ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<parse_long_long>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x28(%rbp)
  jne    L65
  mov    -0x30(%rbp),%rax
  jmp    L62
L65:
  mov    D13(%rip),%rax
  movq   $0x0,(%rax)
  lea    -0x18(%rbp),%rcx
  mov    -0x28(%rbp),%rax
  mov    $0x0,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtoll@plt>
  mov    %rax,-0x10(%rbp)
  mov    D13(%rip),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L58
  mov    -0x30(%rbp),%rax
  jmp    L62
L58:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L6
  mov    -0x18(%rbp),%rax
  cmp    %rax,-0x28(%rbp)
  jne    L66
L6:
  mov    -0x30(%rbp),%rax
  jmp    L62
L66:
  mov    -0x10(%rbp),%rax
L62:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L76
  call   <__stack_chk_fail@plt>
L76:
  leave
  ret
```

assistant: ```c
long long parse_long_long(const char *value, long long default_value) {
  char *endptr;
  long long parsed;
  if (!value)
    return default_value;
  errno = 0;
  parsed = strtoll(value, &endptr, 0);
  if (errno != 0)
    return default_value;
  if (*endptr != '\0' || value == endptr)
    return default_value;
  return parsed;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_long_long>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rsi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L33
  mov    %rdi,%rbx
  mov    D3(%rip),%r12
  movq   $0x0,(%r12)
  mov    %rsp,%rsi
  mov    $0x0,%edx
  call   <strtoll@plt>
  cmpq   $0x0,(%r12)
  jne    L33
  mov    (%rsp),%rdx
  cmpb   $0x0,(%rdx)
  jne    L33
  cmp    %rbx,%rdx
  cmovne %rax,%rbp
L33:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  mov    %rbp,%rax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long parse_long_long(const char *value, long long default_value) {
  char *endptr;
  long long parsed;
  if (!value)
    return default_value;
  errno = 0;
  parsed = strtoll(value, &endptr, 0);
  if (errno != 0)
    return default_value;
  if (*endptr != '\0' || value == endptr)
    return default_value;
  return parsed;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<parse_long_long>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L44
  mov    D0(%rip),%r12
  xor    %edx,%edx
  mov    %rsp,%rsi
  mov    %rdi,%rbx
  movq   $0x0,(%r12)
  call   <strtoll@plt>
  cmpq   $0x0,(%r12)
  jne    L44
  mov    (%rsp),%rdx
  cmpb   $0x0,(%rdx)
  jne    L44
  cmp    %rbx,%rdx
  jne    L17
  nopl   (%rax)
L44:
  mov    %rbp,%rax
L17:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L2
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long parse_long_long(const char *value, long long default_value) {
  char *endptr;
  long long parsed;
  if (!value)
    return default_value;
  errno = 0;
  parsed = strtoll(value, &endptr, 0);
  if (errno != 0)
    return default_value;
  if (*endptr != '\0' || value == endptr)
    return default_value;
  return parsed;
}
```
user: What is the c source code of the assembly code below?

```asm
<parse_long_long>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %rdi,%rdi
  je     L4
  mov    D0(%rip),%r12
  xor    %edx,%edx
  mov    %rsp,%rsi
  mov    %rdi,%rbx
  movq   $0x0,(%r12)
  call   <strtoll@plt>
  cmpq   $0x0,(%r12)
  jne    L4
  mov    (%rsp),%rdx
  cmpb   $0x0,(%rdx)
  jne    L4
  cmp    %rbx,%rdx
  jne    L32
  nopl   (%rax)
L4:
  mov    %rbp,%rax
L32:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L6
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
long long parse_long_long(const char *value, long long default_value) {
  char *endptr;
  long long parsed;
  if (!value)
    return default_value;
  errno = 0;
  parsed = strtoll(value, &endptr, 0);
  if (errno != 0)
    return default_value;
  if (*endptr != '\0' || value == endptr)
    return default_value;
  return parsed;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<_be_free_str>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L8
  mov    -0x8(%rbp),%rax
  sub    $0x8,%rax
  mov    %rax,%rdi
  call   <free@plt>
L8:
  nop
  leave
  ret
```

