ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<sortp>:
  endbr64
  cmp    $0x1,%esi
  jle    L25
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  movslq %esi,%r14
  mov    $0x0,%ebp
  mov    $0x0,%r15d
  jmp    L29
L1:
  mov    %rax,%rbx
L30:
  movslq %ebp,%rax
  mov    (%r12,%rax,8),%rsi
  mov    (%r12,%rbx,8),%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  cmovg  %ebx,%ebp
  lea    0x1(%rbx),%rax
  cmp    %rbx,%r13
  jne    L1
L55:
  movslq %ebp,%rax
  lea    (%r12,%rax,8),%rax
  mov    (%rax),%rdx
  mov    -0x8(%r12,%r14,8),%rcx
  mov    %rcx,(%rax)
  mov    %rdx,-0x8(%r12,%r14,8)
  sub    $0x1,%r14
  cmp    $0x1,%r14d
  jle    L33
L29:
  lea    -0x1(%r14),%r13d
  mov    %r15,%rbx
  test   %r14d,%r14d
  jg     L30
  jmp    L55
L33:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L25:
  ret
```

assistant: ```c
void sortp(char *str[], int n) {
  int maxpos = 0, i, size;
  char *tmp;
  for (size = n; size > 1; size--) {
    for (i = 0; i < size; i++)
      if (strcmp(str[i], str[maxpos]) > 0)
        maxpos = i;
    tmp = str[maxpos];
    str[maxpos] = str[size - 1];
    str[size - 1] = tmp;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<sortp>:
  endbr64
  cmp    $0x1,%esi
  jle    L14
  push   %r15
  mov    %rdi,%r15
  push   %r14
  movslq %esi,%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
L29:
  mov    %r15,%r12
  xor    %ebp,%ebp
  nopl   (%rax)
L25:
  movslq %r13d,%rax
  mov    (%r12),%rdi
  lea    (%r15,%rax,8),%rbx
  mov    (%rbx),%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  cmovg  %r12,%rbx
  cmovg  %ebp,%r13d
  add    $0x1,%ebp
  add    $0x8,%r12
  cmp    %r14d,%ebp
  jl     L25
  mov    -0x8(%r15,%r14,8),%rdx
  mov    (%rbx),%rax
  mov    %rdx,(%rbx)
  mov    %rax,-0x8(%r15,%r14,8)
  sub    $0x1,%r14
  cmp    $0x1,%r14d
  jg     L29
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L14:
  ret
```

assistant: ```c
void sortp(char *str[], int n) {
  int maxpos = 0, i, size;
  char *tmp;
  for (size = n; size > 1; size--) {
    for (i = 0; i < size; i++)
      if (strcmp(str[i], str[maxpos]) > 0)
        maxpos = i;
    tmp = str[maxpos];
    str[maxpos] = str[size - 1];
    str[size - 1] = tmp;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<sortp>:
  endbr64
  cmp    $0x1,%esi
  jle    L12
  push   %r15
  mov    %rdi,%r15
  push   %r14
  movslq %esi,%r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
L16:
  mov    %r15,%r12
  xor    %ebp,%ebp
  nopl   (%rax)
L26:
  movslq %r13d,%rax
  mov    (%r12),%rdi
  lea    (%r15,%rax,8),%rbx
  mov    (%rbx),%rsi
  call   <strcmp@plt>
  test   %rax,%rax
  cmovg  %r12,%rbx
  cmovg  %ebp,%r13d
  add    $0x1,%ebp
  add    $0x8,%r12
  cmp    %r14d,%ebp
  jl     L26
  mov    -0x8(%r15,%r14,8),%rdx
  mov    (%rbx),%rax
  mov    %rdx,(%rbx)
  mov    %rax,-0x8(%r15,%r14,8)
  sub    $0x1,%r14
  cmp    $0x1,%r14d
  jg     L16
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  ret
```

assistant: ```c
void sortp(char *str[], int n) {
  int maxpos = 0, i, size;
  char *tmp;
  for (size = n; size > 1; size--) {
    for (i = 0; i < size; i++)
      if (strcmp(str[i], str[maxpos]) > 0)
        maxpos = i;
    tmp = str[maxpos];
    str[maxpos] = str[size - 1];
    str[size - 1] = tmp;
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  jmp    L8
L26:
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <dump_str@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <dump_str@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <sum@plt>
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <dump_str@plt>
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <dump_str@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  lea    D59(%rip),%rax
  mov    %rax,%rdi
  call   <dump_str@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <fac@plt>
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <dump_str@plt>
  addl   $0x1,-0x4(%rbp)
L8:
  cmpl   $0x9,-0x4(%rbp)
  jle    L26
L29:
  jmp    L29
```

