ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<getMozLibPath>:
  endbr64
  push   %r12
  mov    $0x6a,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L9
  mov    %r12,%rax
  pop    %r12
  ret
  xchg   %ax,%ax
L9:
  lea    D6(%rip),%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    %r12,%rax
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
type: string

tool: D6: /usr/lib/:/usr/lib/x86_64-linux-gnu/:/usr/lib/x86_64/:/usr/lib/x86_64-linux/:     /usr/lib64/:/usr/lib32/

assistant: ```c
char *getMozLibPath() {
  unsigned long pathSize = 256;
  const char *path = "/usr/lib/:/usr/lib/x86_64-linux-gnu/:/usr/lib/x86_64/:/usr/lib/x86_64-linux/:     /usr/lib64/:/usr/lib32/";
  char *mozDir = (char *)malloc(strlen(path) + 1);
  if (mozDir) {
    strcpy(mozDir, path);
  }
  return mozDir;
}
```
user: What is the c source code of the assembly code below:

```asm
<ezThreeFourths>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x24(%rbp)
  movl   $0x80000000,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  or     $0x3,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x24(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  sar    $0x2,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  and    -0x14(%rbp),%eax
  cmp    %eax,-0x18(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int ezThreeFourths(int x) {
  int tmin = 0x80 << 24;
  int mask = tmin | 0x3;
  int mult3 = x + x + x;
  int init_z = mult3 >> 2;
  int add1 = !!((mult3 & mask) ^ tmin);
  int z = init_z + add1;
  return z;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<ezThreeFourths>:
  endbr64
  lea    (%rdi,%rdi,2),%edx
  mov    %edx,%eax
  and    $0x80000003,%eax
  cmp    $0x80000000,%eax
  setne  %al
  movzbl %al,%eax
  sar    $0x2,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
int ezThreeFourths(int x) {
  int tmin = 0x80 << 24;
  int mask = tmin | 0x3;
  int mult3 = x + x + x;
  int init_z = mult3 >> 2;
  int add1 = !!((mult3 & mask) ^ tmin);
  int z = init_z + add1;
  return z;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ezThreeFourths>:
  endbr64
  lea    (%rdi,%rdi,2),%edx
  mov    %edx,%eax
  and    $0x80000003,%eax
  cmp    $0x80000000,%eax
  setne  %al
  sar    $0x2,%edx
  movzbl %al,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int ezThreeFourths(int x) {
  int tmin = 0x80 << 24;
  int mask = tmin | 0x3;
  int mult3 = x + x + x;
  int init_z = mult3 >> 2;
  int add1 = !!((mult3 & mask) ^ tmin);
  int z = init_z + add1;
  return z;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ezThreeFourths>:
  endbr64
  lea    (%rdi,%rdi,2),%edx
  mov    %edx,%eax
  and    $0x80000003,%eax
  cmp    $0x80000000,%eax
  setne  %al
  sar    $0x2,%edx
  movzbl %al,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
int ezThreeFourths(int x) {
  int tmin = 0x80 << 24;
  int mask = tmin | 0x3;
  int mult3 = x + x + x;
  int init_z = mult3 >> 2;
  int add1 = !!((mult3 & mask) ^ tmin);
  int z = init_z + add1;
  return z;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<mp_mul_cmul>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    -0x48(%rbp),%rax
  movsd  (%rax),%xmm1
  mov    -0x40(%rbp),%rax
  movsd  (%rax),%xmm0
  addsd  %xmm1,%xmm0
  mov    -0x48(%rbp),%rax
  movsd  %xmm0,(%rax)
  mov    -0x48(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm1
  mov    -0x40(%rbp),%rax
  add    $0x8,%rax
  movsd  (%rax),%xmm0
  mov    -0x48(%rbp),%rax
  add    $0x8,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  mov    -0x48(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm1