assistant: ```c
void _be_free_str(char *str) {
  if (str)
    free(str - sizeof(long long));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<_be_free_str>:
  endbr64
  test   %rdi,%rdi
  je     L9
  sub    $0x8,%rsp
  sub    $0x8,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  ret
L9:
  ret
```

assistant: ```c
void _be_free_str(char *str) {
  if (str)
    free(str - sizeof(long long));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_be_free_str>:
  endbr64
  test   %rdi,%rdi
  je     L2
  sub    $0x8,%rdi
  jmp    1050 <free@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  ret
```

assistant: ```c
void _be_free_str(char *str) {
  if (str)
    free(str - sizeof(long long));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<_be_free_str>:
  endbr64
  test   %rdi,%rdi
  je     L0
  sub    $0x8,%rdi
  jmp    1050 <free@plt>
  nopw   0x0(%rax,%rax,1)
L0:
  ret
```

assistant: ```c
void _be_free_str(char *str) {
  if (str)
    free(str - sizeof(long long));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sum_n_recur>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x1,-0x4(%rbp)
  jne    L4
  mov    $0x1,%eax
  jmp    L16
L4:
  mov    -0x4(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <sum_n_recur>
  mov    -0x4(%rbp),%edx
  add    %edx,%eax
L16:
  leave
  ret
```

assistant: ```c
int sum_n_recur(int n) {
  if (n == 1)
    return 1;
  return n + sum_n_recur(n - 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<sum_n_recur>:
  endbr64
  mov    %edi,%eax
  cmp    $0x1,%edi
  je     L6
  push   %rbx
  mov    %edi,%ebx
  lea    -0x1(%rdi),%edi
  call   <sum_n_recur>
  add    %ebx,%eax
  pop    %rbx
  ret
L6:
  ret
```

assistant: ```c
int sum_n_recur(int n) {
  if (n == 1)
    return 1;
  return n + sum_n_recur(n - 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<sum_n_recur>:
  endbr64
  xor    %eax,%eax
  cmp    $0x1,%edi
  je     L15
  nopl   0x0(%rax,%rax,1)
L9:
  mov    %edi,%edx
  sub    $0x1,%edi
  add    %edx,%eax
  cmp    $0x1,%edi
  jne    L9
  add    $0x1,%eax
  ret
L15:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int sum_n_recur(int n) {
  if (n == 1)
    return 1;
  return n + sum_n_recur(n - 1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<sum_n_recur>:
  endbr64
  cmp    $0x1,%edi
  je     L66
  lea    -0x2(%rdi),%eax
  lea    -0x1(%rdi),%ecx
  cmp    $0x9,%eax
  jbe    L26
  movd   %edi,%xmm4
  mov    %ecx,%edx
  movdqa D7(%rip),%xmm3
  xor    %eax,%eax
  pshufd $0x0,%xmm4,%xmm1
  paddd  D2(%rip),%xmm1
  shr    $0x2,%edx
  pxor   %xmm0,%xmm0
L111:
  movdqa %xmm1,%xmm2
  add    $0x1,%eax
  paddd  %xmm3,%xmm1
  paddd  %xmm2,%xmm0
  cmp    %eax,%edx
  jne    L111
  movdqa %xmm0,%xmm1
  mov    %ecx,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%edx
  paddd  %xmm1,%xmm0
  sub    %edx,%edi
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %edx,%ecx
  je     L48
  lea    -0x1(%rdi),%ecx
L90:
  add    %edi,%eax
  cmp    $0x1,%ecx
  je     L48
  lea    -0x2(%rdi),%edx
  add    %ecx,%eax
  cmp    $0x3,%edi
  je     L48
  lea    -0x3(%rdi),%ecx
  add    %edx,%eax
  cmp    $0x4,%edi
  je     L48
  lea    -0x4(%rdi),%edx
  add    %ecx,%eax
  cmp    $0x5,%edi
  je     L48
  lea    -0x5(%rdi),%ecx
  add    %edx,%eax
  cmp    $0x6,%edi
  je     L48
  lea    -0x6(%rdi),%edx
  add    %ecx,%eax
  cmp    $0x7,%edi
  je     L48
  lea    -0x7(%rdi),%ecx
  add    %edx,%eax
  cmp    $0x8,%edi
  je     L48
  lea    -0x8(%rdi),%edx
  add    %ecx,%eax
  cmp    $0x9,%edi
  je     L48
  add    %edx,%eax
  sub    $0x9,%edi
  lea    (%rax,%rdi,1),%edx
  cmp    $0x1,%edi
  cmovne %edx,%eax
L48:
  add    $0x1,%eax
  ret
  xchg   %ax,%ax
L66:
  mov    $0x1,%eax
  ret
L26:
  xor    %eax,%eax
  jmp    L90
```