tool_calls: name: unpack_data
arguments: tag: D4
type_infered: string
name: unpack_data
arguments: tag: D59
type_infered: string
name: unpack_data
arguments: tag: D50
type_infered: string
name: unpack_data
arguments: tag: D34
type_infered: string

tool: D4: sum(

tool: D59: ) = 

tool: D50: 


tool: D34: fac(

assistant: ```c
void main(void) {
  uint32_t i;
  for (i = 0; i < 10; i++) {
    dump_str("sum(");
    dump_uint32_t(i);
    dump_str(") = ");
    dump_uint32_t(sum(i));
    dump_str("\n");
    dump_str("fac(");
    dump_uint32_t(i);
    dump_str(") = ");
    dump_uint32_t(fac(i));
    dump_str("\n");
  }
  for (;;)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%ebx
  lea    D11(%rip),%r14
  lea    D34(%rip),%r12
  lea    D10(%rip),%rbp
  lea    D39(%rip),%r13
L3:
  mov    %r14,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <dump_uint32_t@plt>
  mov    %r12,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <sum@plt>
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  mov    %rbp,%rdi
  call   <dump_str@plt>
  mov    %r13,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <dump_uint32_t@plt>
  mov    %r12,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <fac@plt>
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  mov    %rbp,%rdi
  call   <dump_str@plt>
  add    $0x1,%ebx
  cmp    $0xa,%ebx
  jne    L3
L16:
  jmp    L16
```

tool_calls: name: retrieve_data
arguments: label: D11
type: string
name: retrieve_data
arguments: label: D34
type: string
name: retrieve_data
arguments: label: D10
type: string
name: retrieve_data
arguments: label: D39
type: string

tool: D11: sum(

tool: D34: ) = 

tool: D10: 


tool: D39: fac(

assistant: ```c
void main(void) {
  uint32_t i;
  for (i = 0; i < 10; i++) {
    dump_str("sum(");
    dump_uint32_t(i);
    dump_str(") = ");
    dump_uint32_t(sum(i));
    dump_str("\n");
    dump_str("fac(");
    dump_uint32_t(i);
    dump_str(") = ");
    dump_uint32_t(fac(i));
    dump_str("\n");
  }
  for (;;)
    ;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r14
  lea    D22(%rip),%r14
  push   %r13
  lea    D9(%rip),%r13
  push   %r12
  lea    D31(%rip),%r12
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L15:
  mov    %r14,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <dump_uint32_t@plt>
  mov    %r12,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <sum@plt>
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  mov    %rbp,%rdi
  call   <dump_str@plt>
  mov    %r13,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <dump_uint32_t@plt>
  mov    %r12,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <fac@plt>
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  mov    %rbp,%rdi
  call   <dump_str@plt>
  cmp    $0xa,%ebx
  jne    L15
L10:
  jmp    L10
```

tool_calls: name: parse_data
arguments: data_tag: D22
type_infered: string
name: parse_data
arguments: data_tag: D9
type_infered: string
name: parse_data
arguments: data_tag: D31
type_infered: string
name: parse_data
arguments: data_tag: D1
type_infered: string

tool: D22: sum(

tool: D9: fac(

tool: D31: ) = 

tool: D1: 


