ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<scale2x>:
  endbr64
  push   %r15
  push   %r14
  mov    %ecx,%r14d
  push   %r13
  mov    %r8d,%r13d
  push   %r12
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x38,%rsp
  mov    0x70(%rsp),%r15d
  mov    0x78(%rsp),%edi
  mov    %rdx,0x18(%rsp)
  mov    %r9d,0x4(%rsp)
  mov    %edi,0x28(%rsp)
  xor    %edi,%edi
  cmp    $0x1,%r15d
  seta   %dil
  call   <assert@plt>
  mov    0x28(%rsp),%edi
  mov    %ebp,%ecx
  mov    0x18(%rsp),%rax
  mov    %rcx,0x8(%rsp)
  lea    (%rbx,%rcx,1),%rsi
  mov    %r13d,%r9d
  add    %ebp,%ebp
  push   %rdi
  mov    0xc(%rsp),%edi
  lea    (%rax,%r14,1),%r8
  mov    %rax,%rcx
  mov    %rax,%rdx
  push   %rdi
  mov    %rbx,%rdi
  mov    %r8,0x20(%rsp)
  mov    %rax,0x28(%rsp)
  call   <stage_scale2x@plt>
  lea    -0x2(%r15),%ecx
  lea    (%rbx,%rbp,1),%rdi
  test   %ecx,%ecx
  mov    %rdi,0x30(%rsp)
  mov    %ecx,0x3c(%rsp)
  pop    %rsi
  pop    %r8
  mov    0x10(%rsp),%r8
  je     L17
  mov    %ecx,%ebx
  lea    (%r14,%r14,1),%ecx
  mov    0x18(%rsp),%r15
  mov    %r13d,%eax
  mov    %rcx,0x10(%rsp)
  mov    %eax,%r12d
  mov    %r15,%r13
  mov    %rdi,%r15
  nopl   0x0(%rax,%rax,1)
L0:
  mov    0x10(%rsp),%rax
  mov    %r13,%rdx
  mov    %r15,%rdi
  mov    %r12d,%r9d
  lea    0x0(%r13,%rax,1),%r8
  mov    0x8(%rsp),%rax
  add    %r14,%r13
  mov    %r13,%rcx
  lea    (%r15,%rax,1),%rsi
  mov    0x28(%rsp),%eax
  add    %rbp,%r15
  push   %rax
  mov    0xc(%rsp),%eax
  push   %rax
  call   <stage_scale2x@plt>
  pop    %rax
  pop    %rdx
  sub    $0x1,%ebx
  jne    L0
  mov    0x2c(%rsp),%eax
  mov    %r14,%rdx
  mov    %r12d,%r13d
  imul   %rax,%rbp
  add    %rbp,0x20(%rsp)
  imul   %rax,%rdx
  add    %rdx,0x18(%rsp)
  mov    0x18(%rsp),%rbx
  lea    (%rbx,%r14,1),%r8
L17:
  mov    0x28(%rsp),%eax
  mov    0x8(%rsp),%rsi
  mov    %r13d,%r9d
  mov    %r8,%rcx
  mov    0x20(%rsp),%rdi
  mov    0x18(%rsp),%rdx
  mov    %eax,0x78(%rsp)
  mov    0x4(%rsp),%eax
  add    %rdi,%rsi
  mov    %eax,0x70(%rsp)
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1070 <stage_scale2x@plt>
```

assistant: ```c
void scale2x(void *void_dst, unsigned dst_slice, const void *void_src, unsigned src_slice, unsigned pixel, unsigned width, unsigned height, unsigned opt) {
  unsigned char *dst = (unsigned char *)void_dst;
  const unsigned char *src = (const unsigned char *)void_src;
  unsigned count;
  assert(height >= 2);
  count = height;
  stage_scale2x((dst + (0) * dst_slice), (dst + (1) * dst_slice), (src + (0) * src_slice), (src + (0) * src_slice), (src + (1) * src_slice), pixel, width, opt);
  dst = (dst + (2) * dst_slice);
  count -= 2;
  while (count) {
    stage_scale2x((dst + (0) * dst_slice), (dst + (1) * dst_slice), (src + (0) * src_slice), (src + (1) * src_slice), (src + (2) * src_slice), pixel, width, opt);
    dst = (dst + (2) * dst_slice);
    src = (src + (1) * src_slice);
    --count;
  }
  stage_scale2x((dst + (0) * dst_slice), (dst + (1) * dst_slice), (src + (0) * src_slice), (src + (1) * src_slice), (src + (1) * src_slice), pixel, width, opt);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_next_mlfq_to_assess>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L4
  movl   $0x3,-0x4(%rbp)
L4:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int get_next_mlfq_to_assess(int mlfq_to_assess) {
  int next_mlfq = (mlfq_to_assess - 1);
  if (next_mlfq < 0) {
    next_mlfq = 4 - 1;
  }
  return next_mlfq;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<get_next_mlfq_to_assess>:
  endbr64
  mov    %edi,%eax
  sub    $0x1,%eax
  mov    $0x3,%edx
  cmovs  %edx,%eax
  ret
```

