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

```asm
<vista_abort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    D18(%rip),%rax
  mov    (%rax),%ebx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <vista_msg@plt>
  mov    %ebx,%esi
  mov    %eax,%edi
  call   <fputs@plt>
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <fflush@plt>
  mov    $0x5d,%edi
  call   <dberr_exit@plt>
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void vista_abort(char *location) {
  fputs(vista_msg(location), stderr);
  fflush(stderr);
  dberr_exit(93);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vista_abort>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D1(%rip),%rbx
  mov    (%rbx),%ebp
  call   <vista_msg@plt>
  mov    %eax,%edi
  mov    %ebp,%esi
  call   <fputs@plt>
  mov    (%rbx),%edi
  call   <fflush@plt>
  mov    $0x5d,%edi
  call   <dberr_exit@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void vista_abort(char *location) {
  fputs(vista_msg(location), stderr);
  fflush(stderr);
  dberr_exit(93);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<vista_abort>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D9(%rip),%rbx
  mov    (%rbx),%ebp
  call   <vista_msg@plt>
  mov    %eax,%edi
  mov    %ebp,%esi
  call   <fputs@plt>
  mov    (%rbx),%edi
  call   <fflush@plt>
  add    $0x8,%rsp
  mov    $0x5d,%edi
  pop    %rbx
  pop    %rbp
  jmp    1090 <dberr_exit@plt>
```

assistant: ```c
void vista_abort(char *location) {
  fputs(vista_msg(location), stderr);
  fflush(stderr);
  dberr_exit(93);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vista_abort>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D5(%rip),%rbx
  mov    (%rbx),%ebp
  call   <vista_msg@plt>
  mov    %eax,%edi
  mov    %ebp,%esi
  call   <fputs@plt>
  mov    (%rbx),%edi
  call   <fflush@plt>
  add    $0x8,%rsp
  mov    $0x5d,%edi
  pop    %rbx
  pop    %rbp
  jmp    1090 <dberr_exit@plt>
```

assistant: ```c
void vista_abort(char *location) {
  fputs(vista_msg(location), stderr);
  fflush(stderr);
  dberr_exit(93);
}
```
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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xfc0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xfbc(%rbp)
  movl   $0x0,-0xfb8(%rbp)
  movl   $0x1f4,-0xfb4(%rbp)
  movl   $0x0,-0xfc0(%rbp)
  jmp    L10
L57:
  mov    -0xfc0(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xfc0(%rbp),%eax
  cltq
  mov    -0xfc0(%rbp),%edx
  mov    %edx,-0xfb0(%rbp,%rax,4)
  addl   $0x1,-0xfc0(%rbp)
L10:
  cmpl   $0x3e7,-0xfc0(%rbp)
  jle    L57
  movl   $0x0,-0xfc0(%rbp)
  jmp    L60
L46:
  mov    -0xfc0(%rbp),%eax
  mov    $0x3e8,%esi
  mov    %eax,%edi
  call   <randint@plt>
  mov    %eax,-0xfbc(%rbp)
  mov    -0xfbc(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xfc0(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,-0xfb8(%rbp)
  mov    -0xfbc(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%edx
  mov    -0xfc0(%rbp),%eax
  cltq
  mov    %edx,-0xfb0(%rbp,%rax,4)
  mov    -0xfbc(%rbp),%eax
  cltq
  mov    -0xfb8(%rbp),%edx
  mov    %edx,-0xfb0(%rbp,%rax,4)
  addl   $0x1,-0xfc0(%rbp)
L60:
  cmpl   $0x3e7,-0xfc0(%rbp)
  jle    L46
  movl   $0x0,-0xfc0(%rbp)
  jmp    L16
L38:
  mov    -0xfc0(%rbp),%eax
  cltq
  mov    -0xfb0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xfc0(%rbp)
L16:
  cmpl   $0x3e7,-0xfc0(%rbp)
  jle    L38
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L47
  call   <__stack_chk_fail@plt>
L47:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D27
type: string

tool: D27: %d


assistant: ```c
int main() {
  int arr[1000];
  int i;
  int r = 0;
  int t = 0;
  int mid = 1000 / 2;
  for (i = 0; i < 1000; i++) {
    printf("%d\n", i);
    arr[i] = i;
  }
  for (i = 0; i < 1000; i++) {
    r = randint(i, 1000);
    printf("%d\n", r);
    t = arr[i];
    arr[i] = arr[r];
    arr[r] = t;
  }
  for (i = 0; i < 1000; i++) {
    printf("%d\n", arr[i]);
  }
}
```
user: Translate the below assembly code into C programming language:

```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    $0x0,%ebx
  lea    D11(%rip),%rbp
L38:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %ebx,(%rsp,%rbx,4)
  add    $0x1,%rbx
  cmp    $0x3e8,%rbx
  jne    L38
  mov    %rsp,%rbp
  mov    %rbp,%r13
  mov    $0x0,%r12d
  lea    D11(%rip),%r14
L34:
  mov    $0x3e8,%esi
  mov    %r12d,%edi
  call   <randint@plt>
  mov    %eax,%ebx
  mov    %eax,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  mov    0x0(%r13),%eax
  movslq %ebx,%rbx
  mov    (%rsp,%rbx,4),%edx
  mov    %edx,0x0(%r13)
  mov    %eax,(%rsp,%rbx,4)
  add    $0x1,%r12d
  add    $0x4,%r13
  cmp    $0x3e8,%r12d
  jne    L34
  lea    0xfa0(%rbp),%r12
  lea    D11(%rip),%rbx
L8:
  mov    0x0(%rbp),%esi
  mov    %rbx,%rdi
  call   <printf@plt>
  add    $0x4,%rbp
  cmp    %r12,%rbp
  jne    L8
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0xfb0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D11
type_infered: string

tool: D11: %d


assistant: ```c
int main() {
  int arr[1000];
  int i;
  int r = 0;
  int t = 0;
  int mid = 1000 / 2;
  for (i = 0; i < 1000; i++) {
    printf("%d\n", i);
    arr[i] = i;
  }
  for (i = 0; i < 1000; i++) {
    r = randint(i, 1000);
    printf("%d\n", r);
    t = arr[i];
    arr[i] = arr[r];
    arr[r] = t;
  }
  for (i = 0; i < 1000; i++) {
    printf("%d\n", arr[i]);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  nopl   0x0(%rax)
L28:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %ebx,(%r12,%rbx,4)
  add    $0x1,%rbx
  cmp    $0x3e8,%rbx
  jne    L28
  mov    %r12,%r13
  mov    %r12,%r15
  xor    %r14d,%r14d
  nopl   0x0(%rax)
L4:
  mov    %r14d,%edi
  mov    $0x3e8,%esi
  add    $0x1,%r14d
  add    $0x4,%r15
  call   <randint@plt>
  mov    %rbp,%rdi
  movslq %eax,%rbx
  mov    %ebx,%esi
  call   <printf@plt>
  mov    -0x4(%r15),%eax
  mov    (%rsp,%rbx,4),%edx
  mov    %edx,-0x4(%r15)
  mov    %eax,(%rsp,%rbx,4)
  cmp    $0x3e8,%r14d
  jne    L4
  add    $0xfa0,%r12
L7:
  mov    0x0(%r13),%esi
  mov    %rbp,%rdi
  add    $0x4,%r13
  call   <printf@plt>
  cmp    %r12,%r13
  jne    L7
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0xfb8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L24:
  call   <__stack_chk_fail@plt>
```

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

tool: D2: %d


assistant: ```c
int main() {
  int arr[1000];
  int i;
  int r = 0;
  int t = 0;
  int mid = 1000 / 2;
  for (i = 0; i < 1000; i++) {
    printf("%d\n", i);
    arr[i] = i;
  }
  for (i = 0; i < 1000; i++) {
    r = randint(i, 1000);
    printf("%d\n", r);
    t = arr[i];
    arr[i] = arr[r];
    arr[r] = t;
  }
  for (i = 0; i < 1000; i++) {
    printf("%d\n", arr[i]);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0xfb8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0xfa8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  nopl   0x0(%rax)
L3:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    %ebx,(%r12,%rbx,4)
  add    $0x1,%rbx
  cmp    $0x3e8,%rbx
  jne    L3
  mov    %r12,%r13
  mov    %r12,%r15
  xor    %r14d,%r14d
  nopl   0x0(%rax)
L27:
  mov    %r14d,%edi
  mov    $0x3e8,%esi
  add    $0x1,%r14d
  add    $0x4,%r15
  call   <randint@plt>
  mov    %rbp,%rdi
  movslq %eax,%rbx
  mov    %ebx,%esi
  call   <printf@plt>
  mov    -0x4(%r15),%eax
  mov    (%rsp,%rbx,4),%edx
  mov    %edx,-0x4(%r15)
  mov    %eax,(%rsp,%rbx,4)
  cmp    $0x3e8,%r14d
  jne    L27
  add    $0xfa0,%r12
L4:
  mov    0x0(%r13),%esi
  mov    %rbp,%rdi
  add    $0x4,%r13
  call   <printf@plt>
  cmp    %r12,%r13
  jne    L4
  mov    0xfa8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0xfb8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D7
datatype: string

tool: D7: %d


assistant: ```c
int main() {
  int arr[1000];
  int i;
  int r = 0;
  int t = 0;
  int mid = 1000 / 2;
  for (i = 0; i < 1000; i++) {
    printf("%d\n", i);
    arr[i] = i;
  }
  for (i = 0; i < 1000; i++) {
    r = randint(i, 1000);
    printf("%d\n", r);
    t = arr[i];
    arr[i] = arr[r];
    arr[r] = t;
  }
  for (i = 0; i < 1000; i++) {
    printf("%d\n", arr[i]);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<b_sc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x14(%rbp),%eax
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  sete   %al
  movzbl %al,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int b_sc(int a, signed char *b, int *c) {
  int d = a - *b;
  *c = d == 0;
  return d;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<b_sc>:
  endbr64
  mov    %rdx,%rcx
  movsbl (%rsi),%edx
  mov    %edi,%eax
  sub    %edx,%eax
  sete   %dl
  movzbl %dl,%edx
  mov    %edx,(%rcx)
  ret
```

