i < n; i++) {
    GLfloat x = 2 * Pi / n * i;
    glVertex2f(R * cos(x), R * sin(x));
  }
  glEnd();
  glFlush();
}
```
user: Translate the below assembly code into C programming language:

```asm
<circle_display>:
  endbr64
  mov    D4(%rip),%rax
  push   %rbx
  mov    $0x14,%ebx
  mov    (%rax),%edi
  call   <glClear@plt>
  mov    D16(%rip),%rax
  mov    (%rax),%edi
  call   <glBegin@plt>
  cs nopw 0x0(%rax,%rax,1)
L2:
  xor    %edi,%edi
  call   <sin@plt>
  xor    %edi,%edi
  call   <cos@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  call   <glVertex2f@plt>
  sub    $0x1,%ebx
  jne    L2
  xor    %eax,%eax
  call   <glEnd@plt>
  xor    %eax,%eax
  pop    %rbx
  jmp    10b0 <glFlush@plt>
```

assistant: ```c
void circle_display(void) {
  const int n = 20;
  const GLfloat R = 0.5f;
  const GLfloat Pi = 3.1415926536f;
  int i;
  glClear(GL_COLOR_BUFFER_BIT);
  glBegin(GL_TRIANGLES);
  for (i = 0; i < n; i++) {
    GLfloat x = 2 * Pi / n * i;
    glVertex2f(R * cos(x), R * sin(x));
  }
  glEnd();
  glFlush();
}
```
user: What is the c source code of the assembly code below?

```asm
<circle_display>:
  endbr64
  mov    D11(%rip),%rax
  push   %rbx
  mov    $0x14,%ebx
  mov    (%rax),%edi
  call   <glClear@plt>
  mov    D7(%rip),%rax
  mov    (%rax),%edi
  call   <glBegin@plt>
  cs nopw 0x0(%rax,%rax,1)
L8:
  xor    %edi,%edi
  call   <sin@plt>
  xor    %edi,%edi
  call   <cos@plt>
  xor    %esi,%esi
  xor    %edi,%edi
  call   <glVertex2f@plt>
  sub    $0x1,%ebx
  jne    L8
  xor    %eax,%eax
  call   <glEnd@plt>
  xor    %eax,%eax
  pop    %rbx
  jmp    10b0 <glFlush@plt>
```

assistant: ```c
void circle_display(void) {
  const int n = 20;
  const GLfloat R = 0.5f;
  const GLfloat Pi = 3.1415926536f;
  int i;
  glClear(GL_COLOR_BUFFER_BIT);
  glBegin(GL_TRIANGLES);
  for (i = 0; i < n; i++) {
    GLfloat x = 2 * Pi / n * i;
    glVertex2f(R * cos(x), R * sin(x));
  }
  glEnd();
  glFlush();
}
```
user: Translate the below assembly code into C programming language:

