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
<expand_ddval>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movzbl (%rdi),%eax
  test   %al,%al
  je     L20
  mov    %rdi,%r14
  xor    %ebx,%ebx
  xor    %r12d,%r12d
  xor    %ecx,%ecx
  lea    0x14(%rsp),%r8
  nopl   0x0(%rax,%rax,1)
L38:
  movslq %ebx,%rdx
  lea    0x0(%r13,%rdx,1),%rbp
  cmp    $0x40,%al
  je     L51
  add    $0x1,%r12d
  mov    %al,0x0(%rbp)
  add    $0x1,%ebx
  movslq %r12d,%rcx
  movzbl (%r14,%rcx,1),%eax
  test   %al,%al
  jne    L38
L43:
  movslq %ebx,%rbx
  add    %rbx,%r13
L20:
  movb   $0x0,0x0(%r13)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L51:
  lea    0x1(%r14,%rcx,1),%rdi
  mov    %r8,%rsi
  mov    %r8,0x8(%rsp)
  call   <lookup_ddval@plt>
  mov    %rbp,%rdi
  mov    %rax,%r15
  mov    %rax,%rsi
  call   <strcpy@plt>
  add    0x14(%rsp),%r12d
  mov    %r15,%rdi
  call   <strlen@plt>
  movslq %r12d,%rcx
  mov    0x8(%rsp),%r8
  add    %eax,%ebx
  movzbl (%r14,%rcx,1),%eax
  test   %al,%al
  jne    L38
  jmp    L43
L28:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void expand_ddval(const char *ddval_raw, char *ddval_expanded) {
  char c, *ddval;
  int i = 0, j = 0, len;
  while ((c = ddval_raw[i]) != '\0') {
    if (c == '@') {
      ddval = lookup_ddval(&ddval_raw[i + 1], &len);
      strcpy(&ddval_expanded[j], ddval);
      i += len;
      j += strlen(ddval);
    } else
      ddval_expanded[j++] = ddval_raw[i++];
  }
  ddval_expanded[j] = 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0x20(%rbp)
  movl   $0x14,-0x1c(%rbp)
  movl   $0x1e,-0x18(%rbp)
  movl   $0x28,-0x14(%rbp)
  movl   $0x32,-0x10(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L20
L9:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%ecx
  mov    -0x28(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x28(%rbp),%rsi
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,%rsi
  lea    -0x28(%rbp),%rax
  mov    %ecx,%r8d
  mov    %edx,%ecx
  mov    %rsi,%rdx
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x28(%rbp),%rax
  add    $0x4,%rax
  mov    %rax,-0x28(%rbp)
  addl   $0x1,-0x2c(%rbp)
L20:
  cmpl   $0x4,-0x2c(%rbp)
  jle    L9
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L23
  call   <__stack_chk_fail@plt>
L23:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
type_guessed: string

tool: D8: %p %p %d %d


assistant: ```c
int main() {
  int num[] = {10, 20, 30, 40, 50};
  int i, *j;
  j = num;
  for (i = 0; i < 5; i++) {
    printf("%p %p %d %d\n", &j, &j[i], *j, num[i]);
    j++;
  }
  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
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0xa,0x10(%rsp)
  movl   $0x14,0x14(%rsp)
  movl   $0x1e,0x18(%rsp)
  movl   $0x28,0x1c(%rsp)
  movl   $0x32,0x20(%rsp)
  lea    0x10(%rsp),%rax
  mov    %rax,0x8(%rsp)
  mov    $0x0,%ebx
  lea    D7(%rip),%rbp
L16:
  mov    0x8(%rsp),%rax
  lea    (%rax,%rbx,1),%rdx
  lea    0x8(%rsp),%rsi
  mov    0x10(%rsp,%rbx,1),%r8d
  mov    (%rax),%ecx
  mov    %rbp,%rdi
  call   <printf@plt>
  addq   $0x4,0x8(%rsp)
  add    $0x4,%rbx
  cmp    $0x14,%rbx
  jne    L16
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L10
  mov    $0x0,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
L10:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D7
type_guessed: string

