ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<to_ipv4>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L21
  mov    %rdi,%rbx
  mov    $0x4,%edi
  mov    %rsp,%r13
  call   <pig_newseg@plt>
  mov    $0x14,%edx
  xor    %esi,%esi
  mov    %r13,%rdi
  movl   $0x0,(%rax)
  mov    %rax,%r12
  call   <memset@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  je     L79
  movabs $0xcccccccccccccccd,%r14
  xor    %edx,%edx
L95:
  movzbl 0x1(%rbx),%ecx
  cmp    $0x2e,%al
  je     L17
L43:
  test   %cl,%cl
  jne    L93
L96:
  mov    %al,(%rsp,%rdx,1)
L30:
  mov    (%r12),%ebp
  mov    %r13,%rdi
  add    $0x1,%rbx
  call   <atoi@plt>
  mov    $0x14,%edx
  xor    %esi,%esi
  mov    %r13,%rdi
  shl    $0x8,%ebp
  or     %eax,%ebp
  mov    %ebp,(%r12)
  call   <memset@plt>
  movzbl (%rbx),%eax
  test   %al,%al
  je     L79
  movzbl 0x1(%rbx),%ecx
  xor    %edx,%edx
  cmp    $0x2e,%al
  jne    L43
L17:
  test   %cl,%cl
  jne    L30
  jmp    L96
  cs nopw 0x0(%rax,%rax,1)
L93:
  lea    0x1(%rdx),%rsi
  mov    %al,(%rsp,%rdx,1)
  add    $0x1,%rbx
  mov    %rsi,%rax
  mul    %r14
  shr    $0x4,%rdx
  lea    (%rdx,%rdx,4),%rax
  shl    $0x2,%rax
  sub    %rax,%rsi
  mov    %ecx,%eax
  mov    %rsi,%rdx
  jmp    L95
  nopw   0x0(%rax,%rax,1)
L21:
  xor    %r12d,%r12d
  nopl   0x0(%rax,%rax,1)
L79:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x20,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L39:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned int *to_ipv4(const char *data) {
  unsigned int *retval = NULL;
  const char *dp = NULL;
  char oct[20];
  size_t o = 0;
  if (data == NULL) {
    return NULL;
  }
  retval = (unsigned int *)pig_newseg(sizeof(unsigned int));
  *retval = 0;
  memset(oct, 0, sizeof(oct));
  for (dp = data; *dp != 0; dp++) {
    if (*dp == '.' || *(dp + 1) == 0) {
      if (*(dp + 1) == 0) {
        oct[o] = *dp;
      }
      *retval = ((*retval) << 8) | atoi(oct);
      o = 0;
      memset(oct, 0, sizeof(oct));
    } else {
      oct[o] = *dp;
      o = (o + 1) % sizeof(oct);
    }
  }
  return retval;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x1,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D43(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  jmp    L23
L10:
  mov    -0x38(%rbp),%rax
  cqto
  shr    $0x3f,%rdx
  add    %rdx,%rax
  and    $0x1,%eax
  sub    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  cmpq   $0x1,-0x10(%rbp)
  jne    L12
  addq   $0x1,-0x20(%rbp)
L12:
  mov    -0x10(%rbp),%rax
  imul   -0x30(%rbp),%rax
  add    %rax,-0x28(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdx
  shr    $0x3f,%rdx
  add    %rdx,%rax
  sar    %rax
  mov    %rax,-0x38(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,-0x30(%rbp)
L23:
  mov    -0x38(%rbp),%rax
  test   %rax,%rax
  jg     L10
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D19
type_infered: string
name: retrieve_data
arguments: label: D43
type_infered: string
name: retrieve_data
arguments: label: D25
type_infered: string
name: retrieve_data
arguments: label: D31
type_infered: string
name: retrieve_data
arguments: label: D28
type_infered: string

