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

```asm
<memcmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L14
L16:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L27
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  jmp    L18
L27:
  subq   $0x1,-0x28(%rbp)
  addq   $0x1,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L14:
  cmpq   $0x0,-0x28(%rbp)
  jne    L16
  mov    $0x0,%eax
L18:
  pop    %rbp
  ret
```

assistant: ```c
int memcmp(void const *p1, void const *p2, long n) {
  char const *pc1 = p1, *pc2 = p2;
  for (; n; n--, pc1++, pc2++)
    if (*pc1 != *pc2)
      return *pc1 - *pc2;
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<memcmp>:
  endbr64
  test   %rdx,%rdx
  je     L22
  mov    $0x0,%eax
L13:
  movzbl (%rdi,%rax,1),%ecx
  movzbl (%rsi,%rax,1),%r8d
  cmp    %r8b,%cl
  jne    L23
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L13
  mov    $0x0,%eax
  ret
L23:
  movsbl %cl,%eax
  movsbl %r8b,%r8d
  sub    %r8d,%eax
  ret
L22:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int memcmp(void const *p1, void const *p2, long n) {
  char const *pc1 = p1, *pc2 = p2;
  for (; n; n--, pc1++, pc2++)
    if (*pc1 != *pc2)
      return *pc1 - *pc2;
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<memcmp>:
  endbr64
  test   %rdx,%rdx
  je     L4
  xor    %eax,%eax
  jmp    L33
  nopl   (%rax)
L24:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L4
L33:
  movzbl (%rdi,%rax,1),%ecx
  movsbl (%rsi,%rax,1),%r8d
  cmp    %r8b,%cl
  je     L24
  movsbl %cl,%eax
  sub    %r8d,%eax
  ret
  xchg   %ax,%ax
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
int memcmp(void const *p1, void const *p2, long n) {
  char const *pc1 = p1, *pc2 = p2;
  for (; n; n--, pc1++, pc2++)
    if (*pc1 != *pc2)
      return *pc1 - *pc2;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<memcmp>:
  endbr64
  test   %rdx,%rdx
  je     L29
  xor    %eax,%eax
  jmp    L4
  nopl   (%rax)
L8:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L29
L4:
  movzbl (%rdi,%rax,1),%ecx
  movsbl (%rsi,%rax,1),%r8d
  cmp    %r8b,%cl
  je     L8
  movsbl %cl,%eax
  sub    %r8d,%eax
  ret
  xchg   %ax,%ax
L29:
  xor    %eax,%eax
  ret
```

assistant: ```c
int memcmp(void const *p1, void const *p2, long n) {
  char const *pc1 = p1, *pc2 = p2;
  for (; n; n--, pc1++, pc2++)
    if (*pc1 != *pc2)
      return *pc1 - *pc2;
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<CCutil_double_perm_quicksort>:
  endbr64
  cmp    $0x1,%edx
  jle    L16
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %edx,%r12d
  mov    (%rdi),%ecx
  lea    -0x1(%rdx),%edx
  mov    %edx,%eax
  shr    $0x1f,%eax
  add    %edx,%eax
  sar    %eax
  cltq
  lea    (%rdi,%rax,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%rdi)
  mov    %ecx,(%rax)
  movslq (%rdi),%rax
  movsd  (%rsi,%rax,8),%xmm1