tool: D7: %p %p %d %d


assistant: ```c
int main() {
  int num[] = {10, 20, 30, 40, 50};
  int i, *j;
  j = num;
  for (i = 0; i < 5; i++) {
    printf("%p %p %d %d\n", &j, &j[i], *j, num[i]);
    j++;
  }
  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
  mov    $0xa,%ecx
  mov    $0xa,%r8d
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0x140000000a,%rax
  lea    0x10(%rsp),%rbp
  movl   $0x32,0x20(%rsp)
  lea    0x8(%rsp),%r13
  mov    %rbp,0x8(%rsp)
  mov    %rbp,%rdx
  mov    %rax,0x10(%rsp)
  movabs $0x280000001e,%rax
  mov    %rax,0x18(%rsp)
L1:
  add    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  cmp    $0x14,%rbx
  je     L28
  mov    0x0(%rbp,%rbx,1),%r8d
  mov    0x4(%rax),%ecx
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L28:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D0
datatype: string

tool: D0: %p %p %d %d


assistant: ```c
int main() {
  int num[] = {10, 20, 30, 40, 50};
  int i, *j;
  j = num;
  for (i = 0; i < 5; i++) {
    printf("%p %p %d %d\n", &j, &j[i], *j, num[i]);
    j++;
  }
  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
  mov    $0xa,%ecx
  push   %r12
  lea    D19(%rip),%r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x38,%rsp
  movdqa D2(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  movl   $0x32,0x20(%rsp)
  lea    0x8(%rsp),%r13
  mov    %rbp,0x8(%rsp)
  mov    %rbp,%rdx
  movaps %xmm0,0x10(%rsp)
L1:
  mov    0x0(%rbp,%rbx,1),%r8d
  add    %rbx,%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,0x8(%rsp)
  cmp    $0x14,%rbx
  je     L11
  mov    0x4(%rax),%ecx
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L11:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x38,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D19
type: string
name: read_data
arguments: label: D2
type: byte[16]

tool: D19: %p %p %d %d


tool: {"D2": [0x0a,0x00,0x00,0x00,0x14,0x00,0x00,0x00,0x1e,0x00,0x00,0x00,0x28,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int num[] = {10, 20, 30, 40, 50};
  int i, *j;
  j = num;
  for (i = 0; i < 5; i++) {
    printf("%p %p %d %d\n", &j, &j[i], *j, num[i]);
    j++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bias_and_round>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  movsd  %xmm0,-0x18(%rbp)
  movsd  %xmm1,-0x20(%rbp)
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%edx
  mov    D7(%rip),%rax
  mov    %edx,%edi
  movq   %rax,%xmm0
  call   <pow@plt>
  mov    %eax,-0xc(%rbp)
  movsd  -0x18(%rbp),%xmm0
  mulsd  -0x20(%rbp),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0xc(%rbp),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x8(%rbp),%xmm1
  movsd  D4(%rip),%xmm0
  addsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <floorf@plt>
  cltd
  idivl  -0xc(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
datatype: f64
name: disassemble_data
arguments: data_tag: D4
datatype: f64

tool: D7: 10

tool: D4: 0.5

assistant: ```c
double bias_and_round(double val, double frac, int precision) {
  int factor = (int)pow(10.0, precision);
  double ret = val * frac;
  return floorf(ret * factor + 0.5) / factor;
}
```
user: What is the c source code of the assembly code below?

```asm
<bias_and_round>:
  endbr64
  push   %r15
  push   %r14
  push   %rbx
  movq   %xmm0,%r14
  movq   %xmm1,%r15
  movsd  D5(%rip),%xmm0
  call   <pow@plt>
  mov    %eax,%ebx
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movq   %r14,%xmm1
  movq   %r15,%xmm2
  mulsd  %xmm2,%xmm1
  mulsd  %xmm1,%xmm0
  addsd  D8(%rip),%xmm0
  call   <floorf@plt>
  cltd
  idiv   %ebx
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  pop    %rbx
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D5
type: f64
name: parse_data
arguments: data_tag: D8
type: f64

tool: D5: 10

tool: D8: 0.5

assistant: ```c
double bias_and_round(double val, double frac, int precision) {
  int factor = (int)pow(10.0, precision);
  double ret = val * frac;
  return floorf(ret * factor + 0.5) / factor;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<bias_and_round>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  movsd  %xmm0,(%rsp)
  movsd  D12(%rip),%xmm0
  movsd  %xmm1,0x8(%rsp)
  call   <pow@plt>
  movsd  (%rsp),%xmm2
  pxor   %xmm0,%xmm0
  mulsd  0x8(%rsp),%xmm2
  cvtsi2sd %eax,%xmm0
  mov    %eax,%ebx
  mulsd  %xmm2,%xmm0
  addsd  D14(%rip),%xmm0
  call   <floorf@plt>
  pxor   %xmm0,%xmm0
  add    $0x10,%rsp
  cltd
  idiv   %ebx
  pop    %rbx
  cvtsi2sd %eax,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: label: D12
type: f64
name: retrieve_data
arguments: label: D14
type: f64

