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

```asm
<fmatrixAeqI>:
  endbr64
  test   %rsi,%rsi
  je     L16
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x0(,%rsi,4),%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L1:
  mov    (%r12,%rbx,2),%rcx
  mov    %rbp,%rdx
  xor    %esi,%esi
  mov    %rcx,%rdi
  call   <memset@plt>
  movl   $0x3f800000,(%rax,%rbx,1)
  add    $0x4,%rbx
  cmp    %rbx,%rbp
  jne    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  ret
```

assistant: ```c
void fmatrixAeqI(float *A[], int16_t rc) {
  float *pAij;
  int8_t i, j;
  for (i = 0; i < rc; i++) {
    pAij = A[i];
    for (j = 0; j < rc; j++) {
      *(pAij++) = 0.0F;
    }
    A[i][i] = 1.0F;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fmatrixAeqI>:
  endbr64
  test   %rsi,%rsi
  je     L3
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x0(,%rsi,4),%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L16:
  mov    (%r12,%rbx,2),%rcx
  mov    %rbp,%rdx
  xor    %esi,%esi
  mov    %rcx,%rdi
  call   <memset@plt>
  movl   $0x3f800000,(%rax,%rbx,1)
  add    $0x4,%rbx
  cmp    %rbx,%rbp
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  ret
```

assistant: ```c
void fmatrixAeqI(float *A[], int16_t rc) {
  float *pAij;
  int8_t i, j;
  for (i = 0; i < rc; i++) {
    pAij = A[i];
    for (j = 0; j < rc; j++) {
      *(pAij++) = 0.0F;
    }
    A[i][i] = 1.0F;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D19
type_infered: string
name: parse_data
arguments: data_label: D13
type_infered: string

tool: D19: Hello World!


tool: D13: Creating new branch featurel successfully!


assistant: ```c
int main() {
  printf("Hello World!\n");
  printf("Creating new branch featurel successfully!\n");
  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    D17(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D17
type_guessed: string
name: unpack_data
arguments: label: D11
type_guessed: string

tool: D17: Hello World!


tool: D11: Creating new branch featurel successfully!


assistant: ```c
int main() {
  printf("Hello World!\n");
  printf("Creating new branch featurel successfully!\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    D16(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
data_type: string
name: extract_symbolic_data
arguments: tag: D16
data_type: string

tool: D7: Hello World!


tool: D16: Creating new branch featurel successfully!


assistant: ```c
int main() {
  printf("Hello World!\n");
  printf("Creating new branch featurel successfully!\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D7
data_type: string
name: parse_data
arguments: data_tag: D15
data_type: string

tool: D7: Hello World!


tool: D15: Creating new branch featurel successfully!


assistant: ```c
int main() {
  printf("Hello World!\n");
  printf("Creating new branch featurel successfully!\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__timedwait>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %edx,-0x40(%rbp)
  mov    %rcx,-0x48(%rbp)
  mov    %r8d,-0x4c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x48(%rbp)
  je     L12
  lea    -0x20(%rbp),%rdx
  mov    -0x40(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <clock_gettime@plt>
  mov    -0x48(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x20(%rbp),%rdx
  sub    %rdx,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x48(%rbp),%rax
  mov    0x8(%rax),%rax
  mov    -0x18(%rbp),%rdx
  sub    %rdx,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  test   %rax,%rax
  jns    L76
  mov    -0x20(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  add    $0x3b9aca00,%rax
  mov    %rax,-0x18(%rbp)
L76:
  mov    -0x20(%rbp),%rax
  test   %rax,%rax
  jns    L12
  mov    D22(%rip),%rax
  mov    (%rax),%eax
  jmp    L85
L12:
  cmpl   $0x0,-0x4c(%rbp)
  je     L11
  movl   $0x80,-0x4c(%rbp)
