ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<type_size>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  lea    D6(%rip),%rbx
  mov    $0x0,%ebp
  lea    D27(%rip),%rdi
L1:
  mov    %r12,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  je     L12
  add    $0x1,%ebp
  add    $0x10,%rbx
  mov    -0x10(%rbx),%rdi
  test   %rdi,%rdi
  jne    L1
  mov    $0x0,%eax
  jmp    L17
L12:
  movslq %ebp,%rbp
  shl    $0x4,%rbp
  lea    D20(%rip),%rax
  mov    0x8(%rax,%rbp,1),%eax
L17:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D27
type: string

tool: D27: u8

assistant: ```c
unsigned int type_size(const char *name) {
  static struct {
    const char *type;
    unsigned int size;
  } table[] = {
      {"u8", 1}, {"u16", 2}, {"u32", 4}, {"u64", 8}, {"s8", 1}, {"s16", 2}, {"s32", 4}, {"s64", 8}, {"char", 1}, {},
  };
  int i;
  for (i = 0; table[i].type; i++) {
    if (!strcmp(table[i].type, name))
      return table[i].size;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<type_size>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  lea    D19(%rip),%rdi
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  lea    D13(%rip),%rbx
  jmp    L19
  nopl   (%rax)
L16:
  mov    (%rbx),%rdi
  add    $0x10,%rbx
  add    $0x1,%ebp
  test   %rdi,%rdi
  je     L20
L19:
  mov    %r12,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L16
  movslq %ebp,%rbp
  lea    D9(%rip),%rax
  pop    %rbx
  shl    $0x4,%rbp
  mov    0x8(%rax,%rbp,1),%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L20:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: data_label: D19
datatype: string

tool: D19: u8

assistant: ```c
unsigned int type_size(const char *name) {
  static struct {
    const char *type;
    unsigned int size;
  } table[] = {
      {"u8", 1}, {"u16", 2}, {"u32", 4}, {"u64", 8}, {"s8", 1}, {"s16", 2}, {"s32", 4}, {"s64", 8}, {"char", 1}, {},
  };
  int i;
  for (i = 0; table[i].type; i++) {
    if (!strcmp(table[i].type, name))
      return table[i].size;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<type_size>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  lea    D19(%rip),%rdi
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  lea    D14(%rip),%rbx
  jmp    L1
  nopl   (%rax)
L15:
  mov    (%rbx),%rdi
  add    $0x10,%rbx
  add    $0x1,%ebp
  test   %rdi,%rdi
  je     L14
L1:
  mov    %r12,%rsi
  call   <strcmp@plt>
  test   %eax,%eax
  jne    L15
  movslq %ebp,%rbp
  lea    D25(%rip),%rax
  pop    %rbx
  shl    $0x4,%rbp
  mov    0x8(%rax,%rbp,1),%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D19
data_type: string

tool: D19: u8

assistant: ```c
unsigned int type_size(const char *name) {
  static struct {
    const char *type;
    unsigned int size;
  } table[] = {
      {"u8", 1}, {"u16", 2}, {"u32", 4}, {"u64", 8}, {"s8", 1}, {"s16", 2}, {"s32", 4}, {"s64", 8}, {"char", 1}, {},
  };
  int i;
  for (i = 0; table[i].type; i++) {
    if (!strcmp(table[i].type, name))
      return table[i].size;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L17
L18:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L17:
  cmpl   $0x9,-0x4(%rbp)
  jle    L18
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_infered: string

tool: D9: %d


assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10; i++) {
    printf("%d\n", i);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%ebx
  lea    D4(%rip),%rbp
L4:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L4
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
datatype: string

tool: D4: %d


assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10; i++) {
    printf("%d\n", i);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L6
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: label: D7
type_infered: string

tool: D7: %d


assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10; i++) {
    printf("%d\n", i);
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L7:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    $0xa,%ebx
  jne    L7
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
data_type: string

tool: D7: %d


assistant: ```c
int main() {
  int i;
  for (i = 0; i < 10; i++) {
    printf("%d\n", i);
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x70(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    -0x74(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x74(%rbp),%edx
  lea    -0x70(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <rotate@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type: string
name: retrieve_data
arguments: data_label: D7
type: string

