ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%r12
  mov    $0x0,%ebx
  lea    D12(%rip),%r14
  lea    D3(%rip),%r13
L108:
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x2a,(%r12)
  je     L3
  add    $0x1,%ebx
  add    $0x4,%r12
  cmp    $0x3e7,%ebx
  jne    L108
  jmp    L10
L24:
  add    $0x4,%rax
  cmp    %rsi,%rax
  je     L78
L86:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L24
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
  jmp    L24
L78:
  sub    $0x1,%ebx
L36:
  cmp    $0xffffffff,%ebx
  je     L52
  mov    %ebx,%eax
  lea    (%rdi,%rax,4),%rsi
  mov    %rbp,%rax
  jmp    L86
L3:
  test   %ebx,%ebx
  je     L52
L10:
  sub    $0x2,%ebx
  lea    0x4(%rbp),%rdi
  jmp    L36
L52:
  mov    (%rsp),%esi
  cmp    $0x2a,%esi
  je     L1
  lea    0x4(%rsp),%rbx
  lea    D2(%rip),%rbp
L103:
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  mov    -0x4(%rbx),%esi
  cmp    $0x2a,%esi
  jne    L103
L1:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L116
  mov    $0x0,%eax
  add    $0xfb0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L116:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D12
data_type: string
name: unpack_data
arguments: label: D3
data_type: string
name: unpack_data
arguments: label: D2
data_type: string

tool: D12: Vavedi chislo - 

tool: D3: %d

tool: D2: 
%d

assistant: ```c
int main() {
  int c[999], i = 0, x, l;
  while (i < 999) {
    printf("Vavedi chislo - ");
    scanf("%d", &c[i]);
    if (c[i] == 42)
      break;
    i++;
  }
  l = i;
  i = 0;
  while (l != 0) {
    while (i != l - 1) {
      if (c[i] > c[i + 1]) {
        x = c[i];
        c[i] = c[i + 1];
        c[i + 1] = x;
      }
      i++;
    }
    i = 0;
    l--;
  }
  i = 0;
  while (c[i] != 42) {
    printf("\n%d", c[i]);
    i++;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D19(%rip),%r14
  push   %r13
  lea    D21(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%r12
  jmp    L68
  xchg   %ax,%ax
L67:
  add    $0x1,%ebx
  add    $0x4,%r12
  cmp    $0x3e7,%ebx
  je     L105
L68:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x2a,(%r12)
  jne    L67
  test   %ebx,%ebx
  je     L104
L105:
  sub    $0x2,%ebx
  lea    0x4(%rbp),%rdi
  cmp    $0xffffffff,%ebx
  je     L104
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %ebx,%eax
  lea    (%rdi,%rax,4),%rsi
  mov    %rbp,%rax
  nopl   0x0(%rax)
L60:
  mov    (%rax),%edx
  mov    0x4(%rax),%ecx
  cmp    %ecx,%edx
  jle    L86
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
L86:
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L60
  sub    $0x1,%ebx
  cmp    $0xffffffff,%ebx
  jne    L13
L104:
  mov    (%rsp),%esi
  lea    0x4(%rsp),%rbx
  lea    D12(%rip),%rbp
  cmp    $0x2a,%esi
  je     L93
  nopl   0x0(%rax,%rax,1)
L97:
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  mov    -0x4(%rbx),%esi
  cmp    $0x2a,%esi
  jne    L97
L93:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  add    $0xfb0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D19
datatype: string
name: retrieve_data
arguments: data_tag: D21
datatype: string
name: retrieve_data
arguments: data_tag: D12
datatype: string

tool: D19: Vavedi chislo - 

tool: D21: %d

tool: D12: 
%d

assistant: ```c
int main() {
  int c[999], i = 0, x, l;
  while (i < 999) {
    printf("Vavedi chislo - ");
    scanf("%d", &c[i]);
    if (c[i] == 42)
      break;
    i++;
  }
  l = i;
  i = 0;
  while (l != 0) {
    while (i != l - 1) {
      if (c[i] > c[i + 1]) {
        x = c[i];
        c[i] = c[i + 1];
        c[i + 1] = x;
      }
      i++;
    }
    i = 0;
    l--;
  }
  i = 0;
  while (c[i] != 42) {
    printf("\n%d", c[i]);
    i++;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D3(%rip),%r14
  push   %r13
  lea    D28(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0xfb0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%r12
  jmp    L18
  xchg   %ax,%ax
L11:
  add    $0x1,%ebx
  add    $0x4,%r12
  cmp    $0x3e7,%ebx
  je     L99
L18:
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  cmpl   $0x2a,(%r12)
  jne    L11
  test   %ebx,%ebx
  je     L14
  sub    $0x1,%ebx
  je     L14
L96:
  lea    0x4(%rbp),%rdi
  nopl   0x0(%rax)
L86:
  lea    -0x1(%rbx),%eax
  mov    %rbp,%rdx
  mov    %rax,%rbx
  lea    (%rdi,%rax,4),%rsi
  nopl   (%rax)
L8:
  movq   (%rdx),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%eax
  movd   %xmm2,%ecx
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jge    L114
  movq   %xmm1,(%rdx)
L114:
  add    $0x4,%rdx
  cmp    %rsi,%rdx
  jne    L8
  test   %ebx,%ebx
  jne    L86
L14:
  mov    (%rsp),%esi
  lea    0x4(%rsp),%rbx
  lea    D16(%rip),%rbp
  cmp    $0x2a,%esi
  je     L47
  nop
L32:
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  mov    -0x4(%rbx),%esi
  cmp    $0x2a,%esi
  jne    L32
L47:
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  add    $0xfb0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L99:
  mov    $0x3e6,%ebx
  jmp    L96
L82:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D3
type_guessed: string
name: parse_data
arguments: tag: D28
type_guessed: string
name: parse_data
arguments: tag: D16
type_guessed: string