  mov    -0x40(%rbp),%rax
  add    $0x10,%rax
  movsd  (%rax),%xmm0
  mov    -0x48(%rbp),%rax
  add    $0x10,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movl   $0x3,-0x24(%rbp)
  jmp    L19
L10:
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  mulsd  -0x8(%rbp),%xmm1
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  movsd  -0x20(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x8(%rbp),%xmm1
  movsd  -0x18(%rbp),%xmm0
  mulsd  -0x10(%rbp),%xmm0
  mov    -0x24(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x2,-0x24(%rbp)
L19:
  mov    -0x24(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L10
  mov    -0x34(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x34(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x34(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void mp_mul_cmul(int nfft, double din[], double dinout[]) {
  int j;
  double xr, xi, yr, yi;
  dinout[0] += din[0];
  dinout[1] *= din[1];
  dinout[2] *= din[2];
  for (j = 3; j < nfft; j += 2) {
    xr = din[j];
    xi = din[j + 1];
    yr = dinout[j];
    yi = dinout[j + 1];
    dinout[j] = xr * yr - xi * yi;
    dinout[j + 1] = xr * yi + xi * yr;
  }
  dinout[nfft + 1] *= din[nfft + 1];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mp_mul_cmul>:
  endbr64
  mov    %rsi,%rcx
  movsd  (%rdx),%xmm0
  addsd  (%rsi),%xmm0
  movsd  %xmm0,(%rdx)
  movsd  0x8(%rdx),%xmm0
  mulsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x8(%rdx)
  movsd  0x10(%rdx),%xmm0
  mulsd  0x10(%rsi),%xmm0
  movsd  %xmm0,0x10(%rdx)
  cmp    $0x3,%edi
  jle    L2
  mov    $0x3,%eax
L14:
  movsd  (%rcx,%rax,8),%xmm0
  movsd  0x8(%rcx,%rax,8),%xmm1
  movsd  (%rdx,%rax,8),%xmm3
  movsd  0x8(%rdx,%rax,8),%xmm4
  movapd %xmm0,%xmm2
  mulsd  %xmm3,%xmm2
  movapd %xmm1,%xmm5
  mulsd  %xmm4,%xmm5
  subsd  %xmm5,%xmm2
  movsd  %xmm2,(%rdx,%rax,8)
  mulsd  %xmm4,%xmm0
  mulsd  %xmm3,%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,0x8(%rdx,%rax,8)
  add    $0x2,%rax
  cmp    %eax,%edi
  jg     L14
L2:
  movslq %edi,%rdi
  add    $0x1,%rdi
  lea    (%rdx,%rdi,8),%rax
  movsd  (%rax),%xmm0
  mulsd  (%rcx,%rdi,8),%xmm0
  movsd  %xmm0,(%rax)
  ret
```

assistant: ```c
void mp_mul_cmul(int nfft, double din[], double dinout[]) {
  int j;
  double xr, xi, yr, yi;
  dinout[0] += din[0];
  dinout[1] *= din[1];
  dinout[2] *= din[2];
  for (j = 3; j < nfft; j += 2) {
    xr = din[j];
    xi = din[j + 1];
    yr = dinout[j];
    yi = dinout[j + 1];
    dinout[j] = xr * yr - xi * yi;
    dinout[j + 1] = xr * yi + xi * yr;
  }
  dinout[nfft + 1] *= din[nfft + 1];
}
```
user: Translate the below assembly code into C programming language:

```asm
<mp_mul_cmul>:
  endbr64
  movsd  (%rdx),%xmm0
  addsd  (%rsi),%xmm0
  movsd  %xmm0,(%rdx)
  movsd  0x8(%rdx),%xmm0
  mulsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x8(%rdx)
  movsd  0x10(%rdx),%xmm0
  mulsd  0x10(%rsi),%xmm0
  movsd  %xmm0,0x10(%rdx)
  cmp    $0x3,%edi
  jle    L19
  mov    $0x3,%eax
  nopl   0x0(%rax,%rax,1)
L12:
  movsd  (%rsi,%rax,8),%xmm0
  movsd  0x8(%rsi,%rax,8),%xmm1
  movsd  (%rdx,%rax,8),%xmm3
  movsd  0x8(%rdx,%rax,8),%xmm4
  movapd %xmm0,%xmm2
  movapd %xmm1,%xmm5
  mulsd  %xmm3,%xmm2
  mulsd  %xmm4,%xmm5
  mulsd  %xmm4,%xmm0
  mulsd  %xmm3,%xmm1
  subsd  %xmm5,%xmm2
  addsd  %xmm1,%xmm0
  movsd  %xmm2,(%rdx,%rax,8)
  movsd  %xmm0,0x8(%rdx,%rax,8)
  add    $0x2,%rax
  cmp    %eax,%edi
  jg     L12
L19:
  movslq %edi,%rdi
  add    $0x1,%rdi
  lea    (%rdx,%rdi,8),%rax
  movsd  (%rax),%xmm0
  mulsd  (%rsi,%rdi,8),%xmm0
  movsd  %xmm0,(%rax)
  ret
```

