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

```asm
<elem_daxmy>:
  endbr64
  mov    %rsi,%rcx
  mov    %edx,%r9d
  sub    $0x1,%edx
  js     L55
  movslq %r9d,%rax
  shl    $0x3,%rax
  lea    -0x10(%rax),%rsi
  lea    -0x8(%rcx,%rax,1),%rax
  lea    (%rdi,%rsi,1),%r10
  cmp    %rax,%r10
  je     L57
  cmp    $0x1,%r9d
  je     L57
  mov    %r9d,%r8d
  movapd %xmm0,%xmm3
  add    %rcx,%rsi
  xor    %eax,%eax
  shr    %r8d
  unpcklpd %xmm3,%xmm3
  neg    %r8
  shl    $0x4,%r8
L31:
  movupd (%r10,%rax,1),%xmm2
  movupd (%rsi,%rax,1),%xmm1
  shufpd $0x1,%xmm2,%xmm2
  mulpd  %xmm3,%xmm2
  shufpd $0x1,%xmm1,%xmm1
  subpd  %xmm2,%xmm1
  shufpd $0x1,%xmm1,%xmm1
  movups %xmm1,(%rsi,%rax,1)
  sub    $0x10,%rax
  cmp    %r8,%rax
  jne    L31
  mov    %r9d,%esi
  mov    %edx,%eax
  and    $0xfffffffe,%esi
  sub    %esi,%eax
  cmp    %esi,%r9d
  je     L55
  cltq
  mulsd  (%rdi,%rax,8),%xmm0
  lea    (%rcx,%rax,8),%rdx
  movsd  (%rdx),%xmm1
  subsd  %xmm0,%xmm1
  movsd  %xmm1,(%rdx)
  ret
  cs nopw 0x0(%rax,%rax,1)
L57:
  movslq %edx,%rax
  nopl   0x0(%rax,%rax,1)
L14:
  movsd  (%rdi,%rax,8),%xmm2
  movsd  (%rcx,%rax,8),%xmm1
  mulsd  %xmm0,%xmm2
  subsd  %xmm2,%xmm1
  movsd  %xmm1,(%rcx,%rax,8)
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L14
L55:
  ret
```