assistant: ```c
int get_next_mlfq_to_assess(int mlfq_to_assess) {
  int next_mlfq = (mlfq_to_assess - 1);
  if (next_mlfq < 0) {
    next_mlfq = 4 - 1;
  }
  return next_mlfq;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<get_next_mlfq_to_assess>:
  endbr64
  mov    %edi,%eax
  mov    $0x3,%edx
  sub    $0x1,%eax
  cmovs  %edx,%eax
  ret
```

assistant: ```c
int get_next_mlfq_to_assess(int mlfq_to_assess) {
  int next_mlfq = (mlfq_to_assess - 1);
  if (next_mlfq < 0) {
    next_mlfq = 4 - 1;
  }
  return next_mlfq;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_next_mlfq_to_assess>:
  endbr64
  mov    %edi,%eax
  mov    $0x3,%edx
  sub    $0x1,%eax
  cmovs  %edx,%eax
  ret
```

assistant: ```c
int get_next_mlfq_to_assess(int mlfq_to_assess) {
  int next_mlfq = (mlfq_to_assess - 1);
  if (next_mlfq < 0) {
    next_mlfq = 4 - 1;
  }
  return next_mlfq;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
L8:
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1b0(%rbp),%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1b0(%rbp),%eax
  cmp    $0x63,%eax
  jg     L8
  movl   $0x0,-0x1ac(%rbp)
  jmp    L39
L28:
  lea    -0x1a0(%rbp),%rdx
  mov    -0x1ac(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x1ac(%rbp)
L39:
  mov    -0x1b0(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L28
  lea    -0x1a0(%rbp),%rax
  mov    %rax,-0x1a8(%rbp)
  mov    -0x1b0(%rbp),%edx
  mov    -0x1a8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sort@plt>
  movl   $0x0,-0x1ac(%rbp)
  jmp    L1
L42:
  mov    -0x1ac(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x1a8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x1ac(%rbp)
L1:
  mov    -0x1b0(%rbp),%eax
  cmp    %eax,-0x1ac(%rbp)
  jl     L42
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L48
  call   <__stack_chk_fail@plt>
L48:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D39
type: string
name: disassemble_data
arguments: data_label: D32
type: string

tool: D39: Unesite n: 

tool: D32: %d

assistant: ```c
int main() {
  int n, i;
  int *p;
  int niz[100];
  do {
    printf("Unesite n: ");
    scanf("%d", &n);
  } while (n >= 100);
  for (i = 0; i < n; i++) {
    scanf("%d", &niz[i]);
  }
  p = &niz[0];
  sort(p, n);
  for (i = 0; i < n; i++)
    printf("%d", *(p + i));
  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
  push   %rbp
  push   %rbx
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    D18(%rip),%r12
  lea    0xc(%rsp),%rbp
  lea    D32(%rip),%rbx
L0:
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    $0x63,%esi
  jg     L0
  test   %esi,%esi
  jle    L20
  lea    0x10(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D32(%rip),%r12
L40:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  mov    0xc(%rsp),%esi
  add    $0x4,%rbp
  cmp    %ebx,%esi
  jg     L40
L20:
  lea    0x10(%rsp),%rdi
  call   <sort@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L30
  mov    $0x0,%ebx
  lea    D32(%rip),%rbp
L29:
  mov    0x10(%rsp,%rbx,4),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,0xc(%rsp)
  jg     L29
L30:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  mov    $0x0,%eax
  add    $0x1b0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D18
type_guessed: string
name: parse_data
arguments: tag: D32
type_guessed: string

tool: D18: Unesite n: 