assistant: ```c
void mp_mul_cmul(int nfft, double din[], double dinout[]) {
  int j;
  double xr, xi, yr, yi;
  dinout[0] += din[0];
  dinout[1] *= din[1];
  dinout[2] *= din[2];
  for (j = 3; j < nfft; j += 2) {
    xr = din[j];
    xi = din[j + 1];
    yr = dinout[j];
    yi = dinout[j + 1];
    dinout[j] = xr * yr - xi * yi;
    dinout[j + 1] = xr * yi + xi * yr;
  }
  dinout[nfft + 1] *= din[nfft + 1];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<mp_mul_cmul>:
  endbr64
  movsd  (%rdx),%xmm0
  addsd  (%rsi),%xmm0
  movsd  %xmm0,(%rdx)
  movsd  0x8(%rdx),%xmm0
  mulsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x8(%rdx)
  movsd  0x10(%rdx),%xmm0
  mulsd  0x10(%rsi),%xmm0
  movsd  %xmm0,0x10(%rdx)
  cmp    $0x3,%edi
  jle    L1
  lea    -0x4(%rdi),%r8d
  mov    $0x18,%eax
  xor    %ecx,%ecx
  shr    %r8d
  add    $0x1,%r8d
  nopl   (%rax)
L12:
  movupd (%rsi,%rax,1),%xmm2
  movupd (%rdx,%rax,1),%xmm0
  add    $0x1,%ecx
  movupd (%rsi,%rax,1),%xmm1
  unpcklpd %xmm2,%xmm2
  mulpd  %xmm0,%xmm2
  unpckhpd %xmm1,%xmm1
  shufpd $0x1,%xmm0,%xmm0
  mulpd  %xmm1,%xmm0
  movapd %xmm2,%xmm1
  subpd  %xmm0,%xmm1
  addpd  %xmm2,%xmm0
  movsd  %xmm1,%xmm0
  movups %xmm0,(%rdx,%rax,1)
  add    $0x10,%rax
  cmp    %ecx,%r8d
  ja     L12
L1:
  movslq %edi,%rdi
  add    $0x1,%rdi
  lea    (%rdx,%rdi,8),%rax
  movsd  (%rax),%xmm0
  mulsd  (%rsi,%rdi,8),%xmm0
  movsd  %xmm0,(%rax)
  ret
```

assistant: ```c
void mp_mul_cmul(int nfft, double din[], double dinout[]) {
  int j;
  double xr, xi, yr, yi;
  dinout[0] += din[0];
  dinout[1] *= din[1];
  dinout[2] *= din[2];
  for (j = 3; j < nfft; j += 2) {
    xr = din[j];
    xi = din[j + 1];
    yr = dinout[j];
    yi = dinout[j + 1];
    dinout[j] = xr * yr - xi * yi;
    dinout[j + 1] = xr * yi + xi * yr;
  }
  dinout[nfft + 1] *= din[nfft + 1];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<next_protos_parse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  mov    %rax,-0x10(%rbp)
  cmpq   $0xfffe,-0x10(%rbp)
  jbe    L47
  mov    $0x0,%eax
  jmp    L79
L47:
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L74
  mov    $0x0,%eax
  jmp    L79
L74:
  movq   $0x0,-0x20(%rbp)
  jmp    L71
L17:
  mov    -0x20(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  je     L77
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2c,%al
  jne    L98
L77:
  mov    -0x20(%rbp),%rax
  sub    -0x18(%rbp),%rax
  cmp    $0xff,%rax
  jbe    L22
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L79
L22:
  mov    -0x20(%rbp),%rax
  mov    %eax,%ecx
  mov    -0x18(%rbp),%rax
  mov    %eax,%esi
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  sub    %esi,%ecx
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  jmp    L56
L98:
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
L56:
  addq   $0x1,-0x20(%rbp)
L71:
  mov    -0x20(%rbp),%rax
  cmp    -0x10(%rbp),%rax
  jbe    L17
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %dx,(%rax)
  mov    -0x8(%rbp),%rax
L79:
  leave
  ret
```