tool: D8: %s

tool: D7: %d

assistant: ```c
int main() {
  char str[100];
  scanf("%s", str);
  int n;
  scanf("%d", &n);
  rotate(str, n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbx
  mov    %rbx,%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  lea    D14(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    %rbx,%rdi
  call   <rotate@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D7
datatype: string
name: read_data
arguments: data_tag: D14
datatype: string

tool: D7: %s

tool: D14: %d

assistant: ```c
int main() {
  char str[100];
  scanf("%s", str);
  int n;
  scanf("%d", &n);
  rotate(str, n);
  return 0;
}
```
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
  lea    D9(%rip),%rdi
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  mov    %rbp,%rsi
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    %rbp,%rdi
  call   <rotate@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  sub    $0xffffffffffffff80,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D9
datatype: string
name: unpack_data
arguments: tag: D15
datatype: string

tool: D9: %s

tool: D15: %d

assistant: ```c
int main() {
  char str[100];
  scanf("%s", str);
  int n;
  scanf("%d", &n);
  rotate(str, n);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D16(%rip),%rdi
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  mov    %rbp,%rsi
  call   <scanf@plt>
  lea    0xc(%rsp),%rsi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    %rbp,%rdi
  call   <rotate@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  sub    $0xffffffffffffff80,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D16
datatype: string
name: disassemble_data
arguments: tag: D15
datatype: string

tool: D16: %s

tool: D15: %d

assistant: ```c
int main() {
  char str[100];
  scanf("%s", str);
  int n;
  scanf("%d", &n);
  rotate(str, n);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<base64_decode>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L63
L65:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x3d,%al
  jne    L7
  mov    $0x0,%eax
  jmp    L4
L7:
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x3,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <base64_unmap@plt>
  movzbl -0xc(%rbp),%eax
  movsbl %al,%eax
  shl    $0x2,%eax
  mov    %eax,%edx
  movzbl -0xb(%rbp),%eax
  sar    $0x4,%al
  and    $0x3,%eax
  or     %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  movzbl -0xb(%rbp),%eax
  movsbl %al,%eax
  shl    $0x4,%eax
  mov    %eax,%ecx
  movzbl -0xa(%rbp),%eax
  sar    $0x2,%al
  and    $0xf,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  or     %ecx,%edx
  mov    %dl,(%rax)
  movzbl -0xa(%rbp),%eax
  movsbl %al,%eax
  shl    $0x6,%eax
  mov    %eax,%ecx
  movzbl -0x9(%rbp),%eax
  and    $0x3f,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  or     %ecx,%edx
  mov    %dl,(%rax)
  addq   $0x3,-0x18(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  cmp    $0x3d,%al
  jne    L37
  mov    $0x0,%eax
  jmp    L4
L37:
  addq   $0x4,-0x20(%rbp)
  addl   $0x4,-0x24(%rbp)
L63:
  mov    -0x24(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L65
  mov    $0x0,%eax
L4:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L42
  call   <__stack_chk_fail@plt>
L42:
  leave
  ret
```

