ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<LAHF>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x9f,%edi
  call   <_c_setbyte@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void LAHF() {
  ;
  _c_setbyte(0x9f);
  ;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<LAHF>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x9f,%edi
  call   <_c_setbyte@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void LAHF() {
  ;
  _c_setbyte(0x9f);
  ;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<LAHF>:
  endbr64
  mov    $0x9f,%edi
  jmp    1050 <_c_setbyte@plt>
```

assistant: ```c
void LAHF() {
  ;
  _c_setbyte(0x9f);
  ;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<LAHF>:
  endbr64
  mov    $0x9f,%edi
  jmp    1050 <_c_setbyte@plt>
```

assistant: ```c
void LAHF() {
  ;
  _c_setbyte(0x9f);
  ;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<vector_sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  pxor   %xmm0,%xmm0
  movss  %xmm0,(%rax)
  movl   $0x0,-0x4(%rbp)
  jmp    L18
L4:
  mov    -0x18(%rbp),%rax
  movss  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movss  (%rax),%xmm0
  addss  %xmm1,%xmm0
  mov    -0x18(%rbp),%rax
  movss  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L18:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L4
  mov    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int vector_sum(float *r, float *a, int n) {
  *r = 0;
  for (int i = 0; i < n; i++)
    *r += a[i];
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vector_sum>:
  endbr64
  movl   $0x0,(%rdi)
  test   %edx,%edx
  jle    L16
  mov    %rsi,%rax
  lea    -0x1(%rdx),%edx
  lea    0x4(%rsi,%rdx,4),%rdx
L9:
  movss  (%rdi),%xmm0
  addss  (%rax),%xmm0
  movss  %xmm0,(%rdi)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L9
L16:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int vector_sum(float *r, float *a, int n) {
  *r = 0;
  for (int i = 0; i < n; i++)
    *r += a[i];
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<vector_sum>:
  endbr64
  movl   $0x0,(%rdi)
  test   %edx,%edx
  jle    L8
  lea    -0x1(%rdx),%eax
  pxor   %xmm0,%xmm0
  lea    0x4(%rsi,%rax,4),%rax
  nopw   0x0(%rax,%rax,1)
L19:
  addss  (%rsi),%xmm0
  add    $0x4,%rsi
  movss  %xmm0,(%rdi)
  cmp    %rax,%rsi
  jne    L19
L8:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int vector_sum(float *r, float *a, int n) {
  *r = 0;
  for (int i = 0; i < n; i++)
    *r += a[i];
  return 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vector_sum>:
  endbr64
  movl   $0x0,(%rdi)
  test   %edx,%edx
  jle    L10
  lea    -0x1(%rdx),%eax
  pxor   %xmm0,%xmm0
  lea    0x4(%rsi,%rax,4),%rax
  nopw   0x0(%rax,%rax,1)
L16:
  addss  (%rsi),%xmm0
  add    $0x4,%rsi
  movss  %xmm0,(%rdi)
  cmp    %rax,%rsi
  jne    L16
L10:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int vector_sum(float *r, float *a, int n) {
  *r = 0;
  for (int i = 0; i < n; i++)
    *r += a[i];
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <pioInit@plt>
  mov    $0x1,%esi
  mov    $0x14,%edi
  call   <pinMode@plt>
  mov    $0x1,%esi
  mov    $0x14,%edi
  call   <digitalWrite@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  pioInit();
  pinMode(20, 1);
  digitalWrite(20, 1);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <pioInit@plt>
  mov    $0x1,%esi
  mov    $0x14,%edi
  call   <pinMode@plt>
  mov    $0x1,%esi
  mov    $0x14,%edi
  call   <digitalWrite@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  pioInit();
  pinMode(20, 1);
  digitalWrite(20, 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <pioInit@plt>
  mov    $0x1,%esi
  mov    $0x14,%edi
  call   <pinMode@plt>
  mov    $0x1,%esi
  mov    $0x14,%edi
  call   <digitalWrite@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  pioInit();
  pinMode(20, 1);
  digitalWrite(20, 1);
}
```
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   <pioInit@plt>
  mov    $0x1,%esi
  mov    $0x14,%edi
  call   <pinMode@plt>
  mov    $0x1,%esi
  mov    $0x14,%edi
  call   <digitalWrite@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  pioInit();
  pinMode(20, 1);
  digitalWrite(20, 1);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<tcp_ack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void tcp_ack(struct Seaper *seap, struct NetFrame *frame) {
  seap;
  frame;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<tcp_ack>:
  endbr64
  ret
```