  mov    %r12d,%edx
  mov    $0x1,%r8d
  jmp    L53
L59:
  movslq %edx,%rax
  lea    -0x4(%rbp,%rax,4),%rax
L32:
  sub    $0x1,%edx
  mov    %rax,%rdi
  mov    (%rax),%ecx
  sub    $0x4,%rax
  movslq %ecx,%rsi
  movsd  (%rbx,%rsi,8),%xmm0
  comisd %xmm1,%xmm0
  ja     L32
  cmp    %r9d,%edx
  jl     L11
  mov    0x0(%rbp,%r8,4),%eax
  mov    %ecx,0x0(%rbp,%r8,4)
  mov    %eax,(%rdi)
L29:
  add    $0x1,%r8
L53:
  mov    %r8d,%r13d
  mov    %r8d,%r9d
  cmp    %r8d,%r12d
  jle    L59
  movslq 0x0(%rbp,%r8,4),%rax
  comisd (%rbx,%rax,8),%xmm1
  ja     L29
  jmp    L59
L11:
  mov    0x0(%rbp),%eax
  mov    %ecx,0x0(%rbp)
  mov    %eax,(%rdi)
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <CCutil_double_perm_quicksort>
  mov    %r12d,%edx
  sub    %r13d,%edx
  movslq %r13d,%r13
  lea    0x0(%rbp,%r13,4),%rdi
  mov    %rbx,%rsi
  mov    $0x0,%eax
  call   <CCutil_double_perm_quicksort>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  ret
```

assistant: ```c
void CCutil_double_perm_quicksort(perm, len, n) int *perm;
double *len;
int n;
{
  int i, j, temp;
  double t;
  if (n <= 1)
    return;
  ((temp) = (perm[0]), (perm[0]) = (perm[(n - 1) / 2]), (perm[(n - 1) / 2]) = (temp));
  i = 0;
  j = n;
  t = len[perm[0]];
  while (1) {
    do
      i++;
    while (i < n && len[perm[i]] < t);
    do
      j--;
    while (len[perm[j]] > t);
    if (j < i)
      break;
    ((temp) = (perm[i]), (perm[i]) = (perm[j]), (perm[j]) = (temp));
  }
  ((temp) = (perm[0]), (perm[0]) = (perm[j]), (perm[j]) = (temp));
  CCutil_double_perm_quicksort(perm, len, j);
  CCutil_double_perm_quicksort(perm + i, len, n - i);
}
```
user: What is the c source code of the assembly code below:

```asm
<CCutil_double_perm_quicksort>:
  endbr64
  cmp    $0x1,%edx
  jle    L61
  push   %r13
  mov    %rdi,%r8
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
L76:
  lea    -0x1(%r13),%eax
  mov    (%r8),%edx
  lea    0x4(%r8),%r12
  xor    %ebp,%ebp
  sar    %eax
  cltq
  lea    (%r8,%rax,4),%rax
  mov    (%rax),%ecx
  mov    %ecx,(%r8)
  mov    %edx,(%rax)
  movslq (%r8),%rax
  mov    %r13d,%edx
  movsd  (%rbx,%rax,8),%xmm1
  nopw   0x0(%rax,%rax,1)
L18:
  add    $0x1,%ebp
  cmp    %ebp,%r13d
  jle    L70
  movslq (%r12),%rax
  comisd (%rbx,%rax,8),%xmm1
  ja     L23
L70:
  movslq %edx,%rax
  lea    -0x4(%r8,%rax,4),%rax
  nopl   0x0(%rax,%rax,1)
L42:
  movslq (%rax),%rsi
  sub    $0x1,%edx
  mov    %rax,%rdi
  sub    $0x4,%rax
  movsd  (%rbx,%rsi,8),%xmm0
  comisd %xmm1,%xmm0
  ja     L42
  cmp    %edx,%ebp
  jg     L66
  mov    (%r12),%eax
  mov    %esi,(%r12)
  mov    %eax,(%rdi)
L23:
  add    $0x4,%r12
  jmp    L18
  nopl   0x0(%rax)
L66:
  mov    (%r8),%eax
  mov    %esi,(%r8)
  sub    %ebp,%r13d
  mov    %rbx,%rsi
  mov    %eax,(%rdi)
  mov    %r8,%rdi
  xor    %eax,%eax
  call   <CCutil_double_perm_quicksort>