assistant: ```c
void elem_daxmy(double a, double *x, double *y, int n) {
  for (n--; n >= 0; n--)
    y[n] -= a * x[n];
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<adpcm_predict_frame>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  shl    $0x8,%eax
  mov    %eax,%esi
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%ecx
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  or     %ecx,%esi
  mov    %esi,%edx
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%eax
  shl    $0x8,%eax
  mov    %eax,%esi
  mov    -0x20(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%ecx
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  or     %ecx,%esi
  mov    %esi,%edx
  mov    %edx,(%rax)
  movl   $0x1,-0x8(%rbp)
  jmp    L1
L10:
  mov    -0x8(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x2c(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  mov    $0x8,%edx
  mov    $0xf0,%esi
  mov    %eax,%edi
  call   <adpcm_predict_sample@plt>
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,%edx
  mov    %edx,(%rax)
  mov    -0x2c(%rbp),%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  mov    $0xc,%edx
  mov    $0xf,%esi
  mov    %eax,%edi
  call   <adpcm_predict_sample@plt>
  mov    %rax,%rcx
  mov    -0x18(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x8(%rbp)
L1:
  cmpl   $0xf,-0x8(%rbp)
  jbe    L10
  nop
  nop
  leave
  ret
```

assistant: ```c
void adpcm_predict_frame(int16_t *dst, const uint8_t *src, const uint8_t *nibbles, unsigned int rshift) {
  unsigned int i;
  *(dst++) = (src[0] << 8) | src[1];
  *(dst++) = (src[2] << 8) | src[3];
  for (i = 1; i < 16; ++i) {
    uint8_t byte = nibbles[i];
    *(dst++) = adpcm_predict_sample(byte, 0xf0, 8, rshift);
    *(dst++) = adpcm_predict_sample(byte, 0x0f, 12, rshift);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<adpcm_predict_frame>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rdx,%r14
  mov    %ecx,%r13d
  mov    (%rsi),%eax
  shl    $0x8,%eax
  or     0x4(%rsi),%eax
  mov    %eax,(%rdi)
  mov    0x8(%rsi),%eax
  shl    $0x8,%eax
  or     0xc(%rsi),%eax
  mov    %eax,0x4(%rdi)
  mov    $0x1,%ebx
L5:
  mov    (%r14,%rbx,4),%ebp
  mov    %r13d,%ecx
  mov    $0x8,%edx
  mov    $0xf0,%esi
  mov    %ebp,%edi
  call   <adpcm_predict_sample@plt>
  mov    %eax,(%r12,%rbx,8)
  mov    %r13d,%ecx
  mov    $0xc,%edx
  mov    $0xf,%esi
  mov    %ebp,%edi
  call   <adpcm_predict_sample@plt>
  mov    %eax,0x4(%r12,%rbx,8)
  add    $0x1,%rbx
  cmp    $0x10,%rbx
  jne    L5
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void adpcm_predict_frame(int16_t *dst, const uint8_t *src, const uint8_t *nibbles, unsigned int rshift) {
  unsigned int i;
  *(dst++) = (src[0] << 8) | src[1];
  *(dst++) = (src[2] << 8) | src[3];
  for (i = 1; i < 16; ++i) {
    uint8_t byte = nibbles[i];
    *(dst++) = adpcm_predict_sample(byte, 0xf0, 8, rshift);
    *(dst++) = adpcm_predict_sample(byte, 0x0f, 12, rshift);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<adpcm_predict_frame>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    (%rsi),%eax
  mov    $0x1,%ebx
  shl    $0x8,%eax
  or     0x4(%rsi),%eax
  mov    %eax,(%rdi)
  mov    0x8(%rsi),%eax
  shl    $0x8,%eax
  or     0xc(%rsi),%eax
  mov    %eax,0x4(%rdi)
L3:
  mov    (%r14,%rbx,4),%ebp
  mov    %r13d,%ecx
  mov    $0x8,%edx
  mov    $0xf0,%esi
  mov    %ebp,%edi
  call   <adpcm_predict_sample@plt>
  mov    %r13d,%ecx
  mov    $0xc,%edx
  mov    %ebp,%edi
  mov    %eax,(%r12,%rbx,8)
  mov    $0xf,%esi
  call   <adpcm_predict_sample@plt>
  mov    %eax,0x4(%r12,%rbx,8)
  add    $0x1,%rbx
  cmp    $0x10,%rbx
  jne    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void adpcm_predict_frame(int16_t *dst, const uint8_t *src, const uint8_t *nibbles, unsigned int rshift) {
  unsigned int i;
  *(dst++) = (src[0] << 8) | src[1];
  *(dst++) = (src[2] << 8) | src[3];
  for (i = 1; i < 16; ++i) {
    uint8_t byte = nibbles[i];
    *(dst++) = adpcm_predict_sample(byte, 0xf0, 8, rshift);
    *(dst++) = adpcm_predict_sample(byte, 0x0f, 12, rshift);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<adpcm_predict_frame>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  mov    (%rsi),%eax
  mov    $0x1,%ebx
  shl    $0x8,%eax
  or     0x4(%rsi),%eax
  mov    %eax,(%rdi)
  mov    0x8(%rsi),%eax
  shl    $0x8,%eax
  or     0xc(%rsi),%eax
  mov    %eax,0x4(%rdi)
L1:
  mov    (%r14,%rbx,4),%ebp
  mov    %r13d,%ecx
  mov    $0x8,%edx
  mov    $0xf0,%esi
  mov    %ebp,%edi
  call   <adpcm_predict_sample@plt>
  mov    %r13d,%ecx
  mov    $0xc,%edx
  mov    %ebp,%edi
  mov    %eax,(%r12,%rbx,8)
  mov    $0xf,%esi
  call   <adpcm_predict_sample@plt>
  mov    %eax,0x4(%r12,%rbx,8)
  add    $0x1,%rbx
  cmp    $0x10,%rbx
  jne    L1
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void adpcm_predict_frame(int16_t *dst, const uint8_t *src, const uint8_t *nibbles, unsigned int rshift) {
  unsigned int i;
  *(dst++) = (src[0] << 8) | src[1];
  *(dst++) = (src[2] << 8) | src[3];
  for (i = 1; i < 16; ++i) {
    uint8_t byte = nibbles[i];
    *(dst++) = adpcm_predict_sample(byte, 0xf0, 8, rshift);
    *(dst++) = adpcm_predict_sample(byte, 0x0f, 12, rshift);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<printn>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jns    L15
  mov    $0x2d,%edi
  call   <putchar@plt>
  negl   -0x14(%rbp)
L15:
  mov    -0x14(%rbp),%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)
  cmpl   $0x0,-0x4(%rbp)
  je     L19
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <printn>
L19:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  lea    0x30(%rdx),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

assistant: ```c
void printn(int n) {
  int a;
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  a = n / 10;
  if (a != 0) {
    printn(a);
  }
  putchar(n % 10 + '0');
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<printn>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   %edi,%edi
  js     L32
L37:
  movslq %ebx,%rdi
  imul   $0x66666667,%rdi,%rdi
  sar    $0x22,%rdi
  mov    %ebx,%eax
  sar    $0x1f,%eax
  sub    %eax,%edi
  jne    L33
L9:
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebx
  lea    0x30(%rbx),%edi
  call   <putchar@plt>
  pop    %rbx
  ret
L32:
  mov    $0x2d,%edi
  call   <putchar@plt>
  neg    %ebx
  jmp    L37
L33:
  call   <printn>
  jmp    L9
```