assistant: ```c
void main(void) {
  uint32_t i;
  for (i = 0; i < 10; i++) {
    dump_str("sum(");
    dump_uint32_t(i);
    dump_str(") = ");
    dump_uint32_t(sum(i));
    dump_str("\n");
    dump_str("fac(");
    dump_uint32_t(i);
    dump_str(") = ");
    dump_uint32_t(fac(i));
    dump_str("\n");
  }
  for (;;)
    ;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  lea    D34(%rip),%r14
  push   %r13
  lea    D25(%rip),%r13
  push   %r12
  lea    D26(%rip),%r12
  push   %rbp
  lea    D16(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %r14,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <dump_uint32_t@plt>
  mov    %r12,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <sum@plt>
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  mov    %rbp,%rdi
  call   <dump_str@plt>
  mov    %r13,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  call   <dump_uint32_t@plt>
  mov    %r12,%rdi
  call   <dump_str@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  call   <fac@plt>
  mov    %eax,%edi
  call   <dump_uint32_t@plt>
  mov    %rbp,%rdi
  call   <dump_str@plt>
  cmp    $0xa,%ebx
  jne    L10
L7:
  jmp    L7
```

tool_calls: name: retrieve_data
arguments: tag: D34
data_type: string
name: retrieve_data
arguments: tag: D25
data_type: string
name: retrieve_data
arguments: tag: D26
data_type: string
name: retrieve_data
arguments: tag: D16
data_type: string

tool: D34: sum(

tool: D25: fac(

tool: D26: ) = 

tool: D16: 


assistant: ```c
void main(void) {
  uint32_t i;
  for (i = 0; i < 10; i++) {
    dump_str("sum(");
    dump_uint32_t(i);
    dump_str(") = ");
    dump_uint32_t(sum(i));
    dump_str("\n");
    dump_str("fac(");
    dump_uint32_t(i);
    dump_str(") = ");
    dump_uint32_t(fac(i));
    dump_str("\n");
  }
  for (;;)
    ;
}
```
user: Translate the below assembly code into C programming language:

```asm
<fd_null_read_n>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    -0x40(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x48(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  jmp    L62
L3:
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rcx
  mov    -0x34(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x24(%rbp)
  cmpl   $0x0,-0x24(%rbp)
  jns    L79
  mov    D10(%rip),%rax
  mov    (%rax),%rdx
  mov    D7(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  jne    L27
  jmp    L62
L27:
  mov    $0xffffffff,%eax
  jmp    L31
L79:
  cmpl   $0x0,-0x24(%rbp)
  je     L99
  mov    -0x24(%rbp),%eax
  cltq
  sub    %rax,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  add    %rax,-0x18(%rbp)
  mov    -0x48(%rbp),%rax
  sub    -0x20(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x10(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memchr@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x48(%rbp),%rax
  sub    -0x20(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  cmp    %rax,-0x8(%rbp)
  ja     L62
  cmpq   $0x0,-0x8(%rbp)
  jne    L25
L62:
  cmpq   $0x0,-0x20(%rbp)
  jne    L3
  jmp    L63
L99:
  nop
  jmp    L63
L25:
  nop
L63:
  mov    -0x48(%rbp),%rax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
L31:
  leave
  ret
```

assistant: ```c
ssize_t fd_null_read_n(int fd, void *buf, size_t n) {
  unsigned char *mp;
  size_t nleft;
  ssize_t nread;
  unsigned char *p;
  unsigned char *q;
  q = p = (unsigned char *)buf;
  nleft = n;
  while (nleft > 0) {
    if ((nread = read(fd, p, nleft)) < 0) {
      if (errno == EINTR)
        continue;
      else
        return (-1);
    } else if (nread == 0) {
      break;
    }
    nleft -= nread;
    p += nread;
    mp = (unsigned char *)memchr(q, '\0', (n - nleft));
    if (mp <= &q[(n - nleft - 1)]) {
      if (mp != NULL)
        break;
    }
  }
  return (n - nleft);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<fd_null_read_n>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%r14
  test   %rdx,%rdx
  je     L99
  mov    %edi,%r13d
  mov    %rsi,%r12
  mov    %rsi,%rbp
  mov    %rdx,%rbx
  jmp    L22
L90:
  mov    D14(%rip),%rdx
  mov    D1(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L21
  mov    $0xffffffff,%eax
  jmp    L13
L73:
  mov    %r15,%rbx
L21:
  test   %rbx,%rbx
  je     L4
L22:
  mov    %rbx,%rdx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <read@plt>
  test   %eax,%eax
  js     L90
  je     L4
  cltq
  mov    %rbx,%r15
  sub    %rax,%r15
  add    %rax,%rbp
  sub    %rbx,%rax
  lea    (%rax,%r14,1),%rbx
  mov    %rbx,%rdx
  mov    $0x0,%esi
  mov    %r12,%rdi
  call   <memchr@plt>
  lea    -0x1(%r12,%rbx,1),%rdx
  cmp    %rax,%rdx
  jb     L73
  test   %rax,%rax
  je     L73
  mov    %r15,%rbx
L4:
  mov    %r14d,%eax
  sub    %ebx,%eax
L13:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L99:
  mov    %rdx,%rbx
  jmp    L4
```

assistant: ```c
ssize_t fd_null_read_n(int fd, void *buf, size_t n) {
  unsigned char *mp;
  size_t nleft;
  ssize_t nread;
  unsigned char *p;
  unsigned char *q;
  q = p = (unsigned char *)buf;
  nleft = n;
  while (nleft > 0) {
    if ((nread = read(fd, p, nleft)) < 0) {
      if (errno == EINTR)
        continue;
      else
        return (-1);
    } else if (nread == 0) {
      break;
    }
    nleft -= nread;
    p += nread;
    mp = (unsigned char *)memchr(q, '\0', (n - nleft));
    if (mp <= &q[(n - nleft - 1)]) {
      if (mp != NULL)
        break;
    }
  }
  return (n - nleft);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fd_null_read_n>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  je     L76
  nopw   0x0(%rax,%rax,1)
L23:
  mov    %r14,%rdx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %eax,%eax
  js     L2
  je     L12
  cltq
  mov    %r14,%r15
  xor    %esi,%esi
  mov    %rbp,%rdi
  sub    %rax,%r15
  add    %rax,%rbx
  sub    %r14,%rax
  lea    (%rax,%r13,1),%r14
  mov    %r14,%rdx
  call   <memchr@plt>
  lea    -0x1(%rbp,%r14,1),%rdx
  cmp    %rax,%rdx
  jb     L55
  test   %rax,%rax
  jne    L3
L55:
  test   %r15,%r15
  je     L76
  mov    %r15,%r14
  jmp    L23
  nopl   0x0(%rax)
L2:
  mov    D9(%rip),%rax
  mov    D17(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L23
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L76:
  xor    %r14d,%r14d
L12:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  sub    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L3:
  mov    %r15d,%r14d
  jmp    L12
```