tool: D12: 10

tool: D14: 0.5

assistant: ```c
double bias_and_round(double val, double frac, int precision) {
  int factor = (int)pow(10.0, precision);
  double ret = val * frac;
  return floorf(ret * factor + 0.5) / factor;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bias_and_round>:
  endbr64
  push   %rbx
  sub    $0x10,%rsp
  movsd  %xmm0,(%rsp)
  movsd  D3(%rip),%xmm0
  movsd  %xmm1,0x8(%rsp)
  call   <pow@plt>
  movsd  (%rsp),%xmm2
  pxor   %xmm0,%xmm0
  mulsd  0x8(%rsp),%xmm2
  cvtsi2sd %eax,%xmm0
  mov    %eax,%ebx
  mulsd  %xmm2,%xmm0
  addsd  D10(%rip),%xmm0
  call   <floorf@plt>
  pxor   %xmm0,%xmm0
  add    $0x10,%rsp
  cltd
  idiv   %ebx
  pop    %rbx
  cvtsi2sd %eax,%xmm0
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D3
datatype: f64
name: unpack_data
arguments: data_tag: D10
datatype: f64

tool: D3: 10

tool: D10: 0.5

assistant: ```c
double bias_and_round(double val, double frac, int precision) {
  int factor = (int)pow(10.0, precision);
  double ret = val * frac;
  return floorf(ret * factor + 0.5) / factor;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %edi,-0x44(%rbp)
  mov    %rsi,-0x50(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x4,-0x20(%rbp)
  movl   $0x5,-0x1c(%rbp)
  movl   $0x2,-0x18(%rbp)
  movl   $0x6,-0x14(%rbp)
  movl   $0x1,-0x10(%rbp)
  movl   $0x5,-0x24(%rbp)
  movl   $0x0,-0x34(%rbp)
  jmp    L30
L76:
  mov    -0x34(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x34(%rbp)
L30:
  mov    -0x34(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L76
  movl   $0x0,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L25
L14:
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x28(%rbp)
  jmp    L33
L16:
  mov    -0x2c(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%edx
  mov    -0x28(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%eax
  cmp    %eax,%edx
  jle    L70
  addl   $0x1,-0x30(%rbp)
L70:
  addl   $0x1,-0x28(%rbp)
L33:
  mov    -0x28(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L16
  addl   $0x1,-0x2c(%rbp)
L25:
  mov    -0x2c(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L14
  mov    -0x30(%rbp),%eax
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L49
  call   <__stack_chk_fail@plt>
L49:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
datatype: string
name: unpack_data
arguments: data_label: D18
datatype: string

tool: D7: %d 

tool: D18: 
The number of inversion parirs are %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a[] = {4, 5, 2, 6, 1};
  int size = sizeof(a) / sizeof(a[0]);
  for (int i = 0; i < size; ++i) {
    printf("%d ", a[i]);
  }
  int count = 0;
  for (int i = 0; i < size; ++i) {
    for (int j = i + 1; j < size; ++j) {
      if (a[i] > a[j])
        count++;
    }
  }
  printf("\nThe number of inversion parirs are %d\n", count);
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x4,(%rsp)
  movl   $0x5,0x4(%rsp)
  movl   $0x2,0x8(%rsp)
  movl   $0x6,0xc(%rsp)
  movl   $0x1,0x10(%rsp)
  mov    %rsp,%rbx
  lea    0x14(%rsp),%r12
  lea    D10(%rip),%rbp
L26:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %rbx,%r12
  jne    L26
  mov    $0x0,%esi
  mov    $0x1,%r8d
  mov    $0x1,%eax
  mov    $0x0,%edi
  jmp    L20
L7:
  add    $0x1,%eax
  cmp    $0x4,%eax
  jg     L23
L20:
  movslq %esi,%rdx
  movslq %eax,%rcx
  mov    (%rsp,%rcx,4),%ebx
  cmp    %ebx,(%rsp,%rdx,4)
  jle    L7
  add    $0x1,%edi
  jmp    L7
L38:
  mov    %edi,%esi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L23:
  lea    0x1(%r8),%eax
  cmp    $0x4,%eax
  jg     L38
  mov    %r8d,%esi
  mov    %eax,%r8d
  jmp    L20
L31:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D10
type: string
name: parse_data
arguments: data_tag: D3
type: string

