ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<basename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L11
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L14
L11:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L14:
  nop
  leave
  ret
```

assistant: ```c
void basename(char *name, char *path) {
  char *slash = strrchr(path, '/');
  if (slash)
    strcpy(name, slash + 1);
  else
    strcpy(name, path);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<basename>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    $0x2f,%esi
  mov    %rbx,%rdi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L11
  lea    0x1(%rax),%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
L19:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L11:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <strcpy@plt>
  jmp    L19
```

assistant: ```c
void basename(char *name, char *path) {
  char *slash = strrchr(path, '/');
  if (slash)
    strcpy(name, slash + 1);
  else
    strcpy(name, path);
}
```
user: What is the c source code of the assembly code below?

```asm
<basename>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x2f,%esi
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <strrchr@plt>
  lea    0x1(%rax),%rsi
  test   %rax,%rax
  je     L6
  add    $0x8,%rsp
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <strcpy@plt>
  cs nopw 0x0(%rax,%rax,1)
L6:
  add    $0x8,%rsp
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <strcpy@plt>
```

assistant: ```c
void basename(char *name, char *path) {
  char *slash = strrchr(path, '/');
  if (slash)
    strcpy(name, slash + 1);
  else
    strcpy(name, path);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<basename>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  mov    $0x2f,%esi
  mov    %rbp,%rdi
  sub    $0x8,%rsp
  call   <strrchr@plt>
  lea    0x1(%rax),%rsi
  test   %rax,%rax
  je     L4
  add    $0x8,%rsp
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <strcpy@plt>
  cs nopw 0x0(%rax,%rax,1)
L4:
  add    $0x8,%rsp
  mov    %rbp,%rsi
  mov    %r12,%rdi
  pop    %rbp
  pop    %r12
  jmp    1060 <strcpy@plt>
```

assistant: ```c
void basename(char *name, char *path) {
  char *slash = strrchr(path, '/');
  if (slash)
    strcpy(name, slash + 1);
  else
    strcpy(name, path);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<crypto_scalarmult>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1e0,%rsp
  mov    %rdi,-0x1c8(%rbp)
  mov    %rsi,-0x1d0(%rbp)
  mov    %rdx,-0x1d8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1b4(%rbp)
  jmp    L2
L67:
  mov    -0x1b4(%rbp),%edx
  mov    -0x1d0(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x1b4(%rbp),%eax
  mov    %dl,-0x30(%rbp,%rax,1)
  addl   $0x1,-0x1b4(%rbp)
L2:
  cmpl   $0x1f,-0x1b4(%rbp)
  jbe    L67
  movzbl -0x30(%rbp),%eax
  and    $0xfffffff8,%eax
  mov    %al,-0x30(%rbp)
  movzbl -0x11(%rbp),%eax
  and    $0x7f,%eax
  mov    %al,-0x11(%rbp)
  movzbl -0x11(%rbp),%eax
  or     $0x40,%eax
  mov    %al,-0x11(%rbp)
  movl   $0x0,-0x1b4(%rbp)
  jmp    L56
L30:
  mov    -0x1b4(%rbp),%edx
  mov    -0x1d8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x1b4(%rbp),%eax
  mov    %edx,-0x1b0(%rbp,%rax,4)
  addl   $0x1,-0x1b4(%rbp)
L56:
  cmpl   $0x1f,-0x1b4(%rbp)
  jbe    L30
  lea    -0x30(%rbp),%rdx
  lea    -0x1b0(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <mainloop@plt>
  lea    -0x1b0(%rbp),%rdx
  sub    $0xffffffffffffff80,%rdx
  lea    -0x1b0(%rbp),%rax
  sub    $0xffffffffffffff80,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <recip@plt>
  lea    -0x1b0(%rbp),%rdx
  sub    $0xffffffffffffff80,%rdx
  lea    -0x1b0(%rbp),%rax
  add    $0x100,%rax
  lea    -0x1b0(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <mult@plt>
  lea    -0x1b0(%rbp),%rax
  add    $0x100,%rax
  mov    %rax,%rdi
  call   <freeze@plt>
  movl   $0x0,-0x1b4(%rbp)
  jmp    L17
L59:
  mov    -0x1b4(%rbp),%eax
  add    $0x40,%eax
  mov    %eax,%eax
  mov    -0x1b0(%rbp,%rax,4),%edx
  mov    -0x1b4(%rbp),%ecx
  mov    -0x1c8(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  addl   $0x1,-0x1b4(%rbp)