assistant: ```c
void tcp_ack(struct Seaper *seap, struct NetFrame *frame) {
  seap;
  frame;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tcp_ack>:
  endbr64
  ret
```

assistant: ```c
void tcp_ack(struct Seaper *seap, struct NetFrame *frame) {
  seap;
  frame;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tcp_ack>:
  endbr64
  ret
```

assistant: ```c
void tcp_ack(struct Seaper *seap, struct NetFrame *frame) {
  seap;
  frame;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<lcm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  imul   -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <gcd@plt>
  mov    %rax,%rcx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%edx
  div    %rcx
  leave
  ret
```

assistant: ```c
unsigned long long int lcm(unsigned long long int a, unsigned long long int b) {
  unsigned long long int ans = (a * b);
  return (ans / gcd(a, b));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<lcm>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  imul   %rsi,%rbx
  call   <gcd@plt>
  mov    %rax,%rcx
  mov    %rbx,%rax
  mov    $0x0,%edx
  div    %rcx
  pop    %rbx
  ret
```

assistant: ```c
unsigned long long int lcm(unsigned long long int a, unsigned long long int b) {
  unsigned long long int ans = (a * b);
  return (ans / gcd(a, b));
}
```
user: Translate the below assembly code into C programming language:

```asm
<lcm>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  imul   %rsi,%rbx
  call   <gcd@plt>
  xor    %edx,%edx
  mov    %rax,%r8
  mov    %rbx,%rax
  pop    %rbx
  div    %r8
  ret
```

assistant: ```c
unsigned long long int lcm(unsigned long long int a, unsigned long long int b) {
  unsigned long long int ans = (a * b);
  return (ans / gcd(a, b));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<lcm>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  imul   %rsi,%rbx
  call   <gcd@plt>
  xor    %edx,%edx
  mov    %rax,%r8
  mov    %rbx,%rax
  pop    %rbx
  div    %r8
  ret
```

assistant: ```c
unsigned long long int lcm(unsigned long long int a, unsigned long long int b) {
  unsigned long long int ans = (a * b);
  return (ans / gcd(a, b));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<xorSwap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  je     L4
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  xor    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  xor    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  xor    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,(%rax)
L4:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void xorSwap(int *v1, int *v2) {
  if (v1 != v2) {
    *v1 ^= *v2;
    *v2 ^= *v1;
    *v1 ^= *v2;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<xorSwap>:
  endbr64
  cmp    %rsi,%rdi
  je     L5
  mov    (%rdi),%eax
  xor    (%rsi),%eax
  mov    %eax,(%rdi)
  xor    (%rsi),%eax
  mov    %eax,(%rsi)
  xor    %eax,(%rdi)
L5:
  ret
```

assistant: ```c
void xorSwap(int *v1, int *v2) {
  if (v1 != v2) {
    *v1 ^= *v2;
    *v2 ^= *v1;
    *v1 ^= *v2;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<xorSwap>:
  endbr64
  cmp    %rsi,%rdi
  je     L1
  mov    (%rdi),%eax
  xor    (%rsi),%eax
  mov    %eax,(%rdi)
  xor    (%rsi),%eax
  mov    %eax,(%rsi)
  xor    %eax,(%rdi)
L1:
  ret
```

