ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<gg_gethostbyname>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    $0x4,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L28
  mov    %rbp,%rdi
  call   <gethostbyname@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L26
  mov    $0x4,%edx
  mov    (%rax),%esi
  mov    %rbx,%rdi
  call   <memcpy@plt>
L0:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L28:
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D5(%rip),%rax
  mov    %edx,(%rax)
  jmp    L0
L26:
  mov    %rbx,%rdi
  call   <free@plt>
  mov    %rbp,%rbx
  jmp    L0
```

assistant: ```c
struct in_addr *gg_gethostbyname(const char *hostname) {
  struct in_addr *addr = NULL;
  struct hostent *hp;
  if (!(addr = malloc(sizeof(struct in_addr)))) {
    errno = ENOMEM;
    goto cleanup;
  }
  if (!(hp = gethostbyname(hostname)))
    goto cleanup;
  memcpy(addr, hp->h_addr, sizeof(struct in_addr));
  return addr;
cleanup:
  if (addr)
    free(addr);
  return NULL;
}
```
user: What is the c source code of the assembly code below?

```asm
<gg_gethostbyname>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x4,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L6
  mov    %rbp,%rdi
  call   <gethostbyname@plt>
  test   %rax,%rax
  je     L18
  mov    (%rax),%esi
  mov    %r12,%rdi
  mov    $0x4,%edx
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %r12,%rdi
  xor    %r12d,%r12d
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L6:
  mov    D2(%rip),%rax
  mov    (%rax),%edx
  mov    D8(%rip),%rax
  mov    %edx,(%rax)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
struct in_addr *gg_gethostbyname(const char *hostname) {
  struct in_addr *addr = NULL;
  struct hostent *hp;
  if (!(addr = malloc(sizeof(struct in_addr)))) {
    errno = ENOMEM;
    goto cleanup;
  }
  if (!(hp = gethostbyname(hostname)))
    goto cleanup;
  memcpy(addr, hp->h_addr, sizeof(struct in_addr));
  return addr;
cleanup:
  if (addr)
    free(addr);
  return NULL;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<gg_gethostbyname>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x4,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L2
  mov    %rbp,%rdi
  call   <gethostbyname@plt>
  test   %rax,%rax
  je     L15
  mov    (%rax),%esi
  mov    %r12,%rdi
  mov    $0x4,%edx
  call   <memcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  mov    %r12,%rdi
  xor    %r12d,%r12d
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L2:
  mov    D3(%rip),%rax
  mov    (%rax),%edx
  mov    D4(%rip),%rax
  mov    %edx,(%rax)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
struct in_addr *gg_gethostbyname(const char *hostname) {
  struct in_addr *addr = NULL;
  struct hostent *hp;
  if (!(addr = malloc(sizeof(struct in_addr)))) {
    errno = ENOMEM;
    goto cleanup;
  }
  if (!(hp = gethostbyname(hostname)))
    goto cleanup;
  memcpy(addr, hp->h_addr, sizeof(struct in_addr));
  return addr;
cleanup:
  if (addr)
    free(addr);
  return NULL;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<uart_read_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L77
L59:
  cmpb   $0x7f,-0x5(%rbp)
  je     L23
  cmpb   $0x8,-0x5(%rbp)
  jne    L3
L23:
  cmpl   $0x0,-0x4(%rbp)
  jle    L77
  subl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  movsbl -0x5(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  movsbl -0x5(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