L17:
  cmpl   $0x1f,-0x1b4(%rbp)
  jbe    L59
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

assistant: ```c
int crypto_scalarmult(unsigned char *q, const unsigned char *n, const unsigned char *p) {
  unsigned int work[96];
  unsigned char e[32];
  unsigned int i;
  for (i = 0; i < 32; ++i)
    e[i] = n[i];
  e[0] &= 248;
  e[31] &= 127;
  e[31] |= 64;
  for (i = 0; i < 32; ++i)
    work[i] = p[i];
  mainloop(work, e);
  recip(work + 32, work + 32);
  mult(work + 64, work, work + 32);
  freeze(work + 64);
  for (i = 0; i < 32; ++i)
    q[i] = work[64 + i];
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<crypto_scalarmult>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
L18:
  movzbl (%rsi,%rax,1),%ecx
  mov    %cl,0x180(%rsp,%rax,1)
  add    $0x1,%rax
  cmp    $0x20,%rax
  jne    L18
  andb   $0xf8,0x180(%rsp)
  movzbl 0x19f(%rsp),%eax
  and    $0x7f,%eax
  or     $0x40,%eax
  mov    %al,0x19f(%rsp)
  mov    $0x0,%eax
L2:
  movzbl (%rdx,%rax,1),%ecx
  mov    %ecx,(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    $0x20,%rax
  jne    L2
  lea    0x180(%rsp),%rsi
  mov    %rsp,%r13
  mov    %r13,%rdi
  call   <mainloop@plt>
  lea    0x80(%rsp),%rbp
  mov    %rbp,%rsi
  mov    %rbp,%rdi
  call   <recip@plt>
  lea    0x100(%rsp),%r12
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <mult@plt>
  mov    %r12,%rdi
  call   <freeze@plt>
  mov    $0x0,%eax
L14:
  mov    0x100(%rsp,%rax,4),%edx
  mov    %dl,(%rbx,%rax,1)
  add    $0x1,%rax
  cmp    $0x20,%rax
  jne    L14
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  mov    $0x0,%eax
  add    $0x1b8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int crypto_scalarmult(unsigned char *q, const unsigned char *n, const unsigned char *p) {
  unsigned int work[96];
  unsigned char e[32];
  unsigned int i;
  for (i = 0; i < 32; ++i)
    e[i] = n[i];
  e[0] &= 248;
  e[31] &= 127;
  e[31] |= 64;
  for (i = 0; i < 32; ++i)
    work[i] = p[i];
  mainloop(work, e);
  recip(work + 32, work + 32);
  mult(work + 64, work, work + 32);
  freeze(work + 64);
  for (i = 0; i < 32; ++i)
    q[i] = work[64 + i];
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<crypto_scalarmult>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x1b8,%rsp
  movdqu 0x10(%rsi),%xmm1
  movdqu (%rsi),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  movaps %xmm1,0x190(%rsp)
  movzbl 0x19f(%rsp),%eax
  movaps %xmm0,0x180(%rsp)
  andb   $0xf8,0x180(%rsp)
  and    $0x7f,%eax
  or     $0x40,%eax
  mov    %al,0x19f(%rsp)
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L8:
  movzbl (%rdx,%rax,1),%ecx
  mov    %ecx,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    $0x20,%rax
  jne    L8
  lea    0x80(%rsp),%r12
  lea    0x180(%rsp),%rsi
  mov    %rbx,%rdi
  call   <mainloop@plt>
  lea    0x100(%rsp),%r13
  mov    %r12,%rsi
  mov    %r12,%rdi
  call   <recip@plt>
  mov    %r12,%rdx
  mov    %rbx,%rsi
  mov    %r13,%rdi
  call   <mult@plt>
  mov    %r13,%rdi
  call   <freeze@plt>
  xor    %eax,%eax
  nopl   0x0(%rax)
L2:
  mov    0x100(%rbx,%rax,4),%edx
  mov    %dl,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  cmp    $0x20,%rax
  jne    L2
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x1b8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int crypto_scalarmult(unsigned char *q, const unsigned char *n, const unsigned char *p) {
  unsigned int work[96];
  unsigned char e[32];
  unsigned int i;
  for (i = 0; i < 32; ++i)
    e[i] = n[i];
  e[0] &= 248;
  e[31] &= 127;
  e[31] |= 64;
  for (i = 0; i < 32; ++i)
    work[i] = p[i];
  mainloop(work, e);
  recip(work + 32, work + 32);
  mult(work + 64, work, work + 32);
  freeze(work + 64);
  for (i = 0; i < 32; ++i)
    q[i] = work[64 + i];
  return 0;
}
```
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
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  jmp    L20
L42:
  cmpl   $0x20,-0x4(%rbp)
  jne    L19
  movl   $0x1,-0x8(%rbp)
  jmp    L20
L19:
  cmpl   $0x0,-0x8(%rbp)
  je     L20
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  movl   $0x0,-0x8(%rbp)
L20:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L42