assistant: ```c
int b_sc(int a, signed char *b, int *c) {
  int d = a - *b;
  *c = d == 0;
  return d;
}
```
user: What is the c source code of the assembly code below:

```asm
<b_sc>:
  endbr64
  mov    %rdx,%r8
  movsbl (%rsi),%edx
  mov    %edi,%eax
  sub    %edx,%eax
  sete   %dl
  movzbl %dl,%edx
  mov    %edx,(%r8)
  ret
```

assistant: ```c
int b_sc(int a, signed char *b, int *c) {
  int d = a - *b;
  *c = d == 0;
  return d;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<b_sc>:
  endbr64
  mov    %rdx,%r8
  movsbl (%rsi),%edx
  mov    %edi,%eax
  sub    %edx,%eax
  sete   %dl
  movzbl %dl,%edx
  mov    %edx,(%r8)
  ret
```

assistant: ```c
int b_sc(int a, signed char *b, int *c) {
  int d = a - *b;
  *c = d == 0;
  return d;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<is_leap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  sub    $0x76b,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L35
  subl   $0x7d0,-0x4(%rbp)
L35:
  addl   $0x76c,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L47
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x5,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L10
  mov    -0x4(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x7,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L47
L10:
  mov    $0x1,%eax
  jmp    L12
L47:
  mov    $0x0,%eax
L12:
  pop    %rbp
  ret
```

assistant: ```c
int is_leap(int y) {
  if (y > INT_MAX - 1900)
    y -= 2000;
  y += 1900;
  return !(y % 4) && ((y % 100) || !(y % 400));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<is_leap>:
  endbr64
  mov    D1(%rip),%rax
  mov    (%rax),%eax
  sub    $0x76b,%eax
  lea    -0x7d0(%rdi),%edx
  cmp    %edi,%eax
  cmovle %edx,%edi
  lea    0x76c(%rdi),%edx
  mov    $0x0,%ecx
  test   $0x3,%dil
  jne    L2
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x25,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x64,%eax,%eax
  mov    $0x1,%ecx
  cmp    %eax,%edx
  jne    L2
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x27,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%eax
  cmp    %eax,%edx
  sete   %cl
  movzbl %cl,%ecx
L2:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int is_leap(int y) {
  if (y > INT_MAX - 1900)
    y -= 2000;
  y += 1900;
  return !(y % 4) && ((y % 100) || !(y % 400));
}
```
user: What is the c source code of the assembly code below:

```asm
<is_leap>:
  endbr64
  mov    D8(%rip),%rax
  lea    -0x7d0(%rdi),%edx
  mov    (%rax),%eax
  sub    $0x76b,%eax
  cmp    %edi,%eax
  cmovle %edx,%edi
  xor    %r8d,%r8d
  lea    0x76c(%rdi),%eax
  and    $0x3,%edi
  jne    L18
  imul   $0xc28f5c29,%eax,%eax
  mov    $0x1,%r8d
  add    $0x51eb850,%eax
  mov    %eax,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  jbe    L14
L18:
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L14:
  ror    $0x4,%eax
  xor    %r8d,%r8d
  cmp    $0xa3d70a,%eax
  setbe  %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_leap(int y) {
  if (y > INT_MAX - 1900)
    y -= 2000;
  y += 1900;
  return !(y % 4) && ((y % 100) || !(y % 400));
}
```
user: Translate the below assembly code into C programming language:

```asm
<is_leap>:
  endbr64
  mov    D7(%rip),%rax
  lea    -0x7d0(%rdi),%edx
  mov    (%rax),%eax
  sub    $0x76b,%eax
  cmp    %edi,%eax
  cmovle %edx,%edi
  xor    %r8d,%r8d
  lea    0x76c(%rdi),%eax
  and    $0x3,%edi
  jne    L5
  imul   $0xc28f5c29,%eax,%eax
  mov    $0x1,%r8d
  add    $0x51eb850,%eax
  mov    %eax,%edx
  ror    $0x2,%edx
  cmp    $0x28f5c28,%edx
  jbe    L8
L5:
  mov    %r8d,%eax
  ret
  nopl   (%rax)
L8:
  ror    $0x4,%eax
  xor    %r8d,%r8d
  cmp    $0xa3d70a,%eax
  setbe  %r8b
  mov    %r8d,%eax
  ret
```

