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

```asm
<print_tag>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L5
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L19
L14:
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L19:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jb     L14
  mov    $0xa,%edi
  call   <putchar@plt>
L5:
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
type_guessed: string
name: extract_symbolic_data
arguments: label: D7
type_guessed: string

tool: D0: Tag: 

tool: D7: %2.2x 

assistant: ```c
void print_tag(const void *out, uint32_t outlen) {
  unsigned i;
  if (out != NULL) {
    printf("Tag: ");
    for (i = 0; i < outlen; ++i) {
      printf("%2.2x ", ((uint8_t *)out)[i]);
    }
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_tag>:
  endbr64
  test   %rdi,%rdi
  je     L29
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r12d
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %r12d,%r12d
  je     L18
  mov    %rbp,%rbx
  lea    -0x1(%r12),%eax
  lea    0x0(%rbp,%rax,4),%r12
  lea    D11(%rip),%rbp
L26:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rax
  add    $0x4,%rbx
  cmp    %r12,%rax
  jne    L26
L18:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L29:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D9
type_infered: string
name: extract_symbolic_data
arguments: data_label: D11
type_infered: string

tool: D9: Tag: 

tool: D11: %2.2x 

assistant: ```c
void print_tag(const void *out, uint32_t outlen) {
  unsigned i;
  if (out != NULL) {
    printf("Tag: ");
    for (i = 0; i < outlen; ++i) {
      printf("%2.2x ", ((uint8_t *)out)[i]);
    }
    printf("\n");
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<print_tag>:
  endbr64
  test   %rdi,%rdi
  je     L15
  push   %r12
  xor    %eax,%eax
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    D13(%rip),%rdi
  call   <printf@plt>
  test   %ebp,%ebp
  je     L19
  lea    -0x1(%rbp),%eax
  lea    D0(%rip),%rbp
  lea    (%rbx,%rax,4),%r12
  nopw   0x0(%rax,%rax,1)
L16:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rax
  add    $0x4,%rbx
  cmp    %r12,%rax
  jne    L16
L19:
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  xchg   %ax,%ax
L15:
  ret
```

tool_calls: name: read_data
arguments: tag: D13
type_guessed: string
name: read_data
arguments: tag: D0
type_guessed: string

tool: D13: Tag: 

tool: D0: %2.2x 

assistant: ```c
void print_tag(const void *out, uint32_t outlen) {
  unsigned i;
  if (out != NULL) {
    printf("Tag: ");
    for (i = 0; i < outlen; ++i) {
      printf("%2.2x ", ((uint8_t *)out)[i]);
    }
    printf("\n");
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_tag>:
  endbr64
  test   %rdi,%rdi
  je     L13
  push   %r12
  xor    %eax,%eax
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  lea    D7(%rip),%rdi
  call   <printf@plt>
  test   %ebp,%ebp
  je     L27
  lea    -0x1(%rbp),%eax
  lea    D19(%rip),%rbp
  lea    (%rbx,%rax,4),%r12
  nopw   0x0(%rax,%rax,1)
L0:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rax
  add    $0x4,%rbx
  cmp    %r12,%rax
  jne    L0
L27:
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  xchg   %ax,%ax
L13:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D7
data_type: string
name: disassemble_data
arguments: data_label: D19
data_type: string

tool: D7: Tag: 

tool: D19: %2.2x 

assistant: ```c
void print_tag(const void *out, uint32_t outlen) {
  unsigned i;
  if (out != NULL) {
    printf("Tag: ");
    for (i = 0; i < outlen; ++i) {
      printf("%2.2x ", ((uint8_t *)out)[i]);
    }
    printf("\n");
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x7f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x7e8(%rbp)
  jmp    L13
L43:
  mov    -0x7e4(%rbp),%eax
  cmp    -0x7e8(%rbp),%eax
  jle    L13
  mov    -0x7e4(%rbp),%eax
  mov    %eax,-0x7e8(%rbp)
  lea    -0x7e0(%rbp),%rdx
  lea    -0x3f0(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <copy@plt>
L13:
  lea    -0x7e0(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <getlinetest@plt>
  mov    %eax,-0x7e4(%rbp)
  cmpl   $0x0,-0x7e4(%rbp)
  jg     L43
  cmpl   $0x0,-0x7e8(%rbp)
  jle    L36
  lea    -0x3f0(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L36:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L40
  call   <__stack_chk_fail@plt>
L40:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type: string