  cmp    $0x1,%r13d
  jle    L53
  mov    %r12,%r8
  jmp    L76
L53:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L61:
  ret
```

assistant: ```c
void CCutil_double_perm_quicksort(perm, len, n) int *perm;
double *len;
int n;
{
  int i, j, temp;
  double t;
  if (n <= 1)
    return;
  ((temp) = (perm[0]), (perm[0]) = (perm[(n - 1) / 2]), (perm[(n - 1) / 2]) = (temp));
  i = 0;
  j = n;
  t = len[perm[0]];
  while (1) {
    do
      i++;
    while (i < n && len[perm[i]] < t);
    do
      j--;
    while (len[perm[j]] > t);
    if (j < i)
      break;
    ((temp) = (perm[i]), (perm[i]) = (perm[j]), (perm[j]) = (temp));
  }
  ((temp) = (perm[0]), (perm[0]) = (perm[j]), (perm[j]) = (temp));
  CCutil_double_perm_quicksort(perm, len, j);
  CCutil_double_perm_quicksort(perm + i, len, n - i);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<CCutil_double_perm_quicksort>:
  endbr64
  cmp    $0x1,%edx
  jle    L15
  push   %r13
  mov    %rdi,%r8
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
L26:
  lea    -0x1(%r13),%eax
  mov    (%r8),%edx
  lea    0x4(%r8),%r12
  xor    %ebp,%ebp
  sar    %eax
  cltq
  lea    (%r8,%rax,4),%rax
  mov    (%rax),%ecx
  mov    %ecx,(%r8)
  mov    %edx,(%rax)
  movslq (%r8),%rax
  mov    %r13d,%edx
  movsd  (%rbx,%rax,8),%xmm1
  nopw   0x0(%rax,%rax,1)
L67:
  add    $0x1,%ebp
  cmp    %ebp,%r13d
  jle    L44
  movslq (%r12),%rax
  comisd (%rbx,%rax,8),%xmm1
  ja     L0
L44:
  movslq %edx,%rax
  lea    -0x4(%r8,%rax,4),%rax
  nopl   0x0(%rax,%rax,1)
L35:
  movslq (%rax),%rsi
  sub    $0x1,%edx
  mov    %rax,%rdi
  sub    $0x4,%rax
  movsd  (%rbx,%rsi,8),%xmm0
  comisd %xmm1,%xmm0
  ja     L35
  cmp    %edx,%ebp
  jg     L56
  mov    (%r12),%eax
  mov    %esi,(%r12)
  mov    %eax,(%rdi)
L0:
  add    $0x4,%r12
  jmp    L67
  nopl   0x0(%rax)
L56:
  mov    (%r8),%eax
  mov    %esi,(%r8)
  sub    %ebp,%r13d
  mov    %rbx,%rsi
  mov    %eax,(%rdi)
  mov    %r8,%rdi
  xor    %eax,%eax
  call   <CCutil_double_perm_quicksort>
  cmp    $0x1,%r13d
  jle    L24
  mov    %r12,%r8
  jmp    L26
L24:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  ret
```

assistant: ```c
void CCutil_double_perm_quicksort(perm, len, n) int *perm;
double *len;
int n;
{
  int i, j, temp;
  double t;
  if (n <= 1)
    return;
  ((temp) = (perm[0]), (perm[0]) = (perm[(n - 1) / 2]), (perm[(n - 1) / 2]) = (temp));
  i = 0;
  j = n;
  t = len[perm[0]];
  while (1) {
    do
      i++;
    while (i < n && len[perm[i]] < t);
    do
      j--;
    while (len[perm[j]] > t);
    if (j < i)
      break;
    ((temp) = (perm[i]), (perm[i]) = (perm[j]), (perm[j]) = (temp));
  }
  ((temp) = (perm[0]), (perm[0]) = (perm[j]), (perm[j]) = (temp));
  CCutil_double_perm_quicksort(perm, len, j);
  CCutil_double_perm_quicksort(perm + i, len, n - i);
}
```
user: What is the c source code of the assembly code below?

