ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strcrypt>:
  endbr64
  push   %r12
  mov    %rdi,%r9
  mov    $0x1,%esi
  mov    $0xcccccccd,%r10d
  lea    D0(%rip),%r8
  sub    $0x30,%rsp
  movdqa D12(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x2e,%eax
  lea    0x12(%rsp),%rdi
  movq   $0x1a4780,(%rsp)
  mov    %ax,0x20(%rsp)
  mov    $0x2e,%eax
  movq   $0xcf9,0x8(%rsp)
  movaps %xmm0,0x10(%rsp)
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %edx,%eax
  imul   %r10,%rax
  shr    $0x22,%rax
  movslq %eax,%r11
  lea    (%rax,%rax,4),%eax
  sub    %eax,%edx
  mov    (%rsp,%r11,8),%rax
  lea    (%rdx,%rdx,2),%ecx
  add    %ecx,%ecx
  shr    %cl,%rax
  and    $0x3f,%eax
  movzbl (%r8,%rax,1),%eax
L15:
  mov    %al,(%rdi,%rsi,1)
  mov    %esi,%edx
  add    $0x1,%rsi
  cmp    $0x9,%rsi
  jne    L17
  lea    0x10(%rsp),%rsi
  mov    %r9,%rdi
  call   <crypt@plt>
  lea    D15(%rip),%rsi
  mov    %rax,%rdx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <sscanf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x30,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D0
data_type: string
name: disassemble_data
arguments: data_tag: D12
data_type: byte[16]
name: disassemble_data
arguments: data_tag: D15
data_type: string

tool: D0: ./0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz

tool: {"D12": [0x24,0x31,0x24,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e]}

tool: D15: $1$.SY4.tn.$%[^:]:

assistant: ```c
char *strcrypt(char *str) {
  unsigned long seed[2];
  char *crypted;
  int i;
  char salt[] = "$1$..............";
  const char *const seedchars = "./0123456789ABCDEFGHIJKLMNOPQRST"
                                "UVWXYZabcdefghijklmnopqrstuvwxyz";
  seed[0] = 1920 * 69 * 13;
  seed[1] = 3321 ^ (seed[0] >> 14 & 0x30000);
  for (i = 0; i < 8; i++)
    salt[3 + i] = seedchars[(seed[i / 5] >> (i % 5) * 6) & 0x3f];
  crypted = crypt(str, salt);
  sscanf(crypted, "$1$.SY4.tn.$%[^:]:", crypted);
  return crypted;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strcrypt>:
  endbr64
  push   %r12
  mov    $0x5953,%edx
  mov    $0x6e74,%ecx
  sub    $0x20,%rsp
  movdqa D8(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  mov    $0x2e,%eax
  movaps %xmm0,(%rsp)
  mov    %ax,0x10(%rsp)
  mov    %dx,0x4(%rsp)
  movb   $0x34,0x6(%rsp)
  mov    %cx,0x8(%rsp)
  call   <crypt@plt>
  lea    D15(%rip),%rsi
  mov    %rax,%rdx
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <sscanf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x20,%rsp
  mov    %r12,%rax
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D8
type_infered: byte[16]
name: unpack_data
arguments: label: D15
type_infered: string

tool: {"D8": [0x24,0x31,0x24,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e,0x2e]}

tool: D15: $1$.SY4.tn.$%[^:]:

assistant: ```c
char *strcrypt(char *str) {
  unsigned long seed[2];
  char *crypted;
  int i;
  char salt[] = "$1$..............";
  const char *const seedchars = "./0123456789ABCDEFGHIJKLMNOPQRST"
                                "UVWXYZabcdefghijklmnopqrstuvwxyz";
  seed[0] = 1920 * 69 * 13;
  seed[1] = 3321 ^ (seed[0] >> 14 & 0x30000);
  for (i = 0; i < 8; i++)
    salt[3 + i] = seedchars[(seed[i / 5] >> (i % 5) * 6) & 0x3f];
  crypted = crypt(str, salt);
  sscanf(crypted, "$1$.SY4.tn.$%[^:]:", crypted);
  return crypted;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dump_board>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L48
L59:
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L57
L18:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  cmp    $0x40,%al
  jle    L68
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  cmp    $0x5a,%al
  jg     L68
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <pstring_red@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L40
L68:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L40:
  addl   $0x1,-0x4(%rbp)
L57:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L18
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x8(%rbp)
L48:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L59
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D18
data_type: string
name: unpack_data
arguments: data_label: D0
data_type: string

