ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fun>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movb   $0x6,-0x22(%rbp)
  movb   $0x2,-0x21(%rbp)
  movl   $0x6c6c6568,-0xe(%rbp)
  movw   $0xa6f,-0xa(%rbp)
  movw   $0x7325,-0x10(%rbp)
  movzbl -0x22(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x20(%rbp)
  movzbl -0x21(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <calloc@plt>
  mov    %rax,-0x18(%rbp)
  movzbl -0x22(%rbp),%edx
  lea    -0xe(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  movzbl -0x21(%rbp),%edx
  lea    -0x10(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strncpy@plt>
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void fun() {
  unsigned char str_len = 6;
  unsigned char fmt_len = 2;
  char str_chars[6] = {'h', 'e', 'l', 'l', 'o', '\n'};
  char fmt_chars[2] = {'%', 's'};
  char *str = (char *)calloc(str_len, sizeof(char));
  char *fmt = (char *)calloc(fmt_len, sizeof(char));
  strncpy(str, str_chars, str_len);
  strncpy(fmt, fmt_chars, fmt_len);
  printf(fmt, str);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<fun>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x6c6c6568,0x2(%rsp)
  movw   $0xa6f,0x6(%rsp)
  movw   $0x7325,(%rsp)
  mov    $0x1,%esi
  mov    $0x6,%edi
  call   <calloc@plt>
  mov    %rax,%rbp
  mov    $0x1,%esi
  mov    $0x2,%edi
  call   <calloc@plt>
  mov    %rax,%rbx
  lea    0x2(%rsp),%rsi
  mov    $0x6,%edx
  mov    %rbp,%rdi
  call   <strncpy@plt>
  mov    %rsp,%rsi
  mov    $0x2,%edx
  mov    %rbx,%rdi
  call   <strncpy@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fun() {
  unsigned char str_len = 6;
  unsigned char fmt_len = 2;
  char str_chars[6] = {'h', 'e', 'l', 'l', 'o', '\n'};
  char fmt_chars[2] = {'%', 's'};
  char *str = (char *)calloc(str_len, sizeof(char));
  char *fmt = (char *)calloc(fmt_len, sizeof(char));
  strncpy(str, str_chars, str_len);
  strncpy(fmt, fmt_chars, fmt_len);
  printf(fmt, str);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<fun>:
  endbr64
  push   %r12
  mov    $0x7325,%edx
  mov    $0x1,%esi
  mov    $0x6,%edi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0xa6f,%eax
  mov    %dx,(%rsp)
  movl   $0x6c6c6568,0x2(%rsp)
  mov    %ax,0x6(%rsp)
  call   <calloc@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  mov    %rax,%r12
  call   <calloc@plt>
  lea    0x2(%rsp),%rsi
  mov    $0x6,%edx
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strncpy@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  mov    $0x2,%edx
  call   <strncpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fun() {
  unsigned char str_len = 6;
  unsigned char fmt_len = 2;
  char str_chars[6] = {'h', 'e', 'l', 'l', 'o', '\n'};
  char fmt_chars[2] = {'%', 's'};
  char *str = (char *)calloc(str_len, sizeof(char));
  char *fmt = (char *)calloc(fmt_len, sizeof(char));
  strncpy(str, str_chars, str_len);
  strncpy(fmt, fmt_chars, fmt_len);
  printf(fmt, str);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<fun>:
  endbr64
  push   %r12
  mov    $0x7325,%edx
  mov    $0x1,%esi
  mov    $0x6,%edi
  push   %rbp
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0xa6f,%eax
  mov    %dx,(%rsp)
  movl   $0x6c6c6568,0x2(%rsp)
  mov    %ax,0x6(%rsp)
  call   <calloc@plt>
  mov    $0x1,%esi
  mov    $0x2,%edi
  mov    %rax,%r12
  call   <calloc@plt>
  lea    0x2(%rsp),%rsi
  mov    $0x6,%edx
  mov    %r12,%rdi
  mov    %rax,%rbp
  call   <strncpy@plt>
  mov    %rsp,%rsi
  mov    %rbp,%rdi
  mov    $0x2,%edx
  call   <strncpy@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x18,%rsp
  pop    %rbp
  pop    %r12
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fun() {
  unsigned char str_len = 6;
  unsigned char fmt_len = 2;
  char str_chars[6] = {'h', 'e', 'l', 'l', 'o', '\n'};
  char fmt_chars[2] = {'%', 's'};
  char *str = (char *)calloc(str_len, sizeof(char));
  char *fmt = (char *)calloc(fmt_len, sizeof(char));
  strncpy(str, str_chars, str_len);
  strncpy(fmt, fmt_chars, fmt_len);
  printf(fmt, str);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D10(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D55
type: string
name: extract_symbolic_data
arguments: data_tag: D29
type: string
name: extract_symbolic_data
arguments: data_tag: D10
type: string
name: extract_symbolic_data
arguments: data_tag: D40
type: string
name: extract_symbolic_data
arguments: data_tag: D30
type: string
name: extract_symbolic_data
arguments: data_tag: D39
type: string