tool: D19: Enter a decimal integer 

tool: D43: %ld

tool: D25: Input number is = %ld


tool: D31: Its binary equivalent is = %ld


tool: D28: No.of 1's in the binary number is = %ld


assistant: ```c
int main(void) {
  long num, decimal_num, remainder, base = 1, binary = 0, no_of_1s = 0;
  printf("Enter a decimal integer \n");
  scanf("%ld", &num);
  decimal_num = num;
  while (num > 0) {
    remainder = num % 2;
    if (remainder == 1) {
      no_of_1s++;
    }
    binary = binary + remainder * base;
    num = num / 2;
    base = base * 10;
  }
  printf("Input number is = %ld\n", decimal_num);
  printf("Its binary equivalent is = %ld\n", binary);
  printf("No.of 1's in the binary number is = %ld\n", no_of_1s);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D25(%rip),%rdi
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    D26(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rdi
  test   %rdi,%rdi
  jle    L10
  mov    %rdi,%rcx
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  mov    $0x1,%edx
  jmp    L2
L38:
  mov    %rax,%rcx
L2:
  mov    %rcx,%rsi
  shr    $0x3f,%rsi
  lea    (%rcx,%rsi,1),%rax
  and    $0x1,%eax
  sub    %rsi,%rax
  cmp    $0x1,%rax
  sete   %sil
  movzbl %sil,%esi
  add    %rsi,%rbx
  imul   %rdx,%rax
  add    %rax,%rbp
  mov    %rcx,%rax
  shr    $0x3f,%rax
  add    %rcx,%rax
  sar    %rax
  lea    (%rdx,%rdx,4),%rdx
  add    %rdx,%rdx
  cmp    $0x1,%rcx
  jg     L38
  mov    %rax,(%rsp)
L30:
  mov    %rdi,%rsi
  lea    D46(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D41(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  jmp    L30
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D25
type: string
name: parse_data
arguments: label: D26
type: string
name: parse_data
arguments: label: D46
type: string
name: parse_data
arguments: label: D19
type: string
name: parse_data
arguments: label: D41
type: string

tool: D25: Enter a decimal integer 

tool: D26: %ld

tool: D46: Input number is = %ld


tool: D19: Its binary equivalent is = %ld


tool: D41: No.of 1's in the binary number is = %ld


assistant: ```c
int main(void) {
  long num, decimal_num, remainder, base = 1, binary = 0, no_of_1s = 0;
  printf("Enter a decimal integer \n");
  scanf("%ld", &num);
  decimal_num = num;
  while (num > 0) {
    remainder = num % 2;
    if (remainder == 1) {
      no_of_1s++;
    }
    binary = binary + remainder * base;
    num = num / 2;
    base = base * 10;
  }
  printf("Input number is = %ld\n", decimal_num);
  printf("Its binary equivalent is = %ld\n", binary);
  printf("No.of 1's in the binary number is = %ld\n", no_of_1s);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  lea    D46(%rip),%rdi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    D34(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rsi
  test   %rsi,%rsi
  jle    L27
  mov    %rsi,%rcx
  xor    %ebp,%ebp
  xor    %r12d,%r12d
  mov    $0x1,%eax
L10:
  mov    %rcx,%rdx
  and    $0x1,%edx
  cmp    $0x1,%rdx
  sbb    $0xffffffffffffffff,%rbp
  imul   %rax,%rdx
  lea    (%rax,%rax,4),%rax
  add    %rax,%rax
  add    %rdx,%r12
  sar    %rcx
  jne    L10
  movq   $0x0,(%rsp)
L6:
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L27:
  xor    %ebp,%ebp
  xor    %r12d,%r12d
  jmp    L6
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D46
datatype: string
name: read_data
arguments: data_tag: D34
datatype: string
name: read_data
arguments: data_tag: D4
datatype: string
name: read_data
arguments: data_tag: D39
datatype: string
name: read_data
arguments: data_tag: D27
datatype: string

tool: D46: Enter a decimal integer 

tool: D34: %ld

tool: D4: Input number is = %ld


tool: D39: Its binary equivalent is = %ld


tool: D27: No.of 1's in the binary number is = %ld


assistant: ```c
int main(void) {
  long num, decimal_num, remainder, base = 1, binary = 0, no_of_1s = 0;
  printf("Enter a decimal integer \n");
  scanf("%ld", &num);
  decimal_num = num;
  while (num > 0) {
    remainder = num % 2;
    if (remainder == 1) {
      no_of_1s++;
    }
    binary = binary + remainder * base;
    num = num / 2;
    base = base * 10;
  }
  printf("Input number is = %ld\n", decimal_num);
  printf("Its binary equivalent is = %ld\n", binary);
  printf("No.of 1's in the binary number is = %ld\n", no_of_1s);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D38(%rip),%rdi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  mov    %rsp,%rsi
  lea    D31(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rsi
  test   %rsi,%rsi
  jle    L19
  mov    %rsi,%rcx
  xor    %ebp,%ebp
  xor    %r12d,%r12d
  mov    $0x1,%eax
L37:
  mov    %rcx,%rdx
  and    $0x1,%edx
  cmp    $0x1,%rdx
  sbb    $0xffffffffffffffff,%rbp
  imul   %rax,%rdx
  lea    (%rax,%rax,4),%rax
  add    %rax,%rax
  add    %rdx,%r12
  sar    %rcx
  jne    L37
  movq   $0x0,(%rsp)
L21:
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  lea    D41(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L19:
  xor    %ebp,%ebp
  xor    %r12d,%r12d
  jmp    L21
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D38
type_infered: string
name: unpack_data
arguments: label: D31
type_infered: string
name: unpack_data
arguments: label: D12
type_infered: string
name: unpack_data
arguments: label: D41
type_infered: string
name: unpack_data
arguments: label: D9
type_infered: string

