ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<hash_str>:
  endbr64
  movsbl (%rdi),%eax
  xor    %r8d,%r8d
  test   %al,%al
  je     L19
  xchg   %ax,%ax
L13:
  mov    %r8d,%edx
  add    $0x1,%rdi
  shl    $0x5,%edx
  sub    %r8d,%edx
  lea    (%rax,%rdx,1),%r8d
  movsbl (%rdi),%eax
  test   %al,%al
  jne    L13
L19:
  mov    %r8d,%eax
  ret
```

assistant: ```c
unsigned int hash_str(char *str) {
  unsigned int hash = 0;
  for (; *str; str++)
    hash = (hash * 32) - hash + *str;
  return hash;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<cl_connect>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %esi,-0x3c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x28(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <gethostbyname@plt>
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  jne    L17
  mov    $0xffffffff,%eax
  jmp    L12
L17:
  lea    -0x14(%rbp),%rax
  mov    $0xc,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x3c(%rbp),%eax
  mov    %eax,%edi
  call   <htons@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L28
  mov    -0x20(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    (%rax),%eax
  mov    %eax,-0x14(%rbp)
  jmp    L40
L28:
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <inet_addr@plt>
  mov    %eax,-0x14(%rbp)
L40:
  mov    D27(%rip),%rax
  mov    (%rax),%ecx
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  mov    $0x0,%edx
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <socket@plt>
  mov    %eax,-0x24(%rbp)
  cmpl   $0xffffffff,-0x24(%rbp)
  jne    L46
  mov    $0xffffffff,%eax
  jmp    L12
L46:
  lea    -0x14(%rbp),%rcx
  mov    -0x24(%rbp),%eax
  mov    $0x4,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <connect@plt>
  cmp    $0xffffffff,%eax
  jne    L0
  mov    $0xffffffff,%eax
  jmp    L12
L0:
  mov    -0x24(%rbp),%eax
L12:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int cl_connect(const char *remote_addr, const uint16_t port) {
  int conn_sock;
  struct sockaddr_in addr;
  struct hostent *resolved;
  int i = 0;
  if ((resolved = gethostbyname(remote_addr)) == NULL)
    return -1;
  memset(&addr, 0, sizeof(struct sockaddr_in));
  addr.sin_family = AF_INET;
  addr.sin_port = htons(port);
  if (resolved->h_addr_list[0]) {
    addr.sin_addr.s_addr = *(uint32_t *)resolved->h_addr_list[i];
  } else {
    addr.sin_addr.s_addr = inet_addr(remote_addr);
  }
  if ((conn_sock = socket(AF_INET, SOCK_STREAM, 0)) == -1)
    return -1;
  if (connect(conn_sock, (struct sockaddr *)(&addr), sizeof(struct sockaddr)) == -1)
    return -1;
  return conn_sock;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cl_connect>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <gethostbyname@plt>
  test   %rax,%rax
  je     L27
  mov    %rax,%rbx
  lea    0xc(%rsp),%rdi
  mov    $0xc,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,0x14(%rsp)
  mov    %r12d,%edi
  call   <htons@plt>
  mov    %eax,0x10(%rsp)
  mov    (%rbx),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L7
  mov    (%rax),%eax
L53:
  mov    %eax,0xc(%rsp)
  mov    $0x0,%edx
  mov    D20(%rip),%rax
  mov    (%rax),%esi
  mov    D17(%rip),%rax
  mov    (%rax),%edi
  call   <socket@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  je     L57
  lea    0xc(%rsp),%rsi
  mov    $0x4,%edx
  mov    %eax,%edi
  call   <connect@plt>
  cmp    $0xffffffff,%eax
  cmove  %eax,%ebx
L57:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    %ebx,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L7:
  mov    %rbp,%rdi
  call   <inet_addr@plt>
  jmp    L53
L27:
  mov    $0xffffffff,%ebx
  jmp    L57
