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

```asm
<main>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  cmp    $0x2,%edi
  jle    L12
L4:
  mov    0x8(%r12),%rdi
  call   <getenv@plt>
  mov    (%r12),%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    0x10(%r12),%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    0x8(%r12),%rsi
  lea    D7(%rip),%rdi
  sub    %eax,%ebx
  xor    %eax,%eax
  add    %ebx,%ebx
  movslq %ebx,%rbx
  lea    0x0(%rbp,%rbx,1),%rdx
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    (%rsi),%rsi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L4
```

tool_calls: name: retrieve_data
arguments: data_label: D7
datatype: string
name: retrieve_data
arguments: data_label: D6
datatype: string

tool: D7: %s true memloc %p


tool: D6: USAGE: %s [env_var] [target prog name]


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 3) {
    printf("USAGE: %s [env_var] [target prog name]\n", argv[0]);
    exit(0);
  }
  char *ptr;
  ptr = getenv(argv[1]);
  ptr += (strlen(argv[0]) - strlen(argv[2])) * 2;
  printf("%s true memloc %p\n", argv[1], ptr);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  cmp    $0x2,%edi
  jle    L6
L1:
  mov    0x8(%r12),%rdi
  call   <getenv@plt>
  mov    (%r12),%rdi
  mov    %rax,%rbp
  call   <strlen@plt>
  mov    0x10(%r12),%rdi
  mov    %eax,%ebx
  call   <strlen@plt>
  mov    0x8(%r12),%rsi
  lea    D16(%rip),%rdi
  sub    %eax,%ebx
  xor    %eax,%eax
  add    %ebx,%ebx
  movslq %ebx,%rbx
  lea    0x0(%rbp,%rbx,1),%rdx
  call   <printf@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L6:
  mov    (%rsi),%rsi
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L1
```

tool_calls: name: disassemble_data
arguments: data_label: D16
type_infered: string
name: disassemble_data
arguments: data_label: D6
type_infered: string

tool: D16: %s true memloc %p


tool: D6: USAGE: %s [env_var] [target prog name]


assistant: ```c
int main(int argc, char *argv[]) {
  if (argc < 3) {
    printf("USAGE: %s [env_var] [target prog name]\n", argv[0]);
    exit(0);
  }
  char *ptr;
  ptr = getenv(argv[1]);
  ptr += (strlen(argv[0]) - strlen(argv[2])) * 2;
  printf("%s true memloc %p\n", argv[1], ptr);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<xorshift1024star_32>:
  endbr64
  mov    0x80(%rdi),%eax
  movslq %eax,%rdx
  mov    (%rdi,%rdx,4),%r9d
  mov    0x4(%rdi,%rdx,4),%esi
  add    $0x2,%eax
  and    $0x1f,%eax
  mov    %eax,0x80(%rdi)
  cltq
  shl    $0x2,%rax
  lea    (%rdi,%rax,1),%r8
  mov    (%r8),%r10d
  lea    0x4(%rdi,%rax,1),%rdi
  mov    (%rdi),%eax
  mov    %eax,%ecx
  shl    $0x1f,%ecx
  mov    %r10d,%edx
  sar    %edx
  or     %edx,%ecx
  xor    %eax,%ecx
  mov    %r10d,%edx
  shl    $0x1f,%edx
  xor    %r10d,%edx
  mov    %r9d,%eax
  sar    $0x1e,%eax
  lea    0x0(,%rsi,4),%r10d
  or     %r10d,%eax
  xor    %r9d,%eax
  xor    %edx,%eax
  sar    $0xb,%edx
  mov    %ecx,%r9d
  shl    $0x15,%r9d
  or     %r9d,%edx
  xor    %edx,%eax
  mov    %eax,(%r8)
  mov    %esi,%edx
  sar    $0x1e,%edx
  xor    %esi,%edx
  xor    %ecx,%edx
  sar    $0xb,%ecx
  xor    %ecx,%edx
  mov    %edx,(%rdi)
  movzwl %ax,%ecx
  sar    $0x10,%eax
  imul   $0x5497fdb5,%eax,%eax
  imul   $0x89d45497,%ecx,%esi
  add    %esi,%eax
  imul   $0xfdb5,%ecx,%ecx
  sar    $0x10,%ecx
  add    %ecx,%eax
  imul   $0xfdb50000,%edx,%edx
  add    %edx,%eax
  ret
```