assistant: ```c
void printn(int n) {
  int a;
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  a = n / 10;
  if (a != 0) {
    printn(a);
  }
  putchar(n % 10 + '0');
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<printn>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %edi,%edi
  js     L39
L54:
  mov    %ebx,%r12d
  mov    $0xcccccccd,%edi
  mov    %r12,%rbp
  imul   %rdi,%rbp
  shr    $0x23,%rbp
  test   %ebp,%ebp
  jne    L67
L69:
  mov    $0xcccccccd,%eax
  add    $0x8,%rsp
  imul   %r12,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebx
  lea    0x30(%rbx),%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <putchar@plt>
  nopl   (%rax)
L67:
  mov    %ebp,%r14d
  mov    %r14,%r13
  imul   %rdi,%r13
  shr    $0x23,%r13
  test   %r13d,%r13d
  jne    L50
L10:
  mov    $0xcccccccd,%eax
  imul   %r14,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebp
  lea    0x30(%rbp),%edi
  call   <putchar@plt>
  jmp    L69
  nopl   0x0(%rax)
L39:
  mov    $0x2d,%edi
  neg    %ebx
  call   <putchar@plt>
  jmp    L54
  nopl   0x0(%rax)
L50:
  mov    %r13d,%r15d
  imul   %r15,%rdi
  shr    $0x23,%rdi
  test   %edi,%edi
  jne    L75
L1:
  mov    $0xcccccccd,%eax
  imul   %rax,%r15
  shr    $0x23,%r15
  lea    (%r15,%r15,4),%eax
  add    %eax,%eax
  sub    %eax,%r13d
  lea    0x30(%r13),%edi
  call   <putchar@plt>
  jmp    L10
L75:
  call   <printn>
  jmp    L1
```

assistant: ```c
void printn(int n) {
  int a;
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  a = n / 10;
  if (a != 0) {
    printn(a);
  }
  putchar(n % 10 + '0');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<printn>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  test   %edi,%edi
  js     L41
L35:
  mov    %ebx,%r12d
  mov    $0xcccccccd,%edi
  mov    %r12,%rbp
  imul   %rdi,%rbp
  shr    $0x23,%rbp
  test   %ebp,%ebp
  jne    L79
L40:
  mov    $0xcccccccd,%eax
  add    $0x8,%rsp
  imul   %r12,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebx
  lea    0x30(%rbx),%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <putchar@plt>
  nopl   (%rax)
L79:
  mov    %ebp,%r14d
  mov    %r14,%r13
  imul   %rdi,%r13
  shr    $0x23,%r13
  test   %r13d,%r13d
  jne    L42
L55:
  mov    $0xcccccccd,%eax
  imul   %r14,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    %eax,%ebp
  lea    0x30(%rbp),%edi
  call   <putchar@plt>
  jmp    L40
  nopl   0x0(%rax)
L41:
  mov    $0x2d,%edi
  neg    %ebx
  call   <putchar@plt>
  jmp    L35
  nopl   0x0(%rax)
L42:
  mov    %r13d,%r15d
  imul   %r15,%rdi
  shr    $0x23,%rdi
  test   %edi,%edi
  jne    L28
L17:
  mov    $0xcccccccd,%eax
  imul   %rax,%r15
  shr    $0x23,%r15
  lea    (%r15,%r15,4),%eax
  add    %eax,%eax
  sub    %eax,%r13d
  lea    0x30(%r13),%edi
  call   <putchar@plt>
  jmp    L55
L28:
  call   <printn>
  jmp    L17
```

