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
<rta_getattr_s64>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <RTA_DATA@plt>
  mov    %eax,%esi
  lea    0x4(%rsp),%rdi
  mov    $0x4,%edx
  call   <memcpy@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__s64 rta_getattr_s64(const struct rtattr *rta) {
  __s64 tmp;
  memcpy(&tmp, RTA_DATA(rta), sizeof(tmp));
  return tmp;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<rta_getattr_s64>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <RTA_DATA@plt>
  lea    0x4(%rsp),%rdi
  mov    $0x4,%edx
  mov    %eax,%esi
  call   <memcpy@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x18,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__s64 rta_getattr_s64(const struct rtattr *rta) {
  __s64 tmp;
  memcpy(&tmp, RTA_DATA(rta), sizeof(tmp));
  return tmp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rta_getattr_s64>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <RTA_DATA@plt>
  lea    0x4(%rsp),%rdi
  mov    $0x4,%edx
  mov    %eax,%esi
  call   <memcpy@plt>
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
__s64 rta_getattr_s64(const struct rtattr *rta) {
  __s64 tmp;
  memcpy(&tmp, RTA_DATA(rta), sizeof(tmp));
  return tmp;
}
```
user: What is the c source code of the assembly code below:

```asm
<foo>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int foo(int x, int y) {
  int res = x + y;
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<foo>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int foo(int x, int y) {
  int res = x + y;
  return res;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<foo>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int foo(int x, int y) {
  int res = x + y;
  return res;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<foo>:
  endbr64
  lea    (%rdi,%rsi,1),%eax
  ret
```

assistant: ```c
int foo(int x, int y) {
  int res = x + y;
  return res;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  jmp    L9
L18:
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <side@plt>
  addl   $0x1,-0xc(%rbp)
L9:
  cmpl   $0x2,-0xc(%rbp)
  jle    L18
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
type_infered: string
name: parse_data
arguments: data_tag: D13
type_infered: string

tool: D3: Enter a number: 

tool: D13: %d

assistant: ```c
int main() {
  int i;
  int num;
  for (i = 0; i < 3; i++) {
    printf("Enter a number: ");
    scanf("%d", &num);
    side(num);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x3,%ebx
  lea    D9(%rip),%r13
  lea    0x4(%rsp),%r12
  lea    D8(%rip),%rbp
L15:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <side@plt>
  sub    $0x1,%ebx
  jne    L15
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D9
type_infered: string
name: read_data
arguments: tag: D8
type_infered: string

tool: D9: Enter a number: 

tool: D8: %d

assistant: ```c
int main() {
  int i;
  int num;
  for (i = 0; i < 3; i++) {
    printf("Enter a number: ");
    scanf("%d", &num);
    side(num);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  lea    D9(%rip),%r13
  push   %r12
  push   %rbp
  lea    D19(%rip),%rbp
  push   %rbx
  mov    $0x3,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
L1:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <side@plt>
  sub    $0x1,%ebx
  jne    L1
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D9
datatype: string
name: read_data
arguments: data_label: D19
datatype: string

tool: D9: Enter a number: 

tool: D19: %d

assistant: ```c
int main() {
  int i;
  int num;
  for (i = 0; i < 3; i++) {
    printf("Enter a number: ");
    scanf("%d", &num);
    side(num);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D15(%rip),%r13
  push   %r12
  push   %rbp
  lea    D1(%rip),%rbp
  push   %rbx
  mov    $0x3,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
L1:
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <side@plt>
  sub    $0x1,%ebx
  jne    L1
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D15
data_type: string
name: extract_symbolic_data
arguments: tag: D1