tool: D32: %d

assistant: ```c
int main() {
  int n, i;
  int *p;
  int niz[100];
  do {
    printf("Unesite n: ");
    scanf("%d", &n);
  } while (n >= 100);
  for (i = 0; i < n; i++) {
    scanf("%d", &niz[i]);
  }
  p = &niz[0];
  sort(p, n);
  for (i = 0; i < n; i++)
    printf("%d", *(p + i));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D11(%rip),%r12
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbx
  nopw   0x0(%rax,%rax,1)
L35:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    $0x63,%esi
  jg     L35
  lea    0x10(%rsp),%r13
  xor    %ebx,%ebx
  mov    %r13,%r12
  test   %esi,%esi
  jle    L6
  nopl   0x0(%rax)
L38:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  add    $0x4,%r12
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    %ebx,%esi
  jg     L38
L6:
  mov    %r13,%rdi
  call   <sort@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L40
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L48:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L48
L40:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x1b8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D11
data_type: string
name: retrieve_data
arguments: label: D5
data_type: string

tool: D11: Unesite n: 

tool: D5: %d

assistant: ```c
int main() {
  int n, i;
  int *p;
  int niz[100];
  do {
    printf("Unesite n: ");
    scanf("%d", &n);
  } while (n >= 100);
  for (i = 0; i < n; i++) {
    scanf("%d", &niz[i]);
  }
  p = &niz[0];
  sort(p, n);
  for (i = 0; i < n; i++)
    printf("%d", *(p + i));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  lea    D10(%rip),%r12
  push   %rbp
  lea    D6(%rip),%rbp
  push   %rbx
  sub    $0x1b8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rbx
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    $0x63,%esi
  jg     L4
  lea    0x10(%rsp),%r13
  xor    %ebx,%ebx
  mov    %r13,%r12
  test   %esi,%esi
  jle    L42
  nopl   0x0(%rax)
L29:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  add    $0x4,%r12
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  cmp    %ebx,%esi
  jg     L29
L42:
  mov    %r13,%rdi
  call   <sort@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L25
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L26:
  mov    0x0(%r13,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L26
L25:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x1b8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D10
datatype: string
name: extract_symbolic_data
arguments: data_tag: D6
datatype: string

tool: D10: Unesite n: 

