ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<xrealloc>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  imul   %rdx,%r13
  mov    %rdi,%r12
  push   %rbp
  mov    %rdx,%rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x8,%rsp
  test   %r13,%r13
  je     L38
L28:
  mov    %rbp,%rax
  mul    %rbx
  jo     L68
L66:
  test   %r12,%r12
  je     L8
  mov    %r12,%rdi
  mov    %r13,%rsi
  call   <realloc@plt>
  mov    %rax,%r12
L76:
  test   %r12,%r12
  je     L69
L18:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L38:
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <panic@plt>
  jmp    L28
  nopl   0x0(%rax,%rax,1)
L8:
  mov    %r13,%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  jmp    L76
L68:
  lea    D9(%rip),%rdi
  xor    %eax,%eax
  call   <panic@plt>
  jmp    L66
L69:
  mov    %r13,%rsi
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  call   <panic@plt>
  jmp    L18
```

tool_calls: name: read_data
arguments: data_label: D4
type_guessed: string
name: read_data
arguments: data_label: D9
type_guessed: string
name: read_data
arguments: data_label: D26
type_guessed: string

tool: D4: xrealloc: zero size


tool: D9: xrealloc: nmemb * size > SIZE_T_MAX


tool: D26: xrealloc: out of memory (size %zu bytes)


assistant: ```c
__attribute__((visibility("hidden"))) void *xrealloc(void *ptr, size_t nmemb, size_t size) {
  void *new_ptr;
  size_t new_size = nmemb * size;
  if (new_size == 0)
    panic("xrealloc: zero size\n");
  if (((size_t)~0) / nmemb < size)
    panic("xrealloc: nmemb * size > SIZE_T_MAX\n");
  if (ptr == NULL)
    new_ptr = malloc(new_size);
  else
    new_ptr = realloc(ptr, new_size);
  if (new_ptr == NULL)
    panic("xrealloc: out of memory (size %zu bytes)\n", new_size);
  return new_ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<vec3_rotateY>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  movss  %xmm0,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x28(%rbp),%rax
  movss  (%rax),%xmm0
  mov    -0x30(%rbp),%rax
  movss  (%rax),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x30(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x18(%rbp)
  movss  -0x18(%rbp),%xmm4
  movss  %xmm4,-0x38(%rbp)
  mov    -0x34(%rbp),%eax
  movd   %eax,%xmm0
  call   <sinf@plt>
  mulss  -0x38(%rbp),%xmm0
  movss  %xmm0,-0x38(%rbp)
  movss  -0x20(%rbp),%xmm5
  movss  %xmm5,-0x3c(%rbp)
  mov    -0x34(%rbp),%eax
  movd   %eax,%xmm0
  call   <cosf@plt>
  mulss  -0x3c(%rbp),%xmm0
  addss  -0x38(%rbp),%xmm0
  movss  %xmm0,-0x14(%rbp)
  movss  -0x1c(%rbp),%xmm0
  movss  %xmm0,-0x10(%rbp)
  movss  -0x18(%rbp),%xmm6
  movss  %xmm6,-0x38(%rbp)
  mov    -0x34(%rbp),%eax
  movd   %eax,%xmm0
  call   <cosf@plt>
  mulss  -0x38(%rbp),%xmm0
  movss  %xmm0,-0x38(%rbp)
  movss  -0x20(%rbp),%xmm3
  movss  %xmm3,-0x3c(%rbp)
  mov    -0x34(%rbp),%eax
  movd   %eax,%xmm0
  call   <sinf@plt>
  movss  -0x3c(%rbp),%xmm1
  mulss  %xmm0,%xmm1
  movss  -0x38(%rbp),%xmm0
  subss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  movss  -0x14(%rbp),%xmm1
  mov    -0x30(%rbp),%rax
  movss  (%rax),%xmm0
  addss  %xmm1,%xmm0
  mov    -0x28(%rbp),%rax
  movss  %xmm0,(%rax)
  movss  -0x10(%rbp),%xmm1
  mov    -0x30(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mov    -0x28(%rbp),%rax
  add    $0x4,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  movss  -0xc(%rbp),%xmm1