```asm
<xrop3_200_1_1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x24(%rbp)
  mov    %al,-0x28(%rbp)
  movl   $0xff,-0x14(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x24(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0x10(%rbp)
  movzbl -0x28(%rbp),%eax
  shl    $0x8,%eax
  mov    %eax,%edx
  movzbl -0x28(%rbp),%eax
  or     %edx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  or     -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  and    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x10(%rbp),%eax
  not    %eax
  and    -0x4(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  not    %eax
  and    %eax,%edx
  mov    -0x14(%rbp),%eax
  and    -0x10(%rbp),%eax
  or     %eax,%edx
  mov    -0x14(%rbp),%eax
  and    -0xc(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned xrop3_200_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  stk2 = D | T;
  stk1 = S & stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: Translate the below assembly code into C programming language:

```asm
<xrop3_200_1_1>:
  endbr64
  or     %edi,%esi
  movzbl %sil,%eax
  ret
```

assistant: ```c
unsigned xrop3_200_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  stk2 = D | T;
  stk1 = S & stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<xrop3_200_1_1>:
  endbr64
  or     %edi,%esi
  movzbl %sil,%eax
  ret
```

assistant: ```c
unsigned xrop3_200_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  stk2 = D | T;
  stk1 = S & stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<xrop3_200_1_1>:
  endbr64
  or     %edi,%esi
  movzbl %sil,%eax
  ret
```

assistant: ```c
unsigned xrop3_200_1_1(unsigned char s, unsigned char t) {
  unsigned D = 0x00ff;
  unsigned S = ((unsigned)s << 8) | s;
  unsigned T = ((unsigned)t << 8) | t;
  unsigned stk1;
  unsigned stk2;
  stk2 = D | T;
  stk1 = S & stk2;
  return (stk1 & (~S) & (~T)) | (D & S) | (D & T);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getInput>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xffffffff,-0x4(%rbp)
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x5(%rbp)
  cmpb   $0x2f,-0x5(%rbp)
  jle    L8
  cmpb   $0x39,-0x5(%rbp)
  jg     L8
  movsbl -0x5(%rbp),%eax
  sub    $0x30,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L68
L8:
  cmpb   $0x60,-0x5(%rbp)
  jle    L68
  cmpb   $0x63,-0x5(%rbp)
  jg     L68
  movsbl -0x5(%rbp),%eax
  sub    $0x57,%eax
  mov    %eax,-0x4(%rbp)
L68:
  nop
L13:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x5(%rbp)
  cmpb   $0xa,-0x5(%rbp)
  je     L20
  mov    D7(%rip),%rax
  movzbl (%rax),%eax
  cmp    %al,-0x5(%rbp)
  jne    L13
L20:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int getInput(void) {
  char ch;
  int ret = -1;
  ch = getchar();
  if (ch >= '0' && ch <= '9') {
    ret = ch - '0';
  } else if (ch >= 'a' && ch <= 'c') {
    ret = ch - 'a' + 10;
  }
  while ((ch = getchar()) != '\n' && ch != EOF)
    ;
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<getInput>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <getchar@plt>
  lea    -0x30(%rax),%edx
  cmp    $0x9,%dl
  ja     L43
  movsbl %al,%eax
  lea    -0x30(%rax),%ebx
L10:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  je     L21
  mov    D0(%rip),%rdx
  cmp    %al,(%rdx)
  jne    L10
L21:
  mov    %ebx,%eax
  pop    %rbx
  ret
L43:
  lea    -0x61(%rax),%edx
  cmp    $0x2,%dl
  ja     L33
  movsbl %al,%eax
  lea    -0x57(%rax),%ebx
  jmp    L10
L33:
  mov    $0xffffffff,%ebx
  jmp    L10
```

assistant: ```c
int getInput(void) {
  char ch;
  int ret = -1;
  ch = getchar();
  if (ch >= '0' && ch <= '9') {
    ret = ch - '0';
  } else if (ch >= 'a' && ch <= 'c') {
    ret = ch - 'a' + 10;
  }
  while ((ch = getchar()) != '\n' && ch != EOF)
    ;