assistant: ```c
uint32_t xorshift1024star_32(uint32_t *s32) {
  int p = s32[32];
  const uint32_t s0l = s32[p];
  const uint32_t s0u = s32[p + 1];
  p = s32[32] = (p + 2) & 31;
  uint32_t s1l = s32[p];
  uint32_t s1u = s32[p + 1];
  s1u ^= (s1u << 31) | (s1l >> 1);
  s1l ^= s1l << 31;
  s32[p] = s1l ^ s0l ^ ((s1l >> 11) | (s1u << 21)) ^ ((s0l >> 30) | (s0u << 2));
  s32[p + 1] = s1u ^ s0u ^ (s1u >> 11) ^ (s0u >> 30);
  uint32_t lo = s32[p] & 0xFFFF;
  return (lo * 64949 >> 16) + lo * 2312393879 + (s32[p] >> 16) * 1419247029 + s32[p + 1] * 4256497664;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<xorshift1024star_32>:
  endbr64
  movslq 0x80(%rdi),%rdx
  mov    %rdx,%rax
  mov    0x4(%rdi,%rdx,4),%esi
  mov    (%rdi,%rdx,4),%r9d
  add    $0x2,%eax
  and    $0x1f,%eax
  mov    %eax,0x80(%rdi)
  cltq
  shl    $0x2,%rax
  lea    (%rdi,%rax,1),%r8
  lea    0x4(%rdi,%rax,1),%rdi
  mov    (%r8),%r10d
  mov    (%rdi),%eax
  mov    %r10d,%edx
  mov    %eax,%ecx
  sar    %edx
  shl    $0x1f,%ecx
  or     %edx,%ecx
  mov    %r10d,%edx
  xor    %eax,%ecx
  shl    $0x1f,%edx
  mov    %r9d,%eax
  xor    %r10d,%edx
  sar    $0x1e,%eax
  lea    0x0(,%rsi,4),%r10d
  or     %r10d,%eax
  xor    %r9d,%eax
  mov    %ecx,%r9d
  xor    %edx,%eax
  shl    $0x15,%r9d
  sar    $0xb,%edx
  or     %r9d,%edx
  xor    %edx,%eax
  mov    %esi,%edx
  sar    $0x1e,%edx
  mov    %eax,(%r8)
  xor    %esi,%edx
  xor    %ecx,%edx
  sar    $0xb,%ecx
  xor    %ecx,%edx
  movzwl %ax,%ecx
  sar    $0x10,%eax
  imul   $0x89d45497,%ecx,%esi
  mov    %edx,(%rdi)
  imul   $0x5497fdb5,%eax,%eax
  imul   $0xfdb5,%ecx,%ecx
  imul   $0xfdb50000,%edx,%edx
  add    %esi,%eax
  sar    $0x10,%ecx
  add    %ecx,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
uint32_t xorshift1024star_32(uint32_t *s32) {
  int p = s32[32];
  const uint32_t s0l = s32[p];
  const uint32_t s0u = s32[p + 1];
  p = s32[32] = (p + 2) & 31;
  uint32_t s1l = s32[p];
  uint32_t s1u = s32[p + 1];
  s1u ^= (s1u << 31) | (s1l >> 1);
  s1l ^= s1l << 31;
  s32[p] = s1l ^ s0l ^ ((s1l >> 11) | (s1u << 21)) ^ ((s0l >> 30) | (s0u << 2));
  s32[p + 1] = s1u ^ s0u ^ (s1u >> 11) ^ (s0u >> 30);
  uint32_t lo = s32[p] & 0xFFFF;
  return (lo * 64949 >> 16) + lo * 2312393879 + (s32[p] >> 16) * 1419247029 + s32[p + 1] * 4256497664;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<xorshift1024star_32>:
  endbr64
  movslq 0x80(%rdi),%rdx
  mov    %rdx,%rax
  mov    0x4(%rdi,%rdx,4),%esi
