ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D26(%rip),%rsi
  lea    D12(%rip),%rdi
  call   <fopen@plt>
  mov    %rax,%r12
  lea    0x6(%rsp),%rbp
  lea    0x2a(%rsp),%r14
  lea    D24(%rip),%r13
L7:
  lea    -0x6(%rbp),%rbx
L18:
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <fscanf@plt>
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L18
  add    $0x6,%rbp
  cmp    %r14,%rbp
  jne    L7
  mov    %rsp,%rdi
  call   <route@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D26
datatype: string
name: disassemble_data
arguments: data_label: D12
datatype: string
name: disassemble_data
arguments: data_label: D24
datatype: string

tool: D26: r

tool: D12: maze_stack.txt

tool: D24: %c

assistant: ```c
int main() {
  int i, j;
  char maze[6][6];
  FILE *fd = fopen("maze_stack.txt", "r");
  for (i = 0; i < 6; i++)
    for (j = 0; j < 6; j++)
      fscanf(fd, "%c", &maze[i][j]);
  route(maze);
  fclose(fd);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D13(%rip),%rsi
  lea    D4(%rip),%rdi
  push   %r13
  lea    D17(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rbp
  lea    0x2a(%rsp),%r14
  call   <fopen@plt>
  mov    %rax,%r12
  nopw   0x0(%rax,%rax,1)
L12:
  lea    -0x6(%rbp),%rbx
  nopl   0x0(%rax)
L4:
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <fscanf@plt>
  cmp    %rbp,%rbx
  jne    L4
  lea    0x6(%rbx),%rbp
  cmp    %r14,%rbp
  jne    L12
  mov    %rsp,%rdi
  call   <route@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D13
type: string
name: extract_symbolic_data
arguments: tag: D4
type: string
name: extract_symbolic_data
arguments: tag: D17
type: string

tool: D13: r

tool: D4: maze_stack.txt

tool: D17: %c

assistant: ```c
int main() {
  int i, j;
  char maze[6][6];
  FILE *fd = fopen("maze_stack.txt", "r");
  for (i = 0; i < 6; i++)
    for (j = 0; j < 6; j++)
      fscanf(fd, "%c", &maze[i][j]);
  route(maze);
  fclose(fd);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r14
  lea    D12(%rip),%rsi
  lea    D28(%rip),%rdi
  push   %r13
  lea    D25(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rbp
  lea    0x2a(%rsp),%r14
  call   <fopen@plt>
  mov    %rax,%r12
  nopw   0x0(%rax,%rax,1)
L28:
  lea    -0x6(%rbp),%rbx
  nopl   0x0(%rax)
L16:
  mov    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x1,%rbx
  call   <fscanf@plt>
  cmp    %rbp,%rbx
  jne    L16
  lea    0x6(%rbx),%rbp
  cmp    %r14,%rbp
  jne    L28
  mov    %rsp,%rdi
  call   <route@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D12
type: string
name: parse_data
arguments: label: D28
type: string
name: parse_data
arguments: label: D25
type: string

tool: D12: r

tool: D28: maze_stack.txt

tool: D25: %c

assistant: ```c
int main() {
  int i, j;
  char maze[6][6];
  FILE *fd = fopen("maze_stack.txt", "r");
  for (i = 0; i < 6; i++)
    for (j = 0; j < 6; j++)
      fscanf(fd, "%c", &maze[i][j]);
  route(maze);
  fclose(fd);
  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
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x6f6c6c6568206968,%rax
  movabs $0x65726120776f6820,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movl   $0x756f7920,-0x10(%rbp)
  movb   $0x0,-0xc(%rbp)
  movl   $0x756f79,-0x24(%rbp)
  lea    -0x24(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strstr@plt>
  mov    %rax,-0x30(%rbp)
  cmpq   $0x0,-0x30(%rbp)
  je     L4
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L19
L4:
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L19:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
type_guessed: string
name: unpack_data
arguments: data_label: D6
type_guessed: string

tool: D7: There exists a substring


tool: D6: There is no substring


assistant: ```c
void main() {
  char haystack[] = "hi hello how are you";
  char needle[] = "you";
  char *pos = strstr(haystack, needle);
  if (pos) {
    printf("There exists a substring\n");
  } else {
    printf("There is no substring\n");
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movabs $0x6f6c6c6568206968,%rax
  movabs $0x65726120776f6820,%rdx
  mov    %rax,0x10(%rsp)
  mov    %rdx,0x18(%rsp)
  movl   $0x756f7920,0x20(%rsp)
  movb   $0x0,0x24(%rsp)
  movl   $0x756f79,0xc(%rsp)
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  call   <strstr@plt>
  test   %rax,%rax
  je     L13
  lea    D12(%rip),%rdi
  call   <printf@plt>
L29:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  add    $0x38,%rsp
  ret
L13:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  jmp    L29
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D12
datatype: string
name: retrieve_data
arguments: data_label: D1
datatype: string