tool: D55:        *


tool: D29:       * 


tool: D10:      *  


tool: D40: *   *   


tool: D30:  * *    


tool: D39:   *     


assistant: ```c
int main(void) {
  printf("       *\n");
  printf("      * \n");
  printf("     *  \n");
  printf("*   *   \n");
  printf(" * *    \n");
  printf("  *     \n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D49(%rip),%rdi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <printf@plt>
  lea    D6(%rip),%rdi
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  lea    D30(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D49
data_type: string
name: read_data
arguments: tag: D5
data_type: string
name: read_data
arguments: tag: D4
data_type: string
name: read_data
arguments: tag: D6
data_type: string
name: read_data
arguments: tag: D39
data_type: string
name: read_data
arguments: tag: D30
data_type: string

tool: D49:        *


tool: D5:       * 


tool: D4:      *  


tool: D6: *   *   


tool: D39:  * *    


tool: D30:   *     


assistant: ```c
int main(void) {
  printf("       *\n");
  printf("      * \n");
  printf("     *  \n");
  printf("*   *   \n");
  printf(" * *    \n");
  printf("  *     \n");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D23(%rip),%rdi
  call   <printf@plt>
  lea    D48(%rip),%rdi
  call   <printf@plt>
  lea    D34(%rip),%rdi
  call   <printf@plt>
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    D24(%rip),%rdi
  call   <printf@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: read_data
arguments: tag: D23
type: string
name: read_data
arguments: tag: D48
type: string
name: read_data
arguments: tag: D34
type: string
name: read_data
arguments: tag: D28
type: string
name: read_data
arguments: tag: D24
type: string
name: read_data
arguments: tag: D1
type: string

tool: D23:        *


tool: D48:       * 


tool: D34:      *  


tool: D28: *   *   


tool: D24:  * *    


tool: D1:   *     


assistant: ```c
int main(void) {
  printf("       *\n");
  printf("      * \n");
  printf("     *  \n");
  printf("*   *   \n");
  printf(" * *    \n");
  printf("  *     \n");
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  lea    D34(%rip),%rdi
  call   <printf@plt>
  lea    D13(%rip),%rdi
  call   <printf@plt>
  lea    D22(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D20(%rip),%rdi
  call   <printf@plt>
  lea    D42(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D34
type_guessed: string
name: parse_data
arguments: data_tag: D13
type_guessed: string
name: parse_data
arguments: data_tag: D22
type_guessed: string
name: parse_data
arguments: data_tag: D15
type_guessed: string
name: parse_data
arguments: data_tag: D20
type_guessed: string
name: parse_data
arguments: data_tag: D42
type_guessed: string

tool: D34:        *


tool: D13:       * 


tool: D22:      *  


tool: D15: *   *   


tool: D20:  * *    


tool: D42:   *     


assistant: ```c
int main(void) {
  printf("       *\n");
  printf("      * \n");
  printf("     *  \n");
  printf("*   *   \n");
  printf(" * *    \n");
  printf("  *     \n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L12
L7:
  mov    -0x8(%rbp),%rax
  add    %rax,-0x10(%rbp)
  addq   $0x1,-0x8(%rbp)
L12:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jle    L7
  mov    -0x10(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
uint32_t sum(uint32_t n) {
  uint32_t s;
  uint32_t i;
  s = 0;
  for (i = 0; i <= n; i++) {
    s += i;
  }
  return s;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sum>:
  endbr64
  test   %rdi,%rdi
  js     L7
  add    $0x1,%rdi
  mov    $0x0,%eax
  mov    $0x0,%edx
L29:
  add    %rax,%rdx
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L29
L0:
  mov    %rdx,%rax
  ret
L7:
  mov    $0x0,%edx
  jmp    L0
```

assistant: ```c
uint32_t sum(uint32_t n) {
  uint32_t s;
  uint32_t i;
  s = 0;
  for (i = 0; i <= n; i++) {
    s += i;
  }
  return s;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sum>:
  endbr64
  test   %rdi,%rdi
  js     L7
  add    $0x1,%rdi
  xor    %eax,%eax
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L14:
  add    %rax,%r8
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L14
  mov    %r8,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
uint32_t sum(uint32_t n) {
  uint32_t s;
  uint32_t i;
  s = 0;
  for (i = 0; i <= n; i++) {
    s += i;
  }
  return s;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<sum>:
  endbr64
  test   %rdi,%rdi
  js     L17
  add    $0x1,%rdi
  xor    %eax,%eax
  xor    %r8d,%r8d
  nopw   0x0(%rax,%rax,1)
L18:
  add    %rax,%r8
  add    $0x1,%rax
  cmp    %rdi,%rax
  jne    L18
  mov    %r8,%rax
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  xor    %r8d,%r8d
  mov    %r8,%rax
  ret
```