tool: D38: Enter a decimal integer 

tool: D31: %ld

tool: D12: Input number is = %ld


tool: D41: Its binary equivalent is = %ld


tool: D9: No.of 1's in the binary number is = %ld


assistant: ```c
int main(void) {
  long num, decimal_num, remainder, base = 1, binary = 0, no_of_1s = 0;
  printf("Enter a decimal integer \n");
  scanf("%ld", &num);
  decimal_num = num;
  while (num > 0) {
    remainder = num % 2;
    if (remainder == 1) {
      no_of_1s++;
    }
    binary = binary + remainder * base;
    num = num / 2;
    base = base * 10;
  }
  printf("Input number is = %ld\n", decimal_num);
  printf("Its binary equivalent is = %ld\n", binary);
  printf("No.of 1's in the binary number is = %ld\n", no_of_1s);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<bad_gettimeofday>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <inc_addr@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  leave
  ret
```

assistant: ```c
int bad_gettimeofday(void *addr) {
  struct timezone *tz = (struct timezone *)inc_addr(addr, 1);
  return gettimeofday((struct timeval *)addr, tz);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bad_gettimeofday>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0x1,%esi
  call   <inc_addr@plt>
  mov    %rax,%rsi
  mov    %rbx,%rdi
  call   <gettimeofday@plt>
  pop    %rbx
  ret
```

assistant: ```c
int bad_gettimeofday(void *addr) {
  struct timezone *tz = (struct timezone *)inc_addr(addr, 1);
  return gettimeofday((struct timeval *)addr, tz);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bad_gettimeofday>:
  endbr64
  push   %rbp
  mov    $0x1,%esi
  mov    %rdi,%rbp
  call   <inc_addr@plt>
  mov    %rbp,%rdi
  pop    %rbp
  mov    %rax,%rsi
  jmp    1060 <gettimeofday@plt>
```

assistant: ```c
int bad_gettimeofday(void *addr) {
  struct timezone *tz = (struct timezone *)inc_addr(addr, 1);
  return gettimeofday((struct timeval *)addr, tz);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<bad_gettimeofday>:
  endbr64
  push   %rbp
  mov    $0x1,%esi
  mov    %rdi,%rbp
  call   <inc_addr@plt>
  mov    %rbp,%rdi
  pop    %rbp
  mov    %rax,%rsi
  jmp    1060 <gettimeofday@plt>
```

assistant: ```c
int bad_gettimeofday(void *addr) {
  struct timezone *tz = (struct timezone *)inc_addr(addr, 1);
  return gettimeofday((struct timeval *)addr, tz);
}
```
user: What is the c source code of the assembly code below?