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int cl_connect(const char *remote_addr, const uint16_t port) {
  int conn_sock;
  struct sockaddr_in addr;
  struct hostent *resolved;
  int i = 0;
  if ((resolved = gethostbyname(remote_addr)) == NULL)
    return -1;
  memset(&addr, 0, sizeof(struct sockaddr_in));
  addr.sin_family = AF_INET;
  addr.sin_port = htons(port);
  if (resolved->h_addr_list[0]) {
    addr.sin_addr.s_addr = *(uint32_t *)resolved->h_addr_list[i];
  } else {
    addr.sin_addr.s_addr = inet_addr(remote_addr);
  }
  if ((conn_sock = socket(AF_INET, SOCK_STREAM, 0)) == -1)
    return -1;
  if (connect(conn_sock, (struct sockaddr *)(&addr), sizeof(struct sockaddr)) == -1)
    return -1;
  return conn_sock;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<cl_connect>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <gethostbyname@plt>
  test   %rax,%rax
  je     L68
  lea    0xc(%rsp),%r13
  mov    $0xc,%edx
  xor    %esi,%esi
  mov    %rax,%r12
  mov    %r13,%rdi
  call   <memset@plt>
  mov    D17(%rip),%rbx
  mov    %r14d,%edi
  mov    (%rbx),%eax
  mov    %eax,0x14(%rsp)
  call   <htons@plt>
  mov    %eax,0x10(%rsp)
  mov    (%r12),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L14
  mov    (%rax),%eax
L18:
  mov    %eax,0xc(%rsp)
  mov    D15(%rip),%rax
  xor    %edx,%edx
  mov    (%rbx),%edi
  mov    (%rax),%esi
  call   <socket@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L68
  mov    $0x4,%edx
  mov    %r13,%rsi
  mov    %eax,%edi
  call   <connect@plt>
  cmp    $0xffffffff,%eax
  je     L68
L7:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x20,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L14:
  mov    %rbp,%rdi
  call   <inet_addr@plt>
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L68:
  mov    $0xffffffff,%r12d
  jmp    L7
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int cl_connect(const char *remote_addr, const uint16_t port) {
  int conn_sock;
  struct sockaddr_in addr;
  struct hostent *resolved;
  int i = 0;
  if ((resolved = gethostbyname(remote_addr)) == NULL)
    return -1;
  memset(&addr, 0, sizeof(struct sockaddr_in));
  addr.sin_family = AF_INET;
  addr.sin_port = htons(port);
  if (resolved->h_addr_list[0]) {
    addr.sin_addr.s_addr = *(uint32_t *)resolved->h_addr_list[i];
  } else {
    addr.sin_addr.s_addr = inet_addr(remote_addr);
  }
  if ((conn_sock = socket(AF_INET, SOCK_STREAM, 0)) == -1)
    return -1;
  if (connect(conn_sock, (struct sockaddr *)(&addr), sizeof(struct sockaddr)) == -1)
    return -1;
  return conn_sock;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<cl_connect>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <gethostbyname@plt>
  test   %rax,%rax
  je     L0
  lea    0xc(%rsp),%r13
  mov    $0xc,%edx
  xor    %esi,%esi
  mov    %rax,%r12
  mov    %r13,%rdi
  call   <memset@plt>
  mov    D19(%rip),%rbx
  mov    %r14d,%edi
  mov    (%rbx),%eax
  mov    %eax,0x14(%rsp)
  call   <htons@plt>
  mov    %eax,0x10(%rsp)
  mov    (%r12),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L19
  mov    (%rax),%eax
L65:
  mov    %eax,0xc(%rsp)
  mov    D3(%rip),%rax
  xor    %edx,%edx
  mov    (%rbx),%edi
  mov    (%rax),%esi
  call   <socket@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L0
  mov    $0x4,%edx
  mov    %r13,%rsi
  mov    %eax,%edi
  call   <connect@plt>
  cmp    $0xffffffff,%eax
  je     L0
L12:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x20,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L19:
  mov    %rbp,%rdi
  call   <inet_addr@plt>
  jmp    L65
  nopw   0x0(%rax,%rax,1)
L0:
  mov    $0xffffffff,%r12d
  jmp    L12
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int cl_connect(const char *remote_addr, const uint16_t port) {
  int conn_sock;
  struct sockaddr_in addr;
  struct hostent *resolved;
  int i = 0;
  if ((resolved = gethostbyname(remote_addr)) == NULL)
    return -1;
  memset(&addr, 0, sizeof(struct sockaddr_in));
  addr.sin_family = AF_INET;
  addr.sin_port = htons(port);
  if (resolved->h_addr_list[0]) {
    addr.sin_addr.s_addr = *(uint32_t *)resolved->h_addr_list[i];
  } else {
    addr.sin_addr.s_addr = inet_addr(remote_addr);
  }
  if ((conn_sock = socket(AF_INET, SOCK_STREAM, 0)) == -1)
    return -1;
  if (connect(conn_sock, (struct sockaddr *)(&addr), sizeof(struct sockaddr)) == -1)
    return -1;
  return conn_sock;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<shaker_sort>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebp
  shr    $0x1f,%ebp
  add    %esi,%ebp
  sar    %ebp
  cmp    $0x1,%esi
  jle    L109
  mov    %rdi,%rbx
  lea    -0x2(%rsi),%r9d
  mov    %r9d,%r11d
  movslq %esi,%rsi
  lea    (%rdi,%rsi,2),%rax
  mov    %rdi,%r10
  mov    $0x0,%r8d
  mov    $0x0,%esi
  lea    0x4(%rdi),%r13
  mov    $0x1,%edi
  lea    -0xc(%rax,%rax,1),%r12
  jmp    L3
L62:
  add    $0x4,%rax
  cmp    %r14,%rax
  je     L63
L32:
  mov    0x4(%rax),%ecx
  mov    (%rax),%edx
  cmp    %edx,%ecx
  jge    L62
  mov    %ecx,(%rax)
  mov    %edx,0x4(%rax)
  mov    %edi,%esi
  jmp    L62
L63:
  cmp    %r15d,%r9d
  jle    L30
  movslq %r9d,%rax
  lea    (%rbx,%rax,4),%rax
  mov    %r12,%r14
  sub    %r10,%r14
  lea    -0x1(%r11),%edx
  shl    $0x2,%rdx
  sub    %rdx,%r14
  jmp    L80
L45:
  sub    $0x4,%rax
  cmp    %r14,%rax
  je     L30
L80:
  mov    -0x4(%rax),%ecx
  mov    (%rax),%edx
  cmp    %edx,%ecx
  jle    L45
  mov    %ecx,(%rax)
  mov    %edx,-0x4(%rax)
  mov    %edi,%esi
  jmp    L45
L30:
  test   %esi,%esi
  je     L109
  add    $0x1,%r8
  sub    $0x1,%r9d
  add    $0x4,%r10
  sub    $0x2,%r11d
  cmp    %r8d,%ebp
  jle    L109
L3:
  mov    %r8d,%r15d
  lea    0x1(%r9),%eax
  cmp    %r8d,%eax
  jle    L63
  mov    %r11d,%eax
  add    %r8,%rax
  lea    0x0(%r13,%rax,4),%r14
  mov    %r10,%rax
  jmp    L32
L109:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void shaker_sort(int a[], int n) {
  int i = 0, j;
  bool swapped = false;
  for (i = 0; i < n / 2; i++) {
    for (j = i; j < n - 1 - i; j++) {
      if (a[j + 1] < a[j]) {
        int temp = a[j];
        a[j] = a[j + 1];
        a[j + 1] = temp;
        swapped = true;
      }
    }
    for (j = n - i - 2; j > i; j--) {
      if (a[j - 1] > a[j]) {
        int temp = a[j];
        a[j] = a[j - 1];
        a[j - 1] = temp;
        swapped = true;
      }
    }
    if (!swapped)
      break;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<shaker_sort>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  shr    $0x1f,%ebx
  add    %esi,%ebx
  sar    %ebx
  cmp    $0x1,%esi
  jle    L35
  lea    -0x2(%rsi),%r8d
  movslq %esi,%rsi
  mov    %rdi,%r11
  mov    %rdi,%r9
  lea    (%rdi,%rsi,2),%rax
  mov    %r8d,%r10d
  xor    %edi,%edi
  xor    %esi,%esi
  lea    0x4(%r11),%r12
  lea    -0xc(%rax,%rax,1),%rbp
  nop
L67:
  lea    0x1(%r8),%eax
  mov    %edi,%r14d
  cmp    %edi,%eax
  jle    L85
  mov    %r10d,%eax
  add    %rdi,%rax
  lea    (%r12,%rax,4),%r13
  mov    %r9,%rax
  nopl   0x0(%rax,%rax,1)
L81:
  mov    0x4(%rax),%ecx
  mov    (%rax),%edx
  cmp    %edx,%ecx
  jge    L44
  mov    %ecx,(%rax)
  mov    $0x1,%esi
  mov    %edx,0x4(%rax)
L44:
  add    $0x4,%rax
  cmp    %r13,%rax
  jne    L81
L85:
  cmp    %r14d,%r8d
  jle    L9