data_type: string

tool: D15: Enter a number: 

tool: D1: %d

assistant: ```c
int main() {
  int i;
  int num;
  for (i = 0; i < 3; i++) {
    printf("Enter a number: ");
    scanf("%d", &num);
    side(num);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getpwnam>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
struct passwd *getpwnam(const char *name) {
  ;
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getpwnam>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
struct passwd *getpwnam(const char *name) {
  ;
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<getpwnam>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
struct passwd *getpwnam(const char *name) {
  ;
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<getpwnam>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
struct passwd *getpwnam(const char *name) {
  ;
  return NULL;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<stty>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x4(%rbp),%eax
  mov    $0x7409,%esi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <_ioctl@plt>
  leave
  ret
```

assistant: ```c
int stty(fildes, argp)
int fildes;
char *argp;
{ return _ioctl(fildes, (('t' << 8) | 9), argp); }
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<stty>:
  endbr64
  sub    $0x8,%rsp
  mov    %rsi,%rdx
  mov    $0x7409,%esi
  mov    $0x0,%eax
  call   <_ioctl@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int stty(fildes, argp)
int fildes;
char *argp;
{ return _ioctl(fildes, (('t' << 8) | 9), argp); }
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<stty>:
  endbr64
  mov    %rsi,%rdx
  xor    %eax,%eax
  mov    $0x7409,%esi
  jmp    1050 <_ioctl@plt>
```

assistant: ```c
int stty(fildes, argp)
int fildes;
char *argp;
{ return _ioctl(fildes, (('t' << 8) | 9), argp); }
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<stty>:
  endbr64
  mov    %rsi,%rdx
  xor    %eax,%eax
  mov    $0x7409,%esi
  jmp    1050 <_ioctl@plt>
```

assistant: ```c
int stty(fildes, argp)
int fildes;
char *argp;
{ return _ioctl(fildes, (('t' << 8) | 9), argp); }
```
user: What would the C source look like for the assembly code provided here:

```asm
<EDirectFadeRowRGBA32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %ecx,-0x38(%rbp)
  jmp    L11
L8:
  mov    -0x28(%rbp),%rax
  add    $0xc,%rax
  mov    (%rax),%eax
  imul   -0x34(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x30(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%ecx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <eblend@plt>
  mov    -0x30(%rbp),%rdx
  mov    %eax,(%rdx)
  mov    -0x30(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  add    $0x4,%rax
  mov    (%rax),%ecx
  mov    -0x30(%rbp),%rax
  lea    0x4(%rax),%rbx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <eblend@plt>
  mov    %eax,(%rbx)
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%ecx
  mov    -0x30(%rbp),%rax
  lea    0x8(%rax),%rbx
  mov    -0x14(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <eblend@plt>
  mov    %eax,(%rbx)
  addq   $0x10,-0x28(%rbp)
  addq   $0x10,-0x30(%rbp)
L11:
  mov    -0x38(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x38(%rbp)
  test   %eax,%eax
  jne    L8
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void EDirectFadeRowRGBA32(u_int8_t *src, u_int8_t *dst, u_int8_t af, int width) {
  while (width--) {
    u_int8_t a = (src[3] * af >> 8);
    dst[0] = eblend(a, src[0], dst[0]);
    dst[1] = eblend(a, src[1], dst[1]);
    dst[2] = eblend(a, src[2], dst[2]);
    src += 4;
    dst += 4;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<EDirectFadeRowRGBA32>:
  endbr64
  test   %ecx,%ecx
  je     L6
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    %edx,%r14d
  mov    %ecx,%ecx
  shl    $0x4,%rcx
  lea    (%rdi,%rcx,1),%r13
L5:
  mov    %r14d,%r12d
  imul   0xc(%rbp),%r12d
  sar    $0x8,%r12d
  mov    (%rbx),%edx
  mov    0x0(%rbp),%esi
  mov    %r12d,%edi
  call   <eblend@plt>
  mov    %eax,(%rbx)
  mov    0x4(%rbx),%edx
  mov    0x4(%rbp),%esi
  mov    %r12d,%edi