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int c;
  int prev_was_blank = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ')
      prev_was_blank = 1;
    else if (prev_was_blank) {
      putchar(' ');
      putchar(c);
      prev_was_blank = 0;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%ebp
  mov    $0x1,%r12d
  jmp    L26
L16:
  mov    %r12d,%ebp
L26:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  mov    D2(%rip),%rax
  cmp    %ebx,(%rax)
  je     L17
  cmp    $0x20,%ebx
  je     L16
  test   %ebp,%ebp
  je     L26
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    %ebx,%edi
  call   <putchar@plt>
  mov    $0x0,%ebp
  jmp    L26
L17:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  int c;
  int prev_was_blank = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ')
      prev_was_blank = 1;
    else if (prev_was_blank) {
      putchar(' ');
      putchar(c);
      prev_was_blank = 0;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  mov    D8(%rip),%rbx
  jmp    L64
  nopl   0x0(%rax)
L83:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    (%rbx),%eax
  je     L45
  xor    %r12d,%r12d
  cmp    $0x20,%eax
  sete   %r12b
L64:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    %eax,(%rbx)
  je     L45
  cmp    $0x20,%ebp
  je     L58
  test   %r12d,%r12d
  je     L83
L55:
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    %ebp,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    (%rbx),%eax
  je     L45
  cmp    $0x20,%eax
  jne    L83
L58:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    (%rbx),%eax
  je     L45
  cmp    $0x20,%eax
  jne    L55
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    %eax,(%rbx)
  je     L45
  cmp    $0x20,%eax
  jne    L55
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L45:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  int c;
  int prev_was_blank = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ')
      prev_was_blank = 1;
    else if (prev_was_blank) {
      putchar(' ');
      putchar(c);
      prev_was_blank = 0;
    }
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  mov    D5(%rip),%rbx
  jmp    L116
  nopl   0x0(%rax)
L13:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    (%rbx),%eax
  je     L34
  xor    %r12d,%r12d
  cmp    $0x20,%eax
  sete   %r12b
L116:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    %eax,(%rbx)
  je     L34
  cmp    $0x20,%ebp
  je     L76
  test   %r12d,%r12d
  je     L13
L48:
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    %ebp,%edi
  call   <putchar@plt>
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    (%rbx),%eax
  je     L34
  cmp    $0x20,%eax
  jne    L13
L76:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    (%rbx),%eax
  je     L34
  cmp    $0x20,%eax
  jne    L48
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  cmp    %eax,(%rbx)
  je     L34
  cmp    $0x20,%eax
  jne    L48
  jmp    L76
  nopl   0x0(%rax,%rax,1)
L34:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int main() {
  int c;
  int prev_was_blank = 0;
  while ((c = getchar()) != EOF) {
    if (c == ' ')
      prev_was_blank = 1;
    else if (prev_was_blank) {
      putchar(' ');
      putchar(c);
      prev_was_blank = 0;
    }
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<rop3_131_1_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%ecx
  mov    %edx,%eax
  mov    %ecx,%edx
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %al,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  or     -0x4(%rbp),%al
  mov    %al,-0x3(%rbp)
  movzbl -0x20(%rbp),%eax
  and    -0x3(%rbp),%al
  mov    %al,-0x2(%rbp)
  movzbl -0x1c(%rbp),%eax
  xor    -0x2(%rbp),%al
  mov    %al,-0x1(%rbp)
  notb   -0x1(%rbp)
  movzbl -0x1c(%rbp),%eax
  not    %eax
  mov    %eax,%edx
  movzbl -0x1(%rbp),%eax
  and    %eax,%edx
  movzbl -0x20(%rbp),%eax
  not    %eax
  and    %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x1c(%rbp),%al
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  and    -0x20(%rbp),%al