assistant: ```c
ssize_t fd_null_read_n(int fd, void *buf, size_t n) {
  unsigned char *mp;
  size_t nleft;
  ssize_t nread;
  unsigned char *p;
  unsigned char *q;
  q = p = (unsigned char *)buf;
  nleft = n;
  while (nleft > 0) {
    if ((nread = read(fd, p, nleft)) < 0) {
      if (errno == EINTR)
        continue;
      else
        return (-1);
    } else if (nread == 0) {
      break;
    }
    nleft -= nread;
    p += nread;
    mp = (unsigned char *)memchr(q, '\0', (n - nleft));
    if (mp <= &q[(n - nleft - 1)]) {
      if (mp != NULL)
        break;
    }
  }
  return (n - nleft);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<fd_null_read_n>:
  endbr64
  push   %r15
  push   %r14
  mov    %rdx,%r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %edi,%r12d
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  test   %rdx,%rdx
  je     L74
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %r14,%rdx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <read@plt>
  test   %eax,%eax
  js     L79
  je     L81
  cltq
  mov    %r14,%r15
  xor    %esi,%esi
  mov    %rbp,%rdi
  sub    %rax,%r15
  add    %rax,%rbx
  sub    %r14,%rax
  lea    (%rax,%r13,1),%r14
  mov    %r14,%rdx
  call   <memchr@plt>
  lea    -0x1(%rbp,%r14,1),%rdx
  cmp    %rax,%rdx
  jb     L56
  test   %rax,%rax
  jne    L78
L56:
  test   %r15,%r15
  je     L74
  mov    %r15,%r14
  jmp    L12
  nopl   0x0(%rax)
L79:
  mov    D10(%rip),%rax
  mov    D4(%rip),%rdx
  mov    (%rax),%rax
  cmp    %rax,(%rdx)
  je     L12
  add    $0x8,%rsp
  mov    $0xffffffff,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L74:
  xor    %r14d,%r14d
L81:
  add    $0x8,%rsp
  mov    %r13d,%eax
  pop    %rbx
  sub    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L78:
  mov    %r15d,%r14d
  jmp    L81
```

assistant: ```c
ssize_t fd_null_read_n(int fd, void *buf, size_t n) {
  unsigned char *mp;
  size_t nleft;
  ssize_t nread;
  unsigned char *p;
  unsigned char *q;
  q = p = (unsigned char *)buf;
  nleft = n;
  while (nleft > 0) {
    if ((nread = read(fd, p, nleft)) < 0) {
      if (errno == EINTR)
        continue;
      else
        return (-1);
    } else if (nread == 0) {
      break;
    }
    nleft -= nread;
    p += nread;
    mp = (unsigned char *)memchr(q, '\0', (n - nleft));
    if (mp <= &q[(n - nleft - 1)]) {
      if (mp != NULL)
        break;
    }
  }
  return (n - nleft);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<wfa_double2timeval>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  cvttsd2si %xmm0,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  pxor   %xmm2,%xmm2
  cvtsi2sd %rax,%xmm2
  movsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  subsd  %xmm2,%xmm1
  movsd  D8(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,0x8(%rax)
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
datatype: byte[8]

tool: {"D8": [0x00,0x00,0x00,0x00,0x80,0x84,0x2e,0x41]}

assistant: ```c
void wfa_double2timeval(struct timeval *tval, double dval) {
  tval->tv_sec = (long int)dval;
  tval->tv_usec = (long int)((dval - tval->tv_sec) * 1000000);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<wfa_double2timeval>:
  endbr64
  cvttsd2si %xmm0,%rax
  mov    %rax,(%rdi)
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  subsd  %xmm1,%xmm0
  mulsd  D3(%rip),%xmm0
  cvttsd2si %xmm0,%rax
  mov    %rax,0x8(%rdi)
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_infered: byte[8]

tool: {"D3": [0x00,0x00,0x00,0x00,0x80,0x84,0x2e,0x41]}

assistant: ```c