```asm
<to_little>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  lea    -0x14(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x8,%eax
  add    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x10,%eax
  add    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  shl    $0x18,%eax
  add    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int to_little(unsigned int a) {
  unsigned char *buf = (unsigned char *)(&a);
  unsigned int w = buf[3];
  w += buf[2] << 8;
  w += buf[1] << 16;
  w += buf[0] << 24;
  return w;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<to_little>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%esi
  shr    $0x18,%esi
  mov    %edi,%edx
  shr    $0x8,%edx
  and    $0xff00,%edx
  movzbl %ah,%ecx
  shl    $0x10,%ecx
  add    %ecx,%edx
  add    %esi,%edx
  shl    $0x18,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
unsigned int to_little(unsigned int a) {
  unsigned char *buf = (unsigned char *)(&a);
  unsigned int w = buf[3];
  w += buf[2] << 8;
  w += buf[1] << 16;
  w += buf[0] << 24;
  return w;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<to_little>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%edx
  mov    %edi,%esi
  shr    $0x8,%edx
  movzbl %ah,%ecx
  shr    $0x18,%esi
  shl    $0x10,%ecx
  and    $0xff00,%edx
  shl    $0x18,%eax
  add    %ecx,%edx
  add    %esi,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
unsigned int to_little(unsigned int a) {
  unsigned char *buf = (unsigned char *)(&a);
  unsigned int w = buf[3];
  w += buf[2] << 8;
  w += buf[1] << 16;
  w += buf[0] << 24;
  return w;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<to_little>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%edx
  mov    %edi,%esi
  shr    $0x8,%edx
  movzbl %ah,%ecx
  shr    $0x18,%esi
  shl    $0x10,%ecx
  and    $0xff00,%edx
  shl    $0x18,%eax
  add    %ecx,%edx
  add    %esi,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
unsigned int to_little(unsigned int a) {
  unsigned char *buf = (unsigned char *)(&a);
  unsigned int w = buf[3];
  w += buf[2] << 8;
  w += buf[1] << 16;
  w += buf[0] << 24;
  return w;
}
```
user: Translate the below assembly code into C programming language:

```asm
<stmmac_get_mac_addr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <readl@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x28(%rbp),%edx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <readl@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  shr    $0x10,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    $0x3,%rax
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  mov    -0x8(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%eax
  shr    $0x8,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  add    $0x5,%rax
  mov    %dl,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void stmmac_get_mac_addr(unsigned long ioaddr, unsigned char *addr, unsigned int high, unsigned int low) {
  unsigned int hi_addr, lo_addr;
  hi_addr = readl(ioaddr + high);
  lo_addr = readl(ioaddr + low);
  addr[0] = lo_addr & 0xff;
  addr[1] = (lo_addr >> 8) & 0xff;
  addr[2] = (lo_addr >> 16) & 0xff;
  addr[3] = (lo_addr >> 24) & 0xff;
  addr[4] = hi_addr & 0xff;
  addr[5] = (hi_addr >> 8) & 0xff;
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<stmmac_get_mac_addr>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbx
  mov    %ecx,%ebp
  mov    %edx,%edx
  lea    (%rdx,%rdi,1),%rdi
  call   <readl@plt>
  mov    %eax,%r12d
  mov    %ebp,%edi
  add    %r13,%rdi
  call   <readl@plt>
  mov    %al,(%rbx)
  movzbl %ah,%ecx
  mov    %cl,0x1(%rbx)
  mov    %eax,%edx
  shr    $0x10,%edx
  mov    %dl,0x2(%rbx)
  shr    $0x18,%eax
  mov    %al,0x3(%rbx)
  mov    %r12b,0x4(%rbx)
  mov    %r12d,%eax
  movzbl %ah,%eax
  mov    %al,0x5(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void stmmac_get_mac_addr(unsigned long ioaddr, unsigned char *addr, unsigned int high, unsigned int low) {
  unsigned int hi_addr, lo_addr;
  hi_addr = readl(ioaddr + high);
  lo_addr = readl(ioaddr + low);
  addr[0] = lo_addr & 0xff;
  addr[1] = (lo_addr >> 8) & 0xff;
  addr[2] = (lo_addr >> 16) & 0xff;
  addr[3] = (lo_addr >> 24) & 0xff;
  addr[4] = hi_addr & 0xff;
  addr[5] = (hi_addr >> 8) & 0xff;
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<stmmac_get_mac_addr>:
  endbr64
  push   %r13
  mov    %edx,%edx
  push   %r12
  mov    %rdi,%r12
  lea    (%rdx,%rdi,1),%rdi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x8,%rsp
  call   <readl@plt>
  mov    %ebx,%edi
  add    %r12,%rdi
  mov    %eax,%r13d
  call   <readl@plt>
  mov    %r13w,0x4(%rbp)
  mov    %eax,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void stmmac_get_mac_addr(unsigned long ioaddr, unsigned char *addr, unsigned int high, unsigned int low) {
  unsigned int hi_addr, lo_addr;
  hi_addr = readl(ioaddr + high);
  lo_addr = readl(ioaddr + low);
  addr[0] = lo_addr & 0xff;
  addr[1] = (lo_addr >> 8) & 0xff;
  addr[2] = (lo_addr >> 16) & 0xff;
  addr[3] = (lo_addr >> 24) & 0xff;
  addr[4] = hi_addr & 0xff;
  addr[5] = (hi_addr >> 8) & 0xff;
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<stmmac_get_mac_addr>:
  endbr64
  push   %r13
  mov    %edx,%edx
  push   %r12
  mov    %rdi,%r12
  lea    (%rdx,%rdi,1),%rdi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x8,%rsp
  call   <readl@plt>
  mov    %ebx,%edi
  add    %r12,%rdi
  mov    %eax,%r13d
  call   <readl@plt>
  mov    %r13w,0x4(%rbp)
  mov    %eax,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void stmmac_get_mac_addr(unsigned long ioaddr, unsigned char *addr, unsigned int high, unsigned int low) {
  unsigned int hi_addr, lo_addr;
  hi_addr = readl(ioaddr + high);
  lo_addr = readl(ioaddr + low);
  addr[0] = lo_addr & 0xff;
  addr[1] = (lo_addr >> 8) & 0xff;
  addr[2] = (lo_addr >> 16) & 0xff;
  addr[3] = (lo_addr >> 24) & 0xff;
  addr[4] = hi_addr & 0xff;
  addr[5] = (hi_addr >> 8) & 0xff;
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<copy_image>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x4000(%rsp),%r11
L38:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L38
  sub    $0x30,%rsp
  mov    %edi,-0x4024(%rbp)
  mov    %esi,-0x4028(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x4018(%rbp)
  movl   $0x0,-0x401c(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    -0x4024(%rbp),%eax
  mov    $0x10000,%esi
  mov    %eax,%edi
  call   <lseek@plt>
  jmp    L18
L44:
  mov    -0x4014(%rbp),%eax
  add    %eax,-0x4018(%rbp)
  mov    -0x4014(%rbp),%edx
  lea    -0x4010(%rbp),%rcx
  mov    -0x4028(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
L18:
  lea    -0x4010(%rbp),%rcx
  mov    -0x4024(%rbp),%eax
  mov    $0x1000,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x4014(%rbp)
  cmpl   $0x0,-0x4014(%rbp)
  jg     L44
  mov    -0x4018(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L26
  addl   $0x1,-0x4018(%rbp)
  lea    -0x401c(%rbp),%rcx
  mov    -0x4028(%rbp),%eax
  mov    $0x1,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
L26:
  mov    -0x4018(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

