ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D12(%rip),%rsi
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D12
type: string
name: parse_data
arguments: data_label: D0
type: string

tool: D12: Hai World!

tool: D0: %s

assistant: ```c
int main() {
  printf("%s", "Hai World!");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D14(%rip),%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D14
data_type: string
name: disassemble_data
arguments: data_tag: D4
data_type: string

tool: D14: Hai World!

tool: D4: %s

assistant: ```c
int main() {
  printf("%s", "Hai World!");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D16(%rip),%rsi
  lea    D17(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D16
data_type: string
name: retrieve_data
arguments: label: D17
data_type: string

tool: D16: Hai World!

tool: D17: %s

assistant: ```c
int main() {
  printf("%s", "Hai World!");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_filename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  jmp    L22
L55:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  je     L47
  subl   $0x1,-0x14(%rbp)
L22:
  cmpl   $0x0,-0x14(%rbp)
  jns    L55
  jmp    L53
L47:
  nop
L53:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    -0x14(%rbp),%eax
  cltq
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x1(%rax),%rcx
  mov    -0x28(%rbp),%rax
  add    %rax,%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x10(%rbp),%rax
  mov    %eax,-0x14(%rbp)
  jmp    L32
L60:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2e,%al
  jne    L4
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L64
L4:
  subl   $0x1,-0x14(%rbp)
L32:
  cmpl   $0x0,-0x14(%rbp)
  jns    L60
L64:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *get_filename(char *full_pathname) {
  int i;
  char *tmp;
  size_t local_path_len;
  for (i = strlen(full_pathname); i >= 0; i--)
    if (full_pathname[i] == '/')
      break;
  local_path_len = strlen(full_pathname) - i;
  tmp = (char *)malloc(local_path_len + 1);
  memcpy(tmp, full_pathname + i + 1, local_path_len + 1);
  for (i = local_path_len; i >= 0; i--) {
    if (tmp[i] == '.') {
      tmp[i] = '\0';
      break;
    }
  }
  return tmp;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_filename>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  js     L57
  cltq
  add    %r12,%rax
L0:
  cmpb   $0x2f,(%rax)
  je     L57
  sub    $0x1,%ebx
  sub    $0x1,%rax
  cmp    $0xffffffff,%ebx
  jne    L0
L57:
  mov    %r12,%rdi
  call   <strlen@plt>
  sub    %ebx,%eax
  mov    %eax,%ebp
  movslq %eax,%r14
  lea    0x1(%r14),%r15
  mov    %r15,%rdi
  call   <malloc@plt>
  mov    %rax,%r13
  movslq %ebx,%rbx
  lea    0x1(%r12,%rbx,1),%rsi
  mov    %r15,%rdx
  mov    %rax,%rdi
  call   <memcpy@plt>
  test   %ebp,%ebp
  js     L41
  lea    (%r14,%r13,1),%rax
  lea    -0x1(%r13,%r14,1),%rdx
  mov    %ebp,%ebp
  sub    %rbp,%rdx
L4:
  cmpb   $0x2e,(%rax)
  je     L18
  sub    $0x1,%rax
  cmp    %rdx,%rax
  jne    L4
  jmp    L41
L18:
  movb   $0x0,(%rax)
L41:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
char *get_filename(char *full_pathname) {
  int i;
  char *tmp;
  size_t local_path_len;
  for (i = strlen(full_pathname); i >= 0; i--)
    if (full_pathname[i] == '/')
      break;
  local_path_len = strlen(full_pathname) - i;
  tmp = (char *)malloc(local_path_len + 1);
  memcpy(tmp, full_pathname + i + 1, local_path_len + 1);
  for (i = local_path_len; i >= 0; i--) {
    if (tmp[i] == '.') {
      tmp[i] = '\0';
      break;
    }
  }
  return tmp;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<get_filename>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%ebp
  movslq %eax,%rbx
  test   %eax,%eax
  jns    L8
  jmp    L70
  nopl   (%rax)
L11:
  lea    -0x1(%rbx),%ebp
  sub    $0x1,%rbx
  cmp    $0xffffffff,%ebx
  je     L79
L8:
  cmpb   $0x2f,(%r12,%rbx,1)
  mov    %ebx,%ebp
  jne    L11
L70:
  mov    %r12,%rdi
  call   <strlen@plt>
  sub    %ebp,%eax
  movslq %eax,%rbp
  lea    0x1(%rbp),%r15
  mov    %r15,%rdi
  call   <malloc@plt>
  lea    0x1(%r12,%rbx,1),%rsi
  mov    %r15,%rdx
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memcpy@plt>
  test   %ebp,%ebp
  js     L5
  lea    -0x1(%r13,%rbp,1),%rdx
  lea    0x0(%r13,%rbp,1),%rax
  sub    %rbp,%rdx
  jmp    L52
  nopl   0x0(%rax)
L89:
  sub    $0x1,%rax
  cmp    %rdx,%rax
  je     L5
L52:
  cmpb   $0x2e,(%rax)
  jne    L89
  movb   $0x0,(%rax)
L5:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L79:
  movslq %ebp,%rbx
  jmp    L70
```

assistant: ```c
char *get_filename(char *full_pathname) {
  int i;
  char *tmp;
  size_t local_path_len;
  for (i = strlen(full_pathname); i >= 0; i--)
    if (full_pathname[i] == '/')
      break;
  local_path_len = strlen(full_pathname) - i;
  tmp = (char *)malloc(local_path_len + 1);
  memcpy(tmp, full_pathname + i + 1, local_path_len + 1);
  for (i = local_path_len; i >= 0; i--) {
    if (tmp[i] == '.') {
      tmp[i] = '\0';
      break;
    }
  }
  return tmp;
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_filename>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%ebp
  movslq %eax,%rbx
  test   %eax,%eax
  jns    L79
  jmp    L4
  nopl   (%rax)
L50:
  lea    -0x1(%rbx),%ebp
  sub    $0x1,%rbx
  cmp    $0xffffffff,%ebx
  je     L53
L79:
  cmpb   $0x2f,(%r12,%rbx,1)
  mov    %ebx,%ebp
  jne    L50
L4:
  mov    %r12,%rdi
  call   <strlen@plt>
  sub    %ebp,%eax
  movslq %eax,%rbp
  lea    0x1(%rbp),%r15
  mov    %r15,%rdi
  call   <malloc@plt>
  lea    0x1(%r12,%rbx,1),%rsi
  mov    %r15,%rdx
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <memcpy@plt>
  test   %ebp,%ebp
  js     L74
  lea    -0x1(%r13,%rbp,1),%rdx
  lea    0x0(%r13,%rbp,1),%rax
  sub    %rbp,%rdx
  jmp    L82
  nopl   0x0(%rax)
L13:
  sub    $0x1,%rax
  cmp    %rdx,%rax
  je     L74
L82:
  cmpb   $0x2e,(%rax)
  jne    L13
  movb   $0x0,(%rax)
L74:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L53:
  movslq %ebp,%rbx
  jmp    L4
```