  mov    (%rdi,%rdx,4),%r10d
  add    $0x2,%eax
  and    $0x1f,%eax
  mov    %eax,0x80(%rdi)
  cltq
  shl    $0x2,%rax
  lea    (%rdi,%rax,1),%r8
  mov    0x4(%rdi,%rax,1),%eax
  lea    0x0(,%rsi,4),%edi
  mov    (%r8),%r9d
  mov    %eax,%ecx
  mov    %r9d,%edx
  shl    $0x1f,%ecx
  sar    %edx
  or     %edx,%ecx
  mov    %r9d,%edx
  xor    %eax,%ecx
  mov    %r10d,%eax
  shl    $0x1f,%edx
  sar    $0x1e,%eax
  xor    %r9d,%edx
  or     %edi,%eax
  mov    %ecx,%edi
  xor    %r10d,%eax
  shl    $0x15,%edi
  xor    %edx,%eax
  sar    $0xb,%edx
  or     %edi,%edx
  xor    %edx,%eax
  mov    %esi,%edx
  sar    $0x1e,%edx
  movd   %eax,%xmm0
  xor    %esi,%edx
  xor    %ecx,%edx
  sar    $0xb,%ecx
  xor    %ecx,%edx
  movzwl %ax,%ecx
  sar    $0x10,%eax
  imul   $0x89d45497,%ecx,%esi
  movd   %edx,%xmm1
  imul   $0x5497fdb5,%eax,%eax
  punpckldq %xmm1,%xmm0
  imul   $0xfdb5,%ecx,%ecx
  movq   %xmm0,(%r8)
  imul   $0xfdb50000,%edx,%edx
  add    %esi,%eax
  sar    $0x10,%ecx
  add    %ecx,%eax
  add    %edx,%eax
  ret
```

assistant: ```c
uint32_t xorshift1024star_32(uint32_t *s32) {
  int p = s32[32];
  const uint32_t s0l = s32[p];
  const uint32_t s0u = s32[p + 1];
  p = s32[32] = (p + 2) & 31;
  uint32_t s1l = s32[p];
  uint32_t s1u = s32[p + 1];
  s1u ^= (s1u << 31) | (s1l >> 1);
  s1l ^= s1l << 31;
  s32[p] = s1l ^ s0l ^ ((s1l >> 11) | (s1u << 21)) ^ ((s0l >> 30) | (s0u << 2));
  s32[p + 1] = s1u ^ s0u ^ (s1u >> 11) ^ (s0u >> 30);
  uint32_t lo = s32[p] & 0xFFFF;
  return (lo * 64949 >> 16) + lo * 2312393879 + (s32[p] >> 16) * 1419247029 + s32[p + 1] * 4256497664;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<read_next_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    $0xa,%edi
  call   <ft_strnew@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0xa,-0x14(%rbp)
  jmp    L17
L36:
  mov    -0x10(%rbp),%rcx
  mov    -0x2c(%rbp),%eax
  mov    $0xa,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0xffffffff,-0x14(%rbp)
  jne    L27
  mov    $0xffffffff,%eax
  jmp    L0
L27:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ft_strjoin@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <ft_strdup@plt>
  mov    -0x28(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L17:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <ft_strchr@plt>
  test   %rax,%rax
  jne    L45
  cmpl   $0xa,-0x14(%rbp)
  je     L36
L45:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x14(%rbp),%eax
L0:
  leave
  ret
```