assistant: ```c
unsigned char *next_protos_parse(unsigned short *outlen, const char *in) {
  size_t i, len, start = 0;
  unsigned char *out;
  len = strlen(in);
  if (len >= 65535)
    return (NULL);
  if ((out = malloc(strlen(in) + 1)) == NULL)
    return (NULL);
  for (i = 0; i <= len; ++i) {
    if (i == len || in[i] == ',') {
      if (i - start > 255) {
        free(out);
        return (NULL);
      }
      out[start] = i - start;
      start = i + 1;
    } else
      out[i + 1] = in[i];
  }
  *outlen = len + 1;
  return (out);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<next_protos_parse>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rsi,%rdi
  call   <strlen@plt>
  movslq %eax,%rbx
  mov    $0x0,%eax
  cmp    $0xfffe,%rbx
  ja     L48
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L48
  mov    $0x0,%esi
  mov    $0x0,%edx
  jmp    L86
L21:
  mov    %rdx,%rcx
  sub    %rsi,%rcx
  cmp    $0xff,%rcx
  ja     L82
  mov    %edx,%ecx
  sub    %esi,%ecx
  mov    %cl,(%rax,%rsi,1)
  lea    0x1(%rdx),%rsi
L52:
  lea    0x1(%rdx),%rcx
  cmp    %rdx,%rbx
  je     L29
  mov    %rcx,%rdx
L86:
  cmp    %rdx,%rbx
  je     L21
  movzbl (%r12,%rdx,1),%ecx
  cmp    $0x2c,%cl
  je     L21
  mov    %cl,0x1(%rax,%rdx,1)
  jmp    L52
L82:
  mov    %rax,%rdi
  call   <free@plt>
  mov    $0x0,%eax
  jmp    L48
L29:
  add    $0x1,%ebx
  mov    %bx,0x0(%rbp)
L48:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned char *next_protos_parse(unsigned short *outlen, const char *in) {
  size_t i, len, start = 0;
  unsigned char *out;
  len = strlen(in);
  if (len >= 65535)
    return (NULL);
  if ((out = malloc(strlen(in) + 1)) == NULL)
    return (NULL);
  for (i = 0; i <= len; ++i) {
    if (i == len || in[i] == ',') {
      if (i - start > 255) {
        free(out);
        return (NULL);
      }
      out[start] = i - start;
      start = i + 1;
    } else
      out[i + 1] = in[i];
  }
  *outlen = len + 1;
  return (out);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<next_protos_parse>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  call   <strlen@plt>
  movslq %eax,%rbx
  cmp    $0xfffe,%rbx
  ja     L76
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L76
  mov    $0x1,%ecx
  xor    %edx,%edx
  xor    %edi,%edi
  jmp    L79
  cs nopw 0x0(%rax,%rax,1)
L1:
  mov    %sil,(%rax,%rcx,1)
  add    $0x1,%rdx
  lea    0x1(%rcx),%rsi
  cmp    %rbx,%rcx
  ja     L6
L52:
  mov    %rsi,%rcx
L79:
  cmp    %rdx,%rbx
  je     L36
  movzbl (%r12,%rdx,1),%esi
  cmp    $0x2c,%sil
  jne    L1
L36:
  mov    %rdx,%rsi
  sub    %rdi,%rsi
  cmp    $0xff,%rsi
  ja     L10
  mov    %edx,%esi
  add    $0x1,%rdx
  sub    %edi,%esi
  mov    %sil,(%rax,%rdi,1)
  mov    %rcx,%rdi
  lea    0x1(%rcx),%rsi
  cmp    %rbx,%rcx
  jbe    L52
L6:
  add    $0x1,%ebx
  mov    %bx,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %rax,%rdi
  call   <free@plt>
L76:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned char *next_protos_parse(unsigned short *outlen, const char *in) {
  size_t i, len, start = 0;