L11:
  movl   $0x0,-0x4c(%rbp)
  cmpq   $0x0,-0x48(%rbp)
  je     L36
  lea    -0x20(%rbp),%rsi
  jmp    L8
L36:
  mov    $0x0,%esi
L8:
  mov    -0x38(%rbp),%rax
  mov    -0x3c(%rbp),%ecx
  mov    -0x4c(%rbp),%edx
  mov    %rsi,%r8
  mov    %rax,%rsi
  mov    $0xf0,%edi
  call   <__syscall@plt>
  neg    %eax
  mov    %eax,-0x24(%rbp)
  mov    D22(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  je     L87
  mov    D21(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x24(%rbp)
  jne    L98
L87:
  mov    -0x24(%rbp),%eax
  jmp    L85
L98:
  mov    $0x0,%eax
L85:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

assistant: ```c
int __timedwait(volatile int *addr, int val, clockid_t clk, const struct timespec *at, int priv) {
  int r;
  struct timespec to;
  if (at) {
    clock_gettime(clk, &to);
    to.tv_sec = at->tv_sec - to.tv_sec;
    if ((to.tv_nsec = at->tv_nsec - to.tv_nsec) < 0) {
      to.tv_sec--;
      to.tv_nsec += 1000000000;
    }
    if (to.tv_sec < 0)
      return ETIMEDOUT;
  }
  if (priv)
    priv = 128;
  priv = 0;
  r = -__syscall(240, (long)addr, 0 | priv, val, at ? (long)&to : 0);
  if (r == ETIMEDOUT || r == EINTR)
    return r;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<__timedwait>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rcx,%rcx
  je     L66
  mov    %rcx,%rbx
  mov    %rsp,%rsi
  mov    %edx,%edi
  call   <clock_gettime@plt>
  mov    (%rbx),%rdx
  sub    (%rsp),%rdx
  mov    %rdx,(%rsp)
  mov    0x8(%rbx),%rax
  sub    0x8(%rsp),%rax
  mov    %rax,0x8(%rsp)
  js     L0
L57:
  mov    %rsp,%r8
  cmpq   $0x0,(%rsp)
  jns    L1
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  jmp    L36
L0:
  sub    $0x1,%rdx
  mov    %rdx,(%rsp)
  add    $0x3b9aca00,%rax
  mov    %rax,0x8(%rsp)
  jmp    L57
L66:
  mov    $0x0,%r8d
L1:
  mov    %r12d,%ecx
  mov    $0x0,%edx
  mov    %rbp,%rsi
  mov    $0xf0,%edi
  call   <__syscall@plt>
  neg    %eax
  mov    D0(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L36
  mov    D20(%rip),%rdx
  cmp    %eax,(%rdx)
  mov    $0x0,%edx
  cmovne %edx,%eax
L36:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int __timedwait(volatile int *addr, int val, clockid_t clk, const struct timespec *at, int priv) {
  int r;
  struct timespec to;
  if (at) {
    clock_gettime(clk, &to);
    to.tv_sec = at->tv_sec - to.tv_sec;
    if ((to.tv_nsec = at->tv_nsec - to.tv_nsec) < 0) {
      to.tv_sec--;
      to.tv_nsec += 1000000000;
    }
    if (to.tv_sec < 0)
      return ETIMEDOUT;
  }
  if (priv)
    priv = 128;
  priv = 0;
  r = -__syscall(240, (long)addr, 0 | priv, val, at ? (long)&to : 0);
  if (r == ETIMEDOUT || r == EINTR)
    return r;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<__timedwait>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rcx,%rcx
  je     L8
  mov    %rsp,%r13
  mov    %rcx,%rbx
  mov    %edx,%edi
  mov    %r13,%rsi
  call   <clock_gettime@plt>
  mov    (%rbx),%rax
  mov    0x8(%rbx),%rdx
  sub    (%rsp),%rax
  sub    0x8(%rsp),%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  js     L24
  mov    %r13,%r8
  test   %rax,%rax
  js     L17
L11:
  xor    %edx,%edx
  mov    %r12d,%ecx
  mov    %rbp,%rsi
  mov    $0xf0,%edi
  call   <__syscall@plt>
  mov    D19(%rip),%rdx
  neg    %eax
  cmp    %eax,(%rdx)
  je     L56
  mov    D27(%rip),%rdx
  cmp    %eax,(%rdx)
  mov    $0x0,%edx
  cmovne %edx,%eax
L56:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L75
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L24:
  sub    $0x1,%rax
  add    $0x3b9aca00,%rdx
  mov    %r13,%r8
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  test   %rax,%rax
  jns    L11
  jmp    L17
  xchg   %ax,%ax
L8:
  xor    %r8d,%r8d
  jmp    L11
  nopl   (%rax)
L17:
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  jmp    L56