tool: D12: There exists a substring


tool: D1: There is no substring


assistant: ```c
void main() {
  char haystack[] = "hi hello how are you";
  char needle[] = "you";
  char *pos = strstr(haystack, needle);
  if (pos) {
    printf("There exists a substring\n");
  } else {
    printf("There is no substring\n");
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  movdqa D1(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  movl   $0x756f7920,0x20(%rsp)
  movb   $0x0,0x24(%rsp)
  movl   $0x756f79,0xc(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <strstr@plt>
  test   %rax,%rax
  je     L22
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  lea    D26(%rip),%rdi
  add    $0x38,%rsp
  jmp    1080 <printf@plt>
L22:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  lea    D19(%rip),%rdi
  add    $0x38,%rsp
  jmp    1080 <printf@plt>
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D1
type_infered: byte[16]
name: parse_data
arguments: data_tag: D26
type_infered: string
name: parse_data
arguments: data_tag: D19
type_infered: string

tool: {"D1": [0x68,0x69,0x20,0x68,0x65,0x6c,0x6c,0x6f,0x20,0x68,0x6f,0x77,0x20,0x61,0x72,0x65]}

tool: D26: There exists a substring


tool: D19: There is no substring


assistant: ```c
void main() {
  char haystack[] = "hi hello how are you";
  char needle[] = "you";
  char *pos = strstr(haystack, needle);
  if (pos) {
    printf("There exists a substring\n");
  } else {
    printf("There is no substring\n");
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  movdqa D0(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x10(%rsp),%rdi
  movl   $0x756f7920,0x20(%rsp)
  movb   $0x0,0x24(%rsp)
  movl   $0x756f79,0xc(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <strstr@plt>
  test   %rax,%rax
  je     L22
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  lea    D3(%rip),%rdi
  add    $0x38,%rsp
  jmp    1080 <printf@plt>
L22:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  lea    D17(%rip),%rdi
  add    $0x38,%rsp
  jmp    1080 <printf@plt>
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D0
data_type: byte[16]
name: read_data
arguments: label: D3
data_type: string
name: read_data
arguments: label: D17
data_type: string

tool: {"D0": [0x68,0x69,0x20,0x68,0x65,0x6c,0x6c,0x6f,0x20,0x68,0x6f,0x77,0x20,0x61,0x72,0x65]}

tool: D3: There exists a substring


tool: D17: There is no substring


assistant: ```c
void main() {
  char haystack[] = "hi hello how are you";
  char needle[] = "you";
  char *pos = strstr(haystack, needle);
  if (pos) {
    printf("There exists a substring\n");
  } else {
    printf("There is no substring\n");
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rowswap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L59
L4:
  mov    -0x20(%rbp),%eax
  imul   -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x1c(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  pxor   %xmm1,%xmm1
  cvtss2sd %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <abs@plt>
  test   %eax,%eax
  jle    L1
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L48
L12:
  mov    -0x20(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x20(%rbp),%eax
  imul   -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x20(%rbp),%eax
  imul   -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x8(%rbp),%eax
  cltq
  add    %rcx,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movss  (%rdx),%xmm0
  movss  %xmm0,(%rax)
  mov    -0x20(%rbp),%eax
  imul   -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movss  -0x4(%rbp),%xmm0
  movss  %xmm0,(%rax)
  addl   $0x1,-0x8(%rbp)
L48:
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L12
  jmp    L18
L1:
  addl   $0x1,-0xc(%rbp)
L59:
  mov    -0x20(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L4
L18:
  mov    $0x1,%eax
  leave
  ret
```

assistant: ```c
int rowswap(float *array, int j, int n) {
  int i, k;
  for (i = j + 1; i < n - 1; i++) {
    if (abs(*(array + (n)*i + j)) > 0.01) {
      for (k = j; k < n; k++) {
        float tmp = *(array + n * j + k);
        *(array + n * j + k) = *(array + n * i + k);
        *(array + n * i + k) = tmp;
      }
      break;
    }
  }
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rowswap>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  lea    0x1(%rsi),%ebp
  lea    -0x1(%rdx),%eax
  cmp    %eax,%ebp
  jge    L40
  mov    %esi,%r15d
  mov    %edx,%r14d
  movslq %edx,%rax
  mov    %rax,(%rsp)
  mov    %ebp,%eax
  imul   %edx,%eax
  mov    %eax,0x14(%rsp)
  movslq %eax,%rbx
  lea    -0x1(%rdx),%r12d
  movslq %esi,%rax
  mov    %rax,0x18(%rsp)
  lea    (%rdi,%rax,4),%r13
L16:
  pxor   %xmm0,%xmm0
  cvtss2sd 0x0(%r13,%rbx,4),%xmm0
  call   <abs@plt>
  test   %eax,%eax
  jg     L33
  add    $0x1,%ebp
  add    (%rsp),%rbx
  cmp    %r12d,%ebp
  jne    L16
L40:
  mov    $0x1,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L33:
  cmp    %r14d,%r15d
  jge    L40
  mov    0x14(%rsp),%edx
  sub    %r14d,%edx
  movslq %edx,%rdx
  mov    0x18(%rsp),%rcx
  add    %rdx,%rcx
  mov    0x8(%rsp),%rsi
  lea    (%rsi,%rcx,4),%rax
  sub    %r15d,%r12d
  add    %rcx,%r12
  lea    0x4(%rsi,%r12,4),%rcx
  sub    %rdx,%rbx
L54:
  movss  (%rax),%xmm0
  movss  (%rax,%rbx,4),%xmm1
  movss  %xmm1,(%rax)
  movss  %xmm0,(%rax,%rbx,4)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L54
  jmp    L40
```