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  movss  (%rax),%xmm0
  mov    -0x28(%rbp),%rax
  add    $0x8,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
void vec3_rotateY(float *dst, float *b, float c) {
  float p[3], r[3];
  p[0] = dst[0] - b[0];
  p[1] = dst[1] - b[1];
  p[2] = dst[2] - b[2];
  r[0] = p[2] * sinf(c) + p[0] * cosf(c);
  r[1] = p[1];
  r[2] = p[2] * cosf(c) - p[0] * sinf(c);
  dst[0] = r[0] + b[0];
  dst[1] = r[1] + b[1];
  dst[2] = r[2] + b[2];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<vec3_rotateY>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  lea    0xc(%rsp),%rdi
  lea    0x8(%rsp),%rsi
  call   <sincosf@plt>
  movss  0x8(%rsp),%xmm5
  movss  0xc(%rsp),%xmm4
  movss  0x0(%rbp),%xmm6
  movss  (%rbx),%xmm1
  subss  %xmm6,%xmm1
  movss  0x4(%rbx),%xmm0
  movaps %xmm0,%xmm2
  subss  0x4(%rbp),%xmm2
  movss  0x8(%rbx),%xmm0
  subss  0x8(%rbp),%xmm0
  movaps %xmm0,%xmm3
  mulss  %xmm4,%xmm3
  movaps %xmm1,%xmm7
  mulss  %xmm5,%xmm7
  addss  %xmm7,%xmm3
  addss  %xmm6,%xmm3
  movss  %xmm3,(%rbx)
  addss  0x4(%rbp),%xmm2
  movss  %xmm2,0x4(%rbx)
  mulss  %xmm5,%xmm0
  mulss  %xmm4,%xmm1
  subss  %xmm1,%xmm0
  addss  0x8(%rbp),%xmm0
  movss  %xmm0,0x8(%rbx)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void vec3_rotateY(float *dst, float *b, float c) {
  float p[3], r[3];
  p[0] = dst[0] - b[0];
  p[1] = dst[1] - b[1];
  p[2] = dst[2] - b[2];
  r[0] = p[2] * sinf(c) + p[0] * cosf(c);
  r[1] = p[1];
  r[2] = p[2] * cosf(c) - p[0] * sinf(c);
  dst[0] = r[0] + b[0];
  dst[1] = r[1] + b[1];
  dst[2] = r[2] + b[2];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vec3_rotateY>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  lea    0xc(%rsp),%rdi
  lea    0x8(%rsp),%rsi
  call   <sincosf@plt>
  movss  0x0(%rbp),%xmm6
  movss  0x4(%rbx),%xmm0
  movss  (%rbx),%xmm1
  movss  0x8(%rsp),%xmm5
  movaps %xmm0,%xmm2
  movss  0x8(%rbx),%xmm0
  subss  0x8(%rbp),%xmm0
  subss  %xmm6,%xmm1
  movss  0xc(%rsp),%xmm4
  subss  0x4(%rbp),%xmm2
  movaps %xmm0,%xmm3
  mulss  %xmm4,%xmm3
  movaps %xmm1,%xmm7
  mulss  %xmm5,%xmm7
  mulss  %xmm5,%xmm0
  mulss  %xmm4,%xmm1
  addss  %xmm7,%xmm3
  subss  %xmm1,%xmm0
  addss  %xmm6,%xmm3
  movss  %xmm3,(%rbx)
  addss  0x4(%rbp),%xmm2
  movss  %xmm2,0x4(%rbx)
  addss  0x8(%rbp),%xmm0
  movss  %xmm0,0x8(%rbx)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void vec3_rotateY(float *dst, float *b, float c) {
  float p[3], r[3];
  p[0] = dst[0] - b[0];
  p[1] = dst[1] - b[1];
  p[2] = dst[2] - b[2];
  r[0] = p[2] * sinf(c) + p[0] * cosf(c);
  r[1] = p[1];
  r[2] = p[2] * cosf(c) - p[0] * sinf(c);
  dst[0] = r[0] + b[0];
  dst[1] = r[1] + b[1];
  dst[2] = r[2] + b[2];
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<vec3_rotateY>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  lea    0xc(%rsp),%rdi
  lea    0x8(%rsp),%rsi
  call   <sincosf@plt>
  movss  0x0(%rbp),%xmm6
  movss  0x4(%rbx),%xmm0
  movss  (%rbx),%xmm1
  movss  0x8(%rsp),%xmm5
  movaps %xmm0,%xmm2
  movss  0x8(%rbx),%xmm0
  subss  0x8(%rbp),%xmm0
  subss  %xmm6,%xmm1
  movss  0xc(%rsp),%xmm4
  subss  0x4(%rbp),%xmm2
  movaps %xmm0,%xmm3
  mulss  %xmm4,%xmm3
  movaps %xmm1,%xmm7
  mulss  %xmm5,%xmm7
  mulss  %xmm5,%xmm0
  mulss  %xmm4,%xmm1
  addss  %xmm7,%xmm3
  subss  %xmm1,%xmm0
  addss  %xmm6,%xmm3
  movss  %xmm3,(%rbx)
  addss  0x4(%rbp),%xmm2