L75:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int __timedwait(volatile int *addr, int val, clockid_t clk, const struct timespec *at, int priv) {
  int r;
  struct timespec to;
  if (at) {
    clock_gettime(clk, &to);
    to.tv_sec = at->tv_sec - to.tv_sec;
    if ((to.tv_nsec = at->tv_nsec - to.tv_nsec) < 0) {
      to.tv_sec--;
      to.tv_nsec += 1000000000;
    }
    if (to.tv_sec < 0)
      return ETIMEDOUT;
  }
  if (priv)
    priv = 128;
  priv = 0;
  r = -__syscall(240, (long)addr, 0 | priv, val, at ? (long)&to : 0);
  if (r == ETIMEDOUT || r == EINTR)
    return r;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mkline>:
  endbr64
  movabs $0x2020202020202020,%rax
  mov    %rax,D16(%rip)
  movb   $0x0,D37(%rip)
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    $0x30,%edi
  mov    %dil,D76(%rip)
  test   %eax,%eax
  je     L12
  movslq %eax,%rcx
  imul   $0x66666667,%rcx,%rcx
  sar    $0x22,%rcx
  cltd
  sub    %edx,%ecx
  lea    (%rcx,%rcx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  add    $0x30,%eax
  mov    %al,D38(%rip)
  test   %ecx,%ecx
  je     L12
  movslq %ecx,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %ecx,%eax
  sar    $0x1f,%eax
  sub    %eax,%edx
  lea    (%rdx,%rdx,4),%eax
  add    %eax,%eax
  sub    %eax,%ecx
  add    $0x30,%ecx
  mov    %cl,D58(%rip)
  test   %edx,%edx
  je     L12
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    $0x30,%edx
  mov    %dl,D66(%rip)
  test   %eax,%eax
  je     L12
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  add    %edx,%edx
  sub    %edx,%eax
  add    $0x30,%eax
  mov    %al,D43(%rip)
L12:
  lea    D16(%rip),%rax
  ret
```

assistant: ```c
char *mkline(n)
register int n;
{
  static char lbuf[9];
  register int i = 2;
  strcpy(lbuf, "        ");
  lbuf[i++] = (n % 10) + '0';
  n /= 10;
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  return lbuf;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<mkline>:
  endbr64
  movabs $0x2020202020202020,%rax
  mov    %edi,%edx
  movb   $0x0,D37(%rip)
  mov    %rax,D6(%rip)
  movslq %edi,%rax
  sar    $0x1f,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    $0x30,%edi
  mov    %dil,D3(%rip)
  test   %eax,%eax
  je     L9
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  add    $0x30,%eax
  mov    %al,D18(%rip)
  test   %edx,%edx
  je     L9
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    $0x30,%edx
  mov    %dl,D10(%rip)
  test   %eax,%eax
  je     L9
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  add    $0x30,%eax
  mov    %al,D34(%rip)
  test   %edx,%edx
  je     L9
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  add    $0x30,%edx
  mov    %dl,D31(%rip)
L9:
  lea    D6(%rip),%rax
  ret
```

assistant: ```c
char *mkline(n)
register int n;
{
  static char lbuf[9];
  register int i = 2;
  strcpy(lbuf, "        ");
  lbuf[i++] = (n % 10) + '0';
  n /= 10;
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  return lbuf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<mkline>:
  endbr64
  movabs $0x2020202020202020,%rax
  mov    %edi,%edx
  movb   $0x0,D29(%rip)
  mov    %rax,D20(%rip)
  movslq %edi,%rax
  sar    $0x1f,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sub    %edx,%eax
  lea    (%rax,%rax,4),%edx
  add    %edx,%edx
  sub    %edx,%edi
  add    $0x30,%edi
  mov    %dil,D70(%rip)
  test   %eax,%eax
  je     L38
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  add    $0x30,%eax
  mov    %al,D13(%rip)
  test   %edx,%edx
  je     L38
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  add    $0x30,%edx
  mov    %dl,D77(%rip)
  test   %eax,%eax
  je     L38
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%eax
  add    $0x30,%eax
  mov    %al,D5(%rip)
  test   %edx,%edx
  je     L38
  movslq %edx,%rax
  mov    %edx,%ecx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%ecx
  sar    $0x22,%rax
