ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<primePart>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rsi,0x8(%rsp)
  mov    %rdx,(%rsp)
  mov    $0x0,%ebx
  cmp    $0x1,%edi
  jle    L17
  mov    %edi,%ebp
  mov    %ecx,%r13d
  movslq %edi,%rax
  lea    (%rsi,%rax,8),%r14
  mov    (%r14),%rax
  test   %rax,%rax
  je     L0
  movslq %ecx,%r13
  mov    (%rax,%r13,4),%ebx
L17:
  mov    %ebx,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L0:
  lea    0x0(,%rcx,4),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  mov    %rax,(%r14)
  mov    %r13d,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    %r13d,%ebx
  test   %r13d,%r13d
  je     L17
  lea    -0x1(%r13),%eax
  mov    %eax,0x1c(%rsp)
  mov    %r13d,%eax
  mov    %rax,0x10(%rsp)
  mov    $0x0,%r12d
  mov    $0x0,%ebx
  jmp    L70
L65:
  cmp    %r12d,%r13d
  je     L17
  movslq %r12d,%rcx
  lea    0x0(,%rcx,4),%rax
  mov    0x1c(%rsp),%edx
  sub    %r12d,%edx
  lea    0x1(%rcx,%rdx,1),%rcx
  shl    $0x2,%rcx
L7:
  mov    (%r14),%rdx
  mov    %ebx,(%rdx,%rax,1)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L7
  jmp    L17
L93:
  mov    (%rsp),%rdx
  mov    0x8(%rsp),%rsi
  call   <primePart>
  add    %eax,%ebx
  mov    (%r14),%rax
  mov    %ebx,(%rax,%r15,1)
L86:
  add    $0x1,%r12
  cmp    0x10(%rsp),%r12
  je     L17
L70:
  mov    %r12d,%ecx
  lea    0x0(,%r12,4),%r15
  mov    (%rsp),%rax
  mov    %ebp,%edi
  sub    (%rax,%r12,4),%edi
  js     L65
  test   %edi,%edi
  jne    L93
  add    $0x1,%ebx
  mov    (%r14),%rax
  mov    %ebx,(%rax,%r15,1)
  jmp    L86
```

assistant: ```c
int primePart(int n, int **ctx, int *prime, int primes_n) {
  if (n <= 1) {
    return 0;
  }
  if (ctx[n] == NULL) {
    ctx[n] = malloc(sizeof(int) * primes_n);
    memset(ctx[n], 0, primes_n);
  } else {
    return ctx[n][primes_n];
  }
  int i, sum;
  for (i = sum = 0; i != primes_n; i++) {
    int r = n - prime[i];
    if (r < 0) {
      break;
    } else if (r == 0) {
      sum += 1;
      ctx[n][i] = sum;
    } else {
      int d = primePart(r, ctx, prime, i);
      sum += d;
      ctx[n][i] = sum;
    }
  }
  for (; i != primes_n; i++) {
    ctx[n][i] = sum;
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<primePart>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  cmp    $0x1,%edi
  jle    L71
  movslq %edi,%rax
  mov    %rdx,%r13
  mov    %edi,%ebx
  mov    %ecx,%r15d
  lea    (%rsi,%rax,8),%r14
  movslq %ecx,%rdx
  mov    (%r14),%rax
  test   %rax,%rax
  je     L41
  mov    (%rax,%rdx,4),%r12d
L38:
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L41:
  lea    0x0(,%rcx,4),%edi
  mov    %rsi,(%rsp)
  call   <malloc@plt>
  xor    %esi,%esi
  mov    %r15d,%edx
  mov    %rax,(%r14)
  mov    %rax,%rdi
  call   <memset@plt>
  test   %r15d,%r15d
  mov    (%rsp),%r9
  jne    L96
  nopl   0x0(%rax)
L71:
  xor    %r12d,%r12d
  jmp    L38
  nopl   (%rax)
L96:
  lea    -0x1(%r15),%eax
  mov    %r15d,%r11d
  xor    %ebp,%ebp
  xor    %r12d,%r12d
  mov    %eax,0x1c(%rsp)
L76:
  mov    %ebp,%ecx
  lea    0x0(,%rbp,4),%r10
  mov    %ebx,%edi
  sub    0x0(%r13,%rbp,4),%edi