tool_calls: name: retrieve_data
arguments: label: D7
data_type: byte[16]
name: retrieve_data
arguments: label: D2
data_type: byte[16]

tool: {"D7": [0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff,0xfc,0xff,0xff,0xff]}

tool: {"D2": [0x00,0x00,0x00,0x00,0xff,0xff,0xff,0xff,0xfe,0xff,0xff,0xff,0xfd,0xff,0xff,0xff]}

assistant: ```c
int sum_n_recur(int n) {
  if (n == 1)
    return 1;
  return n + sum_n_recur(n - 1);
}
```
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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x18(%rbp)
  movl   $0x7,-0xc(%rbp)
  lea    -0x10(%rbp),%rdx
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%eax
  mov    -0xc(%rbp),%ecx
  lea    -0x18(%rbp),%rdx
  mov    %eax,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D23
type: string
name: extract_symbolic_data
arguments: data_tag: D3
type: string
name: extract_symbolic_data
arguments: data_tag: D39
type: string
name: extract_symbolic_data
arguments: data_tag: D31
type: string

tool: D23: The number of bytes written up to this point X%n is being stored in count_two.


tool: D3: count_one: %d


tool: D39: count_two: %d


tool: D31: A is %d and is at %08x.  B is %x.


assistant: ```c
int main() {
  int A = 5, B = 7, count_one, count_two;
  printf("The number of bytes written up to this point X%n is being stored in count_two.\n", &count_one, &count_two);
  printf("count_one: %d\n", count_one);
  printf("count_two: %d\n", count_two);
  printf("A is %d and is at %08x.  B is %x.\n", A, &A, B);
  exit(0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x5,0xc(%rsp)
  lea    0x14(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x10(%rsp),%esi
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%esi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rdx
  mov    $0x7,%ecx
  mov    0xc(%rsp),%esi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
```

tool_calls: name: read_data
arguments: label: D4
type_guessed: string
name: read_data
arguments: label: D19
type_guessed: string
name: read_data
arguments: label: D18
type_guessed: string
name: read_data
arguments: label: D16
type_guessed: string

tool: D4: The number of bytes written up to this point X%n is being stored in count_two.


tool: D19: count_one: %d


tool: D18: count_two: %d


tool: D16: A is %d and is at %08x.  B is %x.


assistant: ```c
int main() {
  int A = 5, B = 7, count_one, count_two;
  printf("The number of bytes written up to this point X%n is being stored in count_two.\n", &count_one, &count_two);
  printf("count_one: %d\n", count_one);
  printf("count_two: %d\n", count_two);
  printf("A is %d and is at %08x.  B is %x.\n", A, &A, B);
  exit(0);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D3(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  movl   $0x5,0xc(%rsp)
  call   <printf@plt>
  mov    0x10(%rsp),%esi
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%esi
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    0xc(%rsp),%rdx
  xor    %eax,%eax
  lea    D31(%rip),%rdi
  mov    $0x7,%ecx
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
```

tool_calls: name: read_data
arguments: tag: D3
data_type: string
name: read_data
arguments: tag: D1
data_type: string
name: read_data
arguments: tag: D27
data_type: string
name: read_data
arguments: tag: D31
data_type: string

tool: D3: The number of bytes written up to this point X%n is being stored in count_two.


tool: D1: count_one: %d


tool: D27: count_two: %d


tool: D31: A is %d and is at %08x.  B is %x.


assistant: ```c
int main() {
  int A = 5, B = 7, count_one, count_two;
  printf("The number of bytes written up to this point X%n is being stored in count_two.\n", &count_one, &count_two);
  printf("count_one: %d\n", count_one);
  printf("count_two: %d\n", count_two);
  printf("A is %d and is at %08x.  B is %x.\n", A, &A, B);
  exit(0);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D9(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  movl   $0x5,0xc(%rsp)
  call   <printf@plt>
  mov    0x10(%rsp),%esi
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%esi
  lea    D21(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    0xc(%rsp),%rdx
  xor    %eax,%eax
  lea    D25(%rip),%rdi
  mov    $0x7,%ecx
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
```

tool_calls: name: parse_data
arguments: label: D9
datatype: string
name: parse_data
arguments: label: D12
datatype: string
name: parse_data
arguments: label: D21
datatype: string
name: parse_data
arguments: label: D25
datatype: string

tool: D9: The number of bytes written up to this point X%n is being stored in count_two.