assistant: ```c
void printn(int n) {
  int a;
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  a = n / 10;
  if (a != 0) {
    printn(a);
  }
  putchar(n % 10 + '0');
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strlen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L17
L3:
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L17:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L3
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t len = 0;
  for (const char *s = str; *s; s++)
    len++;
  return len;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strlen>:
  endbr64
  cmpb   $0x0,(%rdi)
  je     L10
  mov    $0x0,%eax
L19:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L19
  ret
L10:
  mov    $0x0,%eax
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t len = 0;
  for (const char *s = str; *s; s++)
    len++;
  return len;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strlen>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  je     L7
  nopl   0x0(%rax,%rax,1)
L2:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L2
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t len = 0;
  for (const char *s = str; *s; s++)
    len++;
  return len;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<strlen>:
  endbr64
  xor    %eax,%eax
  cmpb   $0x0,(%rdi)
  je     L7
  nopl   0x0(%rax,%rax,1)
L8:
  add    $0x1,%rax
  cmpb   $0x0,(%rdi,%rax,1)
  jne    L8
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  ret
```

assistant: ```c
size_t strlen(const char *str) {
  size_t len = 0;
  for (const char *s = str; *s; s++)
    len++;
  return len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<jpc_floorlog2>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  setg   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L18:
  sarl   -0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L2:
  cmpl   $0x1,-0x14(%rbp)
  jg     L18
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int jpc_floorlog2(int x) {
  int y;
  assert(x > 0);
  y = 0;
  while (x > 1) {
    x >>= 1;
    ++y;
  }
  return y;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<jpc_floorlog2>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   %edi,%edi
  setg   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  cmp    $0x1,%ebx
  jle    L29
  mov    $0x0,%eax
L22:
  sar    %ebx
  add    $0x1,%eax
  cmp    $0x1,%ebx
  jg     L22
L13:
  pop    %rbx
  ret
L29:
  mov    $0x0,%eax
  jmp    L13
```

assistant: ```c
int jpc_floorlog2(int x) {
  int y;
  assert(x > 0);
  y = 0;
  while (x > 1) {
    x >>= 1;
    ++y;
  }
  return y;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<jpc_floorlog2>:
  endbr64
  test   %edi,%edi
  push   %rbx
  mov    %edi,%ebx
  setg   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  xor    %eax,%eax
  cmp    $0x1,%ebx
  jle    L12
  nopl   (%rax)
L19:
  sar    %ebx
  add    $0x1,%eax
  cmp    $0x1,%ebx
  jne    L19
L12:
  pop    %rbx
  ret
```

assistant: ```c
int jpc_floorlog2(int x) {
  int y;
  assert(x > 0);
  y = 0;
  while (x > 1) {
    x >>= 1;
    ++y;
  }
  return y;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<jpc_floorlog2>:
  endbr64
  test   %edi,%edi
  push   %rbx
  mov    %edi,%ebx
  setg   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  xor    %eax,%eax
  cmp    $0x1,%ebx
  jle    L10
  nopl   (%rax)
L11:
  sar    %ebx
  add    $0x1,%eax
  cmp    $0x1,%ebx
  jne    L11
L10:
  pop    %rbx
  ret
```

assistant: ```c
int jpc_floorlog2(int x) {
  int y;
  assert(x > 0);
  y = 0;
  while (x > 1) {
    x >>= 1;
    ++y;
  }
  return y;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<route_handler>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    $0x7d3,%edi
  call   <create_server@plt>
  mov    %eax,-0x8(%rbp)
  cmpl   $0x0,-0x8(%rbp)
  jns    L1
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x0,%eax
  jmp    L44
L1:
  mov    -0x8(%rbp),%eax
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %eax,%edi
  call   <accept@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L41
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x0,%eax
  jmp    L44
L41:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <handle_route_request@plt>
  jmp    L1
L44:
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D1
datatype: string
name: read_data
arguments: tag: D4
datatype: string

tool: D1: could not create server : 

tool: D4: accept error : 

assistant: ```c
void *route_handler(void *t) {
  int listenfd, connfd;
  if ((listenfd = create_server(2003)) < 0) {
    perror("could not create server : ");
    return NULL;
  }
  for (;;) {
    if ((connfd = accept(listenfd, (struct sockaddr *)NULL, NULL)) < 0) {
      perror("accept error : ");
      return NULL;
    }
    handle_route_request(connfd);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<route_handler>:
  endbr64
  push   %rbx
  mov    $0x7d3,%edi
  call   <create_server@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jns    L17
  lea    D6(%rip),%rdi
  call   <perror@plt>
  jmp    L23
L12:
  call   <handle_route_request@plt>
L17:
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %ebx,%edi
  call   <accept@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jns    L12
  lea    D17(%rip),%rdi
  call   <perror@plt>
L23:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
type_guessed: string
name: retrieve_data
arguments: data_tag: D17
type_guessed: string