tool: D18:     

tool: D0:  %c 

assistant: ```c
void dump_board(char board[40][40], int nrows, int ncols) {
  int i, j;
  printf("\n");
  for (i = 0; i < nrows; i++) {
    printf("    ");
    for (j = 0; j < ncols; j++) {
      if (board[i][j] >= 'A' && board[i][j] <= 'Z') {
        printf(" ");
        pstring_red(board[i][j]);
        printf(" ");
      } else {
        printf(" %c ", board[i][j]);
      }
    }
    printf("\n");
  }
  printf("\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dump_board>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %esi,%ebx
  mov    %edx,%r13d
  mov    $0xa,%edi
  call   <putchar@plt>
  test   %ebx,%ebx
  jle    L74
  lea    -0x1(%r13),%eax
  mov    %rax,0x8(%rsp)
  lea    0x1(%r12,%rax,1),%rbp
  lea    0x29(%r12,%rax,1),%rdx
  lea    -0x1(%rbx),%eax
  lea    (%rax,%rax,4),%rax
  lea    (%rdx,%rax,8),%r15
  lea    D3(%rip),%r14
  lea    D0(%rip),%r12
  jmp    L55
L12:
  movsbl %sil,%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L64:
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  je     L4
L59:
  movzbl (%rbx),%esi
  lea    -0x41(%rsi),%eax
  cmp    $0x19,%al
  ja     L12
  mov    $0x20,%edi
  call   <putchar@plt>
  movsbl (%rbx),%edi
  call   <pstring_red@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  jmp    L64
L4:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x28,%rbp
  cmp    %r15,%rbp
  je     L74
L55:
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  test   %r13d,%r13d
  jle    L4
  mov    %rbp,%rbx
  sub    0x8(%rsp),%rbx
  sub    $0x1,%rbx
  jmp    L59
L74:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
data_type: string
name: extract_symbolic_data
arguments: data_label: D0
data_type: string

tool: D3:     