tool: D12: count_one: %d


tool: D21: count_two: %d


tool: D25: A is %d and is at %08x.  B is %x.


assistant: ```c
int main() {
  int A = 5, B = 7, count_one, count_two;
  printf("The number of bytes written up to this point X%n is being stored in count_two.\n", &count_one, &count_two);
  printf("count_one: %d\n", count_one);
  printf("count_two: %d\n", count_two);
  printf("A is %d and is at %08x.  B is %x.\n", A, &A, B);
  exit(0);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<makeWcharsFromUtf8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x18(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L100
L33:
  mov    -0x38(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  lea    -0x20(%rbp),%rdx
  lea    -0x28(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <convertUtf8ToWchar@plt>
  mov    %eax,-0x2c(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x2c(%rbp)
  je     L61
  cmpl   $0x0,-0x2c(%rbp)
  je     L29
  cmpq   $0x0,-0x40(%rbp)
  je     L44
  mov    -0x10(%rbp),%rax
  cmp    -0x48(%rbp),%rax
  je     L10
  mov    -0x40(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    -0x2c(%rbp),%edx
  mov    %dl,(%rax)
L44:
  addq   $0x1,-0x10(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x18(%rbp)
L100:
  cmpq   $0x0,-0x18(%rbp)
  jne    L33
  jmp    L24
L61:
  nop
  jmp    L24
L29:
  nop
  jmp    L24
L10:
  nop
L24:
  cmpq   $0x0,-0x40(%rbp)
  je     L111
  mov    -0x10(%rbp),%rax
  cmp    -0x48(%rbp),%rax
  jae    L111
  mov    -0x40(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L111:
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L80
  call   <__stack_chk_fail@plt>
L80:
  leave
  ret
```

assistant: ```c
size_t makeWcharsFromUtf8(const char *text, unsigned char *characters, size_t size) {
  size_t length = strlen(text);
  size_t count = 0;
  while (length > 0) {
    const char *utf8 = text;
    size_t utfs = length;
    int character = convertUtf8ToWchar(&utf8, &utfs);
    if (character == EOF)
      break;
    if (!character)
      break;
    if (characters) {
      if (count == size)
        break;
      characters[count] = character;
    }
    count += 1;
    text = utf8;
    length = utfs;
  }
  if (characters && (count < size))
    characters[count] = 0;
  return count;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<makeWcharsFromUtf8>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,%r14
  mov    %rsi,%rbp
  mov    %rdx,%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  test   %rax,%rax
  je     L20
  mov    %rax,%rdx
  mov    $0x0,%ebx
  lea    0x10(%rsp),%r12
  jmp    L96
L58:
  add    $0x1,%rbx
  mov    0x8(%rsp),%r14
  mov    0x10(%rsp),%rdx
  test   %rdx,%rdx
  je     L90
L96:
  mov    %r14,0x8(%rsp)
  mov    %rdx,0x10(%rsp)
  lea    0x8(%rsp),%rdi
  mov    %r12,%rsi
  call   <convertUtf8ToWchar@plt>
  mov    D2(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L90
  test   %eax,%eax
  je     L90
  test   %rbp,%rbp
  je     L58
  cmp    %rbx,%r13
  je     L90
  mov    %al,0x0(%rbp,%rbx,1)
  jmp    L58
L20:
  mov    %rax,%rbx
L90:
  test   %rbp,%rbp
  je     L106
  cmp    %rbx,%r13
  jbe    L106
  movb   $0x0,0x0(%rbp,%rbx,1)
L106:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  mov    %rbx,%rax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L42:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t makeWcharsFromUtf8(const char *text, unsigned char *characters, size_t size) {
  size_t length = strlen(text);
  size_t count = 0;
  while (length > 0) {
    const char *utf8 = text;
    size_t utfs = length;
    int character = convertUtf8ToWchar(&utf8, &utfs);
    if (character == EOF)
      break;
    if (!character)
      break;
    if (characters) {
      if (count == size)
        break;
      characters[count] = character;
    }
    count += 1;
    text = utf8;
    length = utfs;
  }
  if (characters && (count < size))
    characters[count] = 0;
  return count;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<makeWcharsFromUtf8>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rdi,0x8(%rsp)