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getInput>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <getchar@plt>
  lea    -0x30(%rax),%edx
  movsbl %dl,%r12d
  cmp    $0x9,%dl
  jbe    L4
  lea    -0x61(%rax),%edx
  cmp    $0x2,%dl
  ja     L12
  sub    $0x57,%eax
  movsbl %al,%r12d
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  je     L5
  nopl   0x0(%rax,%rax,1)
L28:
  mov    D8(%rip),%rdx
  cmp    %al,(%rdx)
  je     L5
L4:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L28
L5:
  mov    %r12d,%eax
  pop    %r12
  ret
L12:
  mov    $0xffffffff,%r12d
  jmp    L4
```

assistant: ```c
int getInput(void) {
  char ch;
  int ret = -1;
  ch = getchar();
  if (ch >= '0' && ch <= '9') {
    ret = ch - '0';
  } else if (ch >= 'a' && ch <= 'c') {
    ret = ch - 'a' + 10;
  }
  while ((ch = getchar()) != '\n' && ch != EOF)
    ;
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getInput>:
  endbr64
  push   %r12
  xor    %eax,%eax
  call   <getchar@plt>
  lea    -0x30(%rax),%edx
  movsbl %dl,%r12d
  cmp    $0x9,%dl
  jbe    L56
  lea    -0x61(%rax),%edx
  cmp    $0x2,%dl
  ja     L3
  sub    $0x57,%eax
  movsbl %al,%r12d
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  je     L19
  nopl   0x0(%rax,%rax,1)
L45:
  mov    D8(%rip),%rdx
  cmp    %al,(%rdx)
  je     L19
L56:
  xor    %eax,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L45
L19:
  mov    %r12d,%eax
  pop    %r12
  ret
L3:
  mov    $0xffffffff,%r12d
  jmp    L56
```

assistant: ```c
int getInput(void) {
  char ch;
  int ret = -1;
  ch = getchar();
  if (ch >= '0' && ch <= '9') {
    ret = ch - '0';
  } else if (ch >= 'a' && ch <= 'c') {
    ret = ch - 'a' + 10;
  }
  while ((ch = getchar()) != '\n' && ch != EOF)
    ;
  return ret;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <kindredSpirits@plt>
  test   %rax,%rax
  je     L1
  lea    D11(%rip),%rax
  jmp    L15
L1:
  lea    D26(%rip),%rax
L15:
  mov    %rax,%rsi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D11
type_infered: string
name: parse_data
arguments: tag: D26
type_infered: string
name: parse_data
arguments: tag: D14
type_infered: string

tool: D11: Success!

tool: D26: fail whale :(

tool: D14: %s


assistant: ```c
int main(void) {
  printf("%s\n", kindredSpirits(NULL, NULL) ? "Success!" : "fail whale :(");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <kindredSpirits@plt>
  test   %rax,%rax
  lea    D2(%rip),%rsi
  lea    D17(%rip),%rax
  cmove  %rax,%rsi
  lea    D19(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D2
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D17
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D19
type_infered: string

tool: D2: Success!

tool: D17: fail whale :(

tool: D19: %s


assistant: ```c
int main(void) {
  printf("%s\n", kindredSpirits(NULL, NULL) ? "Success!" : "fail whale :(");
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  call   <kindredSpirits@plt>
  lea    D1(%rip),%rsi
  lea    D5(%rip),%rdi
  test   %rax,%rax
  lea    D7(%rip),%rax
  cmove  %rax,%rsi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_infered: string
name: unpack_data
arguments: data_label: D5
type_infered: string
name: unpack_data
arguments: data_label: D7
type_infered: string

tool: D1: Success!

tool: D5: %s


tool: D7: fail whale :(

assistant: ```c
int main(void) {
  printf("%s\n", kindredSpirits(NULL, NULL) ? "Success!" : "fail whale :(");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %esi,%esi
  xor    %edi,%edi
  call   <kindredSpirits@plt>
  lea    D23(%rip),%rsi
  lea    D24(%rip),%rdi
  test   %rax,%rax
  lea    D12(%rip),%rax
  cmove  %rax,%rsi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: data_label: D23
type: string
name: read_data
arguments: data_label: D24
type: string
name: read_data
arguments: data_label: D12
type: string

tool: D23: Success!