  movss  %xmm2,0x4(%rbx)
  addss  0x8(%rbp),%xmm0
  movss  %xmm0,0x8(%rbx)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void vec3_rotateY(float *dst, float *b, float c) {
  float p[3], r[3];
  p[0] = dst[0] - b[0];
  p[1] = dst[1] - b[1];
  p[2] = dst[2] - b[2];
  r[0] = p[2] * sinf(c) + p[0] * cosf(c);
  r[1] = p[1];
  r[2] = p[2] * cosf(c) - p[0] * sinf(c);
  dst[0] = r[0] + b[0];
  dst[1] = r[1] + b[1];
  dst[2] = r[2] + b[2];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  movq   $0x1,-0x10(%rbp)
  jmp    L16
L24:
  lea    -0x28(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x28(%rbp),%rax
  imul   -0x20(%rbp),%rax
  mov    %rax,%rcx
  movabs $0x89705f3ee868eb67,%rdx
  mov    %rcx,%rax
  imul   %rdx
  lea    (%rdx,%rcx,1),%rax
  sar    $0x1d,%rax
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  imul   $0x3b9aca01,%rax,%rdx
  mov    %rcx,%rax
  sub    %rdx,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <gcd@plt>
  mov    %rax,-0x18(%rbp)
  addq   $0x1,-0x10(%rbp)
L16:
  mov    -0x30(%rbp),%rax
  cmp    %rax,-0x10(%rbp)
  jl     L24
  mov    -0x18(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    $0x3b9aca01,%edx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <modPower@plt>
  mov    %eax,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D18
type_infered: string

tool: D18: %lld

assistant: ```c
int main() {
  long long int prod, tgcd, i, j, sum, temp, c, num, ele;
  scanf("%lld", &num);
  scanf("%lld", &ele);
  tgcd = ele;
  prod = ele;
  for (i = 1; i < num; i++) {
    scanf("%lld", &ele);
    prod = (prod * ele) % 1000000001;
    tgcd = gcd(tgcd, ele);
  }
  printf("%lld", modPower(prod, tgcd, 1000000001));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    D9(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%rbx
  cmpq   $0x1,0x8(%rsp)
  jle    L8
  mov    %rbx,%rbp
  mov    $0x1,%r12d
  lea    0x10(%rsp),%r15
  lea    D9(%rip),%r14
  movabs $0x89705f3ee868eb67,%r13
L17:
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%rsi
  mov    %rbx,%rcx
  imul   %rsi,%rcx
  mov    %rcx,%rax
  imul   %r13
  lea    (%rdx,%rcx,1),%rbx
  sar    $0x1d,%rbx
  mov    %rcx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rbx
  imul   $0x3b9aca01,%rbx,%rax
  sub    %rax,%rcx
  mov    %rcx,%rbx
  mov    %rbp,%rdi
  call   <gcd@plt>
  mov    %rax,%rbp
  add    $0x1,%r12
  cmp    %r12,0x8(%rsp)
  jg     L17
L30:
  mov    $0x3b9aca01,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <modPower@plt>
  mov    %eax,%esi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L8:
  mov    %rbx,%rbp
  jmp    L30
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D9
datatype: string

tool: D9: %lld

assistant: ```c
int main() {
  long long int prod, tgcd, i, j, sum, temp, c, num, ele;
  scanf("%lld", &num);
  scanf("%lld", &ele);
  tgcd = ele;
  prod = ele;
  for (i = 1; i < num; i++) {
    scanf("%lld", &ele);
    prod = (prod * ele) % 1000000001;
    tgcd = gcd(tgcd, ele);
  }
  printf("%lld", modPower(prod, tgcd, 1000000001));
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  lea    D6(%rip),%r15
  push   %r14
  mov    %r15,%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rbx
  call   <scanf@plt>
  mov    %rbx,%rsi
  mov    %r15,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%rbp
  cmpq   $0x1,0x8(%rsp)
  mov    %rbp,%r13
  jle    L11
  movabs $0x89705f3ee868eb67,%r12