tool: D0: %s

assistant: ```c
main() {
  int len;
  int max;
  char line[1000];
  char longest[1000];
  max = 0;
  while ((len = getlinetest(line, 1000)) > 0)
    if (len > max) {
      max = len;
      copy(longest, line);
    }
  if (max > 0)
    printf("%s", longest);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x7e0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x7d8(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebp
  mov    %rsp,%r12
L4:
  mov    $0x3e8,%esi
  mov    %r12,%rdi
  call   <getlinetest@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L39
  cmp    %ebx,%ebp
  jge    L4
  lea    0x3f0(%rsp),%rdi
  mov    %r12,%rsi
  call   <copy@plt>
  mov    %ebx,%ebp
  jmp    L4
L39:
  test   %ebp,%ebp
  jg     L25
L26:
  mov    0x7d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L50
  mov    $0x0,%eax
  add    $0x7e0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  lea    0x3f0(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L26
L50:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D3
datatype: string

tool: D3: %s

assistant: ```c
main() {
  int len;
  int max;
  char line[1000];
  char longest[1000];
  max = 0;
  while ((len = getlinetest(line, 1000)) > 0)
    if (len > max) {
      max = len;
      copy(longest, line);
    }
  if (max > 0)
    printf("%s", longest);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x7e8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x7d8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x3f0(%rsp),%r13
  jmp    L58
  nopl   0x0(%rax,%rax,1)
L53:
  cmp    %ebx,%ebp
  jl     L38
L58:
  mov    $0x3e8,%esi
  mov    %r12,%rdi
  call   <getlinetest@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L53
  test   %ebp,%ebp
  jne    L24
L33:
  mov    0x7d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L41
  add    $0x7e8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L38:
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %ebx,%ebp
  call   <copy@plt>
  jmp    L58
L24:
  lea    0x3f0(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  jmp    L33
L41:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D6
type: string

tool: D6: %s

assistant: ```c
main() {
  int len;
  int max;
  char line[1000];
  char longest[1000];
  max = 0;
  while ((len = getlinetest(line, 1000)) > 0)
    if (len > max) {
      max = len;
      copy(longest, line);
    }
  if (max > 0)
    printf("%s", longest);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x7e8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x7d8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x3f0(%rsp),%r13
  jmp    L65
  nopl   0x0(%rax,%rax,1)
L40:
  cmp    %ebx,%ebp
  jl     L5
L65:
  mov    $0x3e8,%esi
  mov    %r12,%rdi
  call   <getlinetest@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jg     L40
  test   %ebp,%ebp
  jne    L4
L34:
  mov    0x7d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x7e8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    %r12,%rsi
  mov    %r13,%rdi
  mov    %ebx,%ebp
  call   <copy@plt>
  jmp    L65
L4:
  lea    0x3f0(%rsp),%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  jmp    L34
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D1
type: string

tool: D1: %s

assistant: ```c
main() {
  int len;
  int max;
  char line[1000];
  char longest[1000];
  max = 0;
  while ((len = getlinetest(line, 1000)) > 0)
    if (len > max) {
      max = len;
      copy(longest, line);
    }
  if (max > 0)
    printf("%s", longest);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<setlimit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  lea    -0x10(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <setrlimit@plt>
  test   %rax,%rax
  jns    L16
  mov    $0x0,%edi
  call   <error@plt>
L16:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
void setlimit(int resource, rlim_t n) {
  struct rlimit limit;
  limit.rlim_cur = limit.rlim_max = n;
  if (setrlimit(resource, &limit) < 0)
    error(NULL);
}
```
user: What is the c source code of the assembly code below?