assistant: ```c
void xorSwap(int *v1, int *v2) {
  if (v1 != v2) {
    *v1 ^= *v2;
    *v2 ^= *v1;
    *v1 ^= *v2;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<xorSwap>:
  endbr64
  cmp    %rsi,%rdi
  je     L9
  mov    (%rdi),%eax
  xor    (%rsi),%eax
  mov    %eax,(%rdi)
  xor    (%rsi),%eax
  mov    %eax,(%rsi)
  xor    %eax,(%rdi)
L9:
  ret
```

assistant: ```c
void xorSwap(int *v1, int *v2) {
  if (v1 != v2) {
    *v1 ^= *v2;
    *v2 ^= *v1;
    *v1 ^= *v2;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<databingen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x18(%rbp)
  mov    %ecx,-0x24(%rbp)
  cmpl   $0x7a,-0x14(%rbp)
  je     L54
  cmpl   $0x7a,-0x14(%rbp)
  jg     L28
  cmpl   $0x72,-0x14(%rbp)
  je     L16
  cmpl   $0x72,-0x14(%rbp)
  jg     L28
  cmpl   $0x6f,-0x14(%rbp)
  je     L103
  cmpl   $0x6f,-0x14(%rbp)
  jg     L28
  cmpl   $0x43,-0x14(%rbp)
  je     L62
  cmpl   $0x63,-0x14(%rbp)
  je     L4
L28:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  mov    $0x55,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L119
L4:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  mov    $0xf0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L119
L62:
  movl   $0x0,-0x4(%rbp)
  jmp    L149
L64:
  mov    -0x24(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  cltd
  shr    $0x1d,%edx
  add    %edx,%eax
  and    $0x7,%eax
  sub    %edx,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  and    $0x7f,%ecx
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L149:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L64
  jmp    L119
L103:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  mov    $0xff,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L119
L54:
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L119
L16:
  movl   $0x0,-0x4(%rbp)
  jmp    L82
L77:
  mov    $0x0,%eax
  call   <rand@plt>
  and    $0x3f,%eax
  or     $0x40,%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x4(%rbp)
L82:
  mov    -0x4(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L77
  nop
L119:
  nop
  leave
  ret
```