  unsigned char *out;
  len = strlen(in);
  if (len >= 65535)
    return (NULL);
  if ((out = malloc(strlen(in) + 1)) == NULL)
    return (NULL);
  for (i = 0; i <= len; ++i) {
    if (i == len || in[i] == ',') {
      if (i - start > 255) {
        free(out);
        return (NULL);
      }
      out[start] = i - start;
      start = i + 1;
    } else
      out[i + 1] = in[i];
  }
  *outlen = len + 1;
  return (out);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<next_protos_parse>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  push   %rbx
  call   <strlen@plt>
  movslq %eax,%rbx
  cmp    $0xfffe,%rbx
  ja     L6
  mov    %r12,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L6
  mov    $0x1,%ecx
  xor    %edx,%edx
  xor    %edi,%edi
  jmp    L11
  cs nopw 0x0(%rax,%rax,1)
L8:
  mov    %sil,(%rax,%rcx,1)
  add    $0x1,%rdx
  lea    0x1(%rcx),%rsi
  cmp    %rcx,%rbx
  jb     L38
L21:
  mov    %rsi,%rcx
L11:
  cmp    %rdx,%rbx
  je     L5
  movzbl (%r12,%rdx,1),%esi
  cmp    $0x2c,%sil
  jne    L8
L5:
  mov    %rdx,%rsi
  sub    %rdi,%rsi
  cmp    $0xff,%rsi
  ja     L60
  mov    %edx,%esi
  add    $0x1,%rdx
  sub    %edi,%esi
  mov    %sil,(%rax,%rdi,1)
  mov    %rcx,%rdi
  lea    0x1(%rcx),%rsi
  cmp    %rcx,%rbx
  jae    L21
L38:
  add    $0x1,%ebx
  mov    %bx,0x0(%rbp)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L60:
  mov    %rax,%rdi
  call   <free@plt>
L6:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
unsigned char *next_protos_parse(unsigned short *outlen, const char *in) {
  size_t i, len, start = 0;
  unsigned char *out;
  len = strlen(in);
  if (len >= 65535)
    return (NULL);
  if ((out = malloc(strlen(in) + 1)) == NULL)
    return (NULL);
  for (i = 0; i <= len; ++i) {
    if (i == len || in[i] == ',') {
      if (i - start > 255) {
        free(out);
        return (NULL);
      }
      out[start] = i - start;
      start = i + 1;
    } else
      out[i + 1] = in[i];
  }
  *outlen = len + 1;
  return (out);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<sky_timestamp_unshift>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0xfffff,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  sar    $0x14,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  imul   $0xf4240,%eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int64_t sky_timestamp_unshift(int64_t value) {
  int64_t usec = value & 0xFFFFF;
  int64_t sec = value >> 20;
  return (sec * 1000000) + usec;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<sky_timestamp_unshift>:
  endbr64
  mov    %edi,%eax
  sar    $0x14,%eax
  imul   $0xf4240,%eax,%eax
  and    $0xfffff,%edi
  add    %edi,%eax
  ret
```

assistant: ```c
int64_t sky_timestamp_unshift(int64_t value) {
  int64_t usec = value & 0xFFFFF;
  int64_t sec = value >> 20;
  return (sec * 1000000) + usec;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sky_timestamp_unshift>:
  endbr64
  mov    %edi,%eax
  and    $0xfffff,%edi
  sar    $0x14,%eax
  imul   $0xf4240,%eax,%eax
  add    %edi,%eax
  ret
```

