ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D76(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D29(%rip),%rdi
  call   <scanf@plt>
  xor    %edi,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  mov    %eax,%ecx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x21,%rdx
  sub    %ecx,%edx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  cmp    $0x4,%eax
  ja     L9
  lea    D64(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  lea    D55(%rip),%rdi
  call   <printf@plt>
L9:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
  lea    D32(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D42(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D13(%rip),%rdi
  call   <printf@plt>
  jmp    L9
  lea    D26(%rip),%rdi
  call   <printf@plt>
  jmp    L9
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D76
type: string
name: disassemble_data
arguments: data_label: D29
type: string
name: disassemble_data
arguments: data_label: D55
type: string
name: disassemble_data
arguments: data_label: D32
type: string
name: disassemble_data
arguments: data_label: D42
type: string
name: disassemble_data
arguments: data_label: D13
type: string
name: disassemble_data
arguments: data_label: D26
type: string

tool: D76: Think of your favorite number and enter it:


tool: D29: %d

tool: D55: Definitely not.


tool: D32: Outcome Unclear.


tool: D42: Absolutely Certain.


tool: D13: Ask again.


tool: D26: The answer is no.


assistant: ```c
int main() {
  int number;
  int turn;
  printf("Think of your favorite number and enter it:\n");
  scanf("%d", &number);
  srand(time(NULL));
  turn = rand() % 5;
  switch (turn) {
  case 0:
    printf("Absolutely Certain.\n");
    break;
  case 1:
    printf("Ask again.\n");
    break;
  case 2:
    printf("The answer is no.\n");
    break;
  case 3:
    printf("Outcome Unclear.\n");
    break;
  case 4:
    printf("Definitely not.\n");
    break;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0xfffffcff,%edi
  call   <test@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  test(-769);
  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>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xfffffcff,%edi
  call   <test@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  test(-769);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xfffffcff,%edi
  call   <test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  test(-769);
  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
  sub    $0x8,%rsp
  mov    $0xfffffcff,%edi
  call   <test@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  test(-769);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<insert_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L48
L37:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L11
L46:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rdx
  add    $0x1,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  subl   $0x1,-0xc(%rbp)
L11:
  cmpl   $0x0,-0xc(%rbp)
  js     L38
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jg     L46
L38:
  mov    -0xc(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x8(%rbp)
L48:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L37
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
int *insert_sort(int array[], int length) {
  int current, j;
  for (int i = 0; i < length; i++) {
    current = array[i];
    j = i - 1;
    while (0 <= j && array[j] < current) {
      array[j + 1] = array[j];
      j--;
    }
    array[j + 1] = current;
  }
  return array;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<insert_sort>:
  endbr64
  test   %esi,%esi
  jle    L34
  mov    (%rdi),%r8d
  mov    %rdi,%r9
  lea    -0x1(%rsi),%r10d
  mov    $0x0,%esi
  mov    $0xffffffff,%eax
  jmp    L59
L21:
  add    $0x1,%rsi
  add    $0x4,%r9
  mov    %edx,%eax
L59:
  mov    %esi,%edx
  cltq
  mov    %r8d,0x4(%rdi,%rax,4)
  cmp    %r10,%rsi
  je     L34
  mov    0x4(%r9),%r8d
  test   %esi,%esi
  js     L21
  mov    %r9,%rax
L24:
  mov    (%rax),%ecx
  cmp    %r8d,%ecx
  jge    L21
  mov    %ecx,0x4(%rax)
  sub    $0x1,%edx
  sub    $0x4,%rax
  cmp    $0xffffffff,%edx
  jne    L24
  jmp    L21
L34:
  mov    %rdi,%rax
  ret
```

assistant: ```c
int *insert_sort(int array[], int length) {
  int current, j;
  for (int i = 0; i < length; i++) {
    current = array[i];
    j = i - 1;
    while (0 <= j && array[j] < current) {
      array[j + 1] = array[j];
      j--;
    }
    array[j + 1] = current;
  }
  return array;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<insert_sort>:
  endbr64
  mov    %rdi,%r8
  test   %esi,%esi
  jle    L32
  mov    $0xffffffff,%ecx
  mov    (%rdi),%edi
  lea    -0x1(%rsi),%r9d
  xor    %esi,%esi
  add    $0x1,%ecx
  movslq %ecx,%rcx
  mov    %edi,(%r8,%rcx,4)
  cmp    %rsi,%r9
  je     L32
  nopw   0x0(%rax,%rax,1)
L48:
  mov    0x4(%r8,%rsi,4),%edi
  mov    %rsi,%rax
  jmp    L50
  nopw   0x0(%rax,%rax,1)
L29:
  mov    %edx,0x4(%r8,%rax,4)
  sub    $0x1,%rax
  mov    %eax,%ecx
  cmp    $0xffffffff,%eax
  je     L12
L50:
  mov    (%r8,%rax,4),%edx
  mov    %eax,%ecx
  cmp    %edi,%edx
  jl     L29
L12:
  add    $0x1,%ecx
  add    $0x1,%rsi
  movslq %ecx,%rcx
  mov    %edi,(%r8,%rcx,4)
  cmp    %rsi,%r9
  jne    L48
L32:
  mov    %r8,%rax
  ret
```