tool: D3: Vavedi chislo - 

tool: D28: %d

tool: D16: 
%d

assistant: ```c
int main() {
  int c[999], i = 0, x, l;
  while (i < 999) {
    printf("Vavedi chislo - ");
    scanf("%d", &c[i]);
    if (c[i] == 42)
      break;
    i++;
  }
  l = i;
  i = 0;
  while (l != 0) {
    while (i != l - 1) {
      if (c[i] > c[i + 1]) {
        x = c[i];
        c[i] = c[i + 1];
        c[i + 1] = x;
      }
      i++;
    }
    i = 0;
    l--;
  }
  i = 0;
  while (c[i] != 42) {
    printf("\n%d", c[i]);
    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
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jg     L6
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L3
L6:
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <get_arguments@plt>
L3:
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D4
datatype: string
name: retrieve_data
arguments: label: D5
datatype: string

tool: D4: 
Welcome to mod!
A litte small useless modulizer
written with <3 in C by willeponken.
There was no arguments specified.
Please re-run mod with arguments.

Available arguments:
-c [start] [end], ex: '-c 1 100' will count all numbers from 1 to 100.
-m [mod], ex: '-m 4 5' will modulize with 4 and 5.



tool: D5: 
Welcome to mod!
A litte small useless modulizer
written with <3 in C by willeponken.


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("\nWelcome to mod!\nA litte small useless modulizer\nwritten with <3 in C by willeponken.\nThere was no arguments specified.\nPlease re-run mod with arguments.\n\nAvailable arguments:\n-c [start] [end], ex: '-c 1 100' will count all numbers from 1 to 100.\n-m [mod], ex: '-m 4 5' will modulize with 4 and 5.\n\n");
  } else {
    printf("\nWelcome to mod!\nA litte small useless modulizer\nwritten with <3 in C by willeponken.\n");
    get_arguments(argc, argv);
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L16
  mov    %edi,%ebx
  mov    %rsi,%rbp
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <get_arguments@plt>
L3:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  lea    D16(%rip),%rdi
  call   <printf@plt>
  jmp    L3
```

tool_calls: name: retrieve_data
arguments: label: D19
type_infered: string
name: retrieve_data
arguments: label: D16
type_infered: string

tool: D19: 
Welcome to mod!
A litte small useless modulizer
written with <3 in C by willeponken.


tool: D16: 
Welcome to mod!
A litte small useless modulizer
written with <3 in C by willeponken.
There was no arguments specified.
Please re-run mod with arguments.

Available arguments:
-c [start] [end], ex: '-c 1 100' will count all numbers from 1 to 100.
-m [mod], ex: '-m 4 5' will modulize with 4 and 5.



assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("\nWelcome to mod!\nA litte small useless modulizer\nwritten with <3 in C by willeponken.\nThere was no arguments specified.\nPlease re-run mod with arguments.\n\nAvailable arguments:\n-c [start] [end], ex: '-c 1 100' will count all numbers from 1 to 100.\n-m [mod], ex: '-m 4 5' will modulize with 4 and 5.\n\n");
  } else {
    printf("\nWelcome to mod!\nA litte small useless modulizer\nwritten with <3 in C by willeponken.\n");
    get_arguments(argc, argv);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L19
  mov    %edi,%ebp
  mov    %rsi,%r12
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <get_arguments@plt>
L1:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L19:
  lea    D7(%rip),%rdi
  call   <printf@plt>
  jmp    L1
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D14
data_type: string
name: extract_symbolic_data
arguments: data_label: D7
data_type: string

tool: D14: 
Welcome to mod!
A litte small useless modulizer
written with <3 in C by willeponken.


tool: D7: 
Welcome to mod!
A litte small useless modulizer
written with <3 in C by willeponken.
There was no arguments specified.
Please re-run mod with arguments.

Available arguments:
-c [start] [end], ex: '-c 1 100' will count all numbers from 1 to 100.
-m [mod], ex: '-m 4 5' will modulize with 4 and 5.



assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("\nWelcome to mod!\nA litte small useless modulizer\nwritten with <3 in C by willeponken.\nThere was no arguments specified.\nPlease re-run mod with arguments.\n\nAvailable arguments:\n-c [start] [end], ex: '-c 1 100' will count all numbers from 1 to 100.\n-m [mod], ex: '-m 4 5' will modulize with 4 and 5.\n\n");
  } else {
    printf("\nWelcome to mod!\nA litte small useless modulizer\nwritten with <3 in C by willeponken.\n");
    get_arguments(argc, argv);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  cmp    $0x1,%edi
  jle    L14
  mov    %edi,%ebp
  mov    %rsi,%r12
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %ebp,%edi
  call   <get_arguments@plt>
L7:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L14:
  lea    D17(%rip),%rdi
  call   <printf@plt>
  jmp    L7
```