assistant: ```c
int is_leap(int y) {
  if (y > INT_MAX - 1900)
    y -= 2000;
  y += 1900;
  return !(y % 4) && ((y % 100) || !(y % 400));
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<allocateAll>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <allocateObjectMemory@plt>
  mov    $0x0,%eax
  call   <allocateMethodLookupCache@plt>
  mov    $0x0,%eax
  call   <allocateFileStreamTable@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void allocateAll() {
  allocateObjectMemory();
  allocateMethodLookupCache();
  allocateFileStreamTable();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<allocateAll>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <allocateObjectMemory@plt>
  mov    $0x0,%eax
  call   <allocateMethodLookupCache@plt>
  mov    $0x0,%eax
  call   <allocateFileStreamTable@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void allocateAll() {
  allocateObjectMemory();
  allocateMethodLookupCache();
  allocateFileStreamTable();
}
```
user: What is the c source code of the assembly code below?

```asm
<allocateAll>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <allocateObjectMemory@plt>
  xor    %eax,%eax
  call   <allocateMethodLookupCache@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <allocateFileStreamTable@plt>
```

assistant: ```c
void allocateAll() {
  allocateObjectMemory();
  allocateMethodLookupCache();
  allocateFileStreamTable();
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<allocateAll>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <allocateObjectMemory@plt>
  xor    %eax,%eax
  call   <allocateMethodLookupCache@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <allocateFileStreamTable@plt>
```

assistant: ```c
void allocateAll() {
  allocateObjectMemory();
  allocateMethodLookupCache();
  allocateFileStreamTable();
}
```
user: What is the c source code of the assembly code below:

```asm
<two_branch>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  cmpl   $0x0,-0x4(%rbp)
  je     L3
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L3:
  movl   $0x1,-0x4(%rbp)
L13:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void two_branch() {
  int x;
  if (x) {
    x = 0;
  } else {
    x = 1;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<two_branch>:
  endbr64
  ret
```

assistant: ```c
void two_branch() {
  int x;
  if (x) {
    x = 0;
  } else {
    x = 1;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<two_branch>:
  endbr64
  ret
```

assistant: ```c
void two_branch() {
  int x;
  if (x) {
    x = 0;
  } else {
    x = 1;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<direct>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x46,%esi
  mov    $0x0,%edi
  call   <motor@plt>
  mov    $0x46,%esi
  mov    $0x1,%edi
  call   <motor@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void direct() {
  motor(0, 70);
  motor(1, 70);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<direct>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x46,%esi
  mov    $0x0,%edi
  call   <motor@plt>
  mov    $0x46,%esi
  mov    $0x1,%edi
  call   <motor@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void direct() {
  motor(0, 70);
  motor(1, 70);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<direct>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x46,%esi
  xor    %edi,%edi
  call   <motor@plt>
  mov    $0x46,%esi
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1050 <motor@plt>
```

assistant: ```c
void direct() {
  motor(0, 70);
  motor(1, 70);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<direct>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x46,%esi
  xor    %edi,%edi
  call   <motor@plt>
  mov    $0x46,%esi
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1050 <motor@plt>
```