assistant: ```c
int read_next_line(char **n_ret, int fd) {
  char *buf;
  int nb_read;
  char *temp;
  buf = ft_strnew(10);
  nb_read = 10;
  while (ft_strchr(*n_ret, '\n') == NULL && nb_read == 10) {
    if ((nb_read = read(fd, buf, 10)) == -1)
      return (-1);
    buf[nb_read] = '\0';
    temp = ft_strjoin(*n_ret, buf);
    free(*n_ret);
    *n_ret = ft_strdup(temp);
    free(temp);
  }
  free(buf);
  return (nb_read);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<read_next_line>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %esi,%r14d
  mov    $0xa,%edi
  call   <ft_strnew@plt>
  mov    %rax,%r13
  mov    $0xa,%ebx
  jmp    L42
L5:
  cltq
  movb   $0x0,0x0(%r13,%rax,1)
  mov    %r13,%rsi
  mov    (%r12),%rdi
  call   <ft_strjoin@plt>
  mov    %rax,%rbp
  mov    (%r12),%rdi
  call   <free@plt>
  mov    %rbp,%rdi
  call   <ft_strdup@plt>
  mov    %rax,(%r12)
  mov    %rbp,%rdi
  call   <free@plt>
L42:
  mov    $0xa,%esi
  mov    (%r12),%rdi
  call   <ft_strchr@plt>
  cmp    $0xa,%ebx
  jne    L9
  test   %rax,%rax
  jne    L9
  mov    $0xa,%edx
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <read@plt>
  mov    %eax,%ebx
  cmp    $0xffffffff,%eax
  jne    L5
  jmp    L47
L9:
  mov    %r13,%rdi
  call   <free@plt>
L47:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int read_next_line(char **n_ret, int fd) {
  char *buf;
  int nb_read;
  char *temp;
  buf = ft_strnew(10);
  nb_read = 10;
  while (ft_strchr(*n_ret, '\n') == NULL && nb_read == 10) {
    if ((nb_read = read(fd, buf, 10)) == -1)
      return (-1);
    buf[nb_read] = '\0';
    temp = ft_strjoin(*n_ret, buf);
    free(*n_ret);
    *n_ret = ft_strdup(temp);
    free(temp);
  }
  free(buf);
  return (nb_read);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_next_line>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    $0xa,%r12d
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0xa,%edi
  call   <ft_strnew@plt>
  mov    %rax,%r13
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L20:
  test   %rax,%rax
  jne    L24
  mov    $0xa,%edx
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <read@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L25
  cltq
  mov    %r13,%rsi
  movb   $0x0,0x0(%r13,%rax,1)
  mov    (%rbx),%rdi
  call   <ft_strjoin@plt>
  mov    (%rbx),%rdi
  mov    %rax,%rbp
  call   <free@plt>
  mov    %rbp,%rdi
  call   <ft_strdup@plt>
  mov    %rbp,%rdi
  mov    %rax,(%rbx)
  call   <free@plt>
L18:
  mov    (%rbx),%rdi
  mov    $0xa,%esi
  call   <ft_strchr@plt>
  cmp    $0xa,%r12d
  je     L20
L24:
  mov    %r13,%rdi
  call   <free@plt>
L25:
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int read_next_line(char **n_ret, int fd) {
  char *buf;
  int nb_read;
  char *temp;
  buf = ft_strnew(10);
  nb_read = 10;
  while (ft_strchr(*n_ret, '\n') == NULL && nb_read == 10) {
    if ((nb_read = read(fd, buf, 10)) == -1)
      return (-1);
    buf[nb_read] = '\0';
    temp = ft_strjoin(*n_ret, buf);
    free(*n_ret);
    *n_ret = ft_strdup(temp);
    free(temp);
  }
  free(buf);
  return (nb_read);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<read_next_line>:
  endbr64
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    $0xa,%r12d
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    $0xa,%edi
  call   <ft_strnew@plt>
  mov    %rax,%r13
  jmp    L30
  nopw   0x0(%rax,%rax,1)
L28:
  test   %rax,%rax
  jne    L2
  mov    $0xa,%edx
  mov    %r13,%rsi
  mov    %r14d,%edi
  call   <read@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L13
  cltq
  mov    %r13,%rsi
  movb   $0x0,0x0(%r13,%rax,1)
  mov    (%rbx),%rdi
  call   <ft_strjoin@plt>
  mov    (%rbx),%rdi
  mov    %rax,%rbp
  call   <free@plt>
  mov    %rbp,%rdi
  call   <ft_strdup@plt>
  mov    %rbp,%rdi
  mov    %rax,(%rbx)
  call   <free@plt>
L30:
  mov    (%rbx),%rdi
  mov    $0xa,%esi
  call   <ft_strchr@plt>
  cmp    $0xa,%r12d
  je     L28
L2:
  mov    %r13,%rdi
  call   <free@plt>
L13:
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int read_next_line(char **n_ret, int fd) {
  char *buf;
  int nb_read;
  char *temp;
  buf = ft_strnew(10);
  nb_read = 10;
  while (ft_strchr(*n_ret, '\n') == NULL && nb_read == 10) {
    if ((nb_read = read(fd, buf, 10)) == -1)
      return (-1);
    buf[nb_read] = '\0';
    temp = ft_strjoin(*n_ret, buf);
    free(*n_ret);
    *n_ret = ft_strdup(temp);
    free(temp);
  }
  free(buf);
  return (nb_read);
}
```
user: What is the c source code of the assembly code below:

```asm
<ft_strsub>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movq   $0x0,-0x10(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L49
  movq   $0x0,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L41
  mov    $0x0,%eax
  jmp    L8
L45:
  mov    -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addq   $0x1,-0x10(%rbp)
L41:
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L45
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
  jmp    L8
L49:
  mov    $0x0,%eax
L8:
  leave
  ret
```

assistant: ```c
char *ft_strsub(char const *s, unsigned int start, size_t len) {
  char *new;
  size_t i;
  i = 0;
  if (s != NULL) {
    new = NULL;
    new = (char *)malloc(sizeof(char) * len + 1);
    if (new == NULL)
      return (NULL);
    while (i < len) {
      new[i] = s[i + start];
      i++;
    }
    new[i] = '\0';
    return (new);
  }
  return (NULL);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_strsub>:
  endbr64
  test   %rdi,%rdi
  je     L25
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %rdx,%rbp
  lea    0x1(%rdx),%edi
  call   <malloc@plt>
  mov    $0x0,%r8d
  test   %rax,%rax
  je     L33
  mov    %rax,%r8
  test   %rbp,%rbp
  je     L26
  lea    (%rax,%rbp,1),%rdi
  mov    %rax,%rdx
  mov    %r12d,%r12d
  sub    %rax,%r12
  lea    (%rbx,%r12,1),%rsi
L4:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rdx)
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L4
L26:
  movb   $0x0,(%rax,%rbp,1)
L33:
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L25:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *ft_strsub(char const *s, unsigned int start, size_t len) {
  char *new;
  size_t i;
  i = 0;
  if (s != NULL) {
    new = NULL;
    new = (char *)malloc(sizeof(char) * len + 1);
    if (new == NULL)
      return (NULL);
    while (i < len) {
      new[i] = s[i + start];
      i++;
    }
    new[i] = '\0';
    return (new);
  }
  return (NULL);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_strsub>:
  endbr64
  test   %rdi,%rdi
  je     L25
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rdx),%edi
  push   %rbx
  mov    %rdx,%rbx
  call   <malloc@plt>
  test   %rax,%rax
  je     L48
  test   %rbx,%rbx
  je     L8
  sub    %rax,%r12
  lea    (%rbx,%rax,1),%rdx
  mov    %rax,%rcx
  lea    0x0(%rbp,%r12,1),%rdi
  nopw   0x0(%rax,%rax,1)
L9:
  movzbl (%rdi,%rcx,1),%esi
  add    $0x1,%rcx
  mov    %sil,-0x1(%rcx)
  cmp    %rdx,%rcx
  jne    L9
L27:
  movb   $0x0,(%rcx)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L48:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  mov    %rax,%rcx
  jmp    L27
  nopl   (%rax)
L25:
  xor    %eax,%eax
  ret
```