tool: D6: could not create server : 

tool: D17: accept error : 

assistant: ```c
void *route_handler(void *t) {
  int listenfd, connfd;
  if ((listenfd = create_server(2003)) < 0) {
    perror("could not create server : ");
    return NULL;
  }
  for (;;) {
    if ((connfd = accept(listenfd, (struct sockaddr *)NULL, NULL)) < 0) {
      perror("accept error : ");
      return NULL;
    }
    handle_route_request(connfd);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<route_handler>:
  endbr64
  push   %rbx
  mov    $0x7d3,%edi
  call   <create_server@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jns    L16
  jmp    L15
  nopw   0x0(%rax,%rax,1)
L12:
  call   <handle_route_request@plt>
L16:
  mov    %ebx,%edi
  xor    %edx,%edx
  xor    %esi,%esi
  call   <accept@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jns    L12
  lea    D13(%rip),%rdi
  call   <perror@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L15:
  lea    D2(%rip),%rdi
  call   <perror@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D13
type_guessed: string
name: retrieve_data
arguments: data_label: D2
type_guessed: string

tool: D13: accept error : 

tool: D2: could not create server : 

assistant: ```c
void *route_handler(void *t) {
  int listenfd, connfd;
  if ((listenfd = create_server(2003)) < 0) {
    perror("could not create server : ");
    return NULL;
  }
  for (;;) {
    if ((connfd = accept(listenfd, (struct sockaddr *)NULL, NULL)) < 0) {
      perror("accept error : ");
      return NULL;
    }
    handle_route_request(connfd);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<route_handler>:
  endbr64
  push   %rbx
  mov    $0x7d3,%edi
  call   <create_server@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jns    L29
  jmp    L20
  nopw   0x0(%rax,%rax,1)
L16:
  call   <handle_route_request@plt>
L29:
  mov    %ebx,%edi
  xor    %edx,%edx
  xor    %esi,%esi
  call   <accept@plt>
  mov    %eax,%edi
  test   %eax,%eax
  jns    L16
  lea    D17(%rip),%rdi
  call   <perror@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
  cs nopw 0x0(%rax,%rax,1)
L20:
  lea    D10(%rip),%rdi
  call   <perror@plt>
  xor    %eax,%eax
  pop    %rbx
  ret
```

tool_calls: name: read_data
arguments: data_tag: D17
data_type: string
name: read_data
arguments: data_tag: D10
data_type: string

tool: D17: accept error : 

tool: D10: could not create server : 

assistant: ```c
void *route_handler(void *t) {
  int listenfd, connfd;
  if ((listenfd = create_server(2003)) < 0) {
    perror("could not create server : ");
    return NULL;
  }
  for (;;) {
    if ((connfd = accept(listenfd, (struct sockaddr *)NULL, NULL)) < 0) {
      perror("accept error : ");
      return NULL;
    }
    handle_route_request(connfd);
  }
}
```
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
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D4
type_infered: string

tool: D4: PASS


assistant: ```c
int main() {
#pragma xmp tasks
  {
#pragma xmp task on p(1 : 4)
      {
#pragma xmp barrier on p
      }
#pragma xmp task on p(5 : 8)
      {
#pragma xmp barrier on p
      }}
#pragma xmp task on p(1)
  {
    printf("PASS\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
data_type: string

tool: D0: PASS


assistant: ```c
int main() {
#pragma xmp tasks
  {
#pragma xmp task on p(1 : 4)
      {
#pragma xmp barrier on p
      }
#pragma xmp task on p(5 : 8)
      {
#pragma xmp barrier on p
      }}
#pragma xmp task on p(1)
  {
    printf("PASS\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type_guessed: string

tool: D6: PASS


assistant: ```c
int main() {
#pragma xmp tasks
  {
#pragma xmp task on p(1 : 4)
      {
#pragma xmp barrier on p
      }
#pragma xmp task on p(5 : 8)
      {
#pragma xmp barrier on p
      }}
#pragma xmp task on p(1)
  {
    printf("PASS\n");
  }
  return 0;
}
```
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
  sub    $0x8,%rsp
  lea    D4(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D4
type: string

tool: D4: PASS


assistant: ```c
int main() {
#pragma xmp tasks
  {
#pragma xmp task on p(1 : 4)
      {
#pragma xmp barrier on p
      }
#pragma xmp task on p(5 : 8)
      {
#pragma xmp barrier on p
      }}
#pragma xmp task on p(1)
  {
    printf("PASS\n");
  }
  return 0;
}
```