tool: D6: %d

assistant: ```c
int main() {
  int n, i;
  int *p;
  int niz[100];
  do {
    printf("Unesite n: ");
    scanf("%d", &n);
  } while (n >= 100);
  for (i = 0; i < n; i++) {
    scanf("%d", &niz[i]);
  }
  p = &niz[0];
  sort(p, n);
  for (i = 0; i < n; i++)
    printf("%d", *(p + i));
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<MultM>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %rdi,-0x68(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %rdx,-0x78(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x58(%rbp)
  jmp    L45
L10:
  movl   $0x0,-0x54(%rbp)
  jmp    L28
L21:
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x70(%rbp),%rax
  mov    -0x54(%rbp),%edx
  movslq %edx,%rdx
  movsd  (%rax,%rdx,8),%xmm0
  mulsd  %xmm0,%xmm1
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  movsd  0x8(%rax),%xmm2
  mov    -0x70(%rbp),%rax
  lea    0x18(%rax),%rdx
  mov    -0x54(%rbp),%eax
  cltq
  movsd  (%rdx,%rax,8),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm0,%xmm1
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  shl    $0x3,%rax
  mov    %rax,%rdx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  movsd  0x10(%rax),%xmm2
  mov    -0x70(%rbp),%rax
  lea    0x30(%rax),%rdx
  mov    -0x54(%rbp),%eax
  cltq
  movsd  (%rdx,%rax,8),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  mov    -0x54(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rcx,%rax
  movsd  %xmm0,-0x50(%rbp,%rax,8)
  addl   $0x1,-0x54(%rbp)
L28:
  cmpl   $0x2,-0x54(%rbp)
  jle    L21
  addl   $0x1,-0x58(%rbp)
L45:
  cmpl   $0x2,-0x58(%rbp)
  jle    L10
  mov    -0x78(%rbp),%rdx
  lea    -0x50(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <CopyM@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
void MultM(double a[3][3], double b[3][3], double result[3][3]) {
  int i, j;
  double at[3][3];
  for (i = 0; i <= 2; i++) {
    for (j = 0; j <= 2; j++)
      at[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j] + a[i][2] * b[2][j];
  }
  CopyM(at, result);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<MultM>:
  endbr64
  sub    $0x58,%rsp
  mov    %rsi,%rcx
  mov    %rdx,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r8
  lea    0x48(%rsp),%r9
  jmp    L9
L37:
  add    $0x18,%rdi
  add    $0x18,%r8
  cmp    %r9,%r8
  je     L10
L9:
  movsd  (%rdi),%xmm4
  movsd  0x8(%rdi),%xmm3
  movsd  0x10(%rdi),%xmm2
  mov    $0x0,%eax
L1:
  movapd %xmm4,%xmm0
  mulsd  (%rcx,%rax,1),%xmm0
  movapd %xmm3,%xmm1
  mulsd  0x18(%rcx,%rax,1),%xmm1
  addsd  %xmm1,%xmm0
  movapd %xmm2,%xmm1
  mulsd  0x30(%rcx,%rax,1),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%r8,%rax,1)
  add    $0x8,%rax
  cmp    $0x18,%rax
  jne    L1
  jmp    L37
L10:
  mov    %rsp,%rdi
  call   <CopyM@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x58,%rsp
  ret
L30:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void MultM(double a[3][3], double b[3][3], double result[3][3]) {
  int i, j;
  double at[3][3];
  for (i = 0; i <= 2; i++) {
    for (j = 0; j <= 2; j++)
      at[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j] + a[i][2] * b[2][j];
  }
  CopyM(at, result);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<MultM>:
  endbr64
  sub    $0x58,%rsp
  mov    %rsi,%rcx
  mov    %rdx,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r10
  lea    0x48(%rsp),%r9
  mov    %r10,%r8
L23:
  movsd  (%rdi),%xmm4
  movsd  0x8(%rdi),%xmm3
  xor    %eax,%eax
  movsd  0x10(%rdi),%xmm2
L27:
  movsd  (%rcx,%rax,1),%xmm0
  movsd  0x18(%rcx,%rax,1),%xmm1
  mulsd  %xmm4,%xmm0
  mulsd  %xmm3,%xmm1