assistant: ```c
void direct() {
  motor(0, 70);
  motor(1, 70);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<unrar_mkpath>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L22
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L28
L22:
  mov    $0xffffffff,%eax
  jmp    L85
L28:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strdup@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L45
  mov    $0xffffffff,%eax
  jmp    L85
L45:
  movl   $0x0,-0x10(%rbp)
  jmp    L49
L39:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jne    L65
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L56
  mov    -0x8(%rbp),%rax
  mov    $0x1ff,%esi
  mov    %rax,%rdi
  call   <mkdir@plt>
  mov    %eax,-0xc(%rbp)
L56:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2f,(%rax)
L65:
  addl   $0x1,-0x10(%rbp)
L49:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L39
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0xc(%rbp),%eax
L85:
  leave
  ret
```

assistant: ```c
int unrar_mkpath(char *path) {
  char *path_tmp;
  int i, ret = 0;
  if (path == NULL || path[0] == '\0')
    return (-1);
  path_tmp = strdup(path);
  if (path_tmp == NULL)
    return (-1);
  for (i = 0; path_tmp[i] != '\0'; i++) {
    if (path_tmp[i] == '/') {
      path_tmp[i] = '\0';
      if (path_tmp[0] != '\0') {
        ret = mkdir(path_tmp, 0777);
      }
      path_tmp[i] = '/';
    }
  }
  free(path_tmp);
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<unrar_mkpath>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L119
  cmpb   $0x0,(%rdi)
  je     L67
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L99
  movzbl (%rax),%eax
  test   %al,%al
  je     L6
  mov    %rbp,%rbx
  mov    $0x0,%r12d
  jmp    L81
L14:
  movb   $0x2f,(%rbx)
L65:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L62
L81:
  cmp    $0x2f,%al
  jne    L65
  movb   $0x0,(%rbx)
  cmpb   $0x0,0x0(%rbp)
  je     L14
  mov    $0x1ff,%esi
  mov    %rbp,%rdi
  call   <mkdir@plt>
  mov    %eax,%r12d
  jmp    L14
L6:
  mov    $0x0,%r12d
L62:
  mov    %rbp,%rdi
  call   <free@plt>
L48:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L119:
  mov    $0xffffffff,%r12d
  jmp    L48
L67:
  mov    $0xffffffff,%r12d
  jmp    L48
L99:
  mov    $0xffffffff,%r12d
  jmp    L48
```

assistant: ```c
int unrar_mkpath(char *path) {
  char *path_tmp;
  int i, ret = 0;
  if (path == NULL || path[0] == '\0')
    return (-1);
  path_tmp = strdup(path);
  if (path_tmp == NULL)
    return (-1);
  for (i = 0; path_tmp[i] != '\0'; i++) {
    if (path_tmp[i] == '/') {
      path_tmp[i] = '\0';
      if (path_tmp[0] != '\0') {
        ret = mkdir(path_tmp, 0777);
      }
      path_tmp[i] = '/';
    }
  }
  free(path_tmp);
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<unrar_mkpath>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L35
  cmpb   $0x0,(%rdi)
  je     L35
  call   <strdup@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L35
  movzbl (%rax),%eax
  test   %al,%al
  je     L50
  mov    %rbp,%rbx
  xor    %r12d,%r12d
  jmp    L53
  nopw   0x0(%rax,%rax,1)
L15:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  je     L116
L53:
  cmp    $0x2f,%al
  jne    L15
  movb   $0x0,(%rbx)
  cmpb   $0x0,0x0(%rbp)
  jne    L85
L92:
  movb   $0x2f,(%rbx)
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbx
  test   %al,%al
  jne    L53
L116:
  mov    %rbp,%rdi
  call   <free@plt>
L4:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L85:
  mov    $0x1ff,%esi
  mov    %rbp,%rdi
  call   <mkdir@plt>
  mov    %eax,%r12d
  jmp    L92
  nopw   0x0(%rax,%rax,1)
L50:
  xor    %r12d,%r12d
  jmp    L116
L35:
  mov    $0xffffffff,%r12d
  jmp    L4
```

assistant: ```c
int unrar_mkpath(char *path) {
  char *path_tmp;
  int i, ret = 0;
  if (path == NULL || path[0] == '\0')
    return (-1);
  path_tmp = strdup(path);
  if (path_tmp == NULL)
    return (-1);
  for (i = 0; path_tmp[i] != '\0'; i++) {
    if (path_tmp[i] == '/') {
      path_tmp[i] = '\0';
      if (path_tmp[0] != '\0') {
        ret = mkdir(path_tmp, 0777);
      }
      path_tmp[i] = '/';
    }
  }
  free(path_tmp);
  return ret;
}
```