  call   <eblend@plt>
  mov    %eax,0x4(%rbx)
  mov    0x8(%rbx),%edx
  mov    0x8(%rbp),%esi
  mov    %r12d,%edi
  call   <eblend@plt>
  mov    %eax,0x8(%rbx)
  add    $0x10,%rbp
  add    $0x10,%rbx
  cmp    %r13,%rbp
  jne    L5
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L6:
  ret
```

assistant: ```c
void EDirectFadeRowRGBA32(u_int8_t *src, u_int8_t *dst, u_int8_t af, int width) {
  while (width--) {
    u_int8_t a = (src[3] * af >> 8);
    dst[0] = eblend(a, src[0], dst[0]);
    dst[1] = eblend(a, src[1], dst[1]);
    dst[2] = eblend(a, src[2], dst[2]);
    src += 4;
    dst += 4;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<EDirectFadeRowRGBA32>:
  endbr64
  test   %ecx,%ecx
  je     L8
  mov    %ecx,%ecx
  push   %r14
  mov    %edx,%r14d
  shl    $0x4,%rcx
  push   %r13
  lea    (%rdi,%rcx,1),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopl   0x0(%rax,%rax,1)
L10:
  mov    0xc(%rbp),%r12d
  mov    (%rbx),%edx
  add    $0x10,%rbp
  add    $0x10,%rbx
  mov    -0x10(%rbp),%esi
  imul   %r14d,%r12d
  sar    $0x8,%r12d
  mov    %r12d,%edi
  call   <eblend@plt>
  mov    -0xc(%rbx),%edx
  mov    %r12d,%edi
  mov    %eax,-0x10(%rbx)
  mov    -0xc(%rbp),%esi
  call   <eblend@plt>
  mov    -0x8(%rbx),%edx
  mov    %r12d,%edi
  mov    %eax,-0xc(%rbx)
  mov    -0x8(%rbp),%esi
  call   <eblend@plt>
  mov    %eax,-0x8(%rbx)
  cmp    %r13,%rbp
  jne    L10
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L8:
  ret
```

assistant: ```c
void EDirectFadeRowRGBA32(u_int8_t *src, u_int8_t *dst, u_int8_t af, int width) {
  while (width--) {
    u_int8_t a = (src[3] * af >> 8);
    dst[0] = eblend(a, src[0], dst[0]);
    dst[1] = eblend(a, src[1], dst[1]);
    dst[2] = eblend(a, src[2], dst[2]);
    src += 4;
    dst += 4;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<EDirectFadeRowRGBA32>:
  endbr64
  test   %ecx,%ecx
  je     L10
  mov    %ecx,%ecx
  push   %r14
  mov    %edx,%r14d
  shl    $0x4,%rcx
  push   %r13
  lea    (%rdi,%rcx,1),%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopl   0x0(%rax,%rax,1)
L12:
  mov    0xc(%rbp),%r12d
  mov    (%rbx),%edx
  add    $0x10,%rbp
  add    $0x10,%rbx
  mov    -0x10(%rbp),%esi
  imul   %r14d,%r12d
  sar    $0x8,%r12d
  mov    %r12d,%edi
  call   <eblend@plt>
  mov    -0xc(%rbx),%edx
  mov    %r12d,%edi
  mov    %eax,-0x10(%rbx)
  mov    -0xc(%rbp),%esi
  call   <eblend@plt>
  mov    -0x8(%rbx),%edx
  mov    %r12d,%edi
  mov    %eax,-0xc(%rbx)
  mov    -0x8(%rbp),%esi
  call   <eblend@plt>
  mov    %eax,-0x8(%rbx)
  cmp    %rbp,%r13
  jne    L12
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L10:
  ret
```

assistant: ```c
void EDirectFadeRowRGBA32(u_int8_t *src, u_int8_t *dst, u_int8_t af, int width) {
  while (width--) {
    u_int8_t a = (src[3] * af >> 8);
    dst[0] = eblend(a, src[0], dst[0]);
    dst[1] = eblend(a, src[1], dst[1]);
    dst[2] = eblend(a, src[2], dst[2]);
    src += 4;
    dst += 4;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<test_collapse>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x10,%r13d
  jmp    L17
L1:
  sub    $0x1,%r13d
  je     L44
L17:
  mov    $0x10,%r12d
L15:
  mov    $0x10,%ebp
L35:
  mov    $0x10,%ebx
L26:
  mov    $0x0,%eax
  call   <foo@plt>
  sub    $0x1,%ebx
  jne    L26
  sub    $0x1,%ebp
  jne    L35
  sub    $0x1,%r12d
  jne    L15
  jmp    L1
L44:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void test_collapse() {
  int i;
#pragma omp parallel
#pragma omp taskloop collapse
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4, , 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4)
  for (int i1 = 0; i1 < 16; ++i1)
    for (int i2 = 0; i2 < 16; ++i2)
      for (int i3 = 0; i3 < 16; ++i3)
        for (int i4 = 0; i4 < 16; ++i4)
          foo();
#pragma omp parallel
#pragma omp taskloop collapse(4, 8)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(2.5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(foo())
