ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<work>:
  endbr64
  test   %edx,%edx
  jle    L4
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %rsi,%r14
  mov    %edx,%r13d
  mov    %rsi,%r12
  mov    %rdi,%rbx
  mov    $0x0,%ebp
L6:
  mov    D28(%rip),%rax
  movq   %rax,%xmm2
  movsd  (%r12),%xmm1
  movsd  (%rbx),%xmm0
  call   <add1@plt>
  movsd  %xmm0,0x8(%rsp)
  mov    D28(%rip),%rax
  movq   %rax,%xmm0
  mov    %ebp,%edx
  mov    %r14,%rsi
  mov    %r15,%rdi
  call   <add2@plt>
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,(%rbx)
  mov    D28(%rip),%rax
  movq   %rax,%xmm0
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <add3@plt>
  movsd  %xmm0,(%rbx)
  add    $0x1,%ebp
  add    $0x8,%r12
  add    $0x8,%rbx
  cmp    %ebp,%r13d
  jne    L6
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L4:
  ret
```

tool_calls: name: read_data
arguments: label: D28
type: f64

tool: D28: 1

assistant: ```c
void work(double *a, double *b, int n) {
  int i;
  double tmp;
#pragma omp simd private(tmp)
  for (i = 0; i < n; i++) {
    tmp = add1(a[i], b[i], 1.0);
    a[i] = add2(a, b, i, 1.0) + tmp;
    a[i] = add3(&a[i], &b[i], 1.0);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<work>:
  endbr64
  test   %edx,%edx
  jle    L2
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  nopl   0x0(%rax,%rax,1)
L11:
  mov    D27(%rip),%rax
  movsd  (%r12),%xmm1
  movsd  (%rbx),%xmm0
  movq   %rax,%xmm2
  call   <add1@plt>
  mov    %ebp,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  mov    D27(%rip),%rax
  movsd  %xmm0,0x8(%rsp)
  add    $0x1,%ebp
  movq   %rax,%xmm0
  call   <add2@plt>
  addsd  0x8(%rsp),%xmm0
  mov    %r12,%rsi
  mov    %rbx,%rdi
  mov    D27(%rip),%rax
  add    $0x8,%r12
  add    $0x8,%rbx
  movsd  %xmm0,-0x8(%rbx)
  movq   %rax,%xmm0
  call   <add3@plt>
  movsd  %xmm0,-0x8(%rbx)
  cmp    %ebp,%r13d
  jne    L11
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  ret
```

tool_calls: name: read_data
arguments: tag: D27
type_guessed: f64

tool: D27: 1

assistant: ```c
void work(double *a, double *b, int n) {
  int i;
  double tmp;
#pragma omp simd private(tmp)
  for (i = 0; i < n; i++) {
    tmp = add1(a[i], b[i], 1.0);
    a[i] = add2(a, b, i, 1.0) + tmp;
    a[i] = add3(&a[i], &b[i], 1.0);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<work>:
  endbr64
  test   %edx,%edx
  jle    L12
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %rdi,%r14
  push   %r13
  mov    %edx,%r13d
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  nopl   0x0(%rax,%rax,1)
L4:
  mov    D7(%rip),%rax
  movsd  (%r12),%xmm1
  movsd  (%rbx),%xmm0
  movq   %rax,%xmm2
  call   <add1@plt>
  mov    %ebp,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  mov    D7(%rip),%rax
  movsd  %xmm0,0x8(%rsp)
  add    $0x1,%ebp
  movq   %rax,%xmm0
  call   <add2@plt>
  addsd  0x8(%rsp),%xmm0
  mov    %r12,%rsi
  mov    %rbx,%rdi
  mov    D7(%rip),%rax
  add    $0x8,%r12
  add    $0x8,%rbx
  movsd  %xmm0,-0x8(%rbx)
  movq   %rax,%xmm0
  call   <add3@plt>
  movsd  %xmm0,-0x8(%rbx)
  cmp    %ebp,%r13d
  jne    L4
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L12:
  ret
```

tool_calls: name: parse_data
arguments: data_label: D7
datatype: f64

tool: D7: 1

assistant: ```c
void work(double *a, double *b, int n) {
  int i;
  double tmp;
#pragma omp simd private(tmp)
  for (i = 0; i < n; i++) {
    tmp = add1(a[i], b[i], 1.0);
    a[i] = add2(a, b, i, 1.0) + tmp;
    a[i] = add3(&a[i], &b[i], 1.0);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<incfill>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L11
L14:
  mov    -0x8(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  addq   $0x8,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  sub    $0x8,%rax
  mov    %rax,-0x18(%rbp)