assistant: ```c
int *insert_sort(int array[], int length) {
  int current, j;
  for (int i = 0; i < length; i++) {
    current = array[i];
    j = i - 1;
    while (0 <= j && array[j] < current) {
      array[j + 1] = array[j];
      j--;
    }
    array[j + 1] = current;
  }
  return array;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<insert_sort>:
  endbr64
  mov    %rdi,%r8
  test   %esi,%esi
  jle    L33
  lea    -0x1(%rsi),%r9d
  xor    %edi,%edi
  cmp    $0x1,%esi
  je     L33
  cs nopw 0x0(%rax,%rax,1)
L19:
  mov    0x4(%r8,%rdi,4),%esi
  mov    %rdi,%rax
  jmp    L0
  nopw   0x0(%rax,%rax,1)
L27:
  mov    %edx,0x4(%r8,%rax,4)
  sub    $0x1,%rax
  mov    %eax,%ecx
  cmp    $0xffffffff,%eax
  je     L7
L0:
  mov    (%r8,%rax,4),%edx
  mov    %eax,%ecx
  cmp    %esi,%edx
  jl     L27
L7:
  add    $0x1,%ecx
  add    $0x1,%rdi
  movslq %ecx,%rcx
  mov    %esi,(%r8,%rcx,4)
  cmp    %r9,%rdi
  jne    L19
L33:
  mov    %r8,%rax
  ret
```

assistant: ```c
int *insert_sort(int array[], int length) {
  int current, j;
  for (int i = 0; i < length; i++) {
    current = array[i];
    j = i - 1;
    while (0 <= j && array[j] < current) {
      array[j + 1] = array[j];
      j--;
    }
    array[j + 1] = current;
  }
  return array;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0xc8,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
type: string

tool: D0: You are %d miles away


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 200;
  printf("You are %d miles away\n", distance);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xc8,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D4: You are %d miles away


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 200;
  printf("You are %d miles away\n", distance);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0xc8,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D8
type: string

tool: D8: You are %d miles away


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 200;
  printf("You are %d miles away\n", distance);
  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
  sub    $0x8,%rsp
  mov    $0xc8,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D4
type_guessed: string

tool: D4: You are %d miles away


assistant: ```c
int main(int argc, char *argv[]) {
  int distance = 200;
  printf("You are %d miles away\n", distance);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<yy_3_MagicWordAutoCorrect>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void yy_3_MagicWordAutoCorrect(char *yytext, int yyleng) {
  ;
  ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<yy_3_MagicWordAutoCorrect>:
  endbr64
  ret
```