assistant: ```c
int64_t sky_timestamp_unshift(int64_t value) {
  int64_t usec = value & 0xFFFFF;
  int64_t sec = value >> 20;
  return (sec * 1000000) + usec;
}
```
user: Translate the below assembly code into C programming language:

```asm
<sky_timestamp_unshift>:
  endbr64
  mov    %edi,%eax
  and    $0xfffff,%edi
  sar    $0x14,%eax
  imul   $0xf4240,%eax,%eax
  add    %edi,%eax
  ret
```

assistant: ```c
int64_t sky_timestamp_unshift(int64_t value) {
  int64_t usec = value & 0xFFFFF;
  int64_t sec = value >> 20;
  return (sec * 1000000) + usec;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<tor_strisprint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L24
L12:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <TOR_ISPRINT@plt>
  test   %eax,%eax
  jne    L9
  mov    $0x0,%eax
  jmp    L17
L9:
  addq   $0x1,-0x8(%rbp)
L24:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L12
  mov    $0x1,%eax
L17:
  leave
  ret
```

assistant: ```c
int tor_strisprint(const char *s) {
  while (*s) {
    if (!TOR_ISPRINT(*s))
      return 0;
    s++;
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<tor_strisprint>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  test   %dil,%dil
  je     L34
L9:
  movsbl %dil,%edi
  call   <TOR_ISPRINT@plt>
  test   %eax,%eax
  je     L23
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L9
  mov    $0x1,%eax
L23:
  pop    %rbx
  ret
L34:
  mov    $0x1,%eax
  jmp    L23
```

assistant: ```c
int tor_strisprint(const char *s) {
  while (*s) {
    if (!TOR_ISPRINT(*s))
      return 0;
    s++;
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<tor_strisprint>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L6
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L14:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L19
L6:
  call   <TOR_ISPRINT@plt>
  test   %eax,%eax
  jne    L14
  pop    %rbx
  ret
L19:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int tor_strisprint(const char *s) {
  while (*s) {
    if (!TOR_ISPRINT(*s))
      return 0;
    s++;
  }
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<tor_strisprint>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  movsbl (%rdi),%edi
  test   %dil,%dil
  jne    L34
  jmp    L17
  nopw   0x0(%rax,%rax,1)
L2:
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  test   %dil,%dil
  je     L17
L34:
  call   <TOR_ISPRINT@plt>
  test   %eax,%eax
  jne    L2
  pop    %rbx
  ret
L17:
  mov    $0x1,%eax
  pop    %rbx
  ret
```

assistant: ```c
int tor_strisprint(const char *s) {
  while (*s) {
    if (!TOR_ISPRINT(*s))
      return 0;
    s++;
  }
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<linux_strlcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L0
  mov    -0x28(%rbp),%rax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
L0:
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

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

tool: D2: error: strlcpy: src too big: %d >= %d


assistant: ```c
size_t linux_strlcpy(char *dst, const char *src, size_t dstsize) {
  size_t n = strlen(src);
  if (n >= dstsize) {
    printf("error: strlcpy: src too big: %d >= %d\n", (int)n, (int)dstsize);
    exit(1);
  }
  strncpy(dst, src, dstsize);
  return n;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<linux_strlcpy>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    %rdx,%rbp
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %rax,%rbx
  cmp    %rbp,%rax
  jae    L19
L13:
  mov    %rbp,%rdx
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <strncpy@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L19:
  mov    %ebp,%edx
  mov    %eax,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L13
```

tool_calls: name: unpack_data
arguments: label: D3
type: string

tool: D3: error: strlcpy: src too big: %d >= %d


assistant: ```c
size_t linux_strlcpy(char *dst, const char *src, size_t dstsize) {
  size_t n = strlen(src);
  if (n >= dstsize) {
    printf("error: strlcpy: src too big: %d >= %d\n", (int)n, (int)dstsize);
    exit(1);
  }
  strncpy(dst, src, dstsize);
  return n;
}
```