  mov    $0x1,%r14d
  cs nopw 0x0(%rax,%rax,1)
L29:
  mov    %rbx,%rsi
  mov    %r15,%rdi
  add    $0x1,%r14
  call   <scanf@plt>
  mov    0x10(%rsp),%rsi
  mov    %rbp,%rcx
  mov    %r13,%rdi
  imul   %rsi,%rcx
  mov    %rcx,%rax
  imul   %r12
  mov    %rcx,%rax
  sar    $0x3f,%rax
  lea    (%rdx,%rcx,1),%rbp
  sar    $0x1d,%rbp
  sub    %rax,%rbp
  imul   $0x3b9aca01,%rbp,%rax
  sub    %rax,%rcx
  mov    %rcx,%rbp
  call   <gcd@plt>
  mov    %rax,%r13
  cmp    %r14,0x8(%rsp)
  jg     L29
L11:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  mov    $0x3b9aca01,%edx
  call   <modPower@plt>
  mov    %r15,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D6
type_guessed: string

tool: D6: %lld

assistant: ```c
int main() {
  long long int prod, tgcd, i, j, sum, temp, c, num, ele;
  scanf("%lld", &num);
  scanf("%lld", &ele);
  tgcd = ele;
  prod = ele;
  for (i = 1; i < num; i++) {
    scanf("%lld", &ele);
    prod = (prod * ele) % 1000000001;
    tgcd = gcd(tgcd, ele);
  }
  printf("%lld", modPower(prod, tgcd, 1000000001));
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  lea    D2(%rip),%r15
  push   %r14
  mov    %r15,%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x8(%rsp),%rsi
  lea    0x10(%rsp),%rbx
  call   <scanf@plt>
  mov    %rbx,%rsi
  mov    %r15,%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%rbp
  cmpq   $0x1,0x8(%rsp)
  mov    %rbp,%r13
  jle    L4
  movabs $0x89705f3ee868eb67,%r12
  mov    $0x1,%r14d
  cs nopw 0x0(%rax,%rax,1)
L22:
  mov    %rbx,%rsi
  mov    %r15,%rdi
  add    $0x1,%r14
  call   <scanf@plt>
  mov    0x10(%rsp),%rsi
  mov    %rbp,%rcx
  mov    %r13,%rdi
  imul   %rsi,%rcx
  mov    %rcx,%rax
  imul   %r12
  mov    %rcx,%rax
  sar    $0x3f,%rax
  lea    (%rdx,%rcx,1),%rbp
  sar    $0x1d,%rbp
  sub    %rax,%rbp
  imul   $0x3b9aca01,%rbp,%rax
  sub    %rax,%rcx
  mov    %rcx,%rbp
  call   <gcd@plt>
  mov    %rax,%r13
  cmp    %r14,0x8(%rsp)
  jg     L22
L4:
  mov    %r13,%rsi
  mov    %rbp,%rdi
  mov    $0x3b9aca01,%edx
  call   <modPower@plt>
  mov    %r15,%rdi
  mov    %eax,%esi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D2
type: string

tool: D2: %lld

assistant: ```c
int main() {
  long long int prod, tgcd, i, j, sum, temp, c, num, ele;
  scanf("%lld", &num);
  scanf("%lld", &ele);
  tgcd = ele;
  prod = ele;
  for (i = 1; i < num; i++) {
    scanf("%lld", &ele);
    prod = (prod * ele) % 1000000001;
    tgcd = gcd(tgcd, ele);
  }
  printf("%lld", modPower(prod, tgcd, 1000000001));
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<version>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <show_version@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void version() {
  show_version();
  exit(0);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<version>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <show_version@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void version() {
  show_version();
  exit(0);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<version>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <show_version@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void version() {
  show_version();
  exit(0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<version>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <show_version@plt>
  xor    %edi,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void version() {