  js     L2
  jne    L90
  mov    (%r14),%rax
  add    $0x1,%r12d
  mov    %r12d,(%rax,%r10,1)
L89:
  add    $0x1,%rbp
  cmp    %r11,%rbp
  jne    L76
  jmp    L38
  cs nopw 0x0(%rax,%rax,1)
L90:
  mov    %r9,%rsi
  mov    %r13,%rdx
  mov    %r10,0x10(%rsp)
  mov    %r11,0x8(%rsp)
  mov    %r9,(%rsp)
  call   <primePart>
  mov    0x10(%rsp),%r10
  mov    0x8(%rsp),%r11
  add    %eax,%r12d
  mov    (%r14),%rax
  mov    (%rsp),%r9
  mov    %r12d,(%rax,%r10,1)
  jmp    L89
  nopl   0x0(%rax,%rax,1)
L2:
  cmp    %ebp,%r15d
  je     L38
  mov    (%r14),%rdx
  mov    0x1c(%rsp),%r14d
  movslq %ebp,%rsi
  sub    %ebp,%r14d
  lea    (%rdx,%rsi,4),%rax
  add    %rsi,%r14
  lea    0x4(%rdx,%r14,4),%rdx
  nopl   0x0(%rax,%rax,1)
L74:
  mov    %r12d,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L74
  jmp    L38
```

assistant: ```c
int primePart(int n, int **ctx, int *prime, int primes_n) {
  if (n <= 1) {
    return 0;
  }
  if (ctx[n] == NULL) {
    ctx[n] = malloc(sizeof(int) * primes_n);
    memset(ctx[n], 0, primes_n);
  } else {
    return ctx[n][primes_n];
  }
  int i, sum;
  for (i = sum = 0; i != primes_n; i++) {
    int r = n - prime[i];
    if (r < 0) {
      break;
    } else if (r == 0) {
      sum += 1;
      ctx[n][i] = sum;
    } else {
      int d = primePart(r, ctx, prime, i);
      sum += d;
      ctx[n][i] = sum;
    }
  }
  for (; i != primes_n; i++) {
    ctx[n][i] = sum;
  }
  return sum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<primePart>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  cmp    $0x1,%edi
  jle    L69
  movslq %edi,%rax
  mov    %rdx,%r12
  mov    %edi,%ebp
  mov    %ecx,%r13d
  lea    (%rsi,%rax,8),%r15
  movslq %ecx,%rdx
  mov    (%r15),%rax
  test   %rax,%rax
  je     L0
  mov    (%rax,%rdx,4),%r14d
L49:
  add    $0x28,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L0:
  lea    0x0(,%rcx,4),%edi
  mov    %rsi,0x8(%rsp)
  xor    %ebx,%ebx
  xor    %r14d,%r14d
  call   <malloc@plt>
  xor    %esi,%esi
  mov    %r13d,%edx
  mov    %rax,(%r15)
  mov    %rax,%rdi
  call   <memset@plt>
  test   %r13d,%r13d
  mov    0x8(%rsp),%r8
  mov    %r13d,%r10d
  jne    L150
  nopl   0x0(%rax,%rax,1)
L69:
  xor    %r14d,%r14d
  jmp    L49
  nopl   (%rax)
L150:
  mov    %ebx,%ecx
  mov    %ebx,%edx
  lea    0x0(,%rbx,4),%r9
  mov    %ebp,%edi
  sub    (%r12,%rbx,4),%edi
  js     L128
  jne    L1
  mov    (%r15),%rax
  add    $0x1,%r14d
  mov    %r14d,(%rax,%r9,1)
L3:
  add    $0x1,%rbx
  cmp    %r10,%rbx
  jne    L150
  jmp    L49
  nop
L1:
  mov    %r8,%rsi
  mov    %r12,%rdx
  mov    %r9,0x18(%rsp)
  mov    %r10,0x10(%rsp)
  mov    %r8,0x8(%rsp)
  call   <primePart>
  mov    0x18(%rsp),%r9
  mov    0x10(%rsp),%r10
  add    %eax,%r14d
  mov    (%r15),%rax
  mov    0x8(%rsp),%r8
  mov    %r14d,(%rax,%r9,1)
  jmp    L3
  nopl   (%rax)
L128:
  cmp    %ebx,%r13d
  je     L49
  mov    %r13d,%edi
  mov    (%r15),%r8
  sub    %ebx,%edi
  lea    -0x1(%rdi),%eax
  cmp    $0x2,%eax
  jbe    L167
  movslq %ebx,%rax
  movd   %r14d,%xmm1
  lea    (%r8,%rax,4),%rcx
  mov    %edi,%eax
  pshufd $0x0,%xmm1,%xmm0
  shr    $0x2,%eax