tool_calls: name: read_data
arguments: data_label: D8
type: string
name: read_data
arguments: data_label: D17
type: string

tool: D8: 
Welcome to mod!
A litte small useless modulizer
written with <3 in C by willeponken.


tool: D17: 
Welcome to mod!
A litte small useless modulizer
written with <3 in C by willeponken.
There was no arguments specified.
Please re-run mod with arguments.

Available arguments:
-c [start] [end], ex: '-c 1 100' will count all numbers from 1 to 100.
-m [mod], ex: '-m 4 5' will modulize with 4 and 5.



assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 2) {
    printf("\nWelcome to mod!\nA litte small useless modulizer\nwritten with <3 in C by willeponken.\nThere was no arguments specified.\nPlease re-run mod with arguments.\n\nAvailable arguments:\n-c [start] [end], ex: '-c 1 100' will count all numbers from 1 to 100.\n-m [mod], ex: '-m 4 5' will modulize with 4 and 5.\n\n");
  } else {
    printf("\nWelcome to mod!\nA litte small useless modulizer\nwritten with <3 in C by willeponken.\n");
    get_arguments(argc, argv);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<string_strndup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x1c(%rbp)
  jle    L6
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  jmp    L7
L6:
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L13
  mov    $0x0,%eax
  jmp    L7
L13:
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L7:
  leave
  ret
```

assistant: ```c
char *string_strndup(const char *string, int length) {
  char *result;
  if ((int)strlen(string) < length)
    return strdup(string);
  result = malloc(length + 1);
  if (!result)
    return NULL;
  memcpy(result, string, length);
  result[length] = '\0';
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<string_strndup>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%ebx
  call   <strlen@plt>
  cmp    %eax,%ebx
  jg     L20
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L22
  mov    %ebx,%edx
  mov    %r12,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movslq %ebx,%rbx
  movb   $0x0,0x0(%rbp,%rbx,1)
L22:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L20:
  mov    %r12,%rdi
  call   <strdup@plt>
  mov    %rax,%rbp
  jmp    L22
```

assistant: ```c
char *string_strndup(const char *string, int length) {
  char *result;
  if ((int)strlen(string) < length)
    return strdup(string);
  result = malloc(length + 1);
  if (!result)
    return NULL;
  memcpy(result, string, length);
  result[length] = '\0';
  return result;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<string_strndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  call   <strlen@plt>
  cmp    %eax,%ebx
  jg     L8
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L2:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L8:
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    10b0 <strdup@plt>
```

assistant: ```c
char *string_strndup(const char *string, int length) {
  char *result;
  if ((int)strlen(string) < length)
    return strdup(string);
  result = malloc(length + 1);
  if (!result)
    return NULL;
  memcpy(result, string, length);
  result[length] = '\0';
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<string_strndup>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movslq %esi,%rbx
  call   <strlen@plt>
  cmp    %eax,%ebx
  jg     L16
  lea    0x1(%rbx),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L14
  mov    %ebx,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  movb   $0x0,(%r12,%rbx,1)
L14:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L16:
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  jmp    10b0 <strdup@plt>
```