  show_version();
  exit(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
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xb46e7cef,-0x6c(%rbp)
L22:
  addl   $0x1,-0x6c(%rbp)
  movl   $0xf65fbb68,-0x68(%rbp)
L87:
  addl   $0x1,-0x68(%rbp)
  movl   $0x4b28b94c,-0x64(%rbp)
  jmp    L89
L23:
  movl   $0x74e258d9,-0x60(%rbp)
L105:
  addl   $0x1,-0x60(%rbp)
  movl   $0x7ee07303,-0x5c(%rbp)
  jmp    L80
L79:
  movl   $0xbbfacf49,-0x58(%rbp)
L99:
  addl   $0x1,-0x58(%rbp)
  lea    -0x30(%rbp),%rdx
  lea    -0x50(%rbp),%rax
  cmp    %rax,%rdx
  jbe    L49
  movl   $0xcae77d64,-0x54(%rbp)
L103:
  cmpl   $0xcae77d67,-0x54(%rbp)
  jl     L103
L49:
  cmpl   $0xbbfacf54,-0x58(%rbp)
  jl     L99
  addl   $0x1,-0x5c(%rbp)
L80:
  cmpl   $0x7ee07308,-0x5c(%rbp)
  jle    L79
  cmpl   $0x74e258e2,-0x60(%rbp)
  jle    L105
  addl   $0x1,-0x64(%rbp)
L89:
  cmpl   $0x4b28b950,-0x64(%rbp)
  jle    L23
  cmpl   $0xf65fbb73,-0x68(%rbp)
  jl     L87
  cmpl   $0xb46e7cfb,-0x6c(%rbp)
  jl     L22
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  leave
  ret
```

assistant: ```c
int main() {
  int var6 = -1267827473;
  do {
    ++var6;
    double s1_115[4], s2_108[4];
    int var8 = -161498264;
    do {
      ++var8;
      int var12 = 1260960076;
      for (; var12 <= 1260960080; ++var12) {
        int var13 = 1960990937;
        do {
          ++var13;
          int var14 = 2128638723;
          for (; var14 <= 2128638728; ++var14) {
            int var22 = -1141190839;
            do {
              ++var22;
              if (s2_108 > s1_115) {
                int var23 = -890798748;
                do {
                  long long e_119[4];
                } while (var23 <= -890798746);
              }
            } while (var22 <= -1141190829);
          }
        } while (var13 <= 1960990946);
      }
    } while (var8 <= -161498254);
  } while (var6 <= -1267827462);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    $0xc,%r10d
  lea    0x20(%rsp),%rcx
  mov    %rsp,%rdx
L65:
  mov    $0xb,%r8d
L22:
  mov    $0x5,%r9d
L54:
  mov    $0xa,%edi
L88:
  mov    $0x6,%esi
L59:
  mov    $0xb,%eax
L85:
  cmp    %rdx,%rcx
  ja     L29
  sub    $0x1,%eax
  jne    L85
  sub    $0x1,%esi
  jne    L59
  sub    $0x1,%edi
  jne    L88
  sub    $0x1,%r9d
  jne    L54
  sub    $0x1,%r8d
  jne    L22
  sub    $0x1,%r10d
  jne    L65
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  mov    $0x0,%eax
  add    $0x58,%rsp
  ret
L29:
  jmp    L29
L64:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int var6 = -1267827473;
  do {
    ++var6;
    double s1_115[4], s2_108[4];
    int var8 = -161498264;
    do {
      ++var8;
      int var12 = 1260960076;
      for (; var12 <= 1260960080; ++var12) {
        int var13 = 1960990937;
        do {
          ++var13;
          int var14 = 2128638723;
          for (; var14 <= 2128638728; ++var14) {
            int var22 = -1141190839;
            do {
              ++var22;
              if (s2_108 > s1_115) {
                int var23 = -890798748;
                do {
                  long long e_119[4];
                } while (var23 <= -890798746);
              }
            } while (var22 <= -1141190829);
          }
        } while (var13 <= 1960990946);
      }
    } while (var8 <= -161498254);
  } while (var6 <= -1267827462);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    $0xc,%r10d