  lea    -0x1(%rax),%esi
  lea    0x10(%rcx),%rax
  shl    $0x4,%rsi
  add    %rax,%rsi
  jmp    L119
  nopw   0x0(%rax,%rax,1)
L34:
  add    $0x10,%rax
L119:
  movups %xmm0,(%rcx)
  mov    %rax,%rcx
  cmp    %rsi,%rax
  jne    L34
  mov    %edi,%eax
  and    $0xfffffffc,%eax
  add    %eax,%edx
  cmp    %eax,%edi
  je     L49
L167:
  movslq %edx,%rax
  mov    %r14d,(%r8,%rax,4)
  lea    0x0(,%rax,4),%rcx
  lea    0x1(%rdx),%eax
  cmp    %r13d,%eax
  je     L49
  add    $0x2,%edx
  mov    %r14d,0x4(%r8,%rcx,1)
  cmp    %edx,%r13d
  je     L49
  mov    %r14d,0x8(%r8,%rcx,1)
  jmp    L49
```

assistant: ```c
int primePart(int n, int **ctx, int *prime, int primes_n) {
  if (n <= 1) {
    return 0;
  }
  if (ctx[n] == NULL) {
    ctx[n] = malloc(sizeof(int) * primes_n);
    memset(ctx[n], 0, primes_n);
  } else {
    return ctx[n][primes_n];
  }
  int i, sum;
  for (i = sum = 0; i != primes_n; i++) {
    int r = n - prime[i];
    if (r < 0) {
      break;
    } else if (r == 0) {
      sum += 1;
      ctx[n][i] = sum;
    } else {
      int d = primePart(r, ctx, prime, i);
      sum += d;
      ctx[n][i] = sum;
    }
  }
  for (; i != primes_n; i++) {
    ctx[n][i] = sum;
  }
  return sum;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<potency_escape_markdown_backticks>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  lea    -0xe(%rdx),%r14
  jmp    L45
L21:
  lea    0x1(%rbx),%rax
  movb   $0x5c,0x0(%rbp,%rbx,1)
  add    $0x2,%rbx
  movzbl 0x0(%r13,%r12,1),%edx
  mov    %dl,0x0(%rbp,%rax,1)
L53:
  add    $0x1,%r12
L45:
  mov    %r13,%rdi
  call   <strlen@plt>
  cmp    %r12,%rax
  jbe    L65
  cmp    %rbx,%r14
  jbe    L65
  movzbl 0x0(%r13,%r12,1),%eax
  cmp    $0x5c,%al
  je     L21
  cmp    $0x60,%al
  je     L54
  mov    %al,0x0(%rbp,%rbx,1)
  lea    0x1(%rbx),%rbx
  jmp    L53
L54:
  movb   $0x20,0x0(%rbp,%rbx,1)
  movb   $0x60,0x1(%rbp,%rbx,1)
  movb   $0x20,0x2(%rbp,%rbx,1)
  movb   $0x60,0x3(%rbp,%rbx,1)
  movb   $0x60,0x4(%rbp,%rbx,1)
  movb   $0x20,0x5(%rbp,%rbx,1)
  movb   $0x60,0x6(%rbp,%rbx,1)
  movb   $0x20,0x7(%rbp,%rbx,1)
  movb   $0x60,0x8(%rbp,%rbx,1)
  movb   $0x60,0x9(%rbp,%rbx,1)
  movb   $0x20,0xa(%rbp,%rbx,1)
  lea    0xc(%rbx),%rax
  movb   $0x60,0xb(%rbp,%rbx,1)
  add    $0xd,%rbx
  movb   $0x20,0x0(%rbp,%rax,1)
  jmp    L53
L65:
  movb   $0x0,0x0(%rbp,%rbx,1)
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *potency_escape_markdown_backticks(const char *unescaped, char *escaped, const size_t escapedLength) {
  size_t i = 0;
  size_t j = 0;
  for (i = 0; (i < strlen(unescaped)) && (j < (escapedLength - 14)); i++) {
    switch (unescaped[i]) {
    case '\\':
      escaped[j++] = '\\';
      escaped[j++] = unescaped[i];
      break;
    case '`':
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      break;
    default:
      escaped[j++] = unescaped[i];
    }
  }
  escaped[j] = 0;
  return &escaped[0];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<potency_escape_markdown_backticks>:
  endbr64
  push   %r14
  lea    -0xe(%rdx),%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  jmp    L14