  jmp    L77
L3:
  movsbl -0x5(%rbp),%eax
  mov    %eax,%edi
  call   <isprint@plt>
  test   %rax,%rax
  je     L77
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x5(%rbp),%eax
  mov    %al,(%rdx)
  movsbl -0x5(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
L77:
  cmpl   $0x62,-0x4(%rbp)
  jg     L62
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,-0x5(%rbp)
  cmpb   $0xa,-0x5(%rbp)
  je     L62
  cmpb   $0xd,-0x5(%rbp)
  jne    L59
L62:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int uart_read_line(char *str) {
  int n = 0;
  char c;
  while (n < 100 - 1 && (c = getchar()) != '\n' && c != '\r') {
    if (c == 0x7F || c == 0x08) {
      if (n > 0) {
        str[--n] = '\0';
        putchar(c);
        putchar(' ');
        putchar(c);
      }
    } else if (isprint(c)) {
      str[n++] = c;
      putchar(c);
    }
  }
  str[n] = '\0';
  putchar('\n');
  return n;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<uart_read_line>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    $0x0,%ebp
  jmp    L13
L6:
  test   %ebp,%ebp
  jg     L98
L53:
  cmp    $0x62,%ebp
  jg     L47
L13:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    $0xd,%al
  je     L47
  cmp    $0xa,%al
  je     L47
  cmp    $0x7f,%bl
  je     L6
  cmp    $0x8,%bl
  je     L6
  movsbl %bl,%r13d
  mov    %r13d,%edi
  call   <isprint@plt>
  test   %rax,%rax
  je     L53
  lea    0x1(%rbp),%r14d
  movslq %ebp,%rbp
  mov    %bl,(%r12,%rbp,1)
  mov    %r13d,%edi
  call   <putchar@plt>
  mov    %r14d,%ebp
  jmp    L53
L98:
  sub    $0x1,%ebp
  movslq %ebp,%rax
  movb   $0x0,(%r12,%rax,1)
  movsbl %bl,%ebx
  mov    %ebx,%edi
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    %ebx,%edi
  call   <putchar@plt>
  jmp    L53
L47:
  movslq %ebp,%rax
  movb   $0x0,(%r12,%rax,1)
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int uart_read_line(char *str) {
  int n = 0;
  char c;
  while (n < 100 - 1 && (c = getchar()) != '\n' && c != '\r') {
    if (c == 0x7F || c == 0x08) {
      if (n > 0) {
        str[--n] = '\0';
        putchar(c);
        putchar(' ');
        putchar(c);
      }
    } else if (isprint(c)) {
      str[n++] = c;
      putchar(c);
    }
  }
  str[n] = '\0';
  putchar('\n');
  return n;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<uart_read_line>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  nopw   0x0(%rax,%rax,1)
L32:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    $0xa,%al
  je     L78
  cmp    $0xd,%al
  je     L78
  cmp    $0x7f,%bl
  je     L25
  cmp    $0x8,%bl
  je     L25
  movsbl %bl,%ebp
  mov    %ebp,%edi
  call   <isprint@plt>
  test   %rax,%rax
  jne    L90
L95:
  cmp    $0x63,%r12d
  jne    L32
  mov    $0x63,%eax
L46:
  movb   $0x0,0x0(%r13,%rax,1)
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L25:
  test   %r12d,%r12d
  je     L32
  sub    $0x1,%r12d
  movsbl %bl,%ebp
  movslq %r12d,%rax
  mov    %ebp,%edi
  movb   $0x0,0x0(%r13,%rax,1)
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    %ebp,%edi
  call   <putchar@plt>
  jmp    L95
  nopw   0x0(%rax,%rax,1)
L90:
  lea    0x1(%r12),%r14d
  movslq %r12d,%r12
  mov    %ebp,%edi
  mov    %bl,0x0(%r13,%r12,1)
  mov    %r14d,%r12d
  call   <putchar@plt>
  jmp    L95
  nopl   0x0(%rax)
L78:
  movslq %r12d,%rax
  jmp    L46
```