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rcx
  mov    %rsp,%rdx
L85:
  mov    $0xb,%r9d
L70:
  mov    $0x5,%r8d
L14:
  mov    $0xa,%edi
L53:
  mov    $0x6,%esi
L17:
  mov    $0xb,%eax
  nopl   0x0(%rax)
L57:
  cmp    %rdx,%rcx
  ja     L62
  sub    $0x1,%eax
  jne    L57
  sub    $0x1,%esi
  jne    L17
  sub    $0x1,%edi
  jne    L53
  sub    $0x1,%r8d
  jne    L14
  sub    $0x1,%r9d
  jne    L70
  sub    $0x1,%r10d
  jne    L85
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L62:
  jmp    L62
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int var6 = -1267827473;
  do {
    ++var6;
    double s1_115[4], s2_108[4];
    int var8 = -161498264;
    do {
      ++var8;
      int var12 = 1260960076;
      for (; var12 <= 1260960080; ++var12) {
        int var13 = 1960990937;
        do {
          ++var13;
          int var14 = 2128638723;
          for (; var14 <= 2128638728; ++var14) {
            int var22 = -1141190839;
            do {
              ++var22;
              if (s2_108 > s1_115) {
                int var23 = -890798748;
                do {
                  long long e_119[4];
                } while (var23 <= -890798746);
              }
            } while (var22 <= -1141190829);
          }
        } while (var13 <= 1960990946);
      }
    } while (var8 <= -161498254);
  } while (var6 <= -1267827462);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x58,%rsp
  mov    $0xc,%edi
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rcx
  mov    %rsp,%rdx
L53:
  mov    $0xb,%esi
L35:
  mov    $0x5,%eax
L27:
  cmp    %rdx,%rcx
  ja     L12
  sub    $0x1,%eax
  jne    L27
  sub    $0x1,%esi
  jne    L35
  sub    $0x1,%edi
  jne    L53
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  xor    %eax,%eax
  add    $0x58,%rsp
  ret
L12:
  jmp    L12
L37:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int var6 = -1267827473;
  do {
    ++var6;
    double s1_115[4], s2_108[4];
    int var8 = -161498264;
    do {
      ++var8;
      int var12 = 1260960076;
      for (; var12 <= 1260960080; ++var12) {
        int var13 = 1960990937;
        do {
          ++var13;
          int var14 = 2128638723;
          for (; var14 <= 2128638728; ++var14) {
            int var22 = -1141190839;
            do {
              ++var22;
              if (s2_108 > s1_115) {
                int var23 = -890798748;
                do {
                  long long e_119[4];
                } while (var23 <= -890798746);
              }
            } while (var22 <= -1141190829);
          }
        } while (var13 <= 1960990946);
      }
    } while (var8 <= -161498254);
  } while (var6 <= -1267827462);
}
```
user: What is the c source code of the assembly code below?

```asm
<I2CMasterIntStatus>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  je     L9
  mov    -0x8(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L17
L9:
  mov    -0x8(%rbp),%rax
  add    $0x14,%rax
  mov    (%rax),%rax
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
L17:
  pop    %rbp
  ret
```

assistant: ```c
bool I2CMasterIntStatus(uint32_t ui32Base, bool bMasked) {
  ;
  if (bMasked) {
    return (((*((volatile uint32_t *)(ui32Base + 0x00000018)))) ? true : false);
  } else {
    return (((*((volatile uint32_t *)(ui32Base + 0x00000014)))) ? true : false);
  }
}
```