  nopl   0x0(%rax)
L64:
  mov    %al,(%rcx)
  mov    %rdx,%rbx
L30:
  add    $0x1,%rbp
L14:
  mov    %r13,%rdi
  call   <strlen@plt>
  lea    (%r12,%rbx,1),%rcx
  cmp    %rbp,%rax
  jbe    L55
  cmp    %rbx,%r14
  jbe    L55
  movzbl 0x0(%r13,%rbp,1),%eax
  lea    0x1(%rbx),%rdx
  lea    (%r12,%rdx,1),%rsi
  cmp    $0x5c,%al
  je     L19
  cmp    $0x60,%al
  jne    L64
  lea    0xc(%rbx),%rax
  movb   $0x20,(%rcx)
  movb   $0x60,(%rsi)
  movb   $0x20,0x2(%r12,%rbx,1)
  movb   $0x60,0x3(%r12,%rbx,1)
  movb   $0x60,0x4(%r12,%rbx,1)
  movb   $0x20,0x5(%r12,%rbx,1)
  movb   $0x60,0x6(%r12,%rbx,1)
  movb   $0x20,0x7(%r12,%rbx,1)
  movb   $0x60,0x8(%r12,%rbx,1)
  movb   $0x60,0x9(%r12,%rbx,1)
  movb   $0x20,0xa(%r12,%rbx,1)
  movb   $0x60,0xb(%r12,%rbx,1)
  add    $0xd,%rbx
  movb   $0x20,(%r12,%rax,1)
  jmp    L30
  nopl   0x0(%rax)
L19:
  movb   $0x5c,(%rcx)
  movzbl 0x0(%r13,%rbp,1),%eax
  add    $0x2,%rbx
  mov    %al,(%rsi)
  jmp    L30
  nopl   0x0(%rax)
L55:
  movb   $0x0,(%rcx)
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *potency_escape_markdown_backticks(const char *unescaped, char *escaped, const size_t escapedLength) {
  size_t i = 0;
  size_t j = 0;
  for (i = 0; (i < strlen(unescaped)) && (j < (escapedLength - 14)); i++) {
    switch (unescaped[i]) {
    case '\\':
      escaped[j++] = '\\';
      escaped[j++] = unescaped[i];
      break;
    case '`':
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      break;
    default:
      escaped[j++] = unescaped[i];
    }
  }
  escaped[j] = 0;
  return &escaped[0];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<potency_escape_markdown_backticks>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  lea    -0xe(%rdx),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  mov    D8(%rip),%r14
  jmp    L8
  nopl   0x0(%rax)
L30:
  mov    %al,(%rcx)
  mov    %rdx,%rbx
L27:
  add    $0x1,%rbp
L8:
  mov    %r15,%rdi
  call   <strlen@plt>
  lea    (%r12,%rbx,1),%rcx
  cmp    %rbp,%rax
  jbe    L26
  cmp    %rbx,%r13
  jbe    L26
  movzbl (%r15,%rbp,1),%eax
  lea    0x1(%rbx),%rdx
  lea    (%r12,%rdx,1),%rsi
  cmp    $0x5c,%al
  je     L61
  cmp    $0x60,%al
  jne    L30
  lea    0xc(%rbx),%rax
  mov    %r14,(%rcx)
  movb   $0x60,0x8(%r12,%rbx,1)
  movb   $0x60,0x9(%r12,%rbx,1)
  movb   $0x20,0xa(%r12,%rbx,1)
  movb   $0x60,0xb(%r12,%rbx,1)
  add    $0xd,%rbx
  movb   $0x20,(%r12,%rax,1)
  jmp    L27
  xchg   %ax,%ax
L61:
  movb   $0x5c,(%rcx)
  movzbl (%r15,%rbp,1),%eax
  add    $0x2,%rbx
  mov    %al,(%rsi)
  jmp    L27
L26:
  movb   $0x0,(%rcx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D8
type: byte[8]

tool: {"D8": [0x20,0x60,0x20,0x60,0x60,0x20,0x60,0x20]}

assistant: ```c
char *potency_escape_markdown_backticks(const char *unescaped, char *escaped, const size_t escapedLength) {
  size_t i = 0;
  size_t j = 0;
  for (i = 0; (i < strlen(unescaped)) && (j < (escapedLength - 14)); i++) {
    switch (unescaped[i]) {
    case '\\':
      escaped[j++] = '\\';
      escaped[j++] = unescaped[i];
      break;
    case '`':
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      escaped[j++] = '`';