void wfa_double2timeval(struct timeval *tval, double dval) {
  tval->tv_sec = (long int)dval;
  tval->tv_usec = (long int)((dval - tval->tv_sec) * 1000000);
}
```
user: Translate the below assembly code into C programming language:

```asm
<wfa_double2timeval>:
  endbr64
  cvttsd2si %xmm0,%rax
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  mov    %rax,(%rdi)
  subsd  %xmm1,%xmm0
  mulsd  D0(%rip),%xmm0
  cvttsd2si %xmm0,%rax
  mov    %rax,0x8(%rdi)
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
datatype: byte[8]

tool: {"D0": [0x00,0x00,0x00,0x00,0x80,0x84,0x2e,0x41]}

assistant: ```c
void wfa_double2timeval(struct timeval *tval, double dval) {
  tval->tv_sec = (long int)dval;
  tval->tv_usec = (long int)((dval - tval->tv_sec) * 1000000);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wfa_double2timeval>:
  endbr64
  cvttsd2si %xmm0,%rax
  pxor   %xmm1,%xmm1
  cvtsi2sd %rax,%xmm1
  subsd  %xmm1,%xmm0
  mulsd  D7(%rip),%xmm0
  cvttsd2si %xmm0,%rdx
  movq   %rax,%xmm0
  movq   %rdx,%xmm2
  punpcklqdq %xmm2,%xmm0
  movups %xmm0,(%rdi)
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
type_guessed: byte[8]

tool: {"D7": [0x00,0x00,0x00,0x00,0x80,0x84,0x2e,0x41]}

assistant: ```c
void wfa_double2timeval(struct timeval *tval, double dval) {
  tval->tv_sec = (long int)dval;
  tval->tv_usec = (long int)((dval - tval->tv_sec) * 1000000);
}
```
user: Translate the below assembly code into C programming language:

```asm
<Suma>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x28(%rbp)
  movsd  %xmm1,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
L20:
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x11(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x11(%rbp),%rax
  mov    %rax,%rdi
  call   <validar_es_flotante@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L26
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L26:
  cmpl   $0x0,-0x1c(%rbp)
  je     L20
  lea    -0x11(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtod@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x28(%rbp)
L38:
  lea    D56(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x11(%rbp),%rax
  mov    %rax,%rsi
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x11(%rbp),%rax
  mov    %rax,%rdi
  call   <validar_es_flotante@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L23
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L23:
  cmpl   $0x0,-0x1c(%rbp)
  je     L38
  lea    -0x11(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtod@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x30(%rbp)
  movsd  -0x28(%rbp),%xmm0
  addsd  -0x30(%rbp),%xmm0
  movq   %xmm0,%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D56
type_infered: string
name: disassemble_data
arguments: data_label: D21
type_infered: string
name: disassemble_data
arguments: data_label: D5
type_infered: string

tool: D56: Enter a number: 

tool: D21: %s

tool: D5: Please only integer or floating.


assistant: ```c
double Suma(double a, double b) {
  char buffer[9];
  bool band = false;
  int enteroDummy = 0;
  do {
    printf("Enter a number: ");
    scanf("%s", buffer);
    if ((band = validar_es_flotante(buffer)) == false)
      printf("Please only integer or floating.\n");
  } while (band == false);
  a = strtod(buffer, NULL);
  do {
    printf("Enter a number: ");
    scanf("%s", buffer);
    if ((band = validar_es_flotante(buffer)) == false)
      printf("Please only integer or floating.\n");
  } while (band == false);
  b = strtod(buffer, NULL);
  return (a + b);
}
```