tool: D0:  %c 

assistant: ```c
void dump_board(char board[40][40], int nrows, int ncols) {
  int i, j;
  printf("\n");
  for (i = 0; i < nrows; i++) {
    printf("    ");
    for (j = 0; j < ncols; j++) {
      if (board[i][j] >= 'A' && board[i][j] <= 'Z') {
        printf(" ");
        pstring_red(board[i][j]);
        printf(" ");
      } else {
        printf(" %c ", board[i][j]);
      }
    }
    printf("\n");
  }
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dump_board>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  mov    $0xa,%edi
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x18,%rsp
  call   <putchar@plt>
  test   %ebp,%ebp
  jle    L5
  lea    -0x1(%r12),%r13d
  lea    -0x1(%rbp),%eax
  lea    0x29(%r14,%r13,1),%rdx
  lea    (%rax,%rax,4),%rax
  lea    (%rdx,%rax,8),%rax
  lea    0x1(%r14,%r13,1),%rbx
  mov    %rax,0x8(%rsp)
  lea    D4(%rip),%r14
  lea    D8(%rip),%rbp
  nopl   0x0(%rax)
L26:
  xor    %eax,%eax
  mov    %r14,%rdi
  call   <printf@plt>
  test   %r12d,%r12d
  jle    L30
  mov    %rbx,%rdx
  sub    %r13,%rdx
  lea    -0x1(%rdx),%r15
  jmp    L23
  nopl   0x0(%rax,%rax,1)
L67:
  mov    $0x20,%edi
  add    $0x1,%r15
  call   <putchar@plt>
  movsbl -0x1(%r15),%edi
  call   <pstring_red@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  cmp    %rbx,%r15
  je     L30
L23:
  movsbl (%r15),%esi
  lea    -0x41(%rsi),%eax
  cmp    $0x19,%al
  jbe    L67
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %rbx,%r15
  jne    L23
L30:
  mov    $0xa,%edi
  add    $0x28,%rbx
  call   <putchar@plt>
  cmp    0x8(%rsp),%rbx
  jne    L26
L5:
  add    $0x18,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <putchar@plt>
```

tool_calls: name: read_data
arguments: data_tag: D4
type_infered: string
name: read_data
arguments: data_tag: D8
type_infered: string

tool: D4:     

tool: D8:  %c 

assistant: ```c
void dump_board(char board[40][40], int nrows, int ncols) {
  int i, j;
  printf("\n");
  for (i = 0; i < nrows; i++) {
    printf("    ");
    for (j = 0; j < ncols; j++) {
      if (board[i][j] >= 'A' && board[i][j] <= 'Z') {
        printf(" ");
        pstring_red(board[i][j]);
        printf(" ");
      } else {
        printf(" %c ", board[i][j]);
      }
    }
    printf("\n");
  }
  printf("\n");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dump_board>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdi,%r14
  mov    $0xa,%edi
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  sub    $0x18,%rsp
  call   <putchar@plt>
  test   %ebp,%ebp
  jle    L13
  lea    -0x1(%r12),%r13d
  lea    -0x1(%rbp),%eax
  lea    0x29(%r14,%r13,1),%rdx
  lea    (%rax,%rax,4),%rax
  lea    (%rdx,%rax,8),%rax
  lea    0x1(%r14,%r13,1),%rbx
  mov    %rax,0x8(%rsp)
  lea    D10(%rip),%r14
  lea    D13(%rip),%rbp
  nopl   0x0(%rax)
L22:
  xor    %eax,%eax
  mov    %r14,%rdi
  call   <printf@plt>
  test   %r12d,%r12d
  jle    L9
  mov    %rbx,%rdx
  sub    %r13,%rdx
  lea    -0x1(%rdx),%r15
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L54:
  mov    $0x20,%edi
  add    $0x1,%r15
  call   <putchar@plt>
  movsbl -0x1(%r15),%edi
  call   <pstring_red@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  cmp    %rbx,%r15
  je     L9
L17:
  movsbl (%r15),%esi
  lea    -0x41(%rsi),%eax
  cmp    $0x19,%al
  jbe    L54
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%r15
  call   <printf@plt>
  cmp    %rbx,%r15
  jne    L17
L9:
  mov    $0xa,%edi
  add    $0x28,%rbx
  call   <putchar@plt>
  cmp    0x8(%rsp),%rbx
  jne    L22
L13:
  add    $0x18,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <putchar@plt>
```

tool_calls: name: read_data
arguments: data_label: D10
type_guessed: string
name: read_data
arguments: data_label: D13
type_guessed: string

tool: D10:     

tool: D13:  %c 

assistant: ```c
void dump_board(char board[40][40], int nrows, int ncols) {
  int i, j;
  printf("\n");
  for (i = 0; i < nrows; i++) {
    printf("    ");
    for (j = 0; j < ncols; j++) {
      if (board[i][j] >= 'A' && board[i][j] <= 'Z') {
        printf(" ");
        pstring_red(board[i][j]);
        printf(" ");
      } else {
        printf(" %c ", board[i][j]);
      }
    }
    printf("\n");
  }
  printf("\n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<kup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movsd  -0x10(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x10(%rbp),%xmm0
  mulsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D4
type: string
name: extract_symbolic_data
arguments: label: D13
type: string

tool: D4: Kupu alinacak bir deger veriniz.