assistant: ```c
int uart_read_line(char *str) {
  int n = 0;
  char c;
  while (n < 100 - 1 && (c = getchar()) != '\n' && c != '\r') {
    if (c == 0x7F || c == 0x08) {
      if (n > 0) {
        str[--n] = '\0';
        putchar(c);
        putchar(' ');
        putchar(c);
      }
    } else if (isprint(c)) {
      str[n++] = c;
      putchar(c);
    }
  }
  str[n] = '\0';
  putchar('\n');
  return n;
}
```
user: Translate the below assembly code into C programming language:

```asm
<uart_read_line>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  nopw   0x0(%rax,%rax,1)
L50:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  cmp    $0xa,%al
  je     L32
  cmp    $0xd,%al
  je     L32
  cmp    $0x7f,%bl
  je     L67
  cmp    $0x8,%bl
  je     L67
  movsbl %bl,%ebp
  mov    %ebp,%edi
  call   <isprint@plt>
  test   %rax,%rax
  jne    L80
L40:
  cmp    $0x63,%r12d
  jne    L50
  mov    $0x63,%eax
L81:
  movb   $0x0,0x0(%r13,%rax,1)
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L67:
  test   %r12d,%r12d
  je     L50
  sub    $0x1,%r12d
  movsbl %bl,%ebp
  movslq %r12d,%rax
  mov    %ebp,%edi
  movb   $0x0,0x0(%r13,%rax,1)
  call   <putchar@plt>
  mov    $0x20,%edi
  call   <putchar@plt>
  mov    %ebp,%edi
  call   <putchar@plt>
  jmp    L40
  nopw   0x0(%rax,%rax,1)
L80:
  lea    0x1(%r12),%r14d
  movslq %r12d,%r12
  mov    %ebp,%edi
  mov    %bl,0x0(%r13,%r12,1)
  mov    %r14d,%r12d
  call   <putchar@plt>
  jmp    L40
  nopl   0x0(%rax)
L32:
  movslq %r12d,%rax
  jmp    L81
```

assistant: ```c
int uart_read_line(char *str) {
  int n = 0;
  char c;
  while (n < 100 - 1 && (c = getchar()) != '\n' && c != '\r') {
    if (c == 0x7F || c == 0x08) {
      if (n > 0) {
        str[--n] = '\0';
        putchar(c);
        putchar(' ');
        putchar(c);
      }
    } else if (isprint(c)) {
      str[n++] = c;
      putchar(c);
    }
  }
  str[n] = '\0';
  putchar('\n');
  return n;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<cg_enter_and_wait_for_frozen>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0xffffffff,-0xc(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <cg_prepare_for_wait@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L24
  mov    -0x4(%rbp),%eax
  jmp    L50
L24:
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <cg_enter@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L60
  movl   $0x0,-0x8(%rbp)
  jmp    L10
L68:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <cg_wait_for@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L21
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <cg_check_frozen@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  addl   $0x1,-0x8(%rbp)
L10:
  cmpl   $0x9,-0x8(%rbp)
  jle    L68
  jmp    L32
L60:
  nop
  jmp    L32
L21:
  nop
L32:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  mov    -0xc(%rbp),%eax
L50:
  leave
  ret
```