assistant: ```c
void databingen(mode, buffer, bsize, offset) int mode;
unsigned char *buffer;
int bsize;
int offset;
{
  int ind;
  switch (mode) {
  default:
  case 'a':
    memset(buffer, 0x55, bsize);
    break;
  case 'c':
    memset(buffer, 0xf0, bsize);
    break;
  case 'C':
    for (ind = 0; ind < bsize; ind++) {
      buffer[ind] = ((offset + ind) % 8 & 0177);
    }
    break;
  case 'o':
    memset(buffer, 0xff, bsize);
    break;
  case 'z':
    memset(buffer, 0x0, bsize);
    break;
  case 'r':
    for (ind = 0; ind < bsize; ind++) {
      buffer[ind] = (rand() & 0177) | 0100;
    }
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<databingen>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rax
  cmp    $0x6f,%edi
  je     L31
  jle    L55
  cmp    $0x72,%edi
  je     L29
  cmp    $0x7a,%edi
  jne    L144
  movslq %edx,%rdx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L48
L55:
  cmp    $0x43,%edi
  je     L70
  cmp    $0x63,%edi
  jne    L144
  movslq %edx,%rdx
  mov    $0xf0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L48
L29:
  test   %edx,%edx
  jle    L48
  mov    %rsi,%rbx
  lea    -0x1(%rdx),%edx
  lea    0x1(%rsi,%rdx,1),%rbp
L105:
  mov    $0x0,%eax
  call   <rand@plt>
  and    $0x3f,%eax
  or     $0x40,%eax
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    %rbp,%rbx
  jne    L105
  jmp    L48
L70:
  test   %edx,%edx
  jle    L48
  lea    -0x1(%rdx),%edx
  lea    0x1(%rsi,%rdx,1),%rdi
  mov    %rsi,%rdx
  sub    %esi,%ecx
L96:
  lea    (%rcx,%rdx,1),%eax
  mov    %eax,%esi
  sar    $0x1f,%esi
  shr    $0x1d,%esi
  add    %esi,%eax
  and    $0x7,%eax
  sub    %esi,%eax
  and    $0x7f,%eax
  mov    %al,(%rdx)
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L96
  jmp    L48
L144:
  movslq %edx,%rdx
  mov    $0x55,%esi
  mov    %rax,%rdi
  call   <memset@plt>
L48:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L31:
  movslq %edx,%rdx
  mov    $0xff,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  jmp    L48
```

assistant: ```c
void databingen(mode, buffer, bsize, offset) int mode;
unsigned char *buffer;
int bsize;
int offset;
{
  int ind;
  switch (mode) {
  default:
  case 'a':
    memset(buffer, 0x55, bsize);
    break;
  case 'c':
    memset(buffer, 0xf0, bsize);
    break;
  case 'C':
    for (ind = 0; ind < bsize; ind++) {
      buffer[ind] = ((offset + ind) % 8 & 0177);
    }
    break;
  case 'o':
    memset(buffer, 0xff, bsize);
    break;
  case 'z':
    memset(buffer, 0x0, bsize);
    break;
  case 'r':
    for (ind = 0; ind < bsize; ind++) {
      buffer[ind] = (rand() & 0177) | 0100;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<databingen>:
  endbr64
  mov    %edi,%eax
  mov    %rsi,%rdi
  cmp    $0x6f,%eax
  je     L41
  jle    L43
  cmp    $0x72,%eax
  je     L62
  cmp    $0x7a,%eax
  jne    L64
  movslq %edx,%rdx
  xor    %esi,%esi
  jmp    1060 <memset@plt>
  nopl   0x0(%rax)
L43:
  cmp    $0x43,%eax
  je     L81
  cmp    $0x63,%eax
  jne    L64
  movslq %edx,%rdx
  mov    $0xf0,%esi
  jmp    1060 <memset@plt>
  nop
L41:
  movslq %edx,%rdx
  mov    $0xff,%esi
  jmp    1060 <memset@plt>
  nopl   (%rax)
L62:
  test   %edx,%edx
  jle    L72
  push   %rbp
  lea    -0x1(%rdx),%eax
  push   %rbx
  lea    0x1(%rsi,%rax,1),%rbp
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  nopl   (%rax)
L83:
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <rand@plt>
  and    $0x3f,%eax
  or     $0x40,%eax
  mov    %al,-0x1(%rbx)
  cmp    %rbp,%rbx
  jne    L83
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L64:
  movslq %edx,%rdx
  mov    $0x55,%esi
  jmp    1060 <memset@plt>
  nopl   (%rax)
L81:
  test   %edx,%edx
  jle    L88
  lea    -0x1(%rdx),%eax
  sub    %edi,%ecx
  lea    0x1(%rsi,%rax,1),%rsi
  cs nopw 0x0(%rax,%rax,1)
L6:
  lea    (%rcx,%rdi,1),%eax
  add    $0x1,%rdi
  cltd
  shr    $0x1d,%edx
  add    %edx,%eax
  and    $0x7,%eax
  sub    %edx,%eax
  and    $0x7f,%eax
  mov    %al,-0x1(%rdi)
  cmp    %rsi,%rdi
  jne    L6
  ret
  xchg   %ax,%ax
L72:
  ret
  nopl   0x0(%rax)
L88:
  ret
```