assistant: ```c
char *string_strndup(const char *string, int length) {
  char *result;
  if ((int)strlen(string) < length)
    return strdup(string);
  result = malloc(length + 1);
  if (!result)
    return NULL;
  memcpy(result, string, length);
  result[length] = '\0';
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_uint8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rcx
  lea    -0xc(%rbp),%rax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
uint8_t get_uint8(const void *p) {
  uint8_t x;
  memcpy((void *)&x, p, sizeof(x));
  return x;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_uint8>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  mov    $0x4,%edx
  call   <memcpy@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint8_t get_uint8(const void *p) {
  uint8_t x;
  memcpy((void *)&x, p, sizeof(x));
  return x;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<get_uint8>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rsi
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <memcpy@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint8_t get_uint8(const void *p) {
  uint8_t x;
  memcpy((void *)&x, p, sizeof(x));
  return x;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_uint8>:
  endbr64
  sub    $0x18,%rsp
  mov    %rdi,%rsi
  mov    $0x4,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <memcpy@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint8_t get_uint8(const void *p) {
  uint8_t x;
  memcpy((void *)&x, p, sizeof(x));
  return x;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<check_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L3
  mov    D19(%rip),%rax
  mov    (%rax),%eax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <fprintf@plt>
  mov    D0(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <exit@plt>
L3:
  nop
  leave
  ret
```

assistant: ```c
void check_error(int cond, const char *str) {
  if (cond == 1) {
    fprintf(stderr, str);
    exit(EXIT_FAILURE);
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<check_error>:
  endbr64
  cmp    $0x1,%edi
  je     L6
  ret
L6:
  sub    $0x8,%rsp
  mov    D14(%rip),%rax
  mov    (%rax),%edi
  call   <fprintf@plt>
  mov    D5(%rip),%rax
  mov    (%rax),%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void check_error(int cond, const char *str) {
  if (cond == 1) {
    fprintf(stderr, str);
    exit(EXIT_FAILURE);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<check_error>:
  endbr64
  cmp    $0x1,%edi
  je     L8
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  sub    $0x8,%rsp
  mov    D13(%rip),%rax
  mov    (%rax),%edi
  call   <fprintf@plt>
  mov    D3(%rip),%rax
  mov    (%rax),%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void check_error(int cond, const char *str) {
  if (cond == 1) {
    fprintf(stderr, str);
    exit(EXIT_FAILURE);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<check_error>:
  endbr64
  cmp    $0x1,%edi
  je     L6
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  sub    $0x8,%rsp
  mov    D13(%rip),%rax
  mov    (%rax),%edi
  call   <fprintf@plt>
  mov    D16(%rip),%rax
  mov    (%rax),%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void check_error(int cond, const char *str) {
  if (cond == 1) {
    fprintf(stderr, str);
    exit(EXIT_FAILURE);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x7,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (15 > 8)
    return 7;
  else
    return 8;
}
```
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
  mov    $0x7,%eax
  ret
```

assistant: ```c
int main() {
  if (15 > 8)
    return 7;
  else
    return 8;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  mov    $0x7,%eax
  ret
```

assistant: ```c
int main() {
  if (15 > 8)
    return 7;
  else
    return 8;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  mov    $0x7,%eax
  ret
```

assistant: ```c
int main() {
  if (15 > 8)
    return 7;
  else
    return 8;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<zskip_left>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L10:
  movsbl -0x5(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  je     L21
  addl   $0x1,-0x4(%rbp)
  jmp    L3
L21:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  jmp    L4
L3:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x5(%rbp)
  cmpb   $0x0,-0x5(%rbp)
  jne    L10
  mov    $0x0,%eax
L4:
  leave
  ret
```

assistant: ```c
char *zskip_left(const char *str, const char *ignores) {
  char ch;
  int i = 0;
  while ((ch = str[i])) {
    if (strchr(ignores, ch)) {
      i++;
      continue;
    }
    return (char *)str + i;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<zskip_left>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  movzbl (%rdi),%esi
  test   %sil,%sil
  je     L35
L19:
  movsbl %sil,%esi
  mov    %rbp,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L10
L2:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  add    $0x1,%rbx
  movzbl (%rbx),%esi
  test   %sil,%sil
  jne    L19
  mov    $0x0,%ebx
  jmp    L2
L35:
  mov    $0x0,%ebx
  jmp    L2
```

assistant: ```c
char *zskip_left(const char *str, const char *ignores) {
  char ch;
  int i = 0;
  while ((ch = str[i])) {
    if (strchr(ignores, ch)) {
      i++;
      continue;
    }
    return (char *)str + i;
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<zskip_left>:
  endbr64
  push   %r12
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  movsbl (%rdi),%esi
  test   %sil,%sil
  je     L36
  mov    %rdi,%r12
L39:
  mov    %rbx,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L21
L4:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
  nopl   0x0(%rax)
L21:
  movsbl 0x1(%r12),%esi
  add    $0x1,%r12
  test   %sil,%sil
  jne    L39
L36:
  xor    %r12d,%r12d
  jmp    L4
```

assistant: ```c
char *zskip_left(const char *str, const char *ignores) {
  char ch;
  int i = 0;
  while ((ch = str[i])) {
    if (strchr(ignores, ch)) {
      i++;
      continue;
    }
    return (char *)str + i;
  }
  return 0;
}
```