  addsd  %xmm1,%xmm0
  movsd  0x30(%rcx,%rax,1),%xmm1
  mulsd  %xmm2,%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,(%r8,%rax,1)
  add    $0x8,%rax
  cmp    $0x18,%rax
  jne    L27
  add    $0x18,%r8
  add    $0x18,%rdi
  cmp    %r8,%r9
  jne    L23
  mov    %r10,%rdi
  call   <CopyM@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x58,%rsp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void MultM(double a[3][3], double b[3][3], double result[3][3]) {
  int i, j;
  double at[3][3];
  for (i = 0; i <= 2; i++) {
    for (j = 0; j <= 2; j++)
      at[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j] + a[i][2] * b[2][j];
  }
  CopyM(at, result);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<MultM>:
  endbr64
  mov    %rsi,%rax
  sub    $0x58,%rsp
  movupd 0x10(%rdi),%xmm1
  mov    %rdx,%rsi
  movupd (%rax),%xmm6
  movupd 0x10(%rax),%xmm0
  movupd 0x20(%rax),%xmm4
  movlpd (%rdi),%xmm1
  movupd 0x18(%rax),%xmm7
  unpcklpd %xmm6,%xmm0
  movupd 0x10(%rax),%xmm10
  movupd 0x20(%rdi),%xmm9
  mov    %fs:0x28,%rdx
  mov    %rdx,0x48(%rsp)
  xor    %edx,%edx
  mulpd  %xmm0,%xmm1
  movapd %xmm4,%xmm8
  movupd (%rdi),%xmm0
  unpckhpd %xmm10,%xmm8
  movupd 0x30(%rax),%xmm5
  movupd 0x8(%rax),%xmm3
  shufpd $0x1,%xmm9,%xmm0
  mulpd  %xmm8,%xmm0
  movapd %xmm9,%xmm8
  movupd 0x38(%rax),%xmm2
  movlpd 0x10(%rdi),%xmm8
  addpd  %xmm0,%xmm1
  movsd  0x40(%rax),%xmm0
  movhpd 0x30(%rax),%xmm0
  mulpd  %xmm8,%xmm0
  addpd  %xmm0,%xmm1
  movupd (%rdi),%xmm0
  unpcklpd %xmm0,%xmm0
  mulpd  %xmm6,%xmm0
  movaps %xmm1,0x10(%rsp)
  movapd %xmm9,%xmm1
  unpcklpd %xmm9,%xmm1
  mulpd  %xmm4,%xmm1
  unpckhpd %xmm4,%xmm4
  movapd %xmm0,%xmm8
  movupd (%rdi),%xmm0
  unpckhpd %xmm0,%xmm0
  mulpd  %xmm7,%xmm0
  addpd  %xmm8,%xmm0
  movupd 0x10(%rdi),%xmm8
  unpcklpd %xmm8,%xmm8
  mulpd  %xmm5,%xmm8
  addpd  %xmm8,%xmm0
  movsd  0x40(%rdi),%xmm8
  movaps %xmm0,(%rsp)
  movupd 0x10(%rdi),%xmm0
  unpckhpd %xmm0,%xmm0
  mulpd  %xmm3,%xmm0
  unpckhpd %xmm3,%xmm3
  addpd  %xmm1,%xmm0
  movapd %xmm9,%xmm1
  unpckhpd %xmm9,%xmm1
  mulpd  %xmm2,%xmm1
  unpckhpd %xmm2,%xmm2
  mulsd  %xmm8,%xmm2
  addpd  %xmm1,%xmm0
  movaps %xmm0,0x20(%rsp)
  movupd 0x30(%rdi),%xmm0
  mov    %rsp,%rdi
  mulsd  %xmm0,%xmm3
  movapd %xmm0,%xmm1
  unpckhpd %xmm0,%xmm1
  mulpd  %xmm7,%xmm1
  movapd %xmm0,%xmm7
  unpcklpd %xmm0,%xmm7
  mulpd  %xmm7,%xmm6
  addpd  %xmm6,%xmm1
  movapd %xmm8,%xmm6
  unpcklpd %xmm6,%xmm6
  mulpd  %xmm6,%xmm5
  addpd  %xmm5,%xmm1
  movapd %xmm0,%xmm5
  unpckhpd %xmm5,%xmm5
  movaps %xmm1,0x30(%rsp)
  movapd %xmm5,%xmm1
  mulsd  %xmm4,%xmm1
  addsd  %xmm3,%xmm1
  movapd %xmm1,%xmm0
  addsd  %xmm2,%xmm0
  movsd  %xmm0,0x40(%rsp)
  call   <CopyM@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x58,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void MultM(double a[3][3], double b[3][3], double result[3][3]) {
  int i, j;
  double at[3][3];
  for (i = 0; i <= 2; i++) {
    for (j = 0; j <= 2; j++)
      at[i][j] = a[i][0] * b[0][j] + a[i][1] * b[1][j] + a[i][2] * b[2][j];
  }
  CopyM(at, result);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<outc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <putc@plt>
  leave
  ret
```

assistant: ```c
int outc(int c, void *cl) {
  FILE *f = cl;
  return putc(c, f);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<outc>:
  endbr64
  sub    $0x8,%rsp
  call   <putc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int outc(int c, void *cl) {
  FILE *f = cl;
  return putc(c, f);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<outc>:
  endbr64
  jmp    1050 <putc@plt>
```