assistant: ```c
uint32_t copy_image(int32_t in_fd, int32_t out_fd) {
  uint8_t buf[0x1000];
  int n;
  uint32_t image_size = 0;
  uint8_t zero = 0;
  lseek(in_fd, 65536, SEEK_SET);
  while ((n = read(in_fd, buf, 0x1000)) > 0) {
    image_size = image_size + n;
    write(out_fd, buf, n);
  }
  if (image_size % 2) {
    image_size++;
    write(out_fd, &zero, 1);
  }
  return image_size;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<copy_image>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L58:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L58
  sub    $0x28,%rsp
  mov    %edi,%r12d
  mov    %esi,%r13d
  mov    %fs:0x28,%rax
  mov    %rax,0x4018(%rsp)
  xor    %eax,%eax
  movl   $0x0,0xc(%rsp)
  mov    D4(%rip),%rax
  mov    (%rax),%edx
  mov    $0x10000,%esi
  call   <lseek@plt>
  mov    $0x0,%ebx
  lea    0x10(%rsp),%rbp
  jmp    L59
L21:
  add    %edx,%ebx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <write@plt>
L59:
  mov    $0x1000,%edx
  mov    %rbp,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  mov    %eax,%edx
  test   %eax,%eax
  jg     L21
  test   $0x1,%bl
  jne    L42
L22:
  mov    0x4018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    %ebx,%eax
  add    $0x4028,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L42:
  add    $0x1,%ebx
  lea    0xc(%rsp),%rsi
  mov    $0x1,%edx
  mov    %r13d,%edi
  call   <write@plt>
  jmp    L22
L11:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t copy_image(int32_t in_fd, int32_t out_fd) {
  uint8_t buf[0x1000];
  int n;
  uint32_t image_size = 0;
  uint8_t zero = 0;
  lseek(in_fd, 65536, SEEK_SET);
  while ((n = read(in_fd, buf, 0x1000)) > 0) {
    image_size = image_size + n;
    write(out_fd, buf, n);
  }
  if (image_size % 2) {
    image_size++;
    write(out_fd, &zero, 1);
  }
  return image_size;
}
```