  sub    %ecx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%edx
  add    $0x30,%edx
  mov    %dl,D7(%rip)
L38:
  lea    D20(%rip),%rax
  ret
```

assistant: ```c
char *mkline(n)
register int n;
{
  static char lbuf[9];
  register int i = 2;
  strcpy(lbuf, "        ");
  lbuf[i++] = (n % 10) + '0';
  n /= 10;
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  if (n != 0) {
    lbuf[i++] = (n % 10) + '0';
    n /= 10;
  }
  return lbuf;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<maxSumIn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq %esi,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L20:
  cmp    %rdx,%rsp
  je     L30
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L20
L30:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L99
  orq    $0x0,-0x8(%rsp,%rax,1)
L99:
  lea    0x3(%rsp),%rbx
  mov    %rbx,%r14
  shr    $0x2,%r14
  and    $0xfffffffffffffffc,%rbx
  lea    0x0(,%r13,4),%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    (%r12),%eax
  mov    %eax,0x0(,%r14,4)
  cmp    $0x1,%r13d
  jle    L137
  mov    %r13d,%edi
  mov    $0x1,%edx
  jmp    L126
L46:
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L36
L26:
  cmp    (%r12,%rax,4),%ecx
  jle    L46
  mov    %ecx,%esi
  add    (%rbx,%rax,4),%esi
  cmp    %esi,(%rbx,%rdx,4)
  jge    L46
  mov    %esi,(%rbx,%rdx,4)
  jmp    L46
L36:
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  je     L113
L126:
  test   %edx,%edx
  jle    L36
  mov    (%r12,%rdx,4),%ecx
  mov    $0x0,%eax
  jmp    L26
L137:
  je     L52
  mov    $0x0,%edx
  jmp    L94
L113:
  mov    $0x0,%eax
  mov    $0x0,%edx
L27:
  mov    (%rbx,%rax,4),%ecx
  cmp    %ecx,%edx
  cmovl  %ecx,%edx
  add    $0x1,%rax
  cmp    %eax,%r13d
  jg     L27
L94:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L162
  mov    %edx,%eax
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L52:
  mov    0x0(,%r14,4),%edx
  test   %edx,%edx
  mov    $0x0,%eax
  cmovs  %eax,%edx
  jmp    L94
L162:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int maxSumIn(int a[], int len) {
  int msis[len];
  memset(msis, 0, sizeof(msis));
  msis[0] = a[0];
  for (int i = 1; i < len; i++) {
    for (int j = 0; j < i; j++) {
      if (a[i] > a[j] && msis[i] < msis[j] + a[i]) {
        msis[i] = msis[j] + a[i];
      }
    }
  }
  int max = 0;
  for (int i = 0; i < len; i++) {
    if (msis[i] > max) {
      max = msis[i];
    }
  }
  return max;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<maxSumIn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  push   %r13
  mov    %esi,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  movslq %esi,%rax
  mov    %rsp,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L151
L128:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L128
L151:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L72
L25:
  lea    0x3(%rsp),%rbx
  lea    0x0(,%r13,4),%edx
  xor    %esi,%esi
  mov    %rbx,%r14
  and    $0xfffffffffffffffc,%rbx
  mov    %rbx,%rdi
  shr    $0x2,%r14
  call   <memset@plt>
  mov    (%r12),%edi
  mov    %edi,0x0(,%r14,4)
  cmp    $0x1,%r13d
  jle    L88
  lea    -0x1(%r13),%r8d
  xor    %edx,%edx
  shl    $0x2,%r8
  nopl   0x0(%rax)
L147:
  mov    0x4(%r12,%rdx,1),%esi
  mov    %edi,%ecx
  xor    %eax,%eax
  jmp    L35
  nopl   0x0(%rax,%rax,1)
L8:
  mov    0x4(%r12,%rax,1),%ecx
  add    $0x4,%rax
L35:
  cmp    %ecx,%esi
  jle    L157
  mov    (%rbx,%rax,1),%ecx
  add    %esi,%ecx
  cmp    %ecx,0x4(%rbx,%rdx,1)
  jge    L157
  mov    %ecx,0x4(%rbx,%rdx,1)
L157:
  cmp    %rax,%rdx
  jne    L8
  add    $0x4,%rdx
  cmp    %r8,%rdx
  jne    L147
  xor    %eax,%eax
  xor    %r8d,%r8d
  jmp    L138
  nopl   0x0(%rax)
L10:
  mov    (%rbx,%rax,4),%edi
L138:
  cmp    %edi,%r8d
  cmovl  %edi,%r8d
  add    $0x1,%rax
  cmp    %eax,%r13d
  jg     L10
L113:
  mov    -0x28(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L143