L11:
  cmpq   $0x0,-0x18(%rbp)
  jg     L14
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void incfill(long len, long adr) {
  unsigned long *p = (unsigned long *)adr;
  while (len > 0) {
    *p = (unsigned long)p;
    p++;
    len -= sizeof(long);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<incfill>:
  endbr64
  mov    %rsi,%rax
  test   %rdi,%rdi
  jle    L8
  sub    $0x1,%rdi
  and    $0xfffffffffffffff8,%rdi
  lea    0x8(%rsi,%rdi,1),%rdx
L9:
  mov    %rax,(%rax)
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L9
L8:
  ret
```

assistant: ```c
void incfill(long len, long adr) {
  unsigned long *p = (unsigned long *)adr;
  while (len > 0) {
    *p = (unsigned long)p;
    p++;
    len -= sizeof(long);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<incfill>:
  endbr64
  test   %rdi,%rdi
  jle    L25
  sub    $0x1,%rdi
  lea    0x8(%rsi),%rax
  and    $0xfffffffffffffff8,%rdi
  lea    (%rax,%rdi,1),%rdx
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L15:
  add    $0x8,%rax
L17:
  mov    %rsi,(%rsi)
  mov    %rax,%rsi
  cmp    %rdx,%rax
  jne    L15
L25:
  ret
```

assistant: ```c
void incfill(long len, long adr) {
  unsigned long *p = (unsigned long *)adr;
  while (len > 0) {
    *p = (unsigned long)p;
    p++;
    len -= sizeof(long);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<incfill>:
  endbr64
  test   %rdi,%rdi
  jle    L30
  sub    $0x1,%rdi
  mov    %rdi,%rcx
  shr    $0x3,%rcx
  add    $0x1,%rcx
  cmp    $0x7,%rdi
  jbe    L16
  mov    %rcx,%rdx
  lea    0x8(%rsi),%rax
  movq   %rsi,%xmm0
  movdqa D8(%rip),%xmm2
  shr    %rdx
  movq   %rax,%xmm3
  mov    %rsi,%rax
  shl    $0x4,%rdx
  punpcklqdq %xmm3,%xmm0
  add    %rsi,%rdx
  nopl   0x0(%rax,%rax,1)
L19:
  movdqa %xmm0,%xmm1
  add    $0x10,%rax
  paddq  %xmm2,%xmm0
  movups %xmm1,-0x10(%rax)
  cmp    %rdx,%rax
  jne    L19
  mov    %rcx,%rax
  and    $0xfffffffffffffffe,%rax
  lea    (%rsi,%rax,8),%rsi
  cmp    %rax,%rcx
  je     L30
L16:
  mov    %rsi,(%rsi)
L30:
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type_infered: byte[16]

tool: {"D8": [0x10,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x10,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
void incfill(long len, long adr) {
  unsigned long *p = (unsigned long *)adr;
  while (len > 0) {
    *p = (unsigned long)p;
    p++;
    len -= sizeof(long);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x1,-0x8(%rbp)
  movl   $0x2710,-0x4(%rbp)
  jmp    L16
L8:
  addl   $0x1,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
L16:
  mov    -0x4(%rbp),%eax
  sub    -0x8(%rbp),%eax
  test   %eax,%eax
  jg     L8
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int j = 1;
  for (int i = 10000; i - j >= 1; i--) {
    j++;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  mov    $0x1388,%eax
L0:
  sub    $0x1,%eax
  jne    L0
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main() {
  int j = 1;
  for (int i = 10000; i - j >= 1; i--) {
    j++;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

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

assistant: ```c
int main() {
  int j = 1;
  for (int i = 10000; i - j >= 1; i--) {
    j++;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

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

assistant: ```c
int main() {
  int j = 1;
  for (int i = 10000; i - j >= 1; i--) {
    j++;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%edi
  call   <plusandminus@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
datatype: string
name: extract_symbolic_data
arguments: data_tag: D3
datatype: string

tool: D5: %d

tool: D3: %d


assistant: ```c
main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", plusandminus(n));
