ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<read_16_be>:
  endbr64
  movzbl (%rdi),%eax
  shl    $0x8,%eax
  movzbl 0x1(%rdi),%edx
  or     %edx,%eax
  ret
```

assistant: ```c
uint16_t read_16_be(unsigned char bytes[2]) {
  uint32_t result = 0;
  for (int i = 0; i < 2; i++)
    result = (result << 8) | bytes[i];
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<read_16_be>:
  endbr64
  movzwl (%rdi),%eax
  rol    $0x8,%ax
  movzwl %ax,%eax
  ret
```

assistant: ```c
uint16_t read_16_be(unsigned char bytes[2]) {
  uint32_t result = 0;
  for (int i = 0; i < 2; i++)
    result = (result << 8) | bytes[i];
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_16_be>:
  endbr64
  movzwl (%rdi),%eax
  rol    $0x8,%ax
  movzwl %ax,%eax
  ret
```

assistant: ```c
uint16_t read_16_be(unsigned char bytes[2]) {
  uint32_t result = 0;
  for (int i = 0; i < 2; i++)
    result = (result << 8) | bytes[i];
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<setbit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int setbit(int num, int position, int new_number) {
  new_number = num | 1 << position;
  return new_number;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<setbit>:
  endbr64
  mov    %edi,%eax
  bts    %esi,%eax
  ret
```

assistant: ```c
int setbit(int num, int position, int new_number) {
  new_number = num | 1 << position;
  return new_number;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<setbit>:
  endbr64
  mov    %edi,%eax
  bts    %esi,%eax
  ret
```

assistant: ```c
int setbit(int num, int position, int new_number) {
  new_number = num | 1 << position;
  return new_number;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<setbit>:
  endbr64
  mov    %edi,%eax
  bts    %esi,%eax
  ret
```

assistant: ```c
int setbit(int num, int position, int new_number) {
  new_number = num | 1 << position;
  return new_number;
}
```
user: What is the c source code of the assembly code below:

```asm
<getstring>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    $0xff,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L33
L2:
  mov    D4(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <getc@plt>
  mov    %al,-0xd(%rbp)
  cmpb   $0xa,-0xd(%rbp)
  jne    L25
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  jmp    L22
L25:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L33:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L2
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L22:
  leave
  ret
```

assistant: ```c
char *getstring(int length) {
  char *buffer = (char *)malloc(255);
  char buff_char;
  for (int i = 0; i < length; i++) {
    buff_char = getc(stdin);
    if (buff_char == '\n') {
      buffer[i] = '\0';
      return buffer;
    } else {
      buffer[i] = buff_char;
    }
  }
  buffer[length] = '\0';
  return buffer;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<getstring>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r15d
  mov    $0xff,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %r15d,%r15d
  jle    L13
  mov    %r15d,%r14d
  mov    $0x0,%ebx
  mov    D6(%rip),%r13
L0:
  mov    0x0(%r13),%edi
  call   <getc@plt>
  cmp    $0xa,%al
  je     L9
  mov    %al,0x0(%rbp,%rbx,1)
  add    $0x1,%rbx
  cmp    %r14,%rbx
  jne    L0
L13:
  movslq %r15d,%r15
  movb   $0x0,0x0(%rbp,%r15,1)
L10:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  movslq %ebx,%r12
  movb   $0x0,0x0(%rbp,%r12,1)
  jmp    L10
```