tool: D13: %lf

assistant: ```c
double kup() {
  double a;
  printf("Kupu alinacak bir deger veriniz.\n");
  scanf("%lf", &a);
  return a * a * a;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<kup>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rsp),%xmm1
  movapd %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
data_type: string
name: disassemble_data
arguments: data_tag: D15
data_type: string

tool: D6: Kupu alinacak bir deger veriniz.


tool: D15: %lf

assistant: ```c
double kup() {
  double a;
  printf("Kupu alinacak bir deger veriniz.\n");
  scanf("%lf", &a);
  return a * a * a;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<kup>:
  endbr64
  sub    $0x18,%rsp
  lea    D2(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D17(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rsp),%xmm1
  movapd %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D2
type_infered: string
name: parse_data
arguments: tag: D17
type_infered: string

tool: D2: Kupu alinacak bir deger veriniz.


tool: D17: %lf

assistant: ```c
double kup() {
  double a;
  printf("Kupu alinacak bir deger veriniz.\n");
  scanf("%lf", &a);
  return a * a * a;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<kup>:
  endbr64
  sub    $0x18,%rsp
  lea    D14(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D2(%rip),%rdi
  call   <scanf@plt>
  movsd  (%rsp),%xmm1
  movapd %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  mulsd  %xmm1,%xmm0
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D14
type_infered: string
name: read_data
arguments: data_tag: D2
type_infered: string

tool: D14: Kupu alinacak bir deger veriniz.


tool: D2: %lf

assistant: ```c
double kup() {
  double a;
  printf("Kupu alinacak bir deger veriniz.\n");
  scanf("%lf", &a);
  return a * a * a;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L1
L10:
  mov    -0x10(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L1:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void strcpy(char *dst, const char *src) {
  while (*src != 0)
    *dst++ = *src++;
  *dst = 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<strcpy>:
  endbr64
  movzbl (%rsi),%eax
  test   %al,%al
  je     L11
L5:
  add    $0x1,%rsi
  add    $0x1,%rdi
  mov    %al,-0x1(%rdi)
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L5
L11:
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void strcpy(char *dst, const char *src) {
  while (*src != 0)
    *dst++ = *src++;
  *dst = 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcpy>:
  endbr64
  jmp    L14
  cs nopw 0x0(%rax,%rax,1)
L16:
  mov    %al,(%rdi)
  add    $0x1,%rdi
  add    $0x1,%rsi
L14:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L16
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void strcpy(char *dst, const char *src) {
  while (*src != 0)
    *dst++ = *src++;
  *dst = 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcpy>:
  endbr64
  jmp    L9
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    %al,(%rdi)
  add    $0x1,%rdi
  add    $0x1,%rsi
L9:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L3
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void strcpy(char *dst, const char *src) {
  while (*src != 0)
    *dst++ = *src++;
  *dst = 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mygetline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L53
L65:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x8(%rbp)
  jge    L29
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
L29:
  addl   $0x1,-0x8(%rbp)
L53:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L9
  cmpl   $0xa,-0x4(%rbp)
  jne    L65
L9:
  cmpl   $0xa,-0x4(%rbp)
  jne    L12
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x8(%rbp)
  jge    L50
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L50:
  addl   $0x1,-0x8(%rbp)
  jmp    L30
L12:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L30:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int mygetline(char s[], int lim) {
  int c, i;
  for (i = 0; (c = getchar()) != EOF && c != '\n'; ++i) {
    if (i < lim - 1)
      s[i] = c;
  }
  if (c == '\n') {
    if (i < lim - 1) {
      s[i] = c;
      s[i + 1] = '\0';
    }
    ++i;
  } else
    s[lim] = '\0';
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<mygetline>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %esi,%r14d
  mov    $0x0,%ebx
  lea    -0x1(%rsi),%r12d
  jmp    L65
L68:
  add    $0x1,%rbx
L65:
  mov    %ebx,%ebp
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    D0(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L77
  cmp    $0xa,%eax
  je     L77
  cmp    %ebx,%r12d
  jle    L68
  mov    %al,0x0(%r13,%rbx,1)
  jmp    L68
L77:
  cmp    $0xa,%eax
  je     L37
  movslq %r14d,%r14
  movb   $0x0,0x0(%r13,%r14,1)
L20:
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L37:
  sub    $0x1,%r14d
  cmp    %ebp,%r14d
  jle    L44
  movslq %ebp,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  movb   $0x0,0x1(%r13,%rax,1)
L44:
  add    $0x1,%ebp
  jmp    L20
```

assistant: ```c
int mygetline(char s[], int lim) {
  int c, i;
  for (i = 0; (c = getchar()) != EOF && c != '\n'; ++i) {
    if (i < lim - 1)
      s[i] = c;
  }
  if (c == '\n') {
    if (i < lim - 1) {
      s[i] = c;
      s[i + 1] = '\0';
    }
    ++i;
  } else
    s[lim] = '\0';
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mygetline>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  movslq %esi,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%r14),%ebx
  sub    $0x8,%rsp
  mov    D0(%rip),%rbp
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L31:
  mov    %ebx,%edx
  cmp    $0xa,%eax
  je     L16
  cmp    %r15d,%ebx
  jle    L57
  mov    %al,0x0(%r13,%r15,1)
L57:
  add    $0x1,%r15
L32:
  xor    %eax,%eax
  mov    %r15d,%r12d
  call   <getchar@plt>
  cmp    %eax,0x0(%rbp)
  jne    L31
  cmp    $0xa,%eax
  je     L48
  movb   $0x0,0x0(%r13,%r14,1)
L68:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L48:
  lea    -0x1(%r14),%edx
  nopl   (%rax)
L16:
  cmp    %r12d,%edx
  jle    L41
  movslq %r12d,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  movb   $0x0,0x1(%r13,%rax,1)
L41:
  add    $0x1,%r12d
  jmp    L68
```