  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    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D12(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <plusandminus@plt>
  mov    %eax,%esi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D12
type_infered: string
name: parse_data
arguments: label: D16
type_infered: string

tool: D12: %d

tool: D16: %d


assistant: ```c
main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", plusandminus(n));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D10(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <plusandminus@plt>
  lea    D16(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D10
datatype: string
name: parse_data
arguments: data_label: D16
datatype: string

tool: D10: %d

tool: D16: %d


assistant: ```c
main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", plusandminus(n));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D5(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  call   <plusandminus@plt>
  lea    D6(%rip),%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D5
type_guessed: string
name: unpack_data
arguments: data_label: D6
type_guessed: string

tool: D5: %d

tool: D6: %d


assistant: ```c
main() {
  int n;
  scanf("%d", &n);
  printf("%d\n", plusandminus(n));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x24a8,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L5
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: tag: D2
type_infered: string

tool: D2: %d 

assistant: ```c
int main() {
  int i = 9384;
  do {
    printf("%d ", i);
    i /= 10;
  } while (i > 0);
  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   %r12
  push   %rbp
  push   %rbx
  mov    $0x4,%ebp
  mov    $0x24a8,%ebx
  lea    D7(%rip),%r12
L6:
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  movslq %ebx,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %ebx,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  mov    %eax,%ebx
  sub    $0x1,%ebp
  jne    L6
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D7
type_guessed: string

tool: D7: %d 

assistant: ```c
int main() {
  int i = 9384;
  do {
    printf("%d ", i);
    i /= 10;
  } while (i > 0);
  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   %r12
  lea    D1(%rip),%r12
  push   %rbp
  mov    $0x4,%ebp
  push   %rbx
  mov    $0x24a8,%ebx
L3:
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  movslq %ebx,%rax
  mov    %ebx,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %eax,%ebx
  sub    $0x1,%ebp
  jne    L3
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D1
type_infered: string

tool: D1: %d 

assistant: ```c
int main() {
  int i = 9384;
  do {
    printf("%d ", i);
    i /= 10;
  } while (i > 0);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  lea    D4(%rip),%r12
  push   %rbp
  mov    $0x4,%ebp
  push   %rbx
  mov    $0x24a8,%ebx
L5:
  mov    %ebx,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  movslq %ebx,%rax
  mov    %ebx,%edx
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%edx
  sar    $0x22,%rax
  sub    %edx,%eax
  mov    %eax,%ebx
  sub    $0x1,%ebp
  jne    L5
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D4
datatype: string

tool: D4: %d 

assistant: ```c
int main() {
  int i = 9384;
  do {
    printf("%d ", i);
    i /= 10;
  } while (i > 0);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<read_from_circular_buffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    $0x0,%eax
  call   <difference@plt>
  mov    %eax,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L28
L10:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%eax
  call   <read_a_char_from_circular_buffer@plt>
  mov    %al,(%rbx)
  subl   $0x1,-0x18(%rbp)
  subl   $0x1,-0x2c(%rbp)
  addl   $0x1,-0x14(%rbp)
L28:
  cmpl   $0x0,-0x18(%rbp)
  jle    L24
  cmpl   $0x0,-0x2c(%rbp)