assistant: ```c
uint32_t sum(uint32_t n) {
  uint32_t s;
  uint32_t i;
  s = 0;
  for (i = 0; i <= n; i++) {
    s += i;
  }
  return s;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<unescape>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  movsbl -0x4(%rbp),%eax
  sub    $0x62,%eax
  cmp    $0x14,%eax
  ja     L59
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D1(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D1(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0xa,%eax
  jmp    L56
  mov    $0x8,%eax
  jmp    L56
  mov    $0xc,%eax
  jmp    L56
  mov    $0xd,%eax
  jmp    L56
  mov    $0x9,%eax
  jmp    L56
  mov    $0xb,%eax
  jmp    L56
L59:
  movzbl -0x4(%rbp),%eax
L56:
  pop    %rbp
  ret
```

assistant: ```c
char unescape(char ec) {
  switch (ec) {
  case 'n':
    return '\n';
  case 'b':
    return '\b';
  case 'f':
    return '\f';
  case 'r':
    return '\r';
  case 't':
    return '\t';
  case 'v':
    return '\v';
  default:
    return ec;
  };
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<unescape>:
  endbr64
  lea    -0x62(%rdi),%eax
  cmp    $0x14,%al
  ja     L6
  movzbl %al,%eax
  lea    D5(%rip),%rdx
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x8,%eax
  ret
  mov    $0xc,%eax
  ret
  mov    $0xd,%eax
  ret
  mov    $0x9,%eax
  ret
  mov    $0xb,%eax
  ret
L6:
  mov    %edi,%eax
  ret
  mov    $0xa,%eax
  ret
```

assistant: ```c
char unescape(char ec) {
  switch (ec) {
  case 'n':
    return '\n';
  case 'b':
    return '\b';
  case 'f':
    return '\f';
  case 'r':
    return '\r';
  case 't':
    return '\t';
  case 'v':
    return '\v';
  default:
    return ec;
  };
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<unescape>:
  endbr64
  lea    -0x62(%rdi),%eax
  cmp    $0x14,%al
  ja     L0
  lea    D0(%rip),%rdx
  movzbl %al,%eax
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nop
L0:
  mov    %edi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0xc,%eax
  ret
  xchg   %ax,%ax
  mov    $0xd,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x9,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0xb,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0xa,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x8,%eax
  ret
```

assistant: ```c
char unescape(char ec) {
  switch (ec) {
  case 'n':
    return '\n';
  case 'b':
    return '\b';
  case 'f':
    return '\f';
  case 'r':
    return '\r';
  case 't':
    return '\t';
  case 'v':
    return '\v';
  default:
    return ec;
  };
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<unescape>:
  endbr64
  lea    -0x62(%rdi),%eax
  cmp    $0x14,%al
  ja     L8
  lea    D3(%rip),%rdx
  movzbl %al,%eax
  movslq (%rdx,%rax,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  nop
L8:
  mov    %edi,%eax
  ret
  nopl   0x0(%rax,%rax,1)
  mov    $0xc,%eax
  ret
  xchg   %ax,%ax
  mov    $0xd,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x9,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0xb,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0xa,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
  mov    $0x8,%eax
  ret
```