assistant: ```c
void databingen(mode, buffer, bsize, offset) int mode;
unsigned char *buffer;
int bsize;
int offset;
{
  int ind;
  switch (mode) {
  default:
  case 'a':
    memset(buffer, 0x55, bsize);
    break;
  case 'c':
    memset(buffer, 0xf0, bsize);
    break;
  case 'C':
    for (ind = 0; ind < bsize; ind++) {
      buffer[ind] = ((offset + ind) % 8 & 0177);
    }
    break;
  case 'o':
    memset(buffer, 0xff, bsize);
    break;
  case 'z':
    memset(buffer, 0x0, bsize);
    break;
  case 'r':
    for (ind = 0; ind < bsize; ind++) {
      buffer[ind] = (rand() & 0177) | 0100;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<TP>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L12
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L7
  mov    -0x8(%rbp),%rax
  movq   $0x1,(%rax)
  mov    -0x10(%rbp),%rax
  movq   $0x0,(%rax)
  jmp    L7
L12:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L26
  mov    -0x8(%rbp),%rax
  movq   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  movq   $0xffffffffffffffff,(%rax)
  jmp    L7
L26:
  mov    -0x8(%rbp),%rax
  movq   $0xffffffffffffffff,(%rax)
  mov    -0x10(%rbp),%rax
  movq   $0x0,(%rax)
L7:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void TP(long long *v0, long long *v1) {
  if (*v0 == 0) {
    if (*v1 == 0) {
      *v0 = 1;
      *v1 = 0;
    }
  } else {
    if (*v1 == 0) {
      *v0 = 0;
      *v1 = -1;
    } else {
      *v0 = -1;
      *v1 = 0;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<TP>:
  endbr64
  cmpq   $0x0,(%rdi)
  jne    L10
  cmpq   $0x0,(%rsi)
  jne    L25
  movq   $0x1,(%rdi)
  movq   $0x0,(%rsi)
  ret
L10:
  cmpq   $0x0,(%rsi)
  jne    L17
  movq   $0x0,(%rdi)
  movq   $0xffffffffffffffff,(%rsi)
  ret
L17:
  movq   $0xffffffffffffffff,(%rdi)
  movq   $0x0,(%rsi)
L25:
  ret
```

assistant: ```c
void TP(long long *v0, long long *v1) {
  if (*v0 == 0) {
    if (*v1 == 0) {
      *v0 = 1;
      *v1 = 0;
    }
  } else {
    if (*v1 == 0) {
      *v0 = 0;
      *v1 = -1;
    } else {
      *v0 = -1;
      *v1 = 0;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<TP>:
  endbr64
  cmpq   $0x0,(%rdi)
  mov    (%rsi),%rax
  jne    L9
  test   %rax,%rax
  jne    L0
  movq   $0x1,(%rdi)
  movq   $0x0,(%rsi)
  ret
  nopl   0x0(%rax)
L9:
  test   %rax,%rax
  je     L5
  movq   $0xffffffffffffffff,(%rdi)
  movq   $0x0,(%rsi)
L0:
  ret
  nopl   0x0(%rax)
L5:
  movq   $0x0,(%rdi)
  movq   $0xffffffffffffffff,(%rsi)
  ret
```

assistant: ```c
void TP(long long *v0, long long *v1) {
  if (*v0 == 0) {
    if (*v1 == 0) {
      *v0 = 1;
      *v1 = 0;
    }
  } else {
    if (*v1 == 0) {
      *v0 = 0;
      *v1 = -1;
    } else {
      *v0 = -1;
      *v1 = 0;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<TP>:
  endbr64
  cmpq   $0x0,(%rdi)
  mov    (%rsi),%rax
  jne    L20
  test   %rax,%rax
  jne    L14
  movq   $0x1,(%rdi)
  movq   $0x0,(%rsi)
  ret
  nopl   0x0(%rax)
L20:
  test   %rax,%rax
  je     L2
  movq   $0xffffffffffffffff,(%rdi)
  movq   $0x0,(%rsi)
L14:
  ret
  nopl   0x0(%rax)
L2:
  movq   $0x0,(%rdi)
  movq   $0xffffffffffffffff,(%rsi)
  ret
```

assistant: ```c
void TP(long long *v0, long long *v1) {
  if (*v0 == 0) {
    if (*v1 == 0) {
      *v0 = 1;
      *v1 = 0;
    }
  } else {
    if (*v1 == 0) {
      *v0 = 0;
      *v1 = -1;
    } else {
      *v0 = -1;
      *v1 = 0;
    }
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bin2hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L29
L0:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  shr    $0x4,%al
  mov    %al,-0x9(%rbp)
  cmpb   $0x9,-0x9(%rbp)
  ja     L55
  movzbl -0x9(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L49
L55:
  movzbl -0x9(%rbp),%eax
  lea    0x57(%rax),%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L49:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  and    $0xf,%eax
  mov    %al,-0x9(%rbp)
  cmpb   $0x9,-0x9(%rbp)
  ja     L42
  movzbl -0x9(%rbp),%eax
  lea    0x30(%rax),%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L30
L42:
  movzbl -0x9(%rbp),%eax
  lea    0x57(%rax),%ecx
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L30:
  addl   $0x1,-0x8(%rbp)
L29:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L0
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int bin2hex(const unsigned char *in, int len, char *out) {
  int n1, n2;
  unsigned char ch;
  for (n1 = 0, n2 = 0; n1 < len; ++n1) {
    ch = in[n1] >> 4;
    if (ch <= 0x09)
      out[n2++] = ch + '0';
    else
      out[n2++] = ch - 10 + 'a';
    ch = in[n1] & 0x0f;
    if (ch <= 0x09)
      out[n2++] = ch + '0';
    else
      out[n2++] = ch - 10 + 'a';
  }
  out[n2] = '\0';
  return n2;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bin2hex>:
  endbr64
  mov    %rdi,%r9
  mov    %esi,%eax
  mov    %rdx,%r8
  test   %esi,%esi
  jle    L41
  mov    %esi,%r10d
  mov    $0x1,%edi
  mov    $0x0,%ecx
  jmp    L30
L31:
  mov    %edi,%esi
  add    $0x57,%edx
  mov    %dl,(%r8,%rcx,2)
  jmp    L63
L47:
  movslq %esi,%rsi
  add    $0x57,%edx
  mov    %dl,(%r8,%rsi,1)
L15:
  add    $0x1,%rcx
  add    $0x2,%edi
  cmp    %r10,%rcx
  je     L46
L30:
  movzbl (%r9,%rcx,1),%esi
  mov    %esi,%edx
  shr    $0x4,%dl
  cmp    $0x9f,%sil
  ja     L31
  mov    %edi,%esi
  add    $0x30,%edx
  mov    %dl,(%r8,%rcx,2)
L63:
  movzbl (%r9,%rcx,1),%edx
  and    $0xf,%edx
  cmp    $0x9,%dl
  ja     L47
  movslq %esi,%rsi
  add    $0x30,%edx
  mov    %dl,(%r8,%rsi,1)
  jmp    L15
L46:
  add    %eax,%eax
L37:
  movslq %eax,%rdx
  movb   $0x0,(%r8,%rdx,1)
  ret
L41:
  mov    $0x0,%eax
  jmp    L37
```