assistant: ```c
int cg_enter_and_wait_for_frozen(const char *cgroup, int pid, bool frozen) {
  int fd, ret = -1;
  int attempts;
  fd = cg_prepare_for_wait(cgroup);
  if (fd < 0)
    return fd;
  ret = cg_enter(cgroup, pid);
  if (ret)
    goto out;
  for (attempts = 0; attempts < 10; attempts++) {
    ret = cg_wait_for(fd);
    if (ret)
      break;
    ret = cg_check_frozen(cgroup, frozen);
    if (ret)
      continue;
  }
out:
  close(fd);
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cg_enter_and_wait_for_frozen>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %esi,%ebx
  mov    %edx,%r14d
  call   <cg_prepare_for_wait@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  js     L2
  mov    %ebx,%esi
  mov    %r13,%rdi
  call   <cg_enter@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L37
  mov    $0xa,%r12d
L29:
  mov    %ebp,%edi
  call   <cg_wait_for@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L37
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <cg_check_frozen@plt>
  mov    %eax,%ebx
  sub    $0x1,%r12d
  jne    L29
L37:
  mov    %ebp,%edi
  call   <close@plt>
  mov    %ebx,%ebp
L2:
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int cg_enter_and_wait_for_frozen(const char *cgroup, int pid, bool frozen) {
  int fd, ret = -1;
  int attempts;
  fd = cg_prepare_for_wait(cgroup);
  if (fd < 0)
    return fd;
  ret = cg_enter(cgroup, pid);
  if (ret)
    goto out;
  for (attempts = 0; attempts < 10; attempts++) {
    ret = cg_wait_for(fd);
    if (ret)
      break;
    ret = cg_check_frozen(cgroup, frozen);
    if (ret)
      continue;
  }
out:
  close(fd);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cg_enter_and_wait_for_frozen>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  call   <cg_prepare_for_wait@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  js     L10
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <cg_enter@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L27
  mov    $0xa,%ebp
  nopl   (%rax)
L5:
  mov    %r12d,%edi
  call   <cg_wait_for@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L27
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <cg_check_frozen@plt>
  mov    %eax,%ebx
  sub    $0x1,%ebp
  jne    L5
L27:
  mov    %r12d,%edi
  mov    %ebx,%r12d
  call   <close@plt>
L10:
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int cg_enter_and_wait_for_frozen(const char *cgroup, int pid, bool frozen) {
  int fd, ret = -1;
  int attempts;
  fd = cg_prepare_for_wait(cgroup);
  if (fd < 0)
    return fd;
  ret = cg_enter(cgroup, pid);
  if (ret)
    goto out;
  for (attempts = 0; attempts < 10; attempts++) {
    ret = cg_wait_for(fd);
    if (ret)
      break;
    ret = cg_check_frozen(cgroup, frozen);
    if (ret)
      continue;
  }
out:
  close(fd);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<cg_enter_and_wait_for_frozen>:
  endbr64
  push   %r14
  mov    %edx,%r14d
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  call   <cg_prepare_for_wait@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  js     L0
  mov    %ebp,%esi
  mov    %r13,%rdi
  call   <cg_enter@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L33
  mov    $0xa,%ebp
  nopl   (%rax)
L19:
  mov    %r12d,%edi
  call   <cg_wait_for@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jne    L33
  mov    %r14d,%esi
  mov    %r13,%rdi
  call   <cg_check_frozen@plt>
  mov    %eax,%ebx
  sub    $0x1,%ebp
  jne    L19
L33:
  mov    %r12d,%edi
  mov    %ebx,%r12d
  call   <close@plt>
L0:
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int cg_enter_and_wait_for_frozen(const char *cgroup, int pid, bool frozen) {
  int fd, ret = -1;
  int attempts;
  fd = cg_prepare_for_wait(cgroup);
  if (fd < 0)
    return fd;
  ret = cg_enter(cgroup, pid);
  if (ret)
    goto out;
  for (attempts = 0; attempts < 10; attempts++) {
    ret = cg_wait_for(fd);
    if (ret)
      break;
    ret = cg_check_frozen(cgroup, frozen);
    if (ret)
      continue;
  }
out:
  close(fd);
  return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<subtract>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rdx
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movsd  -0x20(%rbp),%xmm0
  movsd  -0x18(%rbp),%xmm1
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mov    -0x20(%rbp),%rax
  movsd  -0x10(%rbp),%xmm1
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
data_type: string
name: disassemble_data
arguments: data_tag: D39
data_type: string
name: disassemble_data
arguments: data_tag: D9
data_type: string
name: disassemble_data
arguments: data_tag: D16
data_type: string

tool: D3: Subtraction!
 please input two numbers:

tool: D39: %lf %lf

tool: D9: %lf - %lf = %lf


tool: D16: ===========================================

assistant: ```c
int subtract() {
  double num1, num2, sub;
  printf("Subtraction!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  sub = num1 - num2;
  printf("%lf - %lf = %lf\n", num1, num2, sub);
  printf("===========================================\n");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<subtract>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D37(%rip),%rdi
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm0
  movsd  0x10(%rsp),%xmm1
  movapd %xmm0,%xmm2
  subsd  %xmm1,%xmm2
  lea    D32(%rip),%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D4
data_type: string
name: extract_symbolic_data
arguments: data_tag: D37
data_type: string
name: extract_symbolic_data
arguments: data_tag: D32
data_type: string
name: extract_symbolic_data
arguments: data_tag: D24
data_type: string

tool: D4: Subtraction!
 please input two numbers:

tool: D37: %lf %lf

tool: D32: %lf - %lf = %lf


tool: D24: ===========================================

assistant: ```c
int subtract() {
  double num1, num2, sub;
  printf("Subtraction!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  sub = num1 - num2;
  printf("%lf - %lf = %lf\n", num1, num2, sub);
  printf("===========================================\n");
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<subtract>:
  endbr64
  sub    $0x28,%rsp
  lea    D16(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D22(%rip),%rdi
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm0
  movsd  0x10(%rsp),%xmm1
  lea    D4(%rip),%rdi
  mov    $0x3,%eax
  movapd %xmm0,%xmm2
  subsd  %xmm1,%xmm2
  call   <printf@plt>
  lea    D2(%rip),%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D16
type: string
name: read_data
arguments: data_tag: D22
type: string
name: read_data
arguments: data_tag: D4
type: string
name: read_data
arguments: data_tag: D2
type: string

tool: D16: Subtraction!
 please input two numbers:

tool: D22: %lf %lf

tool: D4: %lf - %lf = %lf


tool: D2: ===========================================

assistant: ```c
int subtract() {
  double num1, num2, sub;
  printf("Subtraction!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  sub = num1 - num2;
  printf("%lf - %lf = %lf\n", num1, num2, sub);
  printf("===========================================\n");
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<subtract>:
  endbr64
  sub    $0x28,%rsp
  lea    D3(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D5(%rip),%rdi
  call   <scanf@plt>
  movsd  0x8(%rsp),%xmm0
  movsd  0x10(%rsp),%xmm1
  lea    D28(%rip),%rdi
  mov    $0x3,%eax
  movapd %xmm0,%xmm2
  subsd  %xmm1,%xmm2
  call   <printf@plt>
  lea    D12(%rip),%rdi
  call   <puts@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D3
type_infered: string
name: parse_data
arguments: tag: D5
type_infered: string
name: parse_data
arguments: tag: D28
type_infered: string
name: parse_data
arguments: tag: D12
type_infered: string

tool: D3: Subtraction!
 please input two numbers:

tool: D5: %lf %lf

tool: D28: %lf - %lf = %lf


tool: D12: ===========================================

assistant: ```c
int subtract() {
  double num1, num2, sub;
  printf("Subtraction!\n please input two numbers:");
  scanf("%lf %lf", &num1, &num2);
  sub = num1 - num2;
  printf("%lf - %lf = %lf\n", num1, num2, sub);
  printf("===========================================\n");
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<resid>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  movsd  %xmm0,-0x50(%rbp)
  movsd  %xmm1,-0x58(%rbp)
  movsd  -0x58(%rbp),%xmm0
  mov    -0x50(%rbp),%rsi
  mov    -0x48(%rbp),%rdx
  mov    -0x40(%rbp),%rcx
  mov    -0x38(%rbp),%rax
  movapd %xmm0,%xmm1
  movq   %rsi,%xmm0
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <sompxy@plt>
  movq   %xmm0,%rbx
  mov    -0x48(%rbp),%rdx
  mov    -0x40(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <sompx@plt>
  movapd %xmm0,%xmm1
  movq   %rbx,%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x20(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x58(%rbp),%xmm1
  movsd  -0x50(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movq   $0x0,-0x30(%rbp)
  jmp    L26
L1:
  movq   $0x0,-0x28(%rbp)
  jmp    L3
L2:
  mov    -0x30(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x30(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mulsd  -0x20(%rbp),%xmm1
  addsd  -0x18(%rbp),%xmm1
  mov    -0x30(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%rax
  shl    $0x3,%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addq   $0x1,-0x28(%rbp)
L3:
  mov    -0x28(%rbp),%rax
  cmp    -0x30(%rbp),%rax
  jle    L2
  addq   $0x1,-0x30(%rbp)
L26:
  mov    -0x30(%rbp),%rax
  cmp    -0x48(%rbp),%rax
  jl     L1
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void resid(double **a, double **b, long stop, double lmoyA, double lmoyB) {
  double coef_b;
  double coef_a;
  long i, j;
  coef_b = sompxy(a, b, stop, lmoyA, lmoyB) / sompx(b, stop);
  coef_a = lmoyA - (coef_b * lmoyB);
  for (i = 0; i < stop; i++)
    for (j = 0; j <= i; j++)
      a[i][j] = a[i][j] - (coef_a + (coef_b * b[i][j]));
}
```