assistant: ```c
int base64_decode(char *in, int inlen, char *out) {
  int i;
  char *in_block;
  char *out_block;
  char temp[4];
  out_block = out;
  in_block = in;
  for (i = 0; i < inlen; i += 4) {
    if (*in_block == '=')
      return 0;
    memcpy(temp, in_block, 4);
    memset(out_block, 0, 3);
    base64_unmap(temp);
    out_block[0] = ((temp[0] << 2) & 0xfc) | ((temp[1] >> 4) & 3);
    out_block[1] = ((temp[1] << 4) & 0xf0) | ((temp[2] >> 2) & 0xf);
    out_block[2] = ((temp[2] << 6) & 0xc0) | ((temp[3]) & 0x3f);
    out_block += 3;
    if (in_block[3] == '=')
      return 0;
    in_block += 4;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<base64_decode>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L21
  mov    %rdi,%rbp
  mov    %rdx,%rbx
  lea    -0x1(%rsi),%eax
  shr    $0x2,%eax
  lea    0x1(%rax),%eax
  mov    %eax,%eax
  lea    (%rdi,%rax,4),%r13
  lea    0x4(%rsp),%r12
L1:
  cmpb   $0x3d,0x0(%rbp)
  je     L21
  mov    0x0(%rbp),%eax
  mov    %eax,0x4(%rsp)
  movw   $0x0,(%rbx)
  movb   $0x0,0x2(%rbx)
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <base64_unmap@plt>
  movzbl 0x5(%rsp),%edx
  movsbl 0x4(%rsp),%ecx
  shl    $0x2,%ecx
  mov    %edx,%eax
  sar    $0x4,%al
  and    $0x3,%eax
  or     %ecx,%eax
  mov    %al,(%rbx)
  movzbl 0x6(%rsp),%eax
  mov    %eax,%ecx
  sar    $0x2,%cl
  and    $0xf,%ecx
  shl    $0x4,%edx
  or     %ecx,%edx
  mov    %dl,0x1(%rbx)
  movzbl 0x7(%rsp),%edx
  and    $0x3f,%edx
  shl    $0x6,%eax
  or     %edx,%eax
  mov    %al,0x2(%rbx)
  add    $0x3,%rbx
  cmpb   $0x3d,0x3(%rbp)
  je     L21
  add    $0x4,%rbp
  cmp    %r13,%rbp
  jne    L1
L21:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int base64_decode(char *in, int inlen, char *out) {
  int i;
  char *in_block;
  char *out_block;
  char temp[4];
  out_block = out;
  in_block = in;
  for (i = 0; i < inlen; i += 4) {
    if (*in_block == '=')
      return 0;
    memcpy(temp, in_block, 4);
    memset(out_block, 0, 3);
    base64_unmap(temp);
    out_block[0] = ((temp[0] << 2) & 0xfc) | ((temp[1] >> 4) & 3);
    out_block[1] = ((temp[1] << 4) & 0xf0) | ((temp[2] >> 2) & 0xf);
    out_block[2] = ((temp[2] << 6) & 0xc0) | ((temp[3]) & 0x3f);
    out_block += 3;
    if (in_block[3] == '=')
      return 0;
    in_block += 4;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<base64_decode>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L6
  lea    -0x1(%rsi),%eax
  mov    %rdi,%r13
  lea    0x4(%rsp),%rbp
  mov    %rdx,%rbx
  shr    $0x2,%eax
  lea    0x1(%rax),%eax
  lea    (%rdi,%rax,4),%r12
  jmp    L11
L51:
  mov    0x0(%r13),%eax
  movb   $0x0,0x2(%rbx)
  mov    %rbp,%rdi
  add    $0x3,%rbx
  mov    %eax,0x4(%rsp)
  xor    %eax,%eax
  mov    %ax,-0x3(%rbx)
  xor    %eax,%eax
  call   <base64_unmap@plt>
  movzbl 0x5(%rsp),%edx
  movsbl 0x4(%rsp),%ecx
  mov    %edx,%eax
  shl    $0x2,%ecx
  sar    $0x4,%al
  shl    $0x4,%edx
  and    $0x3,%eax
  or     %ecx,%eax
  mov    %al,-0x3(%rbx)
  movzbl 0x6(%rsp),%eax
  mov    %eax,%ecx
  shl    $0x6,%eax
  sar    $0x2,%cl
  and    $0xf,%ecx
  or     %ecx,%edx
  mov    %dl,-0x2(%rbx)
  movzbl 0x7(%rsp),%edx
  and    $0x3f,%edx
  or     %edx,%eax
  mov    %al,-0x1(%rbx)
  cmpb   $0x3d,0x3(%r13)
  je     L6
  add    $0x4,%r13
  cmp    %r12,%r13
  je     L6
L11:
  cmpb   $0x3d,0x0(%r13)
  jne    L51
L6:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int base64_decode(char *in, int inlen, char *out) {
  int i;
  char *in_block;
  char *out_block;
  char temp[4];
  out_block = out;
  in_block = in;
  for (i = 0; i < inlen; i += 4) {
    if (*in_block == '=')
      return 0;
    memcpy(temp, in_block, 4);
    memset(out_block, 0, 3);
    base64_unmap(temp);
    out_block[0] = ((temp[0] << 2) & 0xfc) | ((temp[1] >> 4) & 3);
    out_block[1] = ((temp[1] << 4) & 0xf0) | ((temp[2] >> 2) & 0xf);
    out_block[2] = ((temp[2] << 6) & 0xc0) | ((temp[3]) & 0x3f);
    out_block += 3;
    if (in_block[3] == '=')
      return 0;
    in_block += 4;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<base64_decode>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L44
  lea    -0x1(%rsi),%eax
  mov    %rdi,%r13
  lea    0x4(%rsp),%rbp
  mov    %rdx,%rbx
  shr    $0x2,%eax
  lea    0x1(%rax),%eax
  lea    (%rdi,%rax,4),%r12
  jmp    L46
L15:
  mov    0x0(%r13),%eax
  movb   $0x0,0x2(%rbx)
  mov    %rbp,%rdi
  add    $0x3,%rbx
  mov    %eax,0x4(%rsp)
  xor    %eax,%eax
  mov    %ax,-0x3(%rbx)
  xor    %eax,%eax
  call   <base64_unmap@plt>
  movzbl 0x5(%rsp),%edx
  movsbl 0x4(%rsp),%ecx
  mov    %edx,%eax
  shl    $0x2,%ecx
  sar    $0x4,%al
  shl    $0x4,%edx
  and    $0x3,%eax
  or     %ecx,%eax
  mov    %al,-0x3(%rbx)
  movzbl 0x6(%rsp),%eax
  mov    %eax,%ecx
  shl    $0x6,%eax
  sar    $0x2,%cl