assistant: ```c
char *getstring(int length) {
  char *buffer = (char *)malloc(255);
  char buff_char;
  for (int i = 0; i < length; i++) {
    buff_char = getc(stdin);
    if (buff_char == '\n') {
      buffer[i] = '\0';
      return buffer;
    } else {
      buffer[i] = buff_char;
    }
  }
  buffer[length] = '\0';
  return buffer;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<getstring>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  movslq %edi,%r12
  mov    $0xff,%edi
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L23
  mov    D2(%rip),%rbx
  mov    %rax,%r14
  lea    (%rax,%r12,1),%rbp
  jmp    L34
  nopl   0x0(%rax)
L15:
  mov    %al,(%r14)
  add    $0x1,%r14
  cmp    %rbp,%r14
  je     L23
L34:
  mov    (%rbx),%edi
  call   <getc@plt>
  cmp    $0xa,%al
  jne    L15
  movb   $0x0,(%r14)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L23:
  movb   $0x0,0x0(%r13,%r12,1)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *getstring(int length) {
  char *buffer = (char *)malloc(255);
  char buff_char;
  for (int i = 0; i < length; i++) {
    buff_char = getc(stdin);
    if (buff_char == '\n') {
      buffer[i] = '\0';
      return buffer;
    } else {
      buffer[i] = buff_char;
    }
  }
  buffer[length] = '\0';
  return buffer;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<getstring>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  movslq %edi,%r12
  mov    $0xff,%edi
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  mov    %rax,%r13
  test   %r12d,%r12d
  jle    L5
  mov    D8(%rip),%rbx
  mov    %rax,%r14
  lea    (%rax,%r12,1),%rbp
  jmp    L14
  nopl   0x0(%rax)
L17:
  mov    %al,(%r14)
  add    $0x1,%r14
  cmp    %rbp,%r14
  je     L5
L14:
  mov    (%rbx),%edi
  call   <getc@plt>
  cmp    $0xa,%al
  jne    L17
  movb   $0x0,(%r14)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nop
L5:
  movb   $0x0,0x0(%r13,%r12,1)
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *getstring(int length) {
  char *buffer = (char *)malloc(255);
  char buff_char;
  for (int i = 0; i < length; i++) {
    buff_char = getc(stdin);
    if (buff_char == '\n') {
      buffer[i] = '\0';
      return buffer;
    } else {
      buffer[i] = buff_char;
    }
  }
  buffer[length] = '\0';
  return buffer;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<b64ton>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x7,%rax
  movzbl (%rax),%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x6,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x5,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x2,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x4,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x3,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x5,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x6,%rax
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%rax
  add    $0x7,%rax
  mov    %dl,(%rax)
  addq   $0x8,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *b64ton(char *data, double val) {
  char *p = (char *)data;
  unsigned char *p2 = (unsigned char *)&val;
  p[0] = p2[7];
  p[1] = p2[6];
  p[2] = p2[5];
  p[3] = p2[4];
  p[4] = p2[3];
  p[5] = p2[2];
  p[6] = p2[1];
  p[7] = p2[0];
  p += 8;
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<b64ton>:
  endbr64
  movq   %xmm0,%rax
  movq   %xmm0,%rdx
  shr    $0x38,%rdx
  mov    %dl,(%rdi)
  movq   %xmm0,%rdx
  shr    $0x30,%rdx
  mov    %dl,0x1(%rdi)
  movq   %xmm0,%rdx
  shr    $0x28,%rdx
  mov    %dl,0x2(%rdi)
  movq   %xmm0,%rdx
  shr    $0x20,%rdx
  mov    %dl,0x3(%rdi)
  movq   %xmm0,%rdx
  shr    $0x18,%rdx
  mov    %dl,0x4(%rdi)
  movq   %xmm0,%rdx
  shr    $0x10,%rdx
  mov    %dl,0x5(%rdi)
  movzbl %ah,%ecx
  mov    %cl,0x6(%rdi)
  mov    %al,0x7(%rdi)
  lea    0x8(%rdi),%rax
  ret
```

assistant: ```c
char *b64ton(char *data, double val) {
  char *p = (char *)data;
  unsigned char *p2 = (unsigned char *)&val;
  p[0] = p2[7];
  p[1] = p2[6];
  p[2] = p2[5];
  p[3] = p2[4];
  p[4] = p2[3];
  p[5] = p2[2];
  p[6] = p2[1];
  p[7] = p2[0];
  p += 8;
  return p;
}
```
user: Translate the below assembly code into C programming language:

```asm
<b64ton>:
  endbr64
  movq   %xmm0,%rdx
  movq   %xmm0,%rax
  shr    $0x38,%rdx
  mov    %ah,0x6(%rdi)
  mov    %dl,(%rdi)
  movq   %xmm0,%rdx
  shr    $0x30,%rdx
  mov    %al,0x7(%rdi)
  lea    0x8(%rdi),%rax
  mov    %dl,0x1(%rdi)
  movq   %xmm0,%rdx
  shr    $0x28,%rdx
  mov    %dl,0x2(%rdi)
  movq   %xmm0,%rdx
  shr    $0x20,%rdx
  mov    %dl,0x3(%rdi)
  movq   %xmm0,%rdx
  shr    $0x18,%rdx
  mov    %dl,0x4(%rdi)
  movq   %xmm0,%rdx
  shr    $0x10,%rdx
  mov    %dl,0x5(%rdi)
  ret
```