  lea    -0x20(%rbp),%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L72:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L25
L88:
  je     L133
  xor    %r8d,%r8d
  jmp    L113
L133:
  xor    %r8d,%r8d
  test   %edi,%edi
  cmovns %edi,%r8d
  jmp    L113
L143:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int maxSumIn(int a[], int len) {
  int msis[len];
  memset(msis, 0, sizeof(msis));
  msis[0] = a[0];
  for (int i = 1; i < len; i++) {
    for (int j = 0; j < i; j++) {
      if (a[i] > a[j] && msis[i] < msis[j] + a[i]) {
        msis[i] = msis[j] + a[i];
      }
    }
  }
  int max = 0;
  for (int i = 0; i < len; i++) {
    if (msis[i] > max) {
      max = msis[i];
    }
  }
  return max;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<AND>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  and    -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int AND(int a, int b) {
  int output = a & b;
  return output;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<AND>:
  endbr64
  mov    %edi,%eax
  and    %esi,%eax
  ret
```

assistant: ```c
int AND(int a, int b) {
  int output = a & b;
  return output;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<AND>:
  endbr64
  mov    %edi,%eax
  and    %esi,%eax
  ret
```

assistant: ```c
int AND(int a, int b) {
  int output = a & b;
  return output;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<AND>:
  endbr64
  mov    %edi,%eax
  and    %esi,%eax
  ret
```

assistant: ```c
int AND(int a, int b) {
  int output = a & b;
  return output;
}
```
user: Translate the below assembly code into C programming language:

```asm
<memchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L3
L35:
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    -0x1c(%rbp),%edx
  cmp    %dl,%al
  jne    L17
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  jmp    L19
L17:
  addq   $0x1,-0x10(%rbp)
L3:
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L35
  mov    $0x0,%eax
L19:
  pop    %rbp
  ret
```

assistant: ```c
void *memchr(const void *s, int c, size_t size) {
  const unsigned char *p = s;
  for (size_t i = 0; i < size; i++) {
    if (p[i] == (unsigned char)c) {
      return (void *)&p[i];
    }
  }
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<memchr>:
  endbr64
  test   %rdx,%rdx
  je     L31
  mov    %rdi,%rax
  add    %rdx,%rdi
L20:
  mov    %rax,%rdx
  cmp    %sil,(%rax)
  je     L39
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L20
  mov    $0x0,%edx
L39:
  mov    %rdx,%rax
  ret
L31:
  mov    $0x0,%edx
  jmp    L39
```

assistant: ```c
void *memchr(const void *s, int c, size_t size) {
  const unsigned char *p = s;
  for (size_t i = 0; i < size; i++) {
    if (p[i] == (unsigned char)c) {
      return (void *)&p[i];
    }
  }
  return NULL;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L36
  add    %rdi,%rdx
  jmp    L13
  nopl   0x0(%rax)
L3:
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L36
L13:
  cmp    %sil,(%rax)
  jne    L3
  ret
  nopw   0x0(%rax,%rax,1)
L36:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memchr(const void *s, int c, size_t size) {
  const unsigned char *p = s;
  for (size_t i = 0; i < size; i++) {
    if (p[i] == (unsigned char)c) {
      return (void *)&p[i];
    }
  }
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<memchr>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L25
  add    %rdi,%rdx
  jmp    L17
  nopl   0x0(%rax)
L39:
  add    $0x1,%rax
  cmp    %rdx,%rax
  je     L25
L17:
  cmp    %sil,(%rax)
  jne    L39
  ret
  nopw   0x0(%rax,%rax,1)
L25:
  xor    %eax,%eax
  ret
```

assistant: ```c
void *memchr(const void *s, int c, size_t size) {
  const unsigned char *p = s;
  for (size_t i = 0; i < size; i++) {
    if (p[i] == (unsigned char)c) {
      return (void *)&p[i];
    }
  }
  return NULL;
}
```