```asm
<setlimit>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %esi,(%rsp)
  mov    %esi,0x4(%rsp)
  mov    %rsp,%rsi
  call   <setrlimit@plt>
  test   %rax,%rax
  js     L12
L0:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x18,%rsp
  ret
L12:
  mov    $0x0,%edi
  call   <error@plt>
  jmp    L0
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void setlimit(int resource, rlim_t n) {
  struct rlimit limit;
  limit.rlim_cur = limit.rlim_max = n;
  if (setrlimit(resource, &limit) < 0)
    error(NULL);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<setlimit>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %esi,(%rsp)
  mov    %esi,0x4(%rsp)
  mov    %rsp,%rsi
  call   <setrlimit@plt>
  test   %rax,%rax
  js     L5
L8:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax)
L5:
  xor    %edi,%edi
  call   <error@plt>
  jmp    L8
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void setlimit(int resource, rlim_t n) {
  struct rlimit limit;
  limit.rlim_cur = limit.rlim_max = n;
  if (setrlimit(resource, &limit) < 0)
    error(NULL);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<setlimit>:
  endbr64
  sub    $0x18,%rsp
  movd   %esi,%xmm1
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  pshufd $0xe0,%xmm1,%xmm0
  mov    %rsp,%rsi
  movq   %xmm0,(%rsp)
  call   <setrlimit@plt>
  test   %rax,%rax
  js     L21
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x18,%rsp
  ret
  nopl   0x0(%rax,%rax,1)
L21:
  xor    %edi,%edi
  call   <error@plt>
  jmp    L9
L24:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void setlimit(int resource, rlim_t n) {
  struct rlimit limit;
  limit.rlim_cur = limit.rlim_max = n;
  if (setrlimit(resource, &limit) < 0)
    error(NULL);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<is_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x2,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  jmp    L24
L33:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L31
  movl   $0x0,-0x8(%rbp)
L31:
  addl   $0x1,-0x4(%rbp)
L24:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L33
  cmpl   $0x0,-0x14(%rbp)
  jg     L7
  movl   $0xffffffff,-0x8(%rbp)
L7:
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int is_prime(int n) {
  int c, i;
  i = 2;
  c = 1;
  while (i < n) {
    if (n % i == 0) {
      c = 0;
    }
    i++;
  }
  if (n <= 0) {
    c = -1;
  }
  return c;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<is_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L26
  mov    $0x2,%ecx
  mov    $0x1,%esi
L23:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  cmove  %edx,%esi
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L23
L15:
  mov    %esi,%eax
  ret
L26:
  test   %edi,%edi
  setg   %sil
  movzbl %sil,%esi
  lea    -0x1(%rsi,%rsi,1),%esi
  jmp    L15
```

assistant: ```c
int is_prime(int n) {
  int c, i;
  i = 2;
  c = 1;
  while (i < n) {
    if (n % i == 0) {
      c = 0;
    }
    i++;
  }
  if (n <= 0) {
    c = -1;
  }
  return c;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<is_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L1
  mov    $0x2,%ecx
  mov    $0x1,%r8d
  xor    %esi,%esi
  cs nopw 0x0(%rax,%rax,1)
L9:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  cmove  %esi,%r8d
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L9
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  xor    %r8d,%r8d
  test   %edi,%edi
  setg   %r8b
  lea    -0x1(%r8,%r8,1),%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_prime(int n) {
  int c, i;
  i = 2;
  c = 1;
  while (i < n) {
    if (n % i == 0) {
      c = 0;
    }
    i++;
  }
  if (n <= 0) {
    c = -1;
  }
  return c;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<is_prime>:
  endbr64
  cmp    $0x2,%edi
  jle    L11
  mov    $0x2,%ecx
  mov    $0x1,%r8d
  xor    %esi,%esi
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  cmove  %esi,%r8d
  add    $0x1,%ecx
  cmp    %ecx,%edi
  jne    L3
  mov    %r8d,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L11:
  xor    %r8d,%r8d
  test   %edi,%edi
  setg   %r8b
  lea    -0x1(%r8,%r8,1),%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_prime(int n) {
  int c, i;
  i = 2;
  c = 1;
  while (i < n) {
    if (n % i == 0) {
      c = 0;
    }
    i++;
  }
  if (n <= 0) {
    c = -1;
  }
  return c;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<aes_shiftRows>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  mov    %rdi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl 0x1(%rax),%ebx
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl 0x5(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  lea    0x5(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl 0x9(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  lea    0x9(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl 0xd(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  add    $0xd,%rax
  mov    %bl,(%rax)
  mov    -0x10(%rbp),%rax
  movzbl 0xa(%rax),%ebx
  mov    -0x10(%rbp),%rax
  lea    0xa(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl 0x2(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  add    $0x2,%rax
  mov    %bl,(%rax)
  mov    -0x10(%rbp),%rax
  movzbl 0x3(%rax),%ebx
  mov    -0x10(%rbp),%rax
  lea    0x3(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl 0xf(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  lea    0xf(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl 0xb(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  lea    0xb(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl 0x7(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  add    $0x7,%rax
  mov    %bl,(%rax)
  mov    -0x10(%rbp),%rax
  movzbl 0xe(%rax),%ebx
  mov    -0x10(%rbp),%rax
  lea    0xe(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl 0x6(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  add    $0x6,%rax
  mov    %bl,(%rax)
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void aes_shiftRows(unsigned char *buf) {
  register unsigned char i, j;
  i = buf[1];
  buf[1] = buf[5];
  buf[5] = buf[9];
  buf[9] = buf[13];
  buf[13] = i;
  i = buf[10];
  buf[10] = buf[2];
  buf[2] = i;
  j = buf[3];
  buf[3] = buf[15];
  buf[15] = buf[11];
  buf[11] = buf[7];
  buf[7] = j;
  j = buf[14];
  buf[14] = buf[6];
  buf[6] = j;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<aes_shiftRows>:
  endbr64
  movzbl 0x1(%rdi),%eax
  movzbl 0x5(%rdi),%edx
  mov    %dl,0x1(%rdi)
  movzbl 0x9(%rdi),%edx
  mov    %dl,0x5(%rdi)
  movzbl 0xd(%rdi),%edx
  mov    %dl,0x9(%rdi)
  mov    %al,0xd(%rdi)
  movzbl 0xa(%rdi),%eax
  movzbl 0x2(%rdi),%edx
  mov    %dl,0xa(%rdi)
  mov    %al,0x2(%rdi)
  movzbl 0x3(%rdi),%eax
  movzbl 0xf(%rdi),%edx
  mov    %dl,0x3(%rdi)
  movzbl 0xb(%rdi),%edx
  mov    %dl,0xf(%rdi)
  movzbl 0x7(%rdi),%edx
  mov    %dl,0xb(%rdi)
  mov    %al,0x7(%rdi)
  movzbl 0xe(%rdi),%eax
  movzbl 0x6(%rdi),%edx
  mov    %dl,0xe(%rdi)
  mov    %al,0x6(%rdi)
  ret
```