assistant: ```c
void yy_3_MagicWordAutoCorrect(char *yytext, int yyleng) {
  ;
  ;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<yy_3_MagicWordAutoCorrect>:
  endbr64
  ret
```

assistant: ```c
void yy_3_MagicWordAutoCorrect(char *yytext, int yyleng) {
  ;
  ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x2c(%rbp)
  jmp    L10
L35:
  mov    -0x30(%rbp),%eax
  mov    %eax,%edi
  call   <poramni@plt>
  mov    %eax,-0x28(%rbp)
  cmpl   $0x4,-0x2c(%rbp)
  jle    L16
  lea    -0x20(%rbp),%rax
  mov    $0x5,%esi
  mov    %rax,%rdi
  call   <max_index@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  cmp    %eax,-0x28(%rbp)
  jge    L10
  mov    -0x24(%rbp),%eax
  cltq
  mov    -0x28(%rbp),%edx
  mov    %edx,-0x20(%rbp,%rax,4)
  jmp    L10
L16:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x28(%rbp),%edx
  mov    %edx,-0x20(%rbp,%rax,4)
  addl   $0x1,-0x2c(%rbp)
L10:
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jne    L35
  mov    -0x2c(%rbp),%edx
  lea    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sort_asc@plt>
  mov    -0x2c(%rbp),%edx
  lea    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <print_arr@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L56
  call   <__stack_chk_fail@plt>
L56:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type_guessed: string

tool: D2: %d

assistant: ```c
int main() {
  int x, k, n, max_i;
  int a[5];
  n = 0;
  while (scanf("%d", &x)) {
    k = poramni(x);
    if (n >= 5) {
      max_i = max_index(a, 5);
      if (a[max_i] > k) {
        a[max_i] = k;
      }
    } else {
      a[n] = k;
      n++;
    }
  }
  sort_asc(a, n);
  print_arr(a, n);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    $0x0,%ebp
  lea    D9(%rip),%r12
  jmp    L11
L55:
  movslq %ebp,%rax
  mov    %ebx,0x10(%rsp,%rax,4)
  add    $0x1,%ebp
L11:
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L27
  mov    0xc(%rsp),%edi
  call   <poramni@plt>
  mov    %eax,%ebx
  cmp    $0x4,%ebp
  jle    L55
  lea    0x10(%rsp),%rdi
  mov    $0x5,%esi
  call   <max_index@plt>
  movslq %eax,%rdx
  cmp    %ebx,0x10(%rsp,%rdx,4)
  jle    L11
  mov    %ebx,0x10(%rsp,%rdx,4)
  jmp    L11
L27:
  lea    0x10(%rsp),%rbx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <sort_asc@plt>
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <print_arr@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L34:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D9
data_type: string

tool: D9: %d

assistant: ```c
int main() {
  int x, k, n, max_i;
  int a[5];
  n = 0;
  while (scanf("%d", &x)) {
    k = poramni(x);
    if (n >= 5) {
      max_i = max_index(a, 5);
      if (a[max_i] > k) {
        a[max_i] = k;
      }
    } else {
      a[n] = k;
      n++;
    }
  }
  sort_asc(a, n);
  print_arr(a, n);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    D7(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r14
  lea    0xc(%rsp),%r13
L40:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L1
L35:
  mov    0xc(%rsp),%edi
  call   <poramni@plt>
  mov    %eax,%ebx
  cmp    $0x4,%ebp
  jg     L21
  movslq %ebp,%rax
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebp
  mov    %ebx,0x10(%rsp,%rax,4)
  call   <scanf@plt>
  test   %rax,%rax
  jne    L35
L1:
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <sort_asc@plt>
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <print_arr@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L21:
  mov    $0x5,%esi
  mov    %r14,%rdi
  call   <max_index@plt>
  cltq
  cmp    %ebx,0x10(%rsp,%rax,4)
  jle    L40
  mov    %ebx,0x10(%rsp,%rax,4)
  jmp    L40
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D7
type: string

tool: D7: %d

assistant: ```c
int main() {
  int x, k, n, max_i;
  int a[5];
  n = 0;
  while (scanf("%d", &x)) {
    k = poramni(x);
    if (n >= 5) {
      max_i = max_index(a, 5);
      if (a[max_i] > k) {
        a[max_i] = k;
      }
    } else {
      a[n] = k;
      n++;
    }
  }
  sort_asc(a, n);
  print_arr(a, n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  lea    D1(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r14
  lea    0xc(%rsp),%r13
L43:
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  je     L12
L33:
  mov    0xc(%rsp),%edi
  call   <poramni@plt>
  mov    %eax,%ebx
  cmp    $0x4,%ebp
  jg     L21
  movslq %ebp,%rax
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebp
  mov    %ebx,0x10(%rsp,%rax,4)
  call   <scanf@plt>
  test   %rax,%rax
  jne    L33
L12:
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <sort_asc@plt>
  mov    %ebp,%esi
  mov    %r14,%rdi
  call   <print_arr@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L21:
  mov    $0x5,%esi
  mov    %r14,%rdi
  call   <max_index@plt>
  cltq
  cmp    %ebx,0x10(%rsp,%rax,4)
  jle    L43
  mov    %ebx,0x10(%rsp,%rax,4)
  jmp    L43
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D1
data_type: string

tool: D1: %d

assistant: ```c
int main() {
  int x, k, n, max_i;
  int a[5];
  n = 0;
  while (scanf("%d", &x)) {
    k = poramni(x);
    if (n >= 5) {
      max_i = max_index(a, 5);
      if (a[max_i] > k) {
        a[max_i] = k;
      }
    } else {
      a[n] = k;
      n++;
    }
  }
  sort_asc(a, n);
  print_arr(a, n);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<encode_uint32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <htonl@plt>
  mov    %eax,-0xc(%rbp)
  lea    -0xc(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
void encode_uint32(uint8_t *dst, const uint32_t src) {
  uint32_t be = htonl(src);
  memcpy(dst, &be, 4);
}
```
user: What is the c source code of the assembly code below:

```asm
<encode_uint32>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,%rbx
  mov    %esi,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <htonl@plt>
  mov    %eax,0x4(%rsp)
  lea    0x4(%rsp),%rsi
  mov    $0x4,%edx
  mov    %rbx,%rdi
  call   <memcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void encode_uint32(uint8_t *dst, const uint32_t src) {
  uint32_t be = htonl(src);
  memcpy(dst, &be, 4);
}
```
user: Translate the below assembly code into C programming language:

```asm
<encode_uint32>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <htonl@plt>
  lea    0x4(%rsp),%rsi
  mov    $0x4,%edx
  mov    %rbp,%rdi
  mov    %eax,0x4(%rsp)
  call   <memcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x10,%rsp
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void encode_uint32(uint8_t *dst, const uint32_t src) {
  uint32_t be = htonl(src);
  memcpy(dst, &be, 4);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<encode_uint32>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <htonl@plt>
  lea    0x4(%rsp),%rsi
  mov    $0x4,%edx
  mov    %rbp,%rdi
  mov    %eax,0x4(%rsp)
  call   <memcpy@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x10,%rsp
  pop    %rbp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void encode_uint32(uint8_t *dst, const uint32_t src) {
  uint32_t be = htonl(src);
  memcpy(dst, &be, 4);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x7b,-0xc(%rbp)
  movl   $0x5,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
L22:
  addl   $0x1,-0xc(%rbp)
L28:
  addl   $0x1,-0x8(%rbp)
L23:
  addl   $0x1,-0x4(%rbp)
  cmpl   $0x3,-0x4(%rbp)
  jle    L23
  cmpl   $0x1,-0x8(%rbp)
  jle    L28
  cmpl   $0x4,-0xc(%rbp)
  jle    L22
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int i = 123, j = 5, k = 1;
  do {
    i++;
    do {
      j++;
      do {
        k++;
      } while (k < 4);
    } while (j < 2);
  } while (i < 5);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int i = 123, j = 5, k = 1;
  do {
    i++;
    do {
      j++;
      do {
        k++;
      } while (k < 4);
    } while (j < 2);
  } while (i < 5);
  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>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int i = 123, j = 5, k = 1;
  do {
    i++;
    do {
      j++;
      do {
        k++;
      } while (k < 4);
    } while (j < 2);
  } while (i < 5);
  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
  xor    %eax,%eax
  ret
```