assistant: ```c
char unescape(char ec) {
  switch (ec) {
  case 'n':
    return '\n';
  case 'b':
    return '\b';
  case 'f':
    return '\f';
  case 'r':
    return '\r';
  case 't':
    return '\t';
  case 'v':
    return '\v';
  default:
    return ec;
  };
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<alloc_3d_float>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %edx,-0x3c(%rbp)
  mov    -0x34(%rbp),%eax
  imul   -0x38(%rbp),%eax
  imul   -0x3c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L10
  cmpq   $0x0,-0x18(%rbp)
  jne    L27
L10:
  mov    $0x0,%eax
  jmp    L36
L27:
  movl   $0x0,-0x24(%rbp)
  jmp    L12
L62:
  mov    -0x38(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x18(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L64
  mov    $0x0,%eax
  jmp    L36
L64:
  movl   $0x0,-0x28(%rbp)
  jmp    L46
L72:
  mov    -0x3c(%rbp),%eax
  imul   -0x38(%rbp),%eax
  imul   -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%eax
  imul   -0x3c(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x24(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x28(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  addl   $0x1,-0x28(%rbp)
L46:
  mov    -0x28(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L72
  addl   $0x1,-0x24(%rbp)
L12:
  mov    -0x24(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L62
  mov    -0x18(%rbp),%rax
L36:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
float ***alloc_3d_float(int ndim1, int ndim2, int ndim3) {
  float *space = malloc(ndim1 * ndim2 * ndim3 * sizeof(float));
  float ***array3 = malloc(ndim1 * sizeof(float **));
  int i, j;
  if (space == NULL || array3 == NULL)
    return NULL;
  for (j = 0; j < ndim1; j++) {
    array3[j] = malloc(ndim2 * sizeof(float *));
    if (array3[j] == NULL)
      return NULL;
    for (i = 0; i < ndim2; i++)
      array3[j][i] = space + j * (ndim3 * ndim2) + i * ndim3;
  }
  return array3;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<alloc_3d_float>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%ebx
  mov    %esi,%r14d
  mov    %edx,%ebp
  imul   %esi,%edi
  imul   %edx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    %rax,%r15
  lea    0x0(,%rbx,8),%edi
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %r15,%r15
  je     L54
  test   %rax,%rax
  je     L54
  test   %ebx,%ebx
  jle    L53
  lea    0x0(,%r14,8),%edi
  mov    %edi,0x8(%rsp)
  mov    %ebp,%edi
  imul   %r14d,%edi
  mov    %edi,0xc(%rsp)
  mov    %rax,%r12
  mov    %ebx,%ebx
  lea    (%rax,%rbx,8),%rax
  mov    %rax,0x10(%rsp)
  movslq %ebp,%rbx
  shl    $0x2,%rbx
  lea    -0x1(%r14),%eax
  lea    0x8(,%rax,8),%rbp
  mov    $0x0,%r13d
  jmp    L17
L66:
  add    $0x8,%r12
  add    0xc(%rsp),%r13d
  cmp    0x10(%rsp),%r12
  je     L53
L17:
  mov    0x8(%rsp),%edi
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %rax,(%r12)
  test   %rax,%rax
  je     L92
  test   %r14d,%r14d
  jle    L66
  movslq %r13d,%rax
  lea    (%r15,%rax,4),%rdx
  mov    $0x0,%eax
L6:
  mov    (%rsi),%rcx
  mov    %rdx,(%rcx,%rax,1)
  add    $0x8,%rax
  add    %rbx,%rdx
  cmp    %rbp,%rax
  jne    L6
  jmp    L66
L54:
  movq   $0x0,0x18(%rsp)
  jmp    L53
L92:
  mov    %rax,0x18(%rsp)
L53:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
float ***alloc_3d_float(int ndim1, int ndim2, int ndim3) {
  float *space = malloc(ndim1 * ndim2 * ndim3 * sizeof(float));
  float ***array3 = malloc(ndim1 * sizeof(float **));
  int i, j;
  if (space == NULL || array3 == NULL)
    return NULL;
  for (j = 0; j < ndim1; j++) {
    array3[j] = malloc(ndim2 * sizeof(float *));
    if (array3[j] == NULL)
      return NULL;
    for (i = 0; i < ndim2; i++)
      array3[j][i] = space + j * (ndim3 * ndim2) + i * ndim3;
  }
  return array3;
}
```
user: Translate the below assembly code into C programming language:

```asm
<alloc_3d_float>:
  endbr64
  push   %r15
  push   %r14
  mov    %edx,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    %ebp,%edi
  push   %rbx
  mov    %esi,%ebx
  imul   %esi,%edi
  sub    $0x28,%rsp
  imul   %edx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  lea    0x0(,%rbp,8),%edi
  mov    %rax,0x10(%rsp)
  mov    %rax,%r15
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %r15,%r15
  je     L77
  test   %rax,%rax
  je     L77
  test   %ebp,%ebp
  jle    L32
  mov    %r14d,%esi
  mov    %rax,%r12
  lea    (%rax,%rbp,8),%rax
  movslq %r14d,%rdx
  imul   %ebx,%esi
  mov    %rax,0x8(%rsp)
  lea    -0x1(%rbx),%ebp
  xor    %r14d,%r14d
  lea    0x0(,%rbx,8),%r13d
  lea    0x0(,%rdx,4),%r15
  add    $0x1,%rbp
  mov    %esi,0x4(%rsp)
  nopl   0x0(%rax,%rax,1)
L78:
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
  test   %rax,%rax
  je     L77
  test   %ebx,%ebx
  jle    L67
  mov    0x10(%rsp),%rcx
  movslq %r14d,%rdx
  lea    (%rax,%rbp,8),%rdi
  lea    (%rcx,%rdx,4),%rdx
  nopl   (%rax)
L33:
  mov    %rdx,(%rax)
  add    $0x8,%rax
  add    %r15,%rdx
  cmp    %rdi,%rax
  jne    L33
L67:
  add    $0x8,%r12
  add    0x4(%rsp),%r14d
  cmp    0x8(%rsp),%r12
  jne    L78
L32:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L77:
  movq   $0x0,0x18(%rsp)
  jmp    L32
```

assistant: ```c
float ***alloc_3d_float(int ndim1, int ndim2, int ndim3) {
  float *space = malloc(ndim1 * ndim2 * ndim3 * sizeof(float));
  float ***array3 = malloc(ndim1 * sizeof(float **));
  int i, j;
  if (space == NULL || array3 == NULL)
    return NULL;
  for (j = 0; j < ndim1; j++) {
    array3[j] = malloc(ndim2 * sizeof(float *));
    if (array3[j] == NULL)
      return NULL;
    for (i = 0; i < ndim2; i++)
      array3[j][i] = space + j * (ndim3 * ndim2) + i * ndim3;
  }
  return array3;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<alloc_3d_float>:
  endbr64
  push   %r15
  push   %r14
  mov    %edx,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  mov    %ebp,%edi
  push   %rbx
  mov    %esi,%ebx
  imul   %esi,%edi
  sub    $0x28,%rsp
  imul   %edx,%edi
  shl    $0x2,%edi
  call   <malloc@plt>
  lea    0x0(,%rbp,8),%edi
  mov    %rax,0x10(%rsp)
  mov    %rax,%r15
  call   <malloc@plt>
  mov    %rax,0x18(%rsp)
  test   %r15,%r15
  je     L14
  test   %rax,%rax
  je     L14
  test   %ebp,%ebp
  jle    L41
  mov    %r14d,%esi
  mov    %rax,%r12
  lea    (%rax,%rbp,8),%rax
  movslq %r14d,%rdx
  imul   %ebx,%esi
  mov    %rax,0x8(%rsp)
  lea    -0x1(%rbx),%ebp
  xor    %r14d,%r14d
  lea    0x0(,%rbx,8),%r13d
  lea    0x0(,%rdx,4),%r15
  add    $0x1,%rbp
  mov    %esi,0x4(%rsp)
  nopl   0x0(%rax,%rax,1)
L59:
  mov    %r13d,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
  test   %rax,%rax
  je     L14
  test   %ebx,%ebx
  jle    L30
  mov    0x10(%rsp),%rcx
  movslq %r14d,%rdx
  lea    (%rax,%rbp,8),%rdi
  lea    (%rcx,%rdx,4),%rdx
  nopl   (%rax)
L73:
  mov    %rdx,(%rax)
  add    $0x8,%rax
  add    %r15,%rdx
  cmp    %rdi,%rax
  jne    L73
L30:
  add    $0x8,%r12
  add    0x4(%rsp),%r14d
  cmp    0x8(%rsp),%r12
  jne    L59
L41:
  mov    0x18(%rsp),%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L14:
  movq   $0x0,0x18(%rsp)
  jmp    L41
```

assistant: ```c
float ***alloc_3d_float(int ndim1, int ndim2, int ndim3) {
  float *space = malloc(ndim1 * ndim2 * ndim3 * sizeof(float));
  float ***array3 = malloc(ndim1 * sizeof(float **));
  int i, j;
  if (space == NULL || array3 == NULL)
    return NULL;
  for (j = 0; j < ndim1; j++) {
    array3[j] = malloc(ndim2 * sizeof(float *));
    if (array3[j] == NULL)
      return NULL;
    for (i = 0; i < ndim2; i++)
      array3[j][i] = space + j * (ndim3 * ndim2) + i * ndim3;
  }
  return array3;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xa,-0x8(%rbp)
  movl   $0x6,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  add    %ecx,%eax
  cmp    $0x8,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int a = 10, b = 6;
  assert(a / 2 + b / 2 == 8);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x1,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  int a = 10, b = 6;
  assert(a / 2 + b / 2 == 8);
  return 0;
}
```