assistant: ```c
char *get_filename(char *full_pathname) {
  int i;
  char *tmp;
  size_t local_path_len;
  for (i = strlen(full_pathname); i >= 0; i--)
    if (full_pathname[i] == '/')
      break;
  local_path_len = strlen(full_pathname) - i;
  tmp = (char *)malloc(local_path_len + 1);
  memcpy(tmp, full_pathname + i + 1, local_path_len + 1);
  for (i = local_path_len; i >= 0; i--) {
    if (tmp[i] == '.') {
      tmp[i] = '\0';
      break;
    }
  }
  return tmp;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<removeLimits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x24(%rbp)
  jmp    L48
L45:
  mov    -0x24(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D43(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <getrlimit@plt>
  mov    -0x20(%rbp),%rdx
  mov    D37(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jge    L7
  mov    D37(%rip),%rax
  mov    (%rax),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D43(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <setrlimit@plt>
  mov    -0x24(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D43(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <getrlimit@plt>
L7:
  mov    -0x18(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  cmp    %rax,%rdx
  jge    L4
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  lea    D43(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  lea    -0x20(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <setrlimit@plt>
L4:
  addl   $0x1,-0x24(%rbp)
L48:
  cmpl   $0x3,-0x24(%rbp)
  jbe    L45
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
void removeLimits() {
  static int res[] = {RLIMIT_CPU, RLIMIT_DATA, RLIMIT_FSIZE, RLIMIT_NPROC};
  for (unsigned i = 0; i < sizeof(res) / sizeof(int); i++) {
    struct rlimit rl;
    getrlimit(res[i], &rl);
    if (rl.rlim_max < RLIM_INFINITY) {
      rl.rlim_max = RLIM_INFINITY;
      setrlimit(res[i], &rl);
      getrlimit(res[i], &rl);
    }
    if (rl.rlim_cur < rl.rlim_max) {
      rl.rlim_cur = rl.rlim_max;
      setrlimit(res[i], &rl);
    }
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<memset>:
  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)
  jmp    L14
L3:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  mov    -0x1c(%rbp),%edx
  mov    %dl,(%rax)
L14:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L3
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *memset(void *s, int c, size_t n) {
  char *to = (char *)s;
  while (n-- > 0)
    *(to++) = c;
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<memset>:
  endbr64
  mov    %rdi,%rax
  test   %rdx,%rdx
  je     L16
  add    %rdi,%rdx
  mov    %rdi,%rcx
L4:
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  cmp    %rcx,%rdx
  jne    L4
L16:
  ret
```

assistant: ```c
void *memset(void *s, int c, size_t n) {
  char *to = (char *)s;
  while (n-- > 0)
    *(to++) = c;
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<memset>:
  endbr64
  test   %rdx,%rdx
  je     L3
  sub    $0x8,%rsp
  movsbl %sil,%esi
  call   <memset@plt>
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memset(void *s, int c, size_t n) {
  char *to = (char *)s;
  while (n-- > 0)
    *(to++) = c;
  return s;
}
```
user: What is the c source code of the assembly code below:

```asm
<memset>:
  endbr64
  test   %rdx,%rdx
  je     L2
  sub    $0x8,%rsp
  movsbl %sil,%esi
  call   <memset@plt>
  add    $0x8,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %rdi,%rax
  ret
```