assistant: ```c
int main() {
  int i = 123, j = 5, k = 1;
  do {
    i++;
    do {
      j++;
      do {
        k++;
      } while (k < 4);
    } while (j < 2);
  } while (i < 5);
  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
<put_word>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <put_char@plt>
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  mov    %eax,%edi
  call   <put_char@plt>
  mov    -0x4(%rbp),%eax
  shr    $0x10,%eax
  mov    %eax,%edi
  call   <put_char@plt>
  mov    -0x4(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%edi
  call   <put_char@plt>
  nop
  leave
  ret
```

assistant: ```c
void put_word(unsigned w) {
  put_char(w);
  put_char(w >> 8);
  put_char(w >> 16);
  put_char(w >> 24);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<put_word>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <put_char@plt>
  mov    %ebx,%edi
  shr    $0x8,%edi
  call   <put_char@plt>
  mov    %ebx,%edi
  shr    $0x10,%edi
  call   <put_char@plt>
  shr    $0x18,%ebx
  mov    %ebx,%edi
  call   <put_char@plt>
  pop    %rbx
  ret
```

assistant: ```c
void put_word(unsigned w) {
  put_char(w);
  put_char(w >> 8);
  put_char(w >> 16);
  put_char(w >> 24);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<put_word>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <put_char@plt>
  mov    %ebx,%edi
  shr    $0x8,%edi
  call   <put_char@plt>
  mov    %ebx,%edi
  shr    $0x18,%ebx
  shr    $0x10,%edi
  call   <put_char@plt>
  mov    %ebx,%edi
  pop    %rbx
  jmp    1050 <put_char@plt>
```

assistant: ```c
void put_word(unsigned w) {
  put_char(w);
  put_char(w >> 8);
  put_char(w >> 16);
  put_char(w >> 24);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<put_word>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  call   <put_char@plt>
  mov    %ebx,%edi
  shr    $0x8,%edi
  call   <put_char@plt>
  mov    %ebx,%edi
  shr    $0x18,%ebx
  shr    $0x10,%edi
  call   <put_char@plt>
  mov    %ebx,%edi
  pop    %rbx
  jmp    1050 <put_char@plt>
```

assistant: ```c
void put_word(unsigned w) {
  put_char(w);
  put_char(w >> 8);
  put_char(w >> 16);
  put_char(w >> 24);
}
```