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(-5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(0)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(5 - 5)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<test_collapse>:
  endbr64
  push   %r13
  mov    $0x10,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
L34:
  mov    $0x10,%r12d
L9:
  mov    $0x10,%ebp
  nop
L27:
  mov    $0x10,%ebx
  nopl   (%rax)
L32:
  xor    %eax,%eax
  call   <foo@plt>
  sub    $0x1,%ebx
  jne    L32
  sub    $0x1,%ebp
  jne    L27
  sub    $0x1,%r12d
  jne    L9
  sub    $0x1,%r13d
  jne    L34
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void test_collapse() {
  int i;
#pragma omp parallel
#pragma omp taskloop collapse
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4, , 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4)
  for (int i1 = 0; i1 < 16; ++i1)
    for (int i2 = 0; i2 < 16; ++i2)
      for (int i3 = 0; i3 < 16; ++i3)
        for (int i4 = 0; i4 < 16; ++i4)
          foo();
#pragma omp parallel
#pragma omp taskloop collapse(4, 8)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(2.5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(foo())
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(-5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(0)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(5 - 5)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<test_collapse>:
  endbr64
  push   %r13
  mov    $0x10,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
L35:
  mov    $0x10,%r12d
L26:
  mov    $0x10,%ebp
  nop
L25:
  mov    $0x10,%ebx
  nopl   (%rax)
L29:
  xor    %eax,%eax
  call   <foo@plt>
  sub    $0x1,%ebx
  jne    L29
  sub    $0x1,%ebp
  jne    L25
  sub    $0x1,%r12d
  jne    L26
  sub    $0x1,%r13d
  jne    L35
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void test_collapse() {
  int i;
#pragma omp parallel
#pragma omp taskloop collapse
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4, , 4)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(4)
  for (int i1 = 0; i1 < 16; ++i1)
    for (int i2 = 0; i2 < 16; ++i2)
      for (int i3 = 0; i3 < 16; ++i3)
        for (int i4 = 0; i4 < 16; ++i4)
          foo();
#pragma omp parallel
#pragma omp taskloop collapse(4, 8)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(2.5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(foo())
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(-5)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(0)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp parallel
#pragma omp taskloop collapse(5 - 5)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<putbitchar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,%eax
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %al,-0x24(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0x1a(%rbp)
  movzbl -0x24(%rbp),%eax
  shl    $0x3,%eax
  cltq
  mov    %rax,-0x18(%rbp)
  addq   $0xffa6e,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L18
L30:
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x1a(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L13
L16:
  mov    -0x18(%rbp),%rax
  mov    %eax,%edx
  mov    $0x7,%eax
  sub    %edx,%eax
  mov    $0x1,%edx
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  mov    %eax,%edx
  movzbl -0x1a(%rbp),%eax
  and    %edx,%eax
  mov    %al,-0x19(%rbp)
  cmpb   $0x0,-0x19(%rbp)
  je     L39
  mov    -0x38(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdx
  mov    -0x30(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rax,%rsi
  mov    $0xdb,%edi
  call   <putasc@plt>
L39:
  addq   $0x1,-0x18(%rbp)
L13:
  cmpq   $0x7,-0x18(%rbp)
  jbe    L16
  addq   $0x1,-0x10(%rbp)
L18:
  cmpq   $0x7,-0x10(%rbp)
  jbe    L30
  nop
  nop
  leave
  ret
```

assistant: ```c
void putbitchar(unsigned char c, long x, long y) {
  unsigned char *pbit;
  unsigned char tbit, cbit = *pbit;
  unsigned long i, j;
  i = c * 8;
  i += 0xFFA6E;
  pbit = (char *)i;
  for (j = 0; j < 8; j++) {
    cbit = *(pbit + j);
    for (i = 0; i < 8; i++) {
      tbit = cbit & (1 << (7 - i));
      if (tbit != 0) {
        putasc(219, x + i, y + j);
      }
    }
  }
}
```