  jg     L10
L24:
  cmpl   $0x1,-0x14(%rbp)
  jle    L27
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L27:
  mov    -0x14(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int read_from_circular_buffer(char *buf, int bytes_to_read) {
  int bytes_avail_to_read = difference();
  int i = 0;
  while (bytes_avail_to_read > 0 && bytes_to_read > 0) {
    buf[i] = read_a_char_from_circular_buffer();
    bytes_avail_to_read--;
    bytes_to_read--;
    i++;
  }
  if (i > 1)
    buf[i] = '\0';
  return i;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<read_from_circular_buffer>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %esi,%r12d
  mov    $0x0,%eax
  call   <difference@plt>
  test   %eax,%eax
  jle    L18
  mov    %eax,%ebp
  test   %r12d,%r12d
  jle    L18
  mov    $0x1,%ebx
L32:
  mov    $0x0,%eax
  call   <read_a_char_from_circular_buffer@plt>
  mov    %al,-0x1(%r13,%rbx,1)
  mov    %ebx,%ecx
  mov    %ebp,%eax
  sub    %ebx,%eax
  mov    %r12d,%edx
  sub    %ebx,%edx
  add    $0x1,%rbx
  test   %eax,%eax
  jle    L0
  test   %edx,%edx
  jg     L32
L0:
  cmp    $0x1,%ecx
  jle    L37
  movslq %ecx,%rax
  movb   $0x0,0x0(%r13,%rax,1)
L37:
  mov    %ecx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L18:
  mov    $0x0,%ecx
  jmp    L37
```

assistant: ```c
int read_from_circular_buffer(char *buf, int bytes_to_read) {
  int bytes_avail_to_read = difference();
  int i = 0;
  while (bytes_avail_to_read > 0 && bytes_to_read > 0) {
    buf[i] = read_a_char_from_circular_buffer();
    bytes_avail_to_read--;
    bytes_to_read--;
    i++;
  }
  if (i > 1)
    buf[i] = '\0';
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<read_from_circular_buffer>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <difference@plt>
  test   %eax,%eax
  jle    L49
  test   %r12d,%r12d
  jle    L49
  mov    %eax,%ebp
  mov    $0x1,%ebx
  nopl   0x0(%rax,%rax,1)
L9:
  xor    %eax,%eax
  call   <read_a_char_from_circular_buffer@plt>
  mov    %r12d,%edx
  mov    %ebx,%r8d
  mov    %al,-0x1(%r13,%rbx,1)
  mov    %ebp,%eax
  sub    %ebx,%edx
  sub    %ebx,%eax
  add    $0x1,%rbx
  test   %eax,%eax
  jle    L12
  test   %edx,%edx
  jg     L9
L12:
  cmp    $0x1,%r8d
  je     L43
  movslq %r8d,%rax
  movb   $0x0,0x0(%r13,%rax,1)
L43:
  add    $0x8,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L49:
  add    $0x8,%rsp
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int read_from_circular_buffer(char *buf, int bytes_to_read) {
  int bytes_avail_to_read = difference();
  int i = 0;
  while (bytes_avail_to_read > 0 && bytes_to_read > 0) {
    buf[i] = read_a_char_from_circular_buffer();
    bytes_avail_to_read--;
    bytes_to_read--;
    i++;
  }
  if (i > 1)
    buf[i] = '\0';
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<read_from_circular_buffer>:
  endbr64
  push   %r13
  xor    %eax,%eax
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <difference@plt>
  test   %eax,%eax
  jle    L2
  test   %r12d,%r12d
  jle    L2
  mov    %eax,%ebp
  mov    $0x1,%ebx
  nopl   0x0(%rax,%rax,1)
L12:
  xor    %eax,%eax
  call   <read_a_char_from_circular_buffer@plt>
  mov    %r12d,%edx
  mov    %ebx,%r8d
  mov    %al,-0x1(%r13,%rbx,1)
  mov    %ebp,%eax
  sub    %ebx,%edx
  sub    %ebx,%eax
  add    $0x1,%rbx
  test   %eax,%eax
  jle    L32
  test   %edx,%edx
  jg     L12
L32:
  cmp    $0x1,%r8d
  je     L43
  movslq %r8d,%rax
  movb   $0x0,0x0(%r13,%rax,1)
L43:
  add    $0x8,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L2:
  add    $0x8,%rsp
  xor    %r8d,%r8d
  pop    %rbx
  mov    %r8d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int read_from_circular_buffer(char *buf, int bytes_to_read) {
  int bytes_avail_to_read = difference();
  int i = 0;
  while (bytes_avail_to_read > 0 && bytes_to_read > 0) {