assistant: ```c
int rowswap(float *array, int j, int n) {
  int i, k;
  for (i = j + 1; i < n - 1; i++) {
    if (abs(*(array + (n)*i + j)) > 0.01) {
      for (k = j; k < n; k++) {
        float tmp = *(array + n * j + k);
        *(array + n * j + k) = *(array + n * i + k);
        *(array + n * i + k) = tmp;
      }
      break;
    }
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<rowswap>:
  endbr64
  push   %r15
  push   %r14
  lea    0x1(%rsi),%r14d
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%rdx),%ebx
  sub    $0x28,%rsp
  mov    %rdi,0x8(%rsp)
  mov    %esi,0x18(%rsp)
  cmp    %ebx,%r14d
  jge    L18
  mov    %r14d,%eax
  movslq %esi,%rcx
  mov    %edx,%r13d
  movslq %edx,%r12
  imul   %edx,%eax
  mov    %rcx,0x10(%rsp)
  lea    (%rdi,%rcx,4),%rbp
  mov    %eax,0x1c(%rsp)
  movslq %eax,%r15
  jmp    L67
  nopl   0x0(%rax)
L13:
  add    $0x1,%r14d
  add    %r12,%r15
  cmp    %ebx,%r14d
  je     L18
L67:
  movss  0x0(%rbp,%r15,4),%xmm1
  pxor   %xmm0,%xmm0
  movss  %xmm1,0x4(%rsp)
  cvtss2sd %xmm1,%xmm0
  call   <abs@plt>
  movss  0x4(%rsp),%xmm1
  test   %eax,%eax
  jle    L13
  mov    0x18(%rsp),%edi
  cmp    %r13d,%edi
  jge    L18
  mov    0x1c(%rsp),%ecx
  mov    0x10(%rsp),%rsi
  sub    %edi,%ebx
  mov    0x8(%rsp),%rdx
  sub    %r13d,%ecx
  movslq %ecx,%rcx
  add    %rcx,%rsi
  add    %rsi,%rbx
  lea    (%rdx,%rsi,4),%rax
  lea    0x4(%rdx,%rbx,4),%rsi
  mov    %r15,%rdx
  sub    %rcx,%rdx
  jmp    L20
  cs nopw 0x0(%rax,%rax,1)
L49:
  movss  (%rax,%rdx,4),%xmm1
L20:
  movss  (%rax),%xmm0
  movss  %xmm1,(%rax)
  movss  %xmm0,(%rax,%rdx,4)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L49
L18:
  add    $0x28,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int rowswap(float *array, int j, int n) {
  int i, k;
  for (i = j + 1; i < n - 1; i++) {
    if (abs(*(array + (n)*i + j)) > 0.01) {
      for (k = j; k < n; k++) {
        float tmp = *(array + n * j + k);
        *(array + n * j + k) = *(array + n * i + k);
        *(array + n * i + k) = tmp;
      }
      break;
    }
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main_bad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    $0x2,%edi
  call   <cond_OK@plt>
  mov    $0x0,%eax
  call   <bar_OK@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <foo_OK@plt>
  add    %eax,%ebx
  mov    $0xf,%edi
  call   <cond_OK@plt>
  add    %eax,%eax
  add    %ebx,%eax
  mov    %eax,-0x20(%rbp)
  mov    $0x0,%eax
  call   <bar_OK@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <foo_OK@plt>
  add    %eax,%ebx
  mov    $0x11,%edi
  call   <cond_OK@plt>
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %ebx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    $0x0,%eax
  call   <bar_OK@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <foo_OK@plt>
  add    %eax,%ebx
  mov    $0xb,%edi
  call   <cond_OK@plt>
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %ebx,%eax
  mov    %eax,-0x18(%rbp)
  mov    $0x0,%eax
  call   <bar_OK@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <foo_OK@plt>
  add    %eax,%ebx
  mov    $0x13,%edi
  call   <cond_OK@plt>
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %ebx,%eax
  mov    %eax,-0x14(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main_bad() {
  int k1, k2, k3, k4;
  cond_OK(2);
  k1 = bar_OK() + foo_OK() + cond_OK(15) * 2;
  k2 = bar_OK() + foo_OK() + cond_OK(17) * 3;