```asm
<recvall>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x28(%rbp)
  mov    %ecx,-0x34(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x18(%rbp)
  mov    -0x28(%rbp),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L42
L75:
  mov    -0x34(%rbp),%ecx
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <recv@plt>
  mov    %eax,-0x18(%rbp)
  cmpl   $0xffffffff,-0x18(%rbp)
  jne    L83
  mov    D16(%rip),%rax
  mov    (%rax),%rdx
  mov    D1(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L83
  mov    $0x186a0,%edi
  call   <usleep@plt>
  jmp    L42
L83:
  cmpl   $0x0,-0x18(%rbp)
  jle    L63
  mov    -0x18(%rbp),%eax
  cltq
  add    %rax,-0x10(%rbp)
  mov    -0x18(%rbp),%eax
  sub    %eax,-0x14(%rbp)
L42:
  cmpl   $0x0,-0x14(%rbp)
  jne    L75
  jmp    L58
L63:
  nop
L58:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  je     L69
  mov    -0x10(%rbp),%rax
  sub    -0x8(%rbp),%rax
  jmp    L19
L69:
  mov    -0x18(%rbp),%eax
L19:
  leave
  ret
```

assistant: ```c
int recvall(int s, void *buf, ssize_t size, int opt) {
  char *p0 = buf, *p;
  ssize_t cb = 0, n = size;
  for (p = p0; n;) {
    cb = recv(s, p, n, opt);
    if (cb == -1 && errno == EAGAIN) {
      usleep(100000);
      continue;
    }
    if (cb <= 0) {
      break;
    }
    p += cb;
    n -= cb;
  }
  return (p - p0 ? p - p0 : cb);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<recvall>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edx,%ebp
  test   %edx,%edx
  je     L61
  mov    %edi,%r13d
  mov    %rsi,%r15
  mov    %ecx,%r14d
  mov    %rsi,%r12
  jmp    L29
L62:
  mov    $0x186a0,%edi
  call   <usleep@plt>
  jmp    L21
L56:
  test   %eax,%eax
  jle    L59
  cltq
  add    %rax,%r12
  sub    %ebx,%ebp
L21:
  test   %ebp,%ebp
  je     L59
L29:
  mov    %r14d,%ecx
  mov    %ebp,%edx
  mov    %r12,%rsi
  mov    %r13d,%edi
  call   <recv@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  jne    L56
  mov    D10(%rip),%rdx
  mov    D12(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L62
L59:
  mov    %r12,%rax
  sub    %r15,%rax
  cmp    %r12,%r15
  cmovne %eax,%ebx
L41:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L61:
  mov    %edx,%ebx
  jmp    L41
```

assistant: ```c
int recvall(int s, void *buf, ssize_t size, int opt) {
  char *p0 = buf, *p;
  ssize_t cb = 0, n = size;
  for (p = p0; n;) {
    cb = recv(s, p, n, opt);
    if (cb == -1 && errno == EAGAIN) {
      usleep(100000);
      continue;
    }
    if (cb <= 0) {
      break;
    }
    p += cb;
    n -= cb;
  }
  return (p - p0 ? p - p0 : cb);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<recvall>:
  endbr64
  test   %edx,%edx
  je     L48
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  jmp    L36
  nop
L62:
  test   %eax,%eax
  jle    L54
  movslq %eax,%rdx
  add    %rdx,%r14
  sub    %eax,%ebx
  je     L54
L36:
  mov    %r13d,%ecx
  mov    %ebx,%edx
  mov    %r14,%rsi
  mov    %ebp,%edi
  call   <recv@plt>
  cmp    $0xffffffff,%eax
  jne    L62
  mov    D11(%rip),%rdx
  mov    D6(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L0
L54:
  mov    %r14,%rdx
  pop    %rbx
  pop    %rbp
  sub    %r12,%rdx
  cmp    %r14,%r12
  pop    %r12
  pop    %r13
  cmovne %edx,%eax
  pop    %r14
  ret
  nopl   (%rax)
L0:
  mov    $0x186a0,%edi
  call   <usleep@plt>
  jmp    L36
L48:
  xor    %eax,%eax
  ret
```

assistant: ```c
int recvall(int s, void *buf, ssize_t size, int opt) {
  char *p0 = buf, *p;
  ssize_t cb = 0, n = size;
  for (p = p0; n;) {
    cb = recv(s, p, n, opt);
    if (cb == -1 && errno == EAGAIN) {
      usleep(100000);
      continue;
    }
    if (cb <= 0) {
      break;
    }
    p += cb;
    n -= cb;
  }
  return (p - p0 ? p - p0 : cb);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<recvall>:
  endbr64
  test   %edx,%edx
  je     L21
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %edx,%ebx
  jmp    L14
  nop
L42:
  test   %eax,%eax
  jle    L48
  movslq %eax,%rdx
  add    %rdx,%r14
  sub    %eax,%ebx
  je     L48
L14:
  mov    %r13d,%ecx
  mov    %ebx,%edx
  mov    %r14,%rsi
  mov    %ebp,%edi
  call   <recv@plt>
  cmp    $0xffffffff,%eax
  jne    L42
  mov    D9(%rip),%rdx
  mov    D12(%rip),%rcx
  mov    (%rdx),%rsi
  cmp    %rsi,(%rcx)
  je     L43
L48:
  mov    %r14,%rdx
  pop    %rbx
  pop    %rbp
  sub    %r12,%rdx
  cmp    %r14,%r12
  pop    %r12
  pop    %r13
  cmovne %edx,%eax
  pop    %r14
  ret
  nopl   (%rax)
L43:
  mov    $0x186a0,%edi
  call   <usleep@plt>
  jmp    L14
L21:
  xor    %eax,%eax
  ret
```