assistant: ```c
char *ft_strsub(char const *s, unsigned int start, size_t len) {
  char *new;
  size_t i;
  i = 0;
  if (s != NULL) {
    new = NULL;
    new = (char *)malloc(sizeof(char) * len + 1);
    if (new == NULL)
      return (NULL);
    while (i < len) {
      new[i] = s[i + start];
      i++;
    }
    new[i] = '\0';
    return (new);
  }
  return (NULL);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<ft_strsub>:
  endbr64
  test   %rdi,%rdi
  je     L166
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x1(%rdx),%edi
  push   %rbx
  mov    %rdx,%rbx
  call   <malloc@plt>
  xor    %r8d,%r8d
  test   %rax,%rax
  je     L66
  mov    %rax,%r8
  test   %rbx,%rbx
  je     L167
  mov    %r12d,%edi
  mov    %rax,%rcx
  lea    -0x1(%rbx),%rdx
  lea    0x1(%rbp,%rdi,1),%rsi
  sub    %rsi,%rcx
  cmp    $0xe,%rcx
  jbe    L107
  cmp    $0x6,%rdx
  jbe    L107
  mov    %rbx,%rsi
  cmp    $0xe,%rdx
  jbe    L191
  mov    %rdi,%rcx
  and    $0xfffffffffffffff0,%rsi
  mov    %rax,%rdx
  sub    %rax,%rcx
  add    %rax,%rsi
  add    %rbp,%rcx
L89:
  movdqu (%rdx,%rcx,1),%xmm0
  add    $0x10,%rdx
  movups %xmm0,-0x10(%rdx)
  cmp    %rdx,%rsi
  jne    L89
  mov    %rbx,%rdx
  and    $0xfffffffffffffff0,%rdx
  test   $0xf,%bl
  je     L154
  mov    %rbx,%rsi
  sub    %rdx,%rsi
  lea    -0x1(%rsi),%rcx
  cmp    $0x6,%rcx
  jbe    L158
L203:
  lea    0x0(%rbp,%rdi,1),%rcx
  mov    (%rcx,%rdx,1),%rdi
  mov    %rdi,(%rax,%rdx,1)
  mov    %rsi,%rdi
  and    $0xfffffffffffffff8,%rdi
  add    %rdi,%rdx
  cmp    %rdi,%rsi
  je     L154
L9:
  movzbl (%rcx,%rdx,1),%esi
  mov    %sil,(%rax,%rdx,1)
  lea    0x1(%rdx),%rsi
  cmp    %rsi,%rbx
  jbe    L154
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,0x1(%rax,%rdx,1)
  lea    0x2(%rdx),%rsi
  cmp    %rsi,%rbx
  jbe    L154
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,0x2(%rax,%rdx,1)
  lea    0x3(%rdx),%rsi
  cmp    %rsi,%rbx
  jbe    L154
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,0x3(%rax,%rdx,1)
  lea    0x4(%rdx),%rsi
  cmp    %rsi,%rbx
  jbe    L154
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,0x4(%rax,%rdx,1)
  lea    0x5(%rdx),%rsi
  cmp    %rsi,%rbx
  jbe    L154
  movzbl (%rcx,%rsi,1),%esi
  mov    %sil,0x5(%rax,%rdx,1)
  lea    0x6(%rdx),%rsi
  cmp    %rsi,%rbx
  jbe    L154
  movzbl (%rcx,%rsi,1),%ecx
  mov    %cl,0x6(%rax,%rdx,1)
L154:
  add    %rax,%rbx
L179:
  movb   $0x0,(%rbx)
L66:
  pop    %rbx
  mov    %r8,%rax
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L167:
  mov    %rax,%rbx
  jmp    L179
  nopl   (%rax)
L166:
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L107:
  mov    %rdi,%rcx
  add    %rax,%rbx
  sub    %rax,%rcx
  lea    0x0(%rbp,%rcx,1),%rdx
  cs nopw 0x0(%rax,%rax,1)
L69:
  movzbl (%rdx,%rax,1),%ecx
  add    $0x1,%rax
  mov    %cl,-0x1(%rax)
  cmp    %rbx,%rax
  jne    L69
  jmp    L179
L191:
  xor    %edx,%edx
  jmp    L203
L158:
  lea    0x0(%rbp,%rdi,1),%rcx
  jmp    L9
```

assistant: ```c
char *ft_strsub(char const *s, unsigned int start, size_t len) {
  char *new;
  size_t i;
  i = 0;
  if (s != NULL) {
    new = NULL;
    new = (char *)malloc(sizeof(char) * len + 1);
    if (new == NULL)
      return (NULL);
    while (i < len) {
      new[i] = s[i + start];
      i++;
    }
    new[i] = '\0';
    return (new);
  }
  return (NULL);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<showprice>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x28(%rbp)
  mov    %edi,-0x2c(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  D22(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x1,-0x14(%rbp)
  jmp    L4
L7:
  movsd  -0x28(%rbp),%xmm0
  mulsd  -0x8(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
  movsd  -0x8(%rbp),%xmm1
  movsd  D1(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
L4:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jle    L7
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D22
type_infered: f64
name: unpack_data
arguments: data_tag: D1
type_infered: f64
name: unpack_data
arguments: data_tag: D11
type_infered: string

tool: D22: 1

tool: D1: 0.95

tool: D11: The total cost will be $%0.2f


assistant: ```c
void showprice(double rate, int nights) {
  int n;
  double total = 0.0, factor = 1.0;
  for (n = 1; n <= nights; n++, factor *= 0.95) {
    total += rate * factor;
  }
  printf("The total cost will be $%0.2f\n", total);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<showprice>:
  endbr64
  sub    $0x8,%rsp
  test   %edi,%edi
  jle    L5
  movapd %xmm0,%xmm3
  add    $0x1,%edi
  movsd  D9(%rip),%xmm1
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  movsd  D3(%rip),%xmm4
L17:
  movapd %xmm3,%xmm2
  mulsd  %xmm1,%xmm2
  addsd  %xmm2,%xmm0
  add    $0x1,%eax
  mulsd  %xmm4,%xmm1
  cmp    %edi,%eax
  jne    L17
L6:
  lea    D16(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
L5:
  pxor   %xmm0,%xmm0
  jmp    L6
```