assistant: ```c
char *b64ton(char *data, double val) {
  char *p = (char *)data;
  unsigned char *p2 = (unsigned char *)&val;
  p[0] = p2[7];
  p[1] = p2[6];
  p[2] = p2[5];
  p[3] = p2[4];
  p[4] = p2[3];
  p[5] = p2[2];
  p[6] = p2[1];
  p[7] = p2[0];
  p += 8;
  return p;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<b64ton>:
  endbr64
  movq   %xmm0,%rcx
  push   %rbx
  movq   %xmm0,%r10
  movq   %xmm0,%r9
  movzbl %cl,%edx
  movzbl %ch,%ebx
  shr    $0x10,%r10
  movq   %xmm0,%r8
  shl    $0x8,%rdx
  movzbl %r10b,%r10d
  shr    $0x18,%r9
  mov    %rdi,%rax
  or     %rbx,%rdx
  movzbl %r9b,%r9d
  shr    $0x20,%r8
  movq   %xmm0,%rdi
  shl    $0x8,%rdx
  movzbl %r8b,%r8d
  shr    $0x28,%rdi
  add    $0x8,%rax
  or     %r10,%rdx
  movq   %xmm0,%rsi
  movzbl %dil,%edi
  shr    $0x38,%rcx
  shl    $0x8,%rdx
  shr    $0x30,%rsi
  or     %r9,%rdx
  movzbl %sil,%esi
  shl    $0x8,%rdx
  or     %r8,%rdx
  shl    $0x8,%rdx
  or     %rdi,%rdx
  shl    $0x8,%rdx
  or     %rsi,%rdx
  shl    $0x8,%rdx
  or     %rcx,%rdx
  mov    %rdx,-0x8(%rax)
  pop    %rbx
  ret
```

assistant: ```c
char *b64ton(char *data, double val) {
  char *p = (char *)data;
  unsigned char *p2 = (unsigned char *)&val;
  p[0] = p2[7];
  p[1] = p2[6];
  p[2] = p2[5];
  p[3] = p2[4];
  p[4] = p2[3];
  p[5] = p2[2];
  p[6] = p2[1];
  p[7] = p2[0];
  p += 8;
  return p;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<elfrw_read_Syminfos>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L30:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <elfrw_read_Syminfo@plt>
  test   %eax,%eax
  je     L12
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L30
  jmp    L38
L12:
  nop
L38:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int elfrw_read_Syminfos(FILE *fp, Elf64_Syminfo *in, int count) {
  int i;
  for (i = 0; i < count; ++i)
    if (!elfrw_read_Syminfo(fp, &in[i]))
      break;
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<elfrw_read_Syminfos>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  jle    L29
  mov    %rdi,%r12
  mov    %edx,%r13d
  mov    %rsi,%rbx
  mov    $0x0,%ebp
L33:
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <elfrw_read_Syminfo@plt>
  test   %eax,%eax
  je     L5
  add    $0x1,%ebp
  add    $0x4,%rbx
  cmp    %ebp,%r13d
  jne    L33
  mov    %r13d,%ebp
L5:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  mov    $0x0,%ebp
  jmp    L5
```