assistant: ```c
int recvall(int s, void *buf, ssize_t size, int opt) {
  char *p0 = buf, *p;
  ssize_t cb = 0, n = size;
  for (p = p0; n;) {
    cb = recv(s, p, n, opt);
    if (cb == -1 && errno == EAGAIN) {
      usleep(100000);
      continue;
    }
    if (cb <= 0) {
      break;
    }
    p += cb;
    n -= cb;
  }
  return (p - p0 ? p - p0 : cb);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mallocMatrizDouble>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L35
  mov    $0x0,%eax
  jmp    L2
L35:
  movl   $0x0,-0xc(%rbp)
  jmp    L19
L14:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  pxor   %xmm0,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0xc(%rbp)
L19:
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L14
  mov    -0x8(%rbp),%rax
L2:
  leave
  ret
```

assistant: ```c
double *mallocMatrizDouble(int cantFila, int cantCol) {
  int i;
  double *datos = ((double *)malloc(cantFila * cantCol * sizeof(double)));
  if (datos == NULL) {
    return NULL;
  } else {
    for (i = 0; i < cantFila * cantCol; i++) {
      datos[i] = 0.0;
    }
  }
  return datos;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mallocMatrizDouble>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  imul   %esi,%ebx
  lea    0x0(,%rbx,8),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L28
  mov    %rax,%rcx
  test   %ebx,%ebx
  jle    L9
  mov    %rax,%rdx
  mov    %ebx,%ebx
  lea    (%rax,%rbx,8),%rax
L29:
  movq   $0x0,(%rdx)
  add    $0x8,%rdx
  cmp    %rax,%rdx
  jne    L29
L9:
  mov    %rcx,%rax
  pop    %rbx
  ret
L28:
  mov    $0x0,%ecx
  jmp    L9
```

assistant: ```c
double *mallocMatrizDouble(int cantFila, int cantCol) {
  int i;
  double *datos = ((double *)malloc(cantFila * cantCol * sizeof(double)));
  if (datos == NULL) {
    return NULL;
  } else {
    for (i = 0; i < cantFila * cantCol; i++) {
      datos[i] = 0.0;
    }
  }
  return datos;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mallocMatrizDouble>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  imul   %esi,%ebx
  sub    $0x8,%rsp
  lea    0x0(,%rbx,8),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L8
  test   %ebx,%ebx
  jle    L8
  movslq %ebx,%rbp
  xor    %esi,%esi
  mov    %rax,%rdi
  lea    0x0(,%rbp,8),%rdx
  call   <memset@plt>
  mov    %rax,%r8
L8:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
double *mallocMatrizDouble(int cantFila, int cantCol) {
  int i;
  double *datos = ((double *)malloc(cantFila * cantCol * sizeof(double)));
  if (datos == NULL) {
    return NULL;
  } else {
    for (i = 0; i < cantFila * cantCol; i++) {
      datos[i] = 0.0;
    }
  }
  return datos;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mallocMatrizDouble>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  imul   %esi,%ebx
  sub    $0x8,%rsp
  lea    0x0(,%rbx,8),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L8
  test   %ebx,%ebx
  jle    L8
  movslq %ebx,%rbp
  xor    %esi,%esi
  mov    %rax,%rdi
  lea    0x0(,%rbp,8),%rdx
  call   <memset@plt>
  mov    %rax,%r8
L8:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
double *mallocMatrizDouble(int cantFila, int cantCol) {
  int i;
  double *datos = ((double *)malloc(cantFila * cantCol * sizeof(double)));
  if (datos == NULL) {
    return NULL;
  } else {
    for (i = 0; i < cantFila * cantCol; i++) {
      datos[i] = 0.0;
    }
  }
  return datos;
}
```