assistant: ```c
void *memset(void *s, int c, size_t n) {
  char *to = (char *)s;
  while (n-- > 0)
    *(to++) = c;
  return s;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<shift>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x18(%rbp)
  je     L98
  movl   $0x1,-0x8(%rbp)
  jmp    L66
L17:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x18(%rbp)
  jne    L38
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L78
L23:
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0x4(%rbp)
L78:
  cmpl   $0x1,-0x4(%rbp)
  jg     L23
  jmp    L60
L38:
  nop
  addl   $0x1,-0x8(%rbp)
L66:
  mov    -0x8(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L17
L60:
  cmpl   $0x0,-0x14(%rbp)
  je     L35
  mov    -0x20(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    -0x18(%rbp),%eax
  mov    %eax,(%rdx)
  jmp    L35
L98:
  nop
  jmp    L81
L35:
  nop
L81:
  pop    %rbp
  ret
```

assistant: ```c
void shift(int n, int ii, int *iact) {
  int j, k;
  if (ii == iact[1])
    return;
  for (j = 1; j <= n; j++) {
    if (ii != iact[j])
      continue;
    for (k = j; k >= 2; k--)
      iact[k] = iact[k - 1];
    break;
  }
  if (n != 0)
    iact[1] = ii;
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<shift>:
  endbr64
  cmp    %esi,0x4(%rdx)
  je     L39
  mov    $0x1,%ecx
  test   %edi,%edi
  jle    L67
L43:
  mov    %ecx,%eax
  cmp    %esi,(%rdx,%rcx,4)
  jne    L71
  cmp    $0x1,%ecx
  jle    L16
  cltq
  lea    0x0(,%rax,4),%rdi
  lea    (%rdx,%rdi,1),%rax
  lea    -0x4(%rdx,%rdi,1),%rdi
  lea    -0x2(%rcx),%ecx
  shl    $0x2,%rcx
  sub    %rcx,%rdi
L3:
  mov    -0x4(%rax),%ecx
  mov    %ecx,(%rax)
  sub    $0x4,%rax
  cmp    %rdi,%rax
  jne    L3
L16:
  mov    %esi,0x4(%rdx)
  jmp    L23
L39:
  ret
L71:
  add    $0x1,%rcx
  cmp    %ecx,%edi
  jge    L43
  jmp    L16
L67:
  jne    L16
L23:
  ret
```

assistant: ```c
void shift(int n, int ii, int *iact) {
  int j, k;
  if (ii == iact[1])
    return;
  for (j = 1; j <= n; j++) {
    if (ii != iact[j])
      continue;
    for (k = j; k >= 2; k--)
      iact[k] = iact[k - 1];
    break;
  }
  if (n != 0)
    iact[1] = ii;
  return;
}
```
user: Translate the below assembly code into C programming language:

```asm
<shift>:
  endbr64
  cmp    %esi,0x4(%rdx)
  je     L43
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  mov    $0x1,%edx
  sub    $0x8,%rsp
  test   %edi,%edi
  jle    L6
L18:
  movslq %edx,%rax
  cmp    %ebp,(%rbx,%rdx,4)
  jne    L53
  cmp    $0x1,%eax
  je     L4
  lea    -0x2(%rax),%edx
  shl    $0x2,%rax
  mov    %rdx,%rdi
  lea    0x4(,%rdx,4),%rdx
  neg    %rdi
  shl    $0x2,%rdi
  lea    -0x4(%rax,%rdi,1),%rsi
  add    %rax,%rdi
  add    %rbx,%rsi
  add    %rbx,%rdi
  call   <memmove@plt>
L4:
  mov    %ebp,0x4(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L53:
  add    $0x1,%rdx
  cmp    %edx,%edi
  jge    L18
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L6:
  jne    L4
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L43:
  ret
```

assistant: ```c
void shift(int n, int ii, int *iact) {
  int j, k;
  if (ii == iact[1])
    return;
  for (j = 1; j <= n; j++) {
    if (ii != iact[j])
      continue;
    for (k = j; k >= 2; k--)
      iact[k] = iact[k - 1];
    break;
  }
  if (n != 0)
    iact[1] = ii;
  return;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<shift>:
  endbr64
  cmp    %esi,0x4(%rdx)
  je     L41
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdx,%rbx
  mov    $0x1,%edx
  sub    $0x8,%rsp
  test   %edi,%edi
  jle    L37
L11:
  movslq %edx,%rax
  cmp    %ebp,(%rbx,%rdx,4)
  jne    L62
  cmp    $0x1,%eax
  je     L17
  lea    -0x2(%rax),%edx
  shl    $0x2,%rax
  mov    %rdx,%rdi
  lea    0x4(,%rdx,4),%rdx
  neg    %rdi
  shl    $0x2,%rdi
  lea    -0x4(%rax,%rdi,1),%rsi
  add    %rax,%rdi
  add    %rbx,%rsi
  add    %rbx,%rdi
  call   <memmove@plt>
L17:
  mov    %ebp,0x4(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L62:
  add    $0x1,%rdx
  cmp    %edx,%edi
  jge    L11
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L37:
  jne    L17
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L41:
  ret
```

assistant: ```c
void shift(int n, int ii, int *iact) {
  int j, k;
  if (ii == iact[1])
    return;
  for (j = 1; j <= n; j++) {
    if (ii != iact[j])
      continue;
    for (k = j; k >= 2; k--)
      iact[k] = iact[k - 1];
    break;
  }
  if (n != 0)
    iact[1] = ii;
  return;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<multiply74>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L18
L3:
  movl   $0x0,-0x3c(%rbp)
  jmp    L35
L28:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L35:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L28
  addl   $0x1,-0x40(%rbp)
L18:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L3
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply74(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<multiply74>:
  endbr64
  test   %edi,%edi
  jle    L5
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L7:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L17:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L17
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L5:
  ret
```

assistant: ```c
void multiply74(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<multiply74>:
  endbr64
  test   %edi,%edi
  jle    L29
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L3:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L5:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L5
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L3
  pop    %rbx
  ret
L29:
  ret
```

assistant: ```c
void multiply74(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<multiply74>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L91
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L99:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L32
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L32
  nopl   0x0(%rax)
L15:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L15
  cmp    -0x24(%rsp),%r11d
  je     L12
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L12
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L12
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L12:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L99
L91:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L32:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L33:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L33
  mov    -0x18(%rsp),%r8
  jmp    L12
```

assistant: ```c
void multiply74(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getUserName>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <getuid@plt>
  mov    %eax,%ebx
  mov    %ebx,%edi
  call   <getpwuid@plt>
  mov    %rax,%rbx
  mov    (%rbx),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *getUserName() {
  register struct passwd *pw;
  register uid_t uid;
  uid = getuid();
  pw = getpwuid(uid);
  return pw->pw_name;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getUserName>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <getuid@plt>
  mov    %eax,%edi
  call   <getpwuid@plt>
  mov    (%rax),%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *getUserName() {
  register struct passwd *pw;
  register uid_t uid;
  uid = getuid();
  pw = getpwuid(uid);
  return pw->pw_name;
}
```