assistant: ```c
void aes_shiftRows(unsigned char *buf) {
  register unsigned char i, j;
  i = buf[1];
  buf[1] = buf[5];
  buf[5] = buf[9];
  buf[9] = buf[13];
  buf[13] = i;
  i = buf[10];
  buf[10] = buf[2];
  buf[2] = i;
  j = buf[3];
  buf[3] = buf[15];
  buf[15] = buf[11];
  buf[11] = buf[7];
  buf[7] = j;
  j = buf[14];
  buf[14] = buf[6];
  buf[6] = j;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<aes_shiftRows>:
  endbr64
  movzbl 0x5(%rdi),%edx
  movzbl 0x1(%rdi),%eax
  mov    %dl,0x1(%rdi)
  movzbl 0x9(%rdi),%edx
  mov    %dl,0x5(%rdi)
  movzbl 0xd(%rdi),%edx
  mov    %al,0xd(%rdi)
  movzbl 0xa(%rdi),%eax
  mov    %dl,0x9(%rdi)
  movzbl 0x2(%rdi),%edx
  mov    %al,0x2(%rdi)
  movzbl 0x3(%rdi),%eax
  mov    %dl,0xa(%rdi)
  movzbl 0xf(%rdi),%edx
  mov    %dl,0x3(%rdi)
  movzbl 0xb(%rdi),%edx
  mov    %dl,0xf(%rdi)
  movzbl 0x7(%rdi),%edx
  mov    %al,0x7(%rdi)
  movzbl 0xe(%rdi),%eax
  mov    %dl,0xb(%rdi)
  movzbl 0x6(%rdi),%edx
  mov    %al,0x6(%rdi)
  mov    %dl,0xe(%rdi)
  ret
```

assistant: ```c
void aes_shiftRows(unsigned char *buf) {
  register unsigned char i, j;
  i = buf[1];
  buf[1] = buf[5];
  buf[5] = buf[9];
  buf[9] = buf[13];
  buf[13] = i;
  i = buf[10];
  buf[10] = buf[2];
  buf[2] = i;
  j = buf[3];
  buf[3] = buf[15];
  buf[15] = buf[11];
  buf[11] = buf[7];
  buf[7] = j;
  j = buf[14];
  buf[14] = buf[6];
  buf[6] = j;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<aes_shiftRows>:
  endbr64
  movzbl 0x5(%rdi),%edx
  movzbl 0x1(%rdi),%eax
  mov    %dl,0x1(%rdi)
  movzbl 0x9(%rdi),%edx
  mov    %dl,0x5(%rdi)
  movzbl 0xd(%rdi),%edx
  mov    %al,0xd(%rdi)
  movzbl 0xa(%rdi),%eax
  mov    %dl,0x9(%rdi)
  movzbl 0x2(%rdi),%edx
  mov    %al,0x2(%rdi)
  movzbl 0x3(%rdi),%eax
  mov    %dl,0xa(%rdi)
  movzbl 0xf(%rdi),%edx
  mov    %dl,0x3(%rdi)
  movzbl 0xb(%rdi),%edx
  mov    %dl,0xf(%rdi)
  movzbl 0x7(%rdi),%edx
  mov    %al,0x7(%rdi)
  movzbl 0xe(%rdi),%eax
  mov    %dl,0xb(%rdi)
  movzbl 0x6(%rdi),%edx
  mov    %al,0x6(%rdi)
  mov    %dl,0xe(%rdi)
  ret
```