tool: D10: %d 

tool: D3: 
The number of inversion parirs are %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a[] = {4, 5, 2, 6, 1};
  int size = sizeof(a) / sizeof(a[0]);
  for (int i = 0; i < size; ++i) {
    printf("%d ", a[i]);
  }
  int count = 0;
  for (int i = 0; i < size; ++i) {
    for (int j = i + 1; j < size; ++j) {
      if (a[i] > a[j])
        count++;
    }
  }
  printf("\nThe number of inversion parirs are %d\n", count);
  printf("\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x4,%esi
  push   %rbp
  lea    D16(%rip),%rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x500000004,%rax
  movl   $0x1,0x10(%rsp)
  mov    %rsp,%rbx
  lea    0x14(%rsp),%r12
  mov    %rax,(%rsp)
  movabs $0x600000002,%rax
  mov    %rax,0x8(%rsp)
L7:
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  je     L75
  mov    (%rbx),%esi
  jmp    L7
  nopw   0x0(%rax,%rax,1)
L75:
  mov    $0x4,%ecx
  mov    $0x1,%eax
  mov    $0x1,%edi
  xor    %esi,%esi
  nopl   0x0(%rax)
L44:
  add    $0x1,%eax
  cmp    $0x5,%eax
  je     L32
L33:
  movslq %eax,%rdx
  cmp    (%rsp,%rdx,4),%ecx
  jle    L44
  add    $0x1,%eax
  add    $0x1,%esi
  cmp    $0x5,%eax
  jne    L33
L32:
  lea    0x1(%rdi),%eax
  cmp    $0x4,%edi
  je     L39
  mov    (%rsp,%rdi,4),%ecx
  movslq %eax,%rdi
  jmp    L33
  nopl   0x0(%rax,%rax,1)
L39:
  lea    D5(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D16
datatype: string
name: read_data
arguments: data_tag: D5
datatype: string

tool: D16: %d 

tool: D5: 
The number of inversion parirs are %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a[] = {4, 5, 2, 6, 1};
  int size = sizeof(a) / sizeof(a[0]);
  for (int i = 0; i < size; ++i) {
    printf("%d ", a[i]);
  }
  int count = 0;
  for (int i = 0; i < size; ++i) {
    for (int j = i + 1; j < size; ++j) {
      if (a[i] > a[j])
        count++;
    }
  }
  printf("\nThe number of inversion parirs are %d\n", count);
  printf("\n");
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  lea    D21(%rip),%rbp
  push   %rbx
  sub    $0x20,%rsp
  movdqa D15(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x10(%rsp)
  mov    %rsp,%rbx
  lea    0x14(%rsp),%r12
  movaps %xmm0,(%rsp)
L3:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L3
  mov    $0x4,%ecx
  mov    $0x1,%eax
  mov    $0x1,%edi
  xor    %esi,%esi
  nopl   (%rax)
L41:
  add    $0x1,%eax
  cmp    $0x5,%eax
  je     L35
L46:
  movslq %eax,%rdx
  cmp    (%rsp,%rdx,4),%ecx
  jle    L41
  add    $0x1,%eax
  add    $0x1,%esi
  cmp    $0x5,%eax
  jne    L46
L35:
  lea    0x1(%rdi),%eax
  cmp    $0x4,%edi
  je     L68
  mov    (%rsp,%rdi,4),%ecx
  movslq %eax,%rdi
  jmp    L46
  nopl   0x0(%rax,%rax,1)
L68:
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D21
type: string
name: read_data
arguments: tag: D15
type: byte[16]
name: read_data
arguments: tag: D7
type: string

tool: D21: %d 

tool: {"D15": [0x04,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

tool: D7: 
The number of inversion parirs are %d


assistant: ```c
int main(int argc, char *argv[]) {
  int a[] = {4, 5, 2, 6, 1};
  int size = sizeof(a) / sizeof(a[0]);
  for (int i = 0; i < size; ++i) {
    printf("%d ", a[i]);
  }
  int count = 0;
  for (int i = 0; i < size; ++i) {
    for (int j = i + 1; j < size; ++j) {
      if (a[i] > a[j])
        count++;
    }
  }
  printf("\nThe number of inversion parirs are %d\n", count);
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<skip>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L15
L12:
  addq   $0x1,-0x8(%rbp)
L15:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L12
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *skip(char *line) {
  while (isspace(*line))
    line++;
  return line;
}
```
user: Translate the below assembly code into C programming language:

```asm
<skip>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  jmp    L6
L5:
  add    $0x1,%rbx
L6:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L5
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *skip(char *line) {
  while (isspace(*line))
    line++;
  return line;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<skip>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L2:
  add    $0x1,%r12
L11:
  movsbl (%r12),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L2
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *skip(char *line) {
  while (isspace(*line))
    line++;
  return line;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<skip>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  jmp    L1
  nopl   0x0(%rax,%rax,1)
L8:
  add    $0x1,%r12
L1:
  movsbl (%r12),%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L8
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *skip(char *line) {
  while (isspace(*line))
    line++;
  return line;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<print_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    %ecx,%eax
  mov    %edi,%edx
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x20(%rbp)
  movq   $0xb8000,-0x8(%rbp)
  cmpb   $0x0,-0x20(%rbp)
  jne    L24
  movb   $0xf,-0x20(%rbp)
L24:
  cmpl   $0x0,-0x18(%rbp)
  js     L32
  cmpl   $0x0,-0x1c(%rbp)
  js     L32
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <get_screen_offset@plt>
  mov    %eax,-0x10(%rbp)
  jmp    L2
L32:
  mov    $0x0,%eax
  call   <get_cursor@plt>
  mov    %eax,-0x10(%rbp)
L2:
  cmpb   $0xa,-0x14(%rbp)
  jne    L0
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x6,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  mov    $0x4f,%edi
  call   <get_screen_offset@plt>
  mov    %eax,-0x10(%rbp)
  jmp    L27
L0:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x14(%rbp),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x20(%rbp),%eax
  mov    %al,(%rdx)
L27:
  addl   $0x2,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <handle_scrolling@plt>
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <set_cursor@plt>
  nop
  leave
  ret
```

assistant: ```c
void print_char(char character, int col, int row, char attribute_byte) {
  unsigned char *vidmem = (unsigned char *)0xb8000;
  if (!attribute_byte) {
    attribute_byte = 0x0f;
  }
  int offset;
  if (col >= 0 && row >= 0) {
    offset = get_screen_offset(col, row);
  } else {
    offset = get_cursor();
  }
  if (character == '\n') {
    int rows = offset / (2 * 80);
    offset = get_screen_offset(80 - 1, rows);
  } else {
    vidmem[offset] = character;
    vidmem[offset + 1] = attribute_byte;
  }
  offset += 2;
  offset = handle_scrolling(offset);
  set_cursor(offset);
}
```
user: What is the c source code of the assembly code below:

```asm
<print_char>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  test   %cl,%cl
  mov    $0xf,%eax
  cmovne %ecx,%eax
  mov    %eax,%ebp
  mov    %esi,%eax
  or     %edx,%eax
  js     L31
  mov    %esi,%edi
  mov    %edx,%esi
  call   <get_screen_offset@plt>
L10:
  cmp    $0xa,%bl
  je     L17
  movslq %eax,%rdx
  mov    %bl,0xb8000(%rdx)
  mov    %bpl,0xb8001(%rdx)
L35:
  lea    0x2(%rax),%edi
  call   <handle_scrolling@plt>
  mov    %eax,%edi
  call   <set_cursor@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L31:
  mov    $0x0,%eax
  call   <get_cursor@plt>
  jmp    L10
L17:
  movslq %eax,%rsi
  imul   $0x66666667,%rsi,%rsi
  sar    $0x26,%rsi
  sar    $0x1f,%eax
  sub    %eax,%esi
  mov    $0x4f,%edi
  call   <get_screen_offset@plt>
  jmp    L35
```

assistant: ```c
void print_char(char character, int col, int row, char attribute_byte) {
  unsigned char *vidmem = (unsigned char *)0xb8000;
  if (!attribute_byte) {
    attribute_byte = 0x0f;
  }
  int offset;
  if (col >= 0 && row >= 0) {
    offset = get_screen_offset(col, row);
  } else {
    offset = get_cursor();
  }
  if (character == '\n') {
    int rows = offset / (2 * 80);
    offset = get_screen_offset(80 - 1, rows);
  } else {
    vidmem[offset] = character;
    vidmem[offset + 1] = attribute_byte;
  }
  offset += 2;
  offset = handle_scrolling(offset);
  set_cursor(offset);
}
```