assistant: ```c
int elfrw_read_Syminfos(FILE *fp, Elf64_Syminfo *in, int count) {
  int i;
  for (i = 0; i < count; ++i)
    if (!elfrw_read_Syminfo(fp, &in[i]))
      break;
  return i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<elfrw_read_Syminfos>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  jle    L39
  mov    %rdi,%rbp
  mov    %edx,%r13d
  mov    %rsi,%rbx
  xor    %r12d,%r12d
  jmp    L11
L45:
  add    $0x1,%r12d
  add    $0x4,%rbx
  cmp    %r12d,%r13d
  je     L40
L11:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <elfrw_read_Syminfo@plt>
  test   %eax,%eax
  jne    L45
L40:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  xor    %r12d,%r12d
  jmp    L40
```

assistant: ```c
int elfrw_read_Syminfos(FILE *fp, Elf64_Syminfo *in, int count) {
  int i;
  for (i = 0; i < count; ++i)
    if (!elfrw_read_Syminfo(fp, &in[i]))
      break;
  return i;
}
```
user: Translate the below assembly code into C programming language:

```asm
<elfrw_read_Syminfos>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %edx,%edx
  jle    L7
  mov    %rdi,%rbp
  mov    %edx,%r13d
  mov    %rsi,%rbx
  xor    %r12d,%r12d
  jmp    L0
L49:
  add    $0x1,%r12d
  add    $0x4,%rbx
  cmp    %r12d,%r13d
  je     L22
L0:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <elfrw_read_Syminfo@plt>
  test   %eax,%eax
  jne    L49
L22:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  xor    %r12d,%r12d
  jmp    L22
```

assistant: ```c
int elfrw_read_Syminfos(FILE *fp, Elf64_Syminfo *in, int count) {
  int i;
  for (i = 0; i < count; ++i)
    if (!elfrw_read_Syminfo(fp, &in[i]))
      break;
  return i;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    D16(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  pop    %rbp
  ret
```

assistant: ```c
time_t time(time_t *t) {
  errno = ENOSYS;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<time>:
  endbr64
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
time_t time(time_t *t) {
  errno = ENOSYS;
  return -1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<time>:
  endbr64
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    D14(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
time_t time(time_t *t) {
  errno = ENOSYS;
  return -1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<time>:
  endbr64
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    D18(%rip),%rax
  mov    %edx,(%rax)
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
time_t time(time_t *t) {
  errno = ENOSYS;
  return -1;
}
```
user: What is the c source code of the assembly code below?

```asm
<xstrdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <xmalloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *xstrdup(const char *str) {
  char *tmp = xmalloc(strlen(str) + 1);
  strcpy(tmp, str);
  return tmp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xstrdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <xmalloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *xstrdup(const char *str) {
  char *tmp = xmalloc(strlen(str) + 1);
  strcpy(tmp, str);
  return tmp;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xstrdup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <xmalloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *xstrdup(const char *str) {
  char *tmp = xmalloc(strlen(str) + 1);
  strcpy(tmp, str);
  return tmp;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2a,%al
  jne    L26
  addq   $0x1,-0x28(%rbp)
L26:
  movl   $0x0,-0x2c(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  jmp    L83
L147:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x26,%al
  jne    L36
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x2c(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x2c(%rbp)
  jmp    L121
L36:
  addl   $0x1,-0x2c(%rbp)
L121:
  addq   $0x1,-0x20(%rbp)
L83:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L89
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2c,%al
  je     L89
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x3b,%al
  je     L89
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x25,%al
  jne    L147
L89:
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  jmp    L101
L129:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x26,%al
  jne    L109
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L44
L134:
  addq   $0x1,-0x18(%rbp)
L44:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L134
  jmp    L52
L109:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    -0x20(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L52:
  addq   $0x1,-0x20(%rbp)
L101:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L104
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2c,%al
  je     L104
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x3b,%al
  je     L104
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x25,%al
  jne    L129
L104:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  char *gecos, *login, *buf;
  char *p;
  char *bp;
  int l;
  bp = buf;
  if (*gecos == '*')
    gecos++;
  l = 0;
  for (p = gecos; *p != '\0' && *p != ',' && *p != ';' && *p != '%'; p++) {
    if (*p == '&')
      l += strlen(login);
    else
      l++;
  }
  for (p = gecos; *p != '\0' && *p != ',' && *p != ';' && *p != '%'; p++) {
    if (*p == '&') {
      (void)strcpy(bp, login);
      while (*bp != '\0')
        bp++;
    } else {
      *bp++ = *p;
    }
  }
  *bp = '\0';
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmpb   $0x2a,0x0
  sete   %bpl
  movzbl %bpl,%ebp
  movzbl 0x0(%rbp),%ecx
  cmp    $0x3b,%cl
  ja     L38
  movabs $0x800102000000001,%rax
  shr    %cl,%rax
  test   $0x1,%al
  jne    L29
  mov    %rbp,%rax
  jmp    L16
L38:
  mov    %rbp,%rax
L16:
  movabs $0x800102000000001,%rsi
L12:
  add    $0x1,%rax
  movzbl (%rax),%edx
  cmp    $0x3b,%dl
  ja     L12
  bt     %rdx,%rsi
  jae    L12
  movabs $0x800102000000001,%r12
L43:
  cmp    $0x26,%cl
  je     L102
  mov    %cl,(%rbx)
  lea    0x1(%rbx),%rbx
L109:
  add    $0x1,%rbp
  movzbl 0x0(%rbp),%ecx
  cmp    $0x3b,%cl
  ja     L43
  bt     %rcx,%r12
  jae    L43
L29:
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L102:
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  cmpb   $0x0,(%rbx)
  je     L109
L40:
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L40
  jmp    L109
```

