ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<bpf_optimize>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x8,%rsp
  mov    (%rdi),%r12
  mov    %r12,%rdi
  call   <opt_init@plt>
  mov    %r12,%rdi
  xor    %esi,%esi
  call   <opt_loop@plt>
  mov    $0x1,%esi
  mov    %r12,%rdi
  call   <opt_loop@plt>
  mov    %r12,%rdi
  call   <intern_blocks@plt>
  mov    %rbp,%rdi
  call   <opt_root@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  jmp    10d0 <opt_cleanup@plt>
```

assistant: ```c
void bpf_optimize(struct block **rootp) {
  struct block *root;
  root = *rootp;
  opt_init(root);
  opt_loop(root, 0);
  opt_loop(root, 1);
  intern_blocks(root);
  opt_root(rootp);
  opt_cleanup();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<factor>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movss  D4(%rip),%xmm0
  movss  %xmm0,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L0
L3:
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x4(%rbp),%xmm0
  movss  -0x8(%rbp),%xmm1
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L3
  movss  -0x8(%rbp),%xmm0
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D4
data_type: f32

tool: D4: 1

assistant: ```c
float factor(int n) {
  float prod = 1.0;
  int i;
  for (i = 1; i <= n; i++)
    prod *= (float)i;
  return prod;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<factor>:
  endbr64
  test   %edi,%edi
  jle    L2
  add    $0x1,%edi
  mov    $0x1,%eax
  movss  D5(%rip),%xmm0
L18:
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  mulss  %xmm1,%xmm0
  add    $0x1,%eax
  cmp    %edi,%eax
  jne    L18
  ret
L2:
  movss  D5(%rip),%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D5
datatype: f32

tool: D5: 1

assistant: ```c
float factor(int n) {
  float prod = 1.0;
  int i;
  for (i = 1; i <= n; i++)
    prod *= (float)i;
  return prod;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<factor>:
  endbr64
  test   %edi,%edi
  jle    L7
  movss  D17(%rip),%xmm0
  add    $0x1,%edi
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L3:
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  add    $0x1,%eax
  mulss  %xmm1,%xmm0
  cmp    %edi,%eax
  jne    L3
  ret
  nopl   0x0(%rax)
L7:
  movss  D17(%rip),%xmm0
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D17
type: f32

tool: D17: 1

assistant: ```c
float factor(int n) {
  float prod = 1.0;
  int i;
  for (i = 1; i <= n; i++)
    prod *= (float)i;
  return prod;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<factor>:
  endbr64
  test   %edi,%edi
  jle    L18
  movss  D8(%rip),%xmm0
  add    $0x1,%edi
  mov    $0x1,%eax
  nopl   0x0(%rax,%rax,1)
L16:
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  add    $0x1,%eax
  mulss  %xmm1,%xmm0
  cmp    %eax,%edi
  jne    L16
  ret
  nopl   0x0(%rax)
L18:
  movss  D8(%rip),%xmm0
  ret
```

tool_calls: name: read_data
arguments: data_tag: D8
type_infered: f32

tool: D8: 1

assistant: ```c
float factor(int n) {
  float prod = 1.0;
  int i;
  for (i = 1; i <= n; i++)
    prod *= (float)i;
  return prod;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x3b9aca00,%edi
  call   <EstimatePi@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movq   %rax,%xmm0
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D9
type_infered: string
name: read_data
arguments: data_tag: D17
type_infered: string

tool: D9: Estimating pi using 1,000,000,000 iterations of summation 4(1 - 1/3 + 1/5 ... 1/(n+3))...

tool: D17: %.15f


assistant: ```c
int main() {
  printf("Estimating pi using 1,000,000,000 iterations of summation 4(1 - 1/3 + 1/5 ... 1/(n+3))...\n");
  double pi = EstimatePi(1000000000);
  printf("%.15f\n", pi);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D7(%rip),%rdi
  call   <puts@plt>
  mov    $0x3b9aca00,%edi
  call   <EstimatePi@plt>
  lea    D17(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
datatype: string
name: unpack_data
arguments: data_label: D17
datatype: string

tool: D7: Estimating pi using 1,000,000,000 iterations of summation 4(1 - 1/3 + 1/5 ... 1/(n+3))...