      escaped[j++] = ' ';
      break;
    default:
      escaped[j++] = unescaped[i];
    }
  }
  escaped[j] = 0;
  return &escaped[0];
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ft_strnstr_return>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L15
L38:
  addq   $0x1,-0x8(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
  mov    $0x1,%eax
  jmp    L5
L15:
  mov    -0x18(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L30
  mov    -0x20(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L30
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L30
  mov    -0x28(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  cmp    %rax,-0x30(%rbp)
  ja     L38
L30:
  mov    $0x0,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int ft_strnstr_return(const char *s1, const char *s2, unsigned long c, unsigned long n) {
  unsigned long count;
  count = 0;
  while (s1[c] && s2[c] && s2[count] == s1[c] && (c++ < n)) {
    count++;
    if (s2[count] == 0x00)
      return (0x01);
  }
  return (0x00);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_strnstr_return>:
  endbr64
  mov    %rdx,%rax
  mov    %rsi,%r8
  sub    %rdx,%r8
L10:
  movzbl (%rdi,%rax,1),%edx
  test   %dl,%dl
  je     L47
  cmpb   $0x0,(%rsi,%rax,1)
  je     L7
  cmp    (%r8,%rax,1),%dl
  jne    L46
  cmp    %rax,%rcx
  jbe    L32
  add    $0x1,%rax
  cmpb   $0x0,(%r8,%rax,1)
  jne    L10
  mov    $0x1,%eax
  ret
L32:
  mov    $0x0,%eax
  ret
L47:
  mov    $0x0,%eax
  ret
L7:
  mov    $0x0,%eax
  ret
L46:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int ft_strnstr_return(const char *s1, const char *s2, unsigned long c, unsigned long n) {
  unsigned long count;
  count = 0;
  while (s1[c] && s2[c] && s2[count] == s1[c] && (c++ < n)) {
    count++;
    if (s2[count] == 0x00)
      return (0x01);
  }
  return (0x00);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<ft_strnstr_return>:
  endbr64
  mov    %rsi,%r8
  sub    %rdx,%r8
  jmp    L14
  nopl   0x0(%rax)
L3:
  cmpb   $0x0,(%rsi,%rdx,1)
  je     L0
  cmp    (%r8,%rdx,1),%al
  jne    L0
  cmp    %rdx,%rcx
  jbe    L0
  add    $0x1,%rdx
  cmpb   $0x0,(%r8,%rdx,1)
  je     L11
L14:
  movzbl (%rdi,%rdx,1),%eax
  test   %al,%al
  jne    L3
L0:
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int ft_strnstr_return(const char *s1, const char *s2, unsigned long c, unsigned long n) {
  unsigned long count;
  count = 0;
  while (s1[c] && s2[c] && s2[count] == s1[c] && (c++ < n)) {
    count++;
    if (s2[count] == 0x00)
      return (0x01);
  }
  return (0x00);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_strnstr_return>:
  endbr64
  mov    %rsi,%r8
  sub    %rdx,%r8
  jmp    L19
  nopl   0x0(%rax)
L40:
  cmpb   $0x0,(%rsi,%rdx,1)
  je     L10
  cmp    (%r8,%rdx,1),%al
  jne    L10
  cmp    %rdx,%rcx
  jbe    L10
  add    $0x1,%rdx
  cmpb   $0x0,(%r8,%rdx,1)
  je     L26
L19:
  movzbl (%rdi,%rdx,1),%eax
  test   %al,%al
  jne    L40
L10:
  xor    %eax,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L26:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int ft_strnstr_return(const char *s1, const char *s2, unsigned long c, unsigned long n) {