  and    $0xf,%ecx
  or     %ecx,%edx
  mov    %dl,-0x2(%rbx)
  movzbl 0x7(%rsp),%edx
  and    $0x3f,%edx
  or     %edx,%eax
  mov    %al,-0x1(%rbx)
  cmpb   $0x3d,0x3(%r13)
  je     L44
  add    $0x4,%r13
  cmp    %r12,%r13
  je     L44
L46:
  cmpb   $0x3d,0x0(%r13)
  jne    L15
L44:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L43:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int base64_decode(char *in, int inlen, char *out) {
  int i;
  char *in_block;
  char *out_block;
  char temp[4];
  out_block = out;
  in_block = in;
  for (i = 0; i < inlen; i += 4) {
    if (*in_block == '=')
      return 0;
    memcpy(temp, in_block, 4);
    memset(out_block, 0, 3);
    base64_unmap(temp);
    out_block[0] = ((temp[0] << 2) & 0xfc) | ((temp[1] >> 4) & 3);
    out_block[1] = ((temp[1] << 4) & 0xf0) | ((temp[2] >> 2) & 0xf);
    out_block[2] = ((temp[2] << 6) & 0xc0) | ((temp[3]) & 0x3f);
    out_block += 3;
    if (in_block[3] == '=')
      return 0;
    in_block += 4;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<milking>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%edx
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L13
L26:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%edx
  sub    %edx,%eax
  cmp    %eax,-0xc(%rbp)
  jge    L10
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    -0x8(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rdx),%edx
  sub    %edx,%eax
  mov    %eax,-0xc(%rbp)
L10:
  addl   $0x1,-0x8(%rbp)
L13:
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L26
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int milking(int start[], int end[], int length) {
  int pivot = 0, longest = end[pivot] - start[pivot];
  for (int i = 0; i < length; i++) {
    if (end[i] - start[i] > longest) {
      longest = end[i] - start[i];
    }
  }
  return longest;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<milking>:
  endbr64
  mov    (%rsi),%eax
  sub    (%rdi),%eax
  test   %edx,%edx
  jle    L3
  mov    %edx,%edx
  mov    $0x0,%ecx
L13:
  mov    (%rsi,%rcx,4),%r8d
  sub    (%rdi,%rcx,4),%r8d
  cmp    %r8d,%eax
  cmovl  %r8d,%eax
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L13
L3:
  ret
```

assistant: ```c
int milking(int start[], int end[], int length) {
  int pivot = 0, longest = end[pivot] - start[pivot];
  for (int i = 0; i < length; i++) {
    if (end[i] - start[i] > longest) {
      longest = end[i] - start[i];
    }
  }
  return longest;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<milking>:
  endbr64
  mov    (%rsi),%eax
  mov    (%rdi),%r9d
  mov    %eax,%r8d
  sub    %r9d,%r8d
  test   %edx,%edx
  jle    L5
  movslq %edx,%rdx
  xor    %ecx,%ecx
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L4:
  mov    (%rsi,%rcx,4),%eax
  mov    (%rdi,%rcx,4),%r9d
L17:
  sub    %r9d,%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L4
L5:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int milking(int start[], int end[], int length) {
  int pivot = 0, longest = end[pivot] - start[pivot];
  for (int i = 0; i < length; i++) {
    if (end[i] - start[i] > longest) {
      longest = end[i] - start[i];
    }
  }
  return longest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<milking>:
  endbr64
  mov    %rdi,%rcx
  mov    (%rsi),%eax
  mov    %edx,%edi
  sub    (%rcx),%eax
  test   %edx,%edx
  jle    L61
  lea    -0x1(%rdx),%edx
  cmp    $0x2,%edx
  jbe    L24
  mov    %edi,%edx
  movd   %eax,%xmm4
  xor    %eax,%eax
  shr    $0x2,%edx
  pshufd $0x0,%xmm4,%xmm2
  shl    $0x4,%rdx
  nopl   (%rax)
L10:
  movdqu (%rsi,%rax,1),%xmm0
  movdqu (%rcx,%rax,1),%xmm3
  add    $0x10,%rax
  psubd  %xmm3,%xmm0
  movdqa %xmm0,%xmm1
  pcmpgtd %xmm2,%xmm1
  pand   %xmm1,%xmm0
  pandn  %xmm2,%xmm1
  movdqa %xmm1,%xmm2
  por    %xmm0,%xmm2
  cmp    %rdx,%rax
  jne    L10
  movdqa %xmm2,%xmm1
  mov    %edi,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%edx
  movdqa %xmm1,%xmm0
  pcmpgtd %xmm2,%xmm0
  pand   %xmm0,%xmm1
  pandn  %xmm2,%xmm0
  por    %xmm1,%xmm0
  movdqa %xmm0,%xmm2
  psrldq $0x4,%xmm2
  movdqa %xmm2,%xmm1
  pcmpgtd %xmm0,%xmm1
  pand   %xmm1,%xmm2
  pandn  %xmm0,%xmm1
  por    %xmm2,%xmm1
  movd   %xmm1,%eax
  test   $0x3,%dil
  je     L66
L2:
  movslq %edx,%r10
  mov    (%rsi,%r10,4),%r8d
  sub    (%rcx,%r10,4),%r8d
  lea    0x0(,%r10,4),%r9
  cmp    %r8d,%eax
  cmovl  %r8d,%eax
  lea    0x1(%rdx),%r8d
  cmp    %r8d,%edi
  jle    L61
  mov    0x4(%rsi,%r9,1),%r8d
  sub    0x4(%rcx,%r9,1),%r8d
  cmp    %r8d,%eax
  cmovl  %r8d,%eax
  add    $0x2,%edx
  cmp    %edx,%edi
  jle    L61
  mov    0x8(%rsi,%r9,1),%edx
  sub    0x8(%rcx,%r9,1),%edx
  cmp    %edx,%eax
  cmovl  %edx,%eax
L61:
  ret
  nopl   0x0(%rax,%rax,1)
L66:
  ret
L24:
  xor    %edx,%edx
  jmp    L2
```