assistant: ```c
int bin2hex(const unsigned char *in, int len, char *out) {
  int n1, n2;
  unsigned char ch;
  for (n1 = 0, n2 = 0; n1 < len; ++n1) {
    ch = in[n1] >> 4;
    if (ch <= 0x09)
      out[n2++] = ch + '0';
    else
      out[n2++] = ch - 10 + 'a';
    ch = in[n1] & 0x0f;
    if (ch <= 0x09)
      out[n2++] = ch + '0';
    else
      out[n2++] = ch - 10 + 'a';
  }
  out[n2] = '\0';
  return n2;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bin2hex>:
  endbr64
  mov    %rdi,%r9
  test   %esi,%esi
  jle    L15
  movslq %esi,%r11
  xor    %ecx,%ecx
L17:
  movzbl (%r9,%rcx,1),%edi
  shr    $0x4,%dil
  lea    0x30(%rdi),%r10d
  cmp    $0x9,%dil
  lea    0x57(%rdi),%r8d
  mov    %r10d,%edi
  cmova  %r8d,%edi
  mov    %dil,(%rdx,%rcx,2)
  movzbl (%r9,%rcx,1),%edi
  and    $0xf,%edi
  lea    0x30(%rdi),%r10d
  cmp    $0x9,%dil
  lea    0x57(%rdi),%r8d
  mov    %r10d,%edi
  cmova  %r8d,%edi
  mov    %dil,0x1(%rdx,%rcx,2)
  add    $0x1,%rcx
  cmp    %r11,%rcx
  jne    L17
  lea    (%rsi,%rsi,1),%eax
  movslq %eax,%rcx
  add    %rcx,%rdx
  movb   $0x0,(%rdx)
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  movb   $0x0,(%rdx)
  xor    %eax,%eax
  ret
```

assistant: ```c
int bin2hex(const unsigned char *in, int len, char *out) {
  int n1, n2;
  unsigned char ch;
  for (n1 = 0, n2 = 0; n1 < len; ++n1) {
    ch = in[n1] >> 4;
    if (ch <= 0x09)
      out[n2++] = ch + '0';
    else
      out[n2++] = ch - 10 + 'a';
    ch = in[n1] & 0x0f;
    if (ch <= 0x09)
      out[n2++] = ch + '0';
    else
      out[n2++] = ch - 10 + 'a';
  }
  out[n2] = '\0';
  return n2;
}
```