assistant: ```c
int mygetline(char s[], int lim) {
  int c, i;
  for (i = 0; (c = getchar()) != EOF && c != '\n'; ++i) {
    if (i < lim - 1)
      s[i] = c;
  }
  if (c == '\n') {
    if (i < lim - 1) {
      s[i] = c;
      s[i + 1] = '\0';
    }
    ++i;
  } else
    s[lim] = '\0';
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mygetline>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  movslq %esi,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%r14),%ebx
  sub    $0x8,%rsp
  mov    D6(%rip),%rbp
  jmp    L30
  nopl   0x0(%rax,%rax,1)
L20:
  mov    %ebx,%edx
  cmp    $0xa,%eax
  je     L37
  cmp    %r15d,%ebx
  jle    L51
  mov    %al,0x0(%r13,%r15,1)
L51:
  add    $0x1,%r15
L30:
  xor    %eax,%eax
  mov    %r15d,%r12d
  call   <getchar@plt>
  cmp    %eax,0x0(%rbp)
  jne    L20
  cmp    $0xa,%eax
  je     L43
  movb   $0x0,0x0(%r13,%r14,1)
L65:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L43:
  lea    -0x1(%r14),%edx
  nopl   (%rax)
L37:
  cmp    %r12d,%edx
  jle    L29
  movslq %r12d,%rax
  movb   $0xa,0x0(%r13,%rax,1)
  movb   $0x0,0x1(%r13,%rax,1)
L29:
  add    $0x1,%r12d
  jmp    L65
```

assistant: ```c
int mygetline(char s[], int lim) {
  int c, i;
  for (i = 0; (c = getchar()) != EOF && c != '\n'; ++i) {
    if (i < lim - 1)
      s[i] = c;
  }
  if (c == '\n') {
    if (i < lim - 1) {
      s[i] = c;
      s[i + 1] = '\0';
    }
    ++i;
  } else
    s[lim] = '\0';
  return i;
}
```