    buf[i] = read_a_char_from_circular_buffer();
    bytes_avail_to_read--;
    bytes_to_read--;
    i++;
  }
  if (i > 1)
    buf[i] = '\0';
  return i;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<inchi_vfprintf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    D0(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,-0x18(%rbp)
  jne    L36
  cmpq   $0x0,-0x20(%rbp)
  je     L36
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L36
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xd,%al
  jne    L36
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <vfprintf@plt>
  mov    %eax,-0x4(%rbp)
  jmp    L34
L36:
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <vfprintf@plt>
  mov    %eax,-0x4(%rbp)
L34:
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int inchi_vfprintf(FILE *f, const char *lpszFormat, va_list argList) {
  int ret = 0;
  if (f == stderr && lpszFormat && lpszFormat[0] && '\r' == lpszFormat[strlen(lpszFormat) - 1]) {
    ret = vfprintf(f, lpszFormat, argList);
  } else {
    ret = vfprintf(f, lpszFormat, argList);
  }
  return ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<inchi_vfprintf>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbx
  mov    %edx,%ebp
  mov    D9(%rip),%rax
  cmp    %rdi,(%rax)
  jne    L16
  test   %rsi,%rsi
  jne    L3
L16:
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <vfprintf@plt>
L8:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L3:
  cmpb   $0x0,(%rsi)
  je     L16
  mov    %rsi,%rdi
  call   <strlen@plt>
  cltq
  cmpb   $0xd,-0x1(%rbx,%rax,1)
  jne    L16
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <vfprintf@plt>
  jmp    L8
```

assistant: ```c
int inchi_vfprintf(FILE *f, const char *lpszFormat, va_list argList) {
  int ret = 0;
  if (f == stderr && lpszFormat && lpszFormat[0] && '\r' == lpszFormat[strlen(lpszFormat) - 1]) {
    ret = vfprintf(f, lpszFormat, argList);
  } else {
    ret = vfprintf(f, lpszFormat, argList);
  }
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inchi_vfprintf>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    D4(%rip),%rax
  cmp    %rdi,(%rax)
  jne    L11
  test   %rsi,%rsi
  jne    L23
L11:
  add    $0x10,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <vfprintf@plt>
  nopw   0x0(%rax,%rax,1)
L23:
  cmpb   $0x0,(%rsi)
  je     L11
  mov    %rsi,%rdi
  mov    %edx,0xc(%rsp)
  mov    %rsi,(%rsp)
  call   <strlen@plt>
  mov    (%rsp),%rsi
  mov    0xc(%rsp),%edx
  mov    %rbp,%rdi
  add    $0x10,%rsp
  pop    %rbp
  jmp    1070 <vfprintf@plt>
```

assistant: ```c
int inchi_vfprintf(FILE *f, const char *lpszFormat, va_list argList) {
  int ret = 0;
  if (f == stderr && lpszFormat && lpszFormat[0] && '\r' == lpszFormat[strlen(lpszFormat) - 1]) {
    ret = vfprintf(f, lpszFormat, argList);
  } else {
    ret = vfprintf(f, lpszFormat, argList);
  }
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<inchi_vfprintf>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x10,%rsp
  mov    D6(%rip),%rax
  cmp    %rdi,(%rax)
  jne    L1
  test   %rsi,%rsi
  jne    L23
L1:
  add    $0x10,%rsp
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <vfprintf@plt>
  nopw   0x0(%rax,%rax,1)
L23:
  cmpb   $0x0,(%rsi)
  je     L1
  mov    %rsi,%rdi
  mov    %edx,0xc(%rsp)
  mov    %rsi,(%rsp)
  call   <strlen@plt>
  mov    (%rsp),%rsi
  mov    0xc(%rsp),%edx
  mov    %rbp,%rdi
  add    $0x10,%rsp
  pop    %rbp
  jmp    1070 <vfprintf@plt>
```

assistant: ```c
int inchi_vfprintf(FILE *f, const char *lpszFormat, va_list argList) {
  int ret = 0;
  if (f == stderr && lpszFormat && lpszFormat[0] && '\r' == lpszFormat[strlen(lpszFormat) - 1]) {
    ret = vfprintf(f, lpszFormat, argList);
  } else {
    ret = vfprintf(f, lpszFormat, argList);
  }
  return ret;
}
```