tool: D17: %.15f


assistant: ```c
int main() {
  printf("Estimating pi using 1,000,000,000 iterations of summation 4(1 - 1/3 + 1/5 ... 1/(n+3))...\n");
  double pi = EstimatePi(1000000000);
  printf("%.15f\n", pi);
  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
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    $0x3b9aca00,%edi
  call   <EstimatePi@plt>
  lea    D12(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D6
type_infered: string
name: retrieve_data
arguments: label: D12
type_infered: string

tool: D6: Estimating pi using 1,000,000,000 iterations of summation 4(1 - 1/3 + 1/5 ... 1/(n+3))...

tool: D12: %.15f


assistant: ```c
int main() {
  printf("Estimating pi using 1,000,000,000 iterations of summation 4(1 - 1/3 + 1/5 ... 1/(n+3))...\n");
  double pi = EstimatePi(1000000000);
  printf("%.15f\n", pi);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D10(%rip),%rdi
  call   <puts@plt>
  mov    $0x3b9aca00,%edi
  call   <EstimatePi@plt>
  lea    D2(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D10
type: string
name: disassemble_data
arguments: label: D2
type: string

tool: D10: Estimating pi using 1,000,000,000 iterations of summation 4(1 - 1/3 + 1/5 ... 1/(n+3))...

tool: D2: %.15f


assistant: ```c
int main() {
  printf("Estimating pi using 1,000,000,000 iterations of summation 4(1 - 1/3 + 1/5 ... 1/(n+3))...\n");
  double pi = EstimatePi(1000000000);
  printf("%.15f\n", pi);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<height>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L21
L1:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L4
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L4:
  addl   $0x1,-0x8(%rbp)
L21:
  cmpl   $0x3,-0x8(%rbp)
  jle    L1
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int height(int *figure) {
  int max = 0;
  for (int i = 0; i < 4; i++) {
    int width = figure[i] / 10;
    if (width > max)
      max = width;
  }
  return max;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<height>:
  endbr64
  lea    0x10(%rdi),%rsi
  mov    $0x0,%ecx
L1:
  mov    (%rdi),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  sar    $0x1f,%edx
  sub    %edx,%eax
  cmp    %eax,%ecx
  cmovl  %eax,%ecx
  add    $0x4,%rdi
  cmp    %rsi,%rdi
  jne    L1
  mov    %ecx,%eax
  ret
```

assistant: ```c
int height(int *figure) {
  int max = 0;
  for (int i = 0; i < 4; i++) {
    int width = figure[i] / 10;
    if (width > max)
      max = width;
  }
  return max;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<height>:
  endbr64
  lea    0x10(%rdi),%rcx
  xor    %r8d,%r8d
L7:
  movslq (%rdi),%rax
  mov    %rax,%rdx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  cmp    %eax,%r8d
  cmovl  %eax,%r8d
  add    $0x4,%rdi
  cmp    %rcx,%rdi
  jne    L7
  mov    %r8d,%eax
  ret
```

assistant: ```c
int height(int *figure) {
  int max = 0;
  for (int i = 0; i < 4; i++) {
    int width = figure[i] / 10;
    if (width > max)
      max = width;
  }
  return max;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<height>:
  endbr64
  movdqu (%rdi),%xmm1
  movdqa D2(%rip),%xmm5
  pxor   %xmm4,%xmm4
  movdqa %xmm4,%xmm6
  movdqa %xmm4,%xmm2
  movdqa %xmm1,%xmm0
  pcmpgtd %xmm5,%xmm6
  punpckldq %xmm1,%xmm0
  pcmpgtd %xmm0,%xmm2
  movdqa %xmm6,%xmm3
  pmuludq %xmm0,%xmm3
  pmuludq %xmm5,%xmm0
  pmuludq %xmm5,%xmm2
  paddq  %xmm3,%xmm2
  movdqa %xmm4,%xmm3
  psllq  $0x20,%xmm2
  paddq  %xmm2,%xmm0
  movdqa %xmm1,%xmm2
  punpckhdq %xmm1,%xmm2
  psrad  $0x1f,%xmm1
  pcmpgtd %xmm2,%xmm3
  pmuludq %xmm2,%xmm6
  pmuludq %xmm5,%xmm2
  pmuludq %xmm5,%xmm3
  paddq  %xmm6,%xmm3
  psllq  $0x20,%xmm3
  paddq  %xmm3,%xmm2
  shufps $0xdd,%xmm2,%xmm0
  psrad  $0x2,%xmm0
  psubd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  pcmpgtd %xmm4,%xmm1
  pand   %xmm1,%xmm0
  movdqa %xmm0,%xmm2
  psrldq $0x8,%xmm2
  movdqa %xmm2,%xmm1
  pcmpgtd %xmm0,%xmm1
  pand   %xmm1,%xmm2
  pandn  %xmm0,%xmm1
  por    %xmm2,%xmm1
  movdqa %xmm1,%xmm2
  psrldq $0x4,%xmm2
  movdqa %xmm2,%xmm0
  pcmpgtd %xmm1,%xmm0
  pand   %xmm0,%xmm2
  pandn  %xmm1,%xmm0
  por    %xmm2,%xmm0
  movd   %xmm0,%eax
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D2
type_guessed: byte[16]