assistant: ```c
int outc(int c, void *cl) {
  FILE *f = cl;
  return putc(c, f);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<outc>:
  endbr64
  jmp    1050 <putc@plt>
```

assistant: ```c
int outc(int c, void *cl) {
  FILE *f = cl;
  return putc(c, f);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<taa_h264_inverse_hadamard_and_dequant_4x4>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r8
  movslq %esi,%rsi
  lea    D4(%rip),%rax
  mov    (%rax,%rsi,4),%edi
  mov    %r8,%rcx
  lea    0x20(%r8),%rbp
  mov    %r8,%rax
L19:
  movswl (%rax),%edx
  movswl 0x4(%rax),%ebx
  movswl 0x2(%rax),%esi
  movswl 0x6(%rax),%r11d
  lea    (%rdx,%rbx,1),%r9d
  lea    (%rsi,%r11,1),%r10d
  lea    (%r10,%r9,1),%r12d
  mov    %r12w,(%rax)
  sub    %ebx,%edx
  sub    %r11d,%esi
  lea    (%rsi,%rdx,1),%r11d
  mov    %r11w,0x2(%rax)
  sub    %esi,%edx
  mov    %dx,0x4(%rax)
  sub    %r10d,%r9d
  mov    %r9w,0x6(%rax)
  add    $0x8,%rax
  cmp    %rbp,%rax
  jne    L19
  add    $0x8,%r8
L16:
  movswl (%rcx),%eax
  movswl 0x10(%rcx),%edx
  lea    (%rax,%rdx,1),%esi
  sub    %edx,%eax
  movswl 0x8(%rcx),%r9d
  movswl 0x18(%rcx),%edx
  mov    %r9d,%r10d
  sub    %edx,%r10d
  add    %edx,%r9d
  lea    (%rsi,%r9,1),%edx
  imul   %edi,%edx
  add    $0x2,%edx
  sar    $0x2,%edx
  mov    %dx,(%rcx)
  lea    (%rax,%r10,1),%edx
  imul   %edi,%edx
  add    $0x2,%edx
  sar    $0x2,%edx
  mov    %dx,0x8(%rcx)
  sub    %r10d,%eax
  mov    %eax,%edx
  imul   %edi,%edx
  add    $0x2,%edx
  sar    $0x2,%edx
  mov    %dx,0x10(%rcx)
  mov    %esi,%eax
  sub    %r9d,%eax
  imul   %edi,%eax
  add    $0x2,%eax
  sar    $0x2,%eax
  mov    %ax,0x18(%rcx)
  add    $0x2,%rcx
  cmp    %r8,%rcx
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D4
type: string

tool: D4: 


assistant: ```c
void taa_h264_inverse_hadamard_and_dequant_4x4(short *block, int qp) {
  int tmp0, tmp1, tmp2, tmp3;
  static const int dequant_coef_dc[52] = {10 << 0, 11 << 0, 13 << 0, 14 << 0, 16 << 0, 18 << 0, 10 << 1, 11 << 1, 13 << 1, 14 << 1, 16 << 1, 18 << 1, 10 << 2, 11 << 2, 13 << 2, 14 << 2, 16 << 2, 18 << 2, 10 << 3, 11 << 3, 13 << 3, 14 << 3, 16 << 3, 18 << 3, 10 << 4, 11 << 4, 13 << 4, 14 << 4, 16 << 4, 18 << 4, 10 << 5, 11 << 5, 13 << 5, 14 << 5, 16 << 5, 18 << 5, 10 << 6, 11 << 6, 13 << 6, 14 << 6, 16 << 6, 18 << 6, 10 << 7, 11 << 7, 13 << 7, 14 << 7, 16 << 7, 18 << 7, 10 << 8, 11 << 8, 13 << 8, 14 << 8};
  int dqc = dequant_coef_dc[qp];
  for (int i = 0; i < 4; i++) {
    tmp0 = block[i * 4 + 0] + block[i * 4 + 2];
    tmp1 = block[i * 4 + 0] - block[i * 4 + 2];
    tmp2 = block[i * 4 + 1] - block[i * 4 + 3];
    tmp3 = block[i * 4 + 1] + block[i * 4 + 3];
    block[i * 4 + 0] = (short)(tmp0 + tmp3);
    block[i * 4 + 1] = (short)(tmp1 + tmp2);
    block[i * 4 + 2] = (short)(tmp1 - tmp2);
    block[i * 4 + 3] = (short)(tmp0 - tmp3);
  }
  for (int i = 0; i < 4; i++) {
    tmp0 = block[0 * 4 + i] + block[2 * 4 + i];
    tmp1 = block[0 * 4 + i] - block[2 * 4 + i];
    tmp2 = block[1 * 4 + i] - block[3 * 4 + i];
    tmp3 = block[1 * 4 + i] + block[3 * 4 + i];
    block[0 * 4 + i] = (short)(((tmp0 + tmp3) * dqc + 2) >> 2);
    block[1 * 4 + i] = (short)(((tmp1 + tmp2) * dqc + 2) >> 2);
    block[2 * 4 + i] = (short)(((tmp1 - tmp2) * dqc + 2) >> 2);
    block[3 * 4 + i] = (short)(((tmp0 - tmp3) * dqc + 2) >> 2);
  }
}
```
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    D29(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0xd(%rbp),%rax
  mov    %rax,%rdi
  call   <gets@plt>
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0xd(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0xd(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D29
type: string
name: unpack_data
arguments: label: D20
type: string
name: unpack_data
arguments: label: D28
type: string

tool: D29: Ente a string, please.

tool: D20: Your string twice:

tool: D28: Done.

assistant: ```c
int main(void) {
  char words[5];
  puts("Ente a string, please.");
  gets(words);
  printf("Your string twice:\n");
  printf("%s\n", words);
  puts(words);
  puts("Done.");
  return 0;
}
```