assistant: ```c
void aes_shiftRows(unsigned char *buf) {
  register unsigned char i, j;
  i = buf[1];
  buf[1] = buf[5];
  buf[5] = buf[9];
  buf[9] = buf[13];
  buf[13] = i;
  i = buf[10];
  buf[10] = buf[2];
  buf[2] = i;
  j = buf[3];
  buf[3] = buf[15];
  buf[15] = buf[11];
  buf[11] = buf[7];
  buf[7] = j;
  j = buf[14];
  buf[14] = buf[6];
  buf[6] = j;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fabs>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movzwl -0x10(%rbp),%eax
  and    $0x7fff,%ax
  mov    %ax,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double fabs(x)
double x;
{
  union {
    double d;
    short i[4];
  } u;
  u.d = x;
  u.i[0] &= 0x7fff;
  return (u.d);
}
```
user: Translate the below assembly code into C programming language:

```asm
<fabs>:
  endbr64
  movq   %xmm0,%rax
  and    $0x7fff,%ax
  movq   %rax,%xmm0
  ret
```

assistant: ```c
double fabs(x)
double x;
{
  union {
    double d;
    short i[4];
  } u;
  u.d = x;
  u.i[0] &= 0x7fff;
  return (u.d);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<fabs>:
  endbr64
  movq   %xmm0,%rax
  and    $0x7fff,%ax
  movq   %rax,%xmm0
  ret
```

assistant: ```c
double fabs(x)
double x;
{
  union {
    double d;
    short i[4];
  } u;
  u.d = x;
  u.i[0] &= 0x7fff;
  return (u.d);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fabs>:
  endbr64
  movq   %xmm0,%rax
  and    $0x7fff,%ax
  movq   %rax,%xmm0
  ret
```

assistant: ```c
double fabs(x)
double x;
{
  union {
    double d;
    short i[4];
  } u;
  u.d = x;
  u.i[0] &= 0x7fff;
  return (u.d);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  mov    $0x16,%edx
  imul   %edx,%eax
  add    $0x1c,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L65
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L9
  movzbl -0x9(%rbp),%eax
  movzbl %al,%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L20
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  jmp    L69
L20:
  movzbl -0x9(%rbp),%eax
  movzbl -0x9(%rbp),%edx
  imul   %edx,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  jmp    L69
L9:
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  movzbl -0x9(%rbp),%edx
  imul   %edx,%eax
  mov    %al,-0x9(%rbp)
  jmp    L69
L65:
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  add    %eax,%eax
  mov    %al,-0x9(%rbp)
L69:
  movzbl -0x9(%rbp),%eax
  lea    0x5(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L29
  call   <__stack_chk_fail@plt>
L29:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned char input[1], unsigned char output[1]) {
  unsigned char state[1];
  {
    state[0UL] = (input[0UL] + 914778474UL) * (unsigned char)22;
    if (state[0UL] & (unsigned char)1) {
      if (state[0UL] & (unsigned char)1) {
        if (state[0UL] & (unsigned char)1) {
          state[0UL] += state[0UL];
          state[0UL] += state[0UL];
        } else {
          state[0UL] *= state[0UL];
          state[0UL] += state[0UL];
        }
      } else {
        state[0UL] += state[0UL];
        state[0UL] *= state[0UL];
      }
    } else {
      state[0UL] += state[0UL];
      state[0UL] += state[0UL];
    }
    output[0UL] = state[0UL] + 634099461UL;
  }
}
```