tool: {"D2": [0x67,0x66,0x66,0x66,0x67,0x66,0x66,0x66,0x67,0x66,0x66,0x66,0x67,0x66,0x66,0x66]}

assistant: ```c
int height(int *figure) {
  int max = 0;
  for (int i = 0; i < 4; i++) {
    int width = figure[i] / 10;
    if (width > max)
      max = width;
  }
  return max;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<arctg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  movd   %eax,%xmm0
  call   <atan@plt>
  movd   %xmm0,%eax
  mov    %eax,-0xc(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0xc(%rbp),%xmm0
  movss  -0x10(%rbp),%xmm1
  pxor   %xmm2,%xmm2
  cvtss2sd %xmm1,%xmm2
  movq   %xmm2,%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D12
type_guessed: string
name: retrieve_data
arguments: data_label: D19
type_guessed: string
name: retrieve_data
arguments: data_label: D11
type_guessed: string

tool: D12: Function arctg(x) - input x: 

tool: D19: %f

tool: D11: acrtg(%f) = %f


assistant: ```c
void arctg(void) {
  float x, result;
  printf("Function arctg(x) - input x: ");
  scanf("%f", &x);
  result = atan(x);
  printf("acrtg(%f) = %f\n", x, result);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<arctg>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D22(%rip),%rdi
  call   <printf@plt>
  lea    0x4(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  call   <atan@plt>
  movaps %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd 0x4(%rsp),%xmm0
  cvtss2sd %xmm1,%xmm1
  lea    D11(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D22
type_infered: string
name: parse_data
arguments: label: D6
type_infered: string
name: parse_data
arguments: label: D11
type_infered: string

tool: D22: Function arctg(x) - input x: 

tool: D6: %f

tool: D11: acrtg(%f) = %f


assistant: ```c
void arctg(void) {
  float x, result;
  printf("Function arctg(x) - input x: ");
  scanf("%f", &x);
  result = atan(x);
  printf("acrtg(%f) = %f\n", x, result);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<arctg>:
  endbr64
  sub    $0x18,%rsp
  lea    D27(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D11(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  call   <atan@plt>
  lea    D0(%rip),%rdi
  mov    $0x2,%eax
  movaps %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd 0x4(%rsp),%xmm0
  cvtss2sd %xmm1,%xmm1
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D27
type_guessed: string
name: unpack_data
arguments: data_tag: D11
type_guessed: string
name: unpack_data
arguments: data_tag: D0
type_guessed: string

tool: D27: Function arctg(x) - input x: 

tool: D11: %f

tool: D0: acrtg(%f) = %f


assistant: ```c
void arctg(void) {
  float x, result;
  printf("Function arctg(x) - input x: ");
  scanf("%f", &x);
  result = atan(x);
  printf("acrtg(%f) = %f\n", x, result);
}
```
user: What is the c source code of the assembly code below:

```asm
<arctg>:
  endbr64
  sub    $0x18,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D17(%rip),%rdi
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  movss  0x4(%rsp),%xmm0
  call   <atan@plt>
  lea    D8(%rip),%rdi
  mov    $0x2,%eax
  movaps %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd 0x4(%rsp),%xmm0
  cvtss2sd %xmm1,%xmm1
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D16
data_type: string
name: parse_data
arguments: label: D17
data_type: string
name: parse_data
arguments: label: D8
data_type: string

tool: D16: Function arctg(x) - input x: 

tool: D17: %f

tool: D8: acrtg(%f) = %f


assistant: ```c
void arctg(void) {
  float x, result;
  printf("Function arctg(x) - input x: ");
  scanf("%f", &x);
  result = atan(x);
  printf("acrtg(%f) = %f\n", x, result);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ckrand>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  add    $0x1c0,%rax
  mov    (%rax),%rax
  mov    -0x8(%rbp),%rdx
  add    $0x1b8,%rdx
  mov    (%rdx),%rdx
  sub    %rdx,%rax
  lea    -0x18(%rax),%rcx
  movabs $0x4a7904a7904a7905,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x4,%rax
  mov    %rcx,%rsi
  sar    $0x3f,%rsi
  sub    %rsi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  test   %rdx,%rdx
  je     L14
  mov    $0x0,%eax
  jmp    L7
L14:
  mov    $0x1,%eax
L7:
  pop    %rbp
  ret
```

assistant: ```c
extern int ckrand(rngstate)
long rngstate[57];
{
  if (((rngstate[57 - 1]) - (rngstate[57 - 2]) - 24) % 55 != 0)
    return (0);
  else
    return (1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ckrand>:
  endbr64
  mov    0x1c0(%rdi),%rcx
  sub    0x1b8(%rdi),%rcx
  sub    $0x18,%rcx
  movabs $0x4a7904a7904a7905,%rdx
  mov    %rcx,%rax
  imul   %rdx
  sar    $0x4,%rdx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rax
  lea    (%rdx,%rax,2),%rax
  lea    (%rax,%rax,4),%rax
  cmp    %rax,%rcx
  sete   %al
  movzbl %al,%eax
  ret
```

assistant: ```c
extern int ckrand(rngstate)
long rngstate[57];
{
  if (((rngstate[57 - 1]) - (rngstate[57 - 2]) - 24) % 55 != 0)
    return (0);
  else
    return (1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ckrand>:
  endbr64
  mov    0x1c0(%rdi),%rax
  sub    0x1b8(%rdi),%rax
  movabs $0x6fb586fb586fb587,%rdx
  sub    $0x18,%rax
  imul   %rdx,%rax
  movabs $0x253c8253c8253c8,%rdx
  add    %rdx,%rax
  movabs $0x4a7904a7904a790,%rdx
  cmp    %rdx,%rax
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
extern int ckrand(rngstate)
long rngstate[57];
{
  if (((rngstate[57 - 1]) - (rngstate[57 - 2]) - 24) % 55 != 0)
    return (0);
  else
    return (1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ckrand>:
  endbr64
  mov    0x1c0(%rdi),%rax
  sub    0x1b8(%rdi),%rax
  movabs $0x6fb586fb586fb587,%rdx
  sub    $0x18,%rax
  imul   %rdx,%rax
  movabs $0x253c8253c8253c8,%rdx
  add    %rdx,%rax
  movabs $0x4a7904a7904a790,%rdx
  cmp    %rdx,%rax
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
extern int ckrand(rngstate)
long rngstate[57];
{
  if (((rngstate[57 - 1]) - (rngstate[57 - 2]) - 24) % 55 != 0)
    return (0);
  else
    return (1);
}
```
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
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <testNew@plt>
  mov    $0x0,%eax
  call   <UnitTest_report@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  testNew();
  UnitTest_report();
}
```
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    $0x0,%eax
  call   <testNew@plt>
  mov    $0x0,%eax
  call   <UnitTest_report@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  testNew();
  UnitTest_report();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <testNew@plt>
  xor    %eax,%eax
  call   <UnitTest_report@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  testNew();
  UnitTest_report();
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <testNew@plt>
  xor    %eax,%eax
  call   <UnitTest_report@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  testNew();
  UnitTest_report();
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<my_system>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <vfork@plt>
  endbr64
  mov    %eax,-0xc(%rbp)
  cmpl   $0xffffffff,-0xc(%rbp)
  jne    L70
  mov    $0x1,%eax
  jmp    L66
L70:
  cmpl   $0x0,-0xc(%rbp)
  jne    L14
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <execv@plt>
  mov    $0x7f,%eax
  jmp    L66
L14:
  lea    -0x10(%rbp),%rcx
  mov    -0xc(%rbp),%eax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <waitpid@plt>
  cmp    $0xffffffff,%eax
  jne    L73
  mov    D19(%rip),%rax
  mov    (%rax),%rdx
  mov    D16(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L14
  mov    $0x1,%eax
  jmp    L66
L73:
  mov    -0x10(%rbp),%eax
L66:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