  or     %edx,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void rop3_131_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  unsigned char stk4;
  stk4 = ~S;
  stk3 = *D | stk4;
  stk2 = T & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rop3_131_1_1>:
  endbr64
  mov    %edx,%eax
  or     %esi,%eax
  mov    %eax,%edx
  not    %edx
  and    (%rdi),%al
  or     %eax,%edx
  mov    %dl,(%rdi)
  ret
```

assistant: ```c
void rop3_131_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  unsigned char stk4;
  stk4 = ~S;
  stk3 = *D | stk4;
  stk2 = T & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<rop3_131_1_1>:
  endbr64
  mov    %edx,%eax
  or     %esi,%eax
  mov    %eax,%edx
  and    (%rdi),%al
  not    %edx
  or     %eax,%edx
  mov    %dl,(%rdi)
  ret
```

assistant: ```c
void rop3_131_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  unsigned char stk4;
  stk4 = ~S;
  stk3 = *D | stk4;
  stk2 = T & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rop3_131_1_1>:
  endbr64
  mov    %edx,%eax
  or     %esi,%eax
  mov    %eax,%edx
  and    (%rdi),%al
  not    %edx
  or     %eax,%edx
  mov    %dl,(%rdi)
  ret
```

assistant: ```c
void rop3_131_1_1(unsigned char *D, unsigned char S, unsigned char T) {
  unsigned char stk1;
  unsigned char stk2;
  unsigned char stk3;
  unsigned char stk4;
  stk4 = ~S;
  stk3 = *D | stk4;
  stk2 = T & stk3;
  stk1 = S ^ stk2;
  stk1 = ~stk1;
  *D = (stk1 & (~S) & (~T)) | (*D & S) | (*D & T);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<upper_power_of_two>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  subq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    %rax
  or     %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x2,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x4,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x8,%rax
  or     %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  shr    $0x10,%rax
  or     %rax,-0x8(%rbp)
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long upper_power_of_two(unsigned long v) {
  v--;
  v |= v >> 1;
  v |= v >> 2;
  v |= v >> 4;
  v |= v >> 8;
  v |= v >> 16;
  v++;
  return v;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<upper_power_of_two>:
  endbr64
  sub    $0x1,%rdi
  mov    %rdi,%rax
  shr    %rax
  or     %rdi,%rax
  mov    %rax,%rdi
  shr    $0x2,%rdi
  or     %rax,%rdi
  mov    %rdi,%rax
  shr    $0x4,%rax
  or     %rdi,%rax
  mov    %rax,%rdx
  shr    $0x8,%rdx
  or     %rax,%rdx
  mov    %rdx,%rax
  shr    $0x10,%rax
  or     %rdx,%rax
  add    $0x1,%rax
  ret
```

assistant: ```c
unsigned long upper_power_of_two(unsigned long v) {
  v--;
  v |= v >> 1;
  v |= v >> 2;
  v |= v >> 4;
  v |= v >> 8;
  v |= v >> 16;
  v++;
  return v;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<upper_power_of_two>:
  endbr64
  sub    $0x1,%rdi
  mov    %rdi,%rax
  shr    %rax
  or     %rdi,%rax
  mov    %rax,%rdi
  shr    $0x2,%rdi
  or     %rax,%rdi
  mov    %rdi,%rax
  shr    $0x4,%rax
  or     %rdi,%rax
  mov    %rax,%rdx
  shr    $0x8,%rdx
  or     %rax,%rdx
  mov    %rdx,%rax
  shr    $0x10,%rax
  or     %rdx,%rax
  add    $0x1,%rax
  ret
```

assistant: ```c
unsigned long upper_power_of_two(unsigned long v) {
  v--;
  v |= v >> 1;
  v |= v >> 2;
  v |= v >> 4;
  v |= v >> 8;
  v |= v >> 16;
  v++;
  return v;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<upper_power_of_two>:
  endbr64
  sub    $0x1,%rdi
  mov    %rdi,%rax
  shr    %rax
  or     %rdi,%rax
  mov    %rax,%rdi
  shr    $0x2,%rdi
  or     %rax,%rdi
  mov    %rdi,%rax
  shr    $0x4,%rax
  or     %rdi,%rax
  mov    %rax,%rdx
  shr    $0x8,%rdx
  or     %rax,%rdx
  mov    %rdx,%rax
  shr    $0x10,%rax
  or     %rdx,%rax
  add    $0x1,%rax
  ret
```

assistant: ```c
unsigned long upper_power_of_two(unsigned long v) {
  v--;
  v |= v >> 1;
  v |= v >> 2;
  v |= v >> 4;
  v |= v >> 8;
  v |= v >> 16;
  v++;
  return v;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<getGroupName>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    D4(%rip),%rax
  cmp    %rax,-0x18(%rbp)
  jne    L23
  mov    D27(%rip),%rax
  test   %rax,%rax
  je     L23
  mov    D27(%rip),%rax