  mov    %rbp,%r13
  lea    -0x1(%r10),%edx
  movslq %r8d,%rax
  sub    %r9,%r13
  shl    $0x2,%rdx
  lea    (%r11,%rax,4),%rax
  sub    %rdx,%r13
  nopl   0x0(%rax)
L71:
  mov    -0x4(%rax),%ecx
  mov    (%rax),%edx
  cmp    %edx,%ecx
  jle    L49
  mov    %ecx,(%rax)
  mov    $0x1,%esi
  mov    %edx,-0x4(%rax)
L49:
  sub    $0x4,%rax
  cmp    %rax,%r13
  jne    L71
L9:
  test   %esi,%esi
  je     L35
  add    $0x1,%rdi
  sub    $0x1,%r8d
  add    $0x4,%r9
  sub    $0x2,%r10d
  cmp    %edi,%ebx
  jg     L67
L35:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void shaker_sort(int a[], int n) {
  int i = 0, j;
  bool swapped = false;
  for (i = 0; i < n / 2; i++) {
    for (j = i; j < n - 1 - i; j++) {
      if (a[j + 1] < a[j]) {
        int temp = a[j];
        a[j] = a[j + 1];
        a[j + 1] = temp;
        swapped = true;
      }
    }
    for (j = n - i - 2; j > i; j--) {
      if (a[j - 1] > a[j]) {
        int temp = a[j];
        a[j] = a[j - 1];
        a[j - 1] = temp;
        swapped = true;
      }
    }
    if (!swapped)
      break;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<shaker_sort>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  shr    $0x1f,%ebx
  add    %esi,%ebx
  sar    %ebx
  cmp    $0x1,%esi
  jle    L39
  lea    -0x2(%rsi),%r8d
  movslq %esi,%rsi
  mov    %rdi,%r11
  mov    %rdi,%r9
  lea    (%rdi,%rsi,2),%rax
  mov    %r8d,%r10d
  xor    %edi,%edi
  xor    %esi,%esi
  lea    0x4(%r11),%r12
  lea    -0x10(%rax,%rax,1),%rbp
  nop
L27:
  lea    0x1(%r8),%eax
  mov    %edi,%r14d
  cmp    %edi,%eax
  jle    L51
  mov    %r10d,%eax
  mov    %r9,%rdx
  add    %rdi,%rax
  lea    (%r12,%rax,4),%r13
  nopl   0x0(%rax,%rax,1)
L78:
  movq   (%rdx),%xmm0
  pshufd $0xe5,%xmm0,%xmm2
  movd   %xmm0,%ecx
  movd   %xmm2,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jle    L12
  movq   %xmm1,(%rdx)
  mov    $0x1,%esi
L12:
  add    $0x4,%rdx
  cmp    %r13,%rdx
  jne    L78
L51:
  cmp    %r14d,%r8d
  jle    L75
  movslq %r8d,%rax
  mov    %rbp,%r13
  lea    -0x4(%r11,%rax,4),%rdx
  lea    -0x1(%r10),%eax
  sub    %r9,%r13
  shl    $0x2,%rax
  sub    %rax,%r13
  nopw   0x0(%rax,%rax,1)
L50:
  movq   (%rdx),%xmm0
  pshufd $0xe5,%xmm0,%xmm3
  movd   %xmm0,%ecx
  movd   %xmm3,%eax
  pshufd $0xe1,%xmm0,%xmm1
  cmp    %eax,%ecx
  jle    L76
  movq   %xmm1,(%rdx)
  mov    $0x1,%esi
L76:
  sub    $0x4,%rdx
  cmp    %r13,%rdx
  jne    L50
L75:
  test   %esi,%esi
  je     L39
  add    $0x1,%rdi
  sub    $0x1,%r8d
  add    $0x4,%r9
  sub    $0x2,%r10d
  cmp    %edi,%ebx
  jg     L27
L39:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void shaker_sort(int a[], int n) {
  int i = 0, j;
  bool swapped = false;
  for (i = 0; i < n / 2; i++) {
    for (j = i; j < n - 1 - i; j++) {
      if (a[j + 1] < a[j]) {
        int temp = a[j];
        a[j] = a[j + 1];
        a[j + 1] = temp;
        swapped = true;
      }
    }
    for (j = n - i - 2; j > i; j--) {
      if (a[j - 1] > a[j]) {
        int temp = a[j];
        a[j] = a[j - 1];
        a[j - 1] = temp;
        swapped = true;
      }
    }
    if (!swapped)
      break;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<pmod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  call   <fmod@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  comisd -0x8(%rbp),%xmm0
  jbe    L6
  movsd  -0x8(%rbp),%xmm0
  addsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
L6:
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double pmod(double x, double y) {
  double z = fmod(x, y);
  if (z < 0)
    z += y;
  return (z);
}
```
user: What is the c source code of the assembly code below:

```asm
<pmod>:
  endbr64
  push   %rbx
  movq   %xmm1,%rbx
  call   <fmod@plt>
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  ja     L0
L1:
  pop    %rbx
  ret
L0:
  movq   %rbx,%xmm2
  addsd  %xmm2,%xmm0
  jmp    L1
```

assistant: ```c
double pmod(double x, double y) {
  double z = fmod(x, y);
  if (z < 0)
    z += y;
  return (z);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<pmod>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm1,0x8(%rsp)
  call   <fmod@plt>
  pxor   %xmm2,%xmm2
  comisd %xmm0,%xmm2
  jbe    L7
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm1,%xmm0
L7:
  add    $0x18,%rsp
  ret
```

assistant: ```c
double pmod(double x, double y) {
  double z = fmod(x, y);
  if (z < 0)
    z += y;
  return (z);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<pmod>:
  endbr64
  sub    $0x18,%rsp
  movsd  %xmm1,0x8(%rsp)
  call   <fmod@plt>
  pxor   %xmm2,%xmm2
  comisd %xmm0,%xmm2
  jbe    L0
  movsd  0x8(%rsp),%xmm1
  addsd  %xmm1,%xmm0
L0:
  add    $0x18,%rsp
  ret
```

assistant: ```c
double pmod(double x, double y) {
  double z = fmod(x, y);
  if (z < 0)
    z += y;
  return (z);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bitread>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  shr    $0x5,%eax
  mov    %eax,%eax
  shl    $0x2,%rax
  add    %rax,-0x18(%rbp)
  andl   $0x1f,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x1c(%rbp),%eax
  mov    %eax,%ecx
  shr    %cl,%edx
  mov    %edx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x20,-0x20(%rbp)
  jne    L24
  cmpl   $0x0,-0x1c(%rbp)
  je     L30
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edx
  mov    $0x20,%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0x4(%rbp)
  jmp    L30
L24:
  mov    -0x1c(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  cmp    $0x20,%eax
  jbe    L27
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edx
  mov    $0x20,%eax
  sub    -0x1c(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,-0x4(%rbp)
L27:
  mov    -0x20(%rbp),%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  sub    $0x1,%eax
  and    %eax,-0x4(%rbp)
L30:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int bitread(unsigned int *e, unsigned int p, unsigned int len) {
  unsigned int answ;
  e += p / 32;
  p %= 32;
  answ = *e >> p;
  if (len == 32) {
    if (p)
      answ |= (*(e + 1)) << (32 - p);
  } else {
    if (p + len > 32)
      answ |= (*(e + 1)) << (32 - p);
    answ &= (1 << len) - 1;
  }
  return answ;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<bitread>:
  endbr64
  mov    %esi,%ecx
  mov    %esi,%eax
  shr    $0x5,%eax
  mov    %eax,%eax
  lea    (%rdi,%rax,4),%rsi
  mov    %ecx,%edi
  and    $0x1f,%edi
  mov    (%rsi),%eax
  shr    %cl,%eax
  cmp    $0x20,%edx
  je     L5
  lea    (%rdi,%rdx,1),%ecx
  cmp    $0x20,%ecx
  jbe    L27
  mov    $0x20,%ecx
  sub    %edi,%ecx
  mov    0x4(%rsi),%esi
  shl    %cl,%esi
  or     %esi,%eax
L27:
  mov    $0x1,%esi
  mov    %edx,%ecx
  shl    %cl,%esi
  sub    $0x1,%esi
  and    %esi,%eax
L24:
  ret
L5:
  test   %edi,%edi
  je     L24
  mov    $0x20,%ecx
  sub    %edi,%ecx
  mov    0x4(%rsi),%edx
  shl    %cl,%edx
  or     %edx,%eax
  ret
```

assistant: ```c
unsigned int bitread(unsigned int *e, unsigned int p, unsigned int len) {
  unsigned int answ;
  e += p / 32;
  p %= 32;
  answ = *e >> p;
  if (len == 32) {
    if (p)
      answ |= (*(e + 1)) << (32 - p);
  } else {
    if (p + len > 32)
      answ |= (*(e + 1)) << (32 - p);
    answ &= (1 << len) - 1;
  }
  return answ;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitread>:
  endbr64
  mov    %esi,%eax
  mov    %esi,%ecx
  shr    $0x5,%eax
  lea    (%rdi,%rax,4),%rsi
  mov    %ecx,%edi
  mov    (%rsi),%eax
  and    $0x1f,%edi
  shr    %cl,%eax
  cmp    $0x20,%edx
  je     L27
  lea    (%rdi,%rdx,1),%ecx
  cmp    $0x20,%ecx
  jbe    L18
  mov    0x4(%rsi),%esi
  mov    $0x20,%ecx
  sub    %edi,%ecx
  shl    %cl,%esi
  or     %esi,%eax
L18:
  mov    $0x1,%esi
  mov    %edx,%ecx
  shl    %cl,%esi
  sub    $0x1,%esi
  and    %esi,%eax
L20:
  ret
  nopw   0x0(%rax,%rax,1)
L27:
  test   %edi,%edi
  je     L20
  mov    0x4(%rsi),%esi
  sub    %edi,%edx
  mov    %edx,%ecx
  shl    %cl,%esi
  or     %esi,%eax
  ret
```

assistant: ```c
unsigned int bitread(unsigned int *e, unsigned int p, unsigned int len) {
  unsigned int answ;
  e += p / 32;
  p %= 32;
  answ = *e >> p;
  if (len == 32) {
    if (p)
      answ |= (*(e + 1)) << (32 - p);
  } else {
    if (p + len > 32)
      answ |= (*(e + 1)) << (32 - p);
    answ &= (1 << len) - 1;
  }
  return answ;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bitread>:
  endbr64
  mov    %esi,%eax
  mov    %esi,%ecx
  shr    $0x5,%eax
  lea    (%rdi,%rax,4),%rsi
  mov    %ecx,%edi
  mov    (%rsi),%eax
  and    $0x1f,%edi
  shr    %cl,%eax
  cmp    $0x20,%edx
  je     L8
  lea    (%rdi,%rdx,1),%ecx
  cmp    $0x20,%ecx
  jbe    L19
  mov    0x4(%rsi),%esi
  mov    $0x20,%ecx
  sub    %edi,%ecx
  shl    %cl,%esi
  or     %esi,%eax
L19:
  mov    $0x1,%esi
  mov    %edx,%ecx
  shl    %cl,%esi
  sub    $0x1,%esi
  and    %esi,%eax
L1:
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  test   %edi,%edi
  je     L1
  mov    0x4(%rsi),%esi
  sub    %edi,%edx
  mov    %edx,%ecx
  shl    %cl,%esi
  or     %esi,%eax
  ret
```

assistant: ```c
unsigned int bitread(unsigned int *e, unsigned int p, unsigned int len) {
  unsigned int answ;
  e += p / 32;
  p %= 32;
  answ = *e >> p;
  if (len == 32) {
    if (p)
      answ |= (*(e + 1)) << (32 - p);
  } else {
    if (p + len > 32)
      answ |= (*(e + 1)) << (32 - p);
    answ &= (1 << len) - 1;
  }
  return answ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<xrealloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L1
  lea    D7(%rip),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <err@plt>
L1:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D7
datatype: string

tool: D7: realloc() failed

assistant: ```c
void *xrealloc(void *p, size_t size) {
  void *ret = realloc(p, size);
  if (ret == NULL)
    err(1, "realloc() failed");
  return ret;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<xrealloc>:
  endbr64
  push   %rbx
  call   <realloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L6
L16:
  mov    %rbx,%rax
  pop    %rbx
  ret
L6:
  lea    D1(%rip),%rsi
  mov    $0x1,%edi
  call   <err@plt>
  jmp    L16
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
datatype: string

tool: D1: realloc() failed

assistant: ```c
void *xrealloc(void *p, size_t size) {
  void *ret = realloc(p, size);
  if (ret == NULL)
    err(1, "realloc() failed");
  return ret;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<xrealloc>:
  endbr64
  push   %r12
  call   <realloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L12
L2:
  mov    %r12,%rax
  pop    %r12
  ret
L12:
  lea    D6(%rip),%rsi
  mov    $0x1,%edi
  call   <err@plt>
  jmp    L2
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
datatype: string

tool: D6: realloc() failed

assistant: ```c
void *xrealloc(void *p, size_t size) {
  void *ret = realloc(p, size);
  if (ret == NULL)
    err(1, "realloc() failed");
  return ret;
}
```