tool_calls: name: parse_data
arguments: data_tag: D9
type: f64
name: parse_data
arguments: data_tag: D3
type: f64
name: parse_data
arguments: data_tag: D16
type: string

tool: D9: 1

tool: D3: 0.95

tool: D16: The total cost will be $%0.2f


assistant: ```c
void showprice(double rate, int nights) {
  int n;
  double total = 0.0, factor = 1.0;
  for (n = 1; n <= nights; n++, factor *= 0.95) {
    total += rate * factor;
  }
  printf("The total cost will be $%0.2f\n", total);
}
```
user: Translate the below assembly code into C programming language:

```asm
<showprice>:
  endbr64
  movapd %xmm0,%xmm3
  test   %edi,%edi
  jle    L9
  movsd  D30(%rip),%xmm1
  add    $0x1,%edi
  pxor   %xmm0,%xmm0
  mov    $0x1,%eax
  movsd  D33(%rip),%xmm4
  nopl   0x0(%rax,%rax,1)
L14:
  movapd %xmm3,%xmm2
  add    $0x1,%eax
  mulsd  %xmm1,%xmm2
  mulsd  %xmm4,%xmm1
  addsd  %xmm2,%xmm0
  cmp    %eax,%edi
  jne    L14
  lea    D13(%rip),%rdi
  jmp    1050 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L9:
  pxor   %xmm0,%xmm0
  lea    D13(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D30
type: f64
name: retrieve_data
arguments: data_label: D33
type: f64
name: retrieve_data
arguments: data_label: D13
type: string

tool: D30: 1

tool: D33: 0.95

tool: D13: The total cost will be $%0.2f


assistant: ```c
void showprice(double rate, int nights) {
  int n;
  double total = 0.0, factor = 1.0;
  for (n = 1; n <= nights; n++, factor *= 0.95) {
    total += rate * factor;
  }
  printf("The total cost will be $%0.2f\n", total);
}
```