assistant: ```c
int main(void) {
  char *gecos, *login, *buf;
  char *p;
  char *bp;
  int l;
  bp = buf;
  if (*gecos == '*')
    gecos++;
  l = 0;
  for (p = gecos; *p != '\0' && *p != ',' && *p != ';' && *p != '%'; p++) {
    if (*p == '&')
      l += strlen(login);
    else
      l++;
  }
  for (p = gecos; *p != '\0' && *p != ',' && *p != ';' && *p != '%'; p++) {
    if (*p == '&') {
      (void)strcpy(bp, login);
      while (*bp != '\0')
        bp++;
    } else {
      *bp++ = *p;
    }
  }
  *bp = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  cmpb   $0x2a,0x0
  sete   %bpl
  movzbl 0x0(%rbp),%ecx
  cmp    $0x3b,%cl
  ja     L78
  movabs $0xf7ffefdffffffffe,%rax
  sar    %cl,%rax
  mov    %rax,%rdx
  mov    %rbp,%rax
  and    $0x1,%dl
  je     L0
L5:
  movabs $0xf7ffefdffffffffe,%rsi
  nopl   (%rax)
L98:
  movzbl 0x1(%rax),%edx
  add    $0x1,%rax
  cmp    $0x3b,%dl
  ja     L98
  bt     %rdx,%rsi
  jb     L98
  movabs $0xf7ffefdffffffffe,%r12
  nopl   (%rax)
L92:
  cmp    $0x26,%cl
  je     L34
  mov    %cl,(%rbx)
  add    $0x1,%rbx
L27:
  movzbl 0x1(%rbp),%ecx
  add    $0x1,%rbp
  cmp    $0x3b,%cl
  ja     L92
  bt     %rcx,%r12
  jb     L92
L0:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L34:
  xor    %esi,%esi
  mov    %rbx,%rdi
  call   <strcpy@plt>
  cmpb   $0x0,(%rbx)
  je     L27
  nop
L36:
  add    $0x1,%rbx
  cmpb   $0x0,(%rbx)
  jne    L36
  jmp    L27
L78:
  mov    %rbp,%rax
  jmp    L5
```

assistant: ```c
int main(void) {
  char *gecos, *login, *buf;
  char *p;
  char *bp;
  int l;
  bp = buf;
  if (*gecos == '*')
    gecos++;
  l = 0;
  for (p = gecos; *p != '\0' && *p != ',' && *p != ';' && *p != '%'; p++) {
    if (*p == '&')
      l += strlen(login);
    else
      l++;
  }
  for (p = gecos; *p != '\0' && *p != ',' && *p != ';' && *p != '%'; p++) {
    if (*p == '&') {
      (void)strcpy(bp, login);
      while (*bp != '\0')
        bp++;
    } else {
      *bp++ = *p;
    }
  }
  *bp = '\0';
}
```