  jmp    L20
L23:
  mov    -0x18(%rbp),%rax
  mov    %rax,D7(%rip)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <getgrgid@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L5
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  jmp    L29
L5:
  mov    $0x0,%eax
L29:
  mov    %rax,D7(%rip)
  mov    D7(%rip),%rax
L20:
  leave
  ret
```

assistant: ```c
const char *getGroupName(gid_t gid) {
  static const char *cachedName;
  static gid_t cachedGid;
  if (gid == cachedGid && cachedName) {
    return cachedName;
  } else {
    cachedGid = gid;
    struct group *group = getgrgid(gid);
    cachedName = group ? group->gr_name : NULL;
    return cachedName;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<getGroupName>:
  endbr64
  cmp    %rdi,D28(%rip)
  jne    L7
  mov    D4(%rip),%rax
  test   %rax,%rax
  je     L7
  ret
L7:
  sub    $0x8,%rsp
  mov    %rdi,D11(%rip)
  call   <getgrgid@plt>
  test   %rax,%rax
  je     L1
  mov    (%rax),%rax
L1:
  mov    %rax,D11(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
const char *getGroupName(gid_t gid) {
  static const char *cachedName;
  static gid_t cachedGid;
  if (gid == cachedGid && cachedName) {
    return cachedName;
  } else {
    cachedGid = gid;
    struct group *group = getgrgid(gid);
    cachedName = group ? group->gr_name : NULL;
    return cachedName;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<getGroupName>:
  endbr64
  cmp    %rdi,D1(%rip)
  jne    L24
  mov    D39(%rip),%rax
  test   %rax,%rax
  je     L24
  ret
  nopw   0x0(%rax,%rax,1)
L24:
  sub    $0x8,%rsp
  mov    %rdi,D28(%rip)
  call   <getgrgid@plt>
  test   %rax,%rax
  je     L0
  mov    (%rax),%rax
L0:
  mov    %rax,D28(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
const char *getGroupName(gid_t gid) {
  static const char *cachedName;
  static gid_t cachedGid;
  if (gid == cachedGid && cachedName) {
    return cachedName;
  } else {
    cachedGid = gid;
    struct group *group = getgrgid(gid);
    cachedName = group ? group->gr_name : NULL;
    return cachedName;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<getGroupName>:
  endbr64
  cmp    %rdi,D33(%rip)
  jne    L11
  mov    D2(%rip),%rax
  test   %rax,%rax
  je     L11
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  sub    $0x8,%rsp
  mov    %rdi,D38(%rip)
  call   <getgrgid@plt>
  test   %rax,%rax
  je     L17
  mov    (%rax),%rax
L17:
  mov    %rax,D38(%rip)
  add    $0x8,%rsp
  ret
```

assistant: ```c
const char *getGroupName(gid_t gid) {
  static const char *cachedName;
  static gid_t cachedGid;
  if (gid == cachedGid && cachedName) {
    return cachedName;
  } else {
    cachedGid = gid;
    struct group *group = getgrgid(gid);
    cachedName = group ? group->gr_name : NULL;
    return cachedName;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ctlogpn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %esi,-0xcc(%rbp)
  mov    %rdx,-0xd8(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L8
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L8:
  mov    -0xcc(%rbp),%edx
  mov    -0xc8(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xd8(%rbp),%rdx
  mov    -0xb4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%edx
  mov    -0xd8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <vprintf@plt>
  mov    -0xb4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

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

tool: D0: %s:%d: 

assistant: ```c
void ctlogpn(char *p, int n, char *fmt, ...) {
  va_list arg;
  printf("%s:%d: ", p, n);
  va_start(arg, fmt);
  vprintf(fmt, arg);
  va_end(arg);
  putchar('\n');
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ctlogpn>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%rbx
  mov    %esi,%edx
  mov    %rdi,%rsi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <vprintf@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D9
datatype: string

tool: D9: %s:%d: 

assistant: ```c
void ctlogpn(char *p, int n, char *fmt, ...) {
  va_list arg;
  printf("%s:%d: ", p, n);
  va_start(arg, fmt);
  vprintf(fmt, arg);
  va_end(arg);
  putchar('\n');
}
```