  unsigned long count;
  count = 0;
  while (s1[c] && s2[c] && s2[count] == s1[c] && (c++ < n)) {
    count++;
    if (s2[count] == 0x00)
      return (0x01);
  }
  return (0x00);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<print_variable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rsi
  mov    -0x8(%rbp),%rax
  mov    %edx,%ecx
  mov    $0x1,%edx
  mov    %rax,%rdi
  call   <fwrite@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

assistant: ```c
void print_variable(const char *var_ptr, size_t var_len) {
  fwrite(var_ptr, var_len, 1, stdout);
  putchar('\n');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_variable>:
  endbr64
  sub    $0x8,%rsp
  mov    D1(%rip),%rax
  mov    (%rax),%ecx
  mov    $0x1,%edx
  call   <fwrite@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void print_variable(const char *var_ptr, size_t var_len) {
  fwrite(var_ptr, var_len, 1, stdout);
  putchar('\n');
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<print_variable>:
  endbr64
  sub    $0x8,%rsp
  mov    D1(%rip),%rax
  mov    $0x1,%edx
  mov    (%rax),%ecx
  call   <fwrite@plt>
  mov    $0xa,%edi
  add    $0x8,%rsp
  jmp    1060 <putchar@plt>
```

assistant: ```c
void print_variable(const char *var_ptr, size_t var_len) {
  fwrite(var_ptr, var_len, 1, stdout);
  putchar('\n');
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<print_variable>:
  endbr64
  sub    $0x8,%rsp
  mov    D4(%rip),%rax
  mov    $0x1,%edx
  mov    (%rax),%ecx
  call   <fwrite@plt>
  mov    $0xa,%edi
  add    $0x8,%rsp
  jmp    1060 <putchar@plt>
```

assistant: ```c
void print_variable(const char *var_ptr, size_t var_len) {
  fwrite(var_ptr, var_len, 1, stdout);
  putchar('\n');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cpuid_ebx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x20(%rbp),%rdx
  mov    -0x24(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <do_cpuid@plt>
  mov    -0x1c(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
uint32_t cpuid_ebx(uint32_t op) {
  uint32_t regs[4];
  do_cpuid(op, regs);
  return regs[1];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cpuid_ebx>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <do_cpuid@plt>
  mov    0x4(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L9
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t cpuid_ebx(uint32_t op) {
  uint32_t regs[4];
  do_cpuid(op, regs);
  return regs[1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cpuid_ebx>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <do_cpuid@plt>
  mov    0x4(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t cpuid_ebx(uint32_t op) {
  uint32_t regs[4];
  do_cpuid(op, regs);
  return regs[1];
}
```
user: Translate the below assembly code into C programming language:

```asm
<cpuid_ebx>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <do_cpuid@plt>
  mov    0x4(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint32_t cpuid_ebx(uint32_t op) {
  uint32_t regs[4];
  do_cpuid(op, regs);
  return regs[1];
}
```