  call   <strlen@plt>
  test   %rax,%rax
  je     L27
  mov    0x8(%rsp),%rdi
  lea    0x20(%rsp),%r13
  lea    0x18(%rsp),%r12
  mov    D1(%rip),%rbp
  jmp    L57
  nopw   0x0(%rax,%rax,1)
L75:
  test   %eax,%eax
  je     L27
  test   %rbx,%rbx
  je     L1
  cmp    %r15,%r14
  je     L27
  mov    %al,(%rbx,%r15,1)
L1:
  mov    0x20(%rsp),%rax
  mov    0x18(%rsp),%rdi
  add    $0x1,%r15
  test   %rax,%rax
  je     L27
L57:
  mov    %rdi,0x18(%rsp)
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %rax,0x20(%rsp)
  call   <convertUtf8ToWchar@plt>
  cmp    %eax,0x0(%rbp)
  jne    L75
L27:
  test   %rbx,%rbx
  je     L82
  cmp    %r15,%r14
  jbe    L82
  movb   $0x0,(%rbx,%r15,1)
L82:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x38,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L55:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t makeWcharsFromUtf8(const char *text, unsigned char *characters, size_t size) {
  size_t length = strlen(text);
  size_t count = 0;
  while (length > 0) {
    const char *utf8 = text;
    size_t utfs = length;
    int character = convertUtf8ToWchar(&utf8, &utfs);
    if (character == EOF)
      break;
    if (!character)
      break;
    if (characters) {
      if (count == size)
        break;
      characters[count] = character;
    }
    count += 1;
    text = utf8;
    length = utfs;
  }
  if (characters && (count < size))
    characters[count] = 0;
  return count;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<makeWcharsFromUtf8>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rdi,0x8(%rsp)
  call   <strlen@plt>
  test   %rax,%rax
  je     L55
  test   %rbx,%rbx
  mov    0x8(%rsp),%rdi
  lea    0x20(%rsp),%r13
  mov    D3(%rip),%rbp
  lea    0x18(%rsp),%r12
  jne    L73
  jmp    L128
  nopw   0x0(%rax,%rax,1)
L75:
  test   %eax,%eax
  je     L55
  cmp    %r15,%r14
  je     L55
  mov    %al,(%rbx,%r15,1)
  mov    0x20(%rsp),%rax
  add    $0x1,%r15
  mov    0x18(%rsp),%rdi
  test   %rax,%rax
  je     L55
L73:
  mov    %rdi,0x18(%rsp)
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %rax,0x20(%rsp)
  call   <convertUtf8ToWchar@plt>
  cmp    %eax,0x0(%rbp)
  jne    L75
L55:
  test   %rbx,%rbx
  je     L98
  cmp    %r15,%r14
  jbe    L98
  movb   $0x0,(%rbx,%r15,1)
L98:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L89
  add    $0x38,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L123:
  test   %eax,%eax
  je     L55
  mov    0x20(%rsp),%rax
  mov    0x18(%rsp),%rdi
  add    $0x1,%r15
  test   %rax,%rax
  je     L98
L128:
  mov    %rdi,0x18(%rsp)
  mov    %r13,%rsi
  mov    %r12,%rdi
  mov    %rax,0x20(%rsp)
  call   <convertUtf8ToWchar@plt>
  cmp    %eax,0x0(%rbp)
  jne    L123
  jmp    L55
L89:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
size_t makeWcharsFromUtf8(const char *text, unsigned char *characters, size_t size) {
  size_t length = strlen(text);
  size_t count = 0;
  while (length > 0) {
    const char *utf8 = text;
    size_t utfs = length;
    int character = convertUtf8ToWchar(&utf8, &utfs);
    if (character == EOF)
      break;
    if (!character)
      break;
    if (characters) {
      if (count == size)
        break;
      characters[count] = character;
    }
    count += 1;
    text = utf8;
    length = utfs;
  }
  if (characters && (count < size))
    characters[count] = 0;
  return count;
}
```
user: What is the c source code of the assembly code below:

```asm
<stbir__support_zero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x4(%rbp)
  pxor   %xmm0,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
float stbir__support_zero(float s) {
  (void)sizeof(s);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<stbir__support_zero>:
  endbr64
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
float stbir__support_zero(float s) {
  (void)sizeof(s);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<stbir__support_zero>:
  endbr64
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
float stbir__support_zero(float s) {
  (void)sizeof(s);
  return 0;
}
```