assistant: ```c
int milking(int start[], int end[], int length) {
  int pivot = 0, longest = end[pivot] - start[pivot];
  for (int i = 0; i < length; i++) {
    if (end[i] - start[i] > longest) {
      longest = end[i] - start[i];
    }
  }
  return longest;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<do_fldl2t>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  fninit
  fldl2t
  mov    -0x8(%rbp),%rax
  fstpl  (%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void do_fldl2t(void *p) {
  asm __volatile__("fninit");
  asm __volatile__("fldl2t");
  asm __volatile__("fstpl (%0)" : : "r"(p) : "memory");
}
```
user: Translate the below assembly code into C programming language:

```asm
<do_fldl2t>:
  endbr64
  fninit
  fldl2t
  fstpl  (%rdi)
  ret
```

assistant: ```c
void do_fldl2t(void *p) {
  asm __volatile__("fninit");
  asm __volatile__("fldl2t");
  asm __volatile__("fstpl (%0)" : : "r"(p) : "memory");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<do_fldl2t>:
  endbr64
  fninit
  fldl2t
  fstpl  (%rdi)
  ret
```

assistant: ```c
void do_fldl2t(void *p) {
  asm __volatile__("fninit");
  asm __volatile__("fldl2t");
  asm __volatile__("fstpl (%0)" : : "r"(p) : "memory");
}
```