tool: D24: %s


tool: D12: fail whale :(

assistant: ```c
int main(void) {
  printf("%s\n", kindredSpirits(NULL, NULL) ? "Success!" : "fail whale :(");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L7
L9:
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x4(%rbp)
  addl   $0x1,-0x8(%rbp)
L7:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L9
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int sum(int ar[], int n) {
  int i;
  int total = 0;
  for (i = 0; i < n; i++)
    total += ar[i];
  return total;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sum>:
  endbr64
  test   %esi,%esi
  jle    L15
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rcx
  mov    $0x0,%edx
L13:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L13
L4:
  mov    %edx,%eax
  ret
L15:
  mov    $0x0,%edx
  jmp    L4
```

assistant: ```c
int sum(int ar[], int n) {
  int i;
  int total = 0;
  for (i = 0; i < n; i++)
    total += ar[i];
  return total;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<sum>:
  endbr64
  test   %esi,%esi
  jle    L17
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L15:
  add    (%rdi),%eax
  add    $0x4,%rdi
  cmp    %rdx,%rdi
  jne    L15
  ret
  nopl   0x0(%rax)
L17:
  xor    %eax,%eax
  ret
```

assistant: ```c
int sum(int ar[], int n) {
  int i;
  int total = 0;
  for (i = 0; i < n; i++)
    total += ar[i];
  return total;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum>:
  endbr64
  mov    %rdi,%rdx
  test   %esi,%esi
  jle    L39
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L52
  mov    %esi,%ecx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L12:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rcx,%rax
  jne    L12
  movdqa %xmm0,%xmm1
  mov    %esi,%ecx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  test   $0x3,%sil
  je     L51
L41:
  movslq %ecx,%rdi
  lea    0x0(,%rdi,4),%r8
  add    (%rdx,%rdi,4),%eax
  lea    0x1(%rcx),%edi
  cmp    %edi,%esi
  jle    L24
  add    $0x2,%ecx
  add    0x4(%rdx,%r8,1),%eax
  cmp    %ecx,%esi
  jle    L24
  add    0x8(%rdx,%r8,1),%eax
  ret
  nopl   0x0(%rax)
L39:
  xor    %eax,%eax
L24:
  ret
  nopl   0x0(%rax,%rax,1)
L51:
  ret
L52:
  xor    %eax,%eax
  xor    %ecx,%ecx
  jmp    L41
```

assistant: ```c
int sum(int ar[], int n) {
  int i;
  int total = 0;
  for (i = 0; i < n; i++)
    total += ar[i];
  return total;
}
```
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
  sub    $0xf0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xd0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x19,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,-0xd8(%rbp)
  lea    -0xe4(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
L53:
  lea    -0xe0(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xe0(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L10
  mov    -0xe0(%rbp),%edx
  mov    -0xd8(%rbp),%eax
  cltq
  mov    %edx,-0xd0(%rbp,%rax,4)
  addl   $0x1,-0xd8(%rbp)
L10:
  mov    -0xe4(%rbp),%eax
  cmp    %eax,-0xd8(%rbp)
  jl     L53
  mov    -0xe4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0xd4(%rbp)
  movl   $0x0,-0xdc(%rbp)
  jmp    L8
L37:
  mov    -0xdc(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%edx
  mov    -0xe4(%rbp),%eax
  sub    -0xdc(%rbp),%eax
  sub    $0x1,%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xdc(%rbp)
L8:
  mov    -0xdc(%rbp),%eax
  cmp    -0xd4(%rbp),%eax
  jl     L37
  mov    -0xe4(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  je     L41
  mov    -0xd4(%rbp),%eax
  cltq
  mov    -0xd0(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L41:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L56
  call   <__stack_chk_fail@plt>
L56:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D23
type: string
name: parse_data
arguments: tag: D16
type: string
name: parse_data
arguments: tag: D39
type: string

tool: D23: %d

tool: D16: 
%d

tool: D39: 
%d


assistant: ```c
int main() {
  int i, n, Num, Odd[50] = {0}, brOdd = 0, Middle;
  scanf("%d", &n);
  do {
    scanf("%d", &Num);
    if (Num % 2 != 0) {
      Odd[brOdd] = Num;
      brOdd++;
    }
  } while (brOdd < n);
  Middle = n / 2;
  for (i = 0; i < Middle; i++) {
    printf("\n%d", Odd[i] - Odd[n - i - 1]);
  }
  if (n % 2 != 0) {
    printf("\n%d\n", Odd[Middle]);
  }
}
```