  k3 = bar_OK() + foo_OK() + cond_OK(11) * 3;
  k4 = bar_OK() + foo_OK() + cond_OK(19) * 3;
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main_bad>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <cond_OK@plt>
  mov    $0x0,%eax
  call   <bar_OK@plt>
  mov    $0x0,%eax
  call   <foo_OK@plt>
  mov    $0xf,%edi
  call   <cond_OK@plt>
  mov    $0x0,%eax
  call   <bar_OK@plt>
  mov    $0x0,%eax
  call   <foo_OK@plt>
  mov    $0x11,%edi
  call   <cond_OK@plt>
  mov    $0x0,%eax
  call   <bar_OK@plt>
  mov    $0x0,%eax
  call   <foo_OK@plt>
  mov    $0xb,%edi
  call   <cond_OK@plt>
  mov    $0x0,%eax
  call   <bar_OK@plt>
  mov    $0x0,%eax
  call   <foo_OK@plt>
  mov    $0x13,%edi
  call   <cond_OK@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main_bad() {
  int k1, k2, k3, k4;
  cond_OK(2);
  k1 = bar_OK() + foo_OK() + cond_OK(15) * 2;
  k2 = bar_OK() + foo_OK() + cond_OK(17) * 3;
  k3 = bar_OK() + foo_OK() + cond_OK(11) * 3;
  k4 = bar_OK() + foo_OK() + cond_OK(19) * 3;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main_bad>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  call   <bar_OK@plt>
  xor    %eax,%eax
  call   <foo_OK@plt>
  mov    $0xf,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  call   <bar_OK@plt>
  xor    %eax,%eax
  call   <foo_OK@plt>
  mov    $0x11,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  call   <bar_OK@plt>
  xor    %eax,%eax
  call   <foo_OK@plt>
  mov    $0xb,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  call   <bar_OK@plt>
  xor    %eax,%eax
  call   <foo_OK@plt>
  mov    $0x13,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main_bad() {
  int k1, k2, k3, k4;
  cond_OK(2);
  k1 = bar_OK() + foo_OK() + cond_OK(15) * 2;
  k2 = bar_OK() + foo_OK() + cond_OK(17) * 3;
  k3 = bar_OK() + foo_OK() + cond_OK(11) * 3;
  k4 = bar_OK() + foo_OK() + cond_OK(19) * 3;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main_bad>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x2,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  call   <bar_OK@plt>
  xor    %eax,%eax
  call   <foo_OK@plt>
  mov    $0xf,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  call   <bar_OK@plt>
  xor    %eax,%eax
  call   <foo_OK@plt>
  mov    $0x11,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  call   <bar_OK@plt>
  xor    %eax,%eax
  call   <foo_OK@plt>
  mov    $0xb,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  call   <bar_OK@plt>
  xor    %eax,%eax
  call   <foo_OK@plt>
  mov    $0x13,%edi
  call   <cond_OK@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main_bad() {
  int k1, k2, k3, k4;
  cond_OK(2);
  k1 = bar_OK() + foo_OK() + cond_OK(15) * 2;
  k2 = bar_OK() + foo_OK() + cond_OK(17) * 3;
  k3 = bar_OK() + foo_OK() + cond_OK(11) * 3;
  k4 = bar_OK() + foo_OK() + cond_OK(19) * 3;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<nibblestr2hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x30(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L128
  cmpq   $0x0,-0x38(%rbp)
  jne    L103
L128:
  mov    $0x0,%eax
  jmp    L94
L103:
  mov    -0x30(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x30(%rbp)
  js     L4
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jg     L4
  mov    -0x30(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L102
L4:
  mov    $0x0,%eax
  jmp    L94
L102:
  movl   $0x0,-0x1c(%rbp)
  jmp    L79
L77:
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  movb   $0x0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L79:
  mov    -0x2c(%rbp),%eax
  sub    -0x14(%rbp),%eax
  cmp    %eax,-0x1c(%rbp)
  jl     L77
  movl   $0x0,-0x1c(%rbp)
  jmp    L31
L120:
  mov    -0x1c(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L3
  mov    -0x38(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x38(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <nibble2num@plt>
  mov    %eax,-0x18(%rbp)
  jmp    L85
L3:
  mov    -0x18(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%ebx
  mov    -0x38(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x38(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <nibble2num@plt>
  add    %ebx,%eax
  mov    %eax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%edx
  mov    %dl,(%rax)
L85:
  addl   $0x1,-0x1c(%rbp)
L31:
  mov    -0x1c(%rbp),%eax
  cmp    -0x30(%rbp),%eax
  jl     L120
  mov    $0x1,%eax
L94:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int nibblestr2hex(char *dst, int dstlen, char *src, int srclen) {
  if (dst == NULL || src == NULL)
    return (0);
  int hexbytes = srclen / 2;
  if (srclen < 0 || hexbytes > dstlen || ((srclen & 0x01) == 1))
    return (0);
  int i;
  for (i = 0; i < (dstlen - hexbytes); i++)
    *(dst++) = 0;
  int val;
  for (i = 0; i < srclen; i++) {
    if ((i & 0x01) == 0) {
      val = nibble2num(*(src++));
    } else {
      val = (val << 4) + nibble2num(*(src++));
      *(dst++) = val;
    }
  }
  return (1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<nibblestr2hex>:
  endbr64
  test   %rdi,%rdi
  je     L113
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rax
  mov    %ecx,%r12d
  test   %rdx,%rdx
  je     L10
  mov    %ecx,%edi
  shr    $0x1f,%edi
  lea    (%rdi,%rcx,1),%ecx
  sar    %ecx
  test   %dil,%dil
  jne    L64
  cmp    %esi,%ecx
  jg     L64
  mov    $0x0,%edi
  mov    %r12d,%ebp
  and    $0x1,%ebp
  jne    L81
  sub    %ecx,%esi
  test   %esi,%esi
  jle    L154
  mov    %esi,%esi
  lea    (%rax,%rsi,1),%r13
L56:
  add    $0x1,%rax
  movb   $0x0,-0x1(%rax)
  cmp    %r13,%rax
  jne    L56
L23:
  test   %r12d,%r12d
  jle    L72
  lea    0x1(%rdx),%rbx
  jmp    L43
L154:
  mov    %rax,%r13
  jmp    L23
L27:
  mov    %r8d,%r14d
  shl    $0x4,%r14d
  movsbl -0x1(%rbx),%edi
  call   <nibble2num@plt>
  lea    (%r14,%rax,1),%r8d
  mov    %r8b,0x0(%r13)
  lea    0x1(%r13),%r13
L94:
  add    $0x1,%ebp
  add    $0x1,%rbx
  cmp    %ebp,%r12d
  je     L42
L43:
  test   $0x1,%bpl
  jne    L27
  movsbl -0x1(%rbx),%edi
  call   <nibble2num@plt>
  mov    %eax,%r8d
  jmp    L94
L42:
  mov    $0x1,%edi
  jmp    L81
L10:
  mov    $0x0,%edi
L81:
  mov    %edi,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L64:
  mov    $0x0,%edi
  jmp    L81
L72:
  mov    $0x1,%edi
  jmp    L81
L113:
  mov    $0x0,%edi
  mov    %edi,%eax
  ret
```

assistant: ```c
int nibblestr2hex(char *dst, int dstlen, char *src, int srclen) {
  if (dst == NULL || src == NULL)
    return (0);
  int hexbytes = srclen / 2;
  if (srclen < 0 || hexbytes > dstlen || ((srclen & 0x01) == 1))
    return (0);
  int i;
  for (i = 0; i < (dstlen - hexbytes); i++)
    *(dst++) = 0;
  int val;
  for (i = 0; i < srclen; i++) {
    if ((i & 0x01) == 0) {
      val = nibble2num(*(src++));
    } else {
      val = (val << 4) + nibble2num(*(src++));
      *(dst++) = val;
    }
  }
  return (1);
}
```