assistant: ```c
int my_system(char *bin, char **argv) {
  int status, pid;
  pid = vfork();
  if (pid == -1)
    return 1;
  if (pid == 0) {
    execv(bin, argv);
    return 127;
  }
  do {
    if (waitpid(pid, &status, 0) == -1) {
      if (errno != EINTR)
        return 1;
    } else
      return status;
  } while (1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<my_system>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,(%rsp)
  mov    %rsi,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <vfork@plt>
  endbr64
  mov    %eax,%ebx
  mov    $0x1,%eax
  cmp    $0xffffffff,%ebx
  je     L49
  lea    0x14(%rsp),%rbp
  test   %ebx,%ebx
  je     L9
L31:
  mov    $0x0,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <waitpid@plt>
  cmp    $0xffffffff,%eax
  jne    L68
  mov    D14(%rip),%rdx
  mov    D18(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L31
  mov    $0x1,%eax
  jmp    L49
L9:
  mov    0x8(%rsp),%rsi
  mov    (%rsp),%rdi
  call   <execv@plt>
  mov    $0x7f,%eax
  jmp    L49
L68:
  mov    0x14(%rsp),%eax
L49:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L22
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L22:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int my_system(char *bin, char **argv) {
  int status, pid;
  pid = vfork();
  if (pid == -1)
    return 1;
  if (pid == 0) {
    execv(bin, argv);
    return 127;
  }
  do {
    if (waitpid(pid, &status, 0) == -1) {
      if (errno != EINTR)
        return 1;
    } else
      return status;
  } while (1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<my_system>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,(%rsp)
  mov    %rsi,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <vfork@plt>
  endbr64
  cmp    $0xffffffff,%eax
  je     L36
  mov    %eax,%ebx
  lea    0x14(%rsp),%rbp
  test   %eax,%eax
  jne    L34
  jmp    L45
  xchg   %ax,%ax
L75:
  mov    D15(%rip),%rax
  mov    D1(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  jne    L36
L34:
  xor    %edx,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <waitpid@plt>
  cmp    $0xffffffff,%eax
  je     L75
  mov    0x14(%rsp),%eax
L52:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L44
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L36:
  mov    $0x1,%eax
  jmp    L52
  nop
L45:
  mov    0x8(%rsp),%rsi
  mov    (%rsp),%rdi
  call   <execv@plt>
  mov    $0x7f,%eax
  jmp    L52
L44:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int my_system(char *bin, char **argv) {
  int status, pid;
  pid = vfork();
  if (pid == -1)
    return 1;
  if (pid == 0) {
    execv(bin, argv);
    return 127;
  }
  do {
    if (waitpid(pid, &status, 0) == -1) {
      if (errno != EINTR)
        return 1;
    } else
      return status;
  } while (1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<my_system>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,(%rsp)
  mov    %rsi,0x8(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <vfork@plt>
  endbr64
  cmp    $0xffffffff,%eax
  je     L9
  mov    %eax,%ebx
  lea    0x14(%rsp),%rbp
  test   %eax,%eax
  jne    L58
  jmp    L49
  xchg   %ax,%ax
L4:
  mov    D14(%rip),%rax
  mov    D17(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  jne    L9
L58:
  xor    %edx,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <waitpid@plt>
  cmp    $0xffffffff,%eax
  je     L4
  mov    0x14(%rsp),%eax
L41:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L10
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  mov    $0x1,%eax
  jmp    L41
  nop
L49:
  mov    0x8(%rsp),%rsi
  mov    (%rsp),%rdi
  call   <execv@plt>
  mov    $0x7f,%eax
  jmp    L41
L10:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int my_system(char *bin, char **argv) {
  int status, pid;
  pid = vfork();
  if (pid == -1)
    return 1;
  if (pid == 0) {
    execv(bin, argv);
    return 127;
  }
  do {
    if (waitpid(pid, &status, 0) == -1) {
      if (errno != EINTR)
        return 1;
    } else
      return status;
  } while (1);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_unlink>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0xffffffff,%eax
  leave
  ret
```

assistant: ```c
int _unlink(const char *n) {
  assert(0);
  return -1;
}
```