ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<conjugate_gradient>:
  endbr64
  push   %r15
  lea    0x0(,%rsi,8),%r15d
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  cmp    %edx,%esi
  jl     L3
  mov    %r15d,%edi
  mov    %r9,0x10(%rsp)
  mov    %r8,0x8(%rsp)
  mov    %ecx,(%rsp)
  call   <malloc@plt>
  mov    %r15d,%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    %r15d,%edi
  mov    %rax,%r14
  call   <malloc@plt>
  push   $0x0
  mov    %r12d,%edx
  mov    %ebp,%esi
  push   %r13
  mov    %rbx,%rdi
  mov    %rax,%r15
  push   %r14
  push   %rax
  mov    0x30(%rsp),%r9
  mov    0x28(%rsp),%r8
  mov    0x20(%rsp),%ecx
  call   <iterate_conjgrad@plt>
  add    $0x20,%rsp
  mov    %eax,%r12d
L29:
  mov    %r15,%rdi
  call   <free@plt>
  mov    %r14,%rdi
  call   <free@plt>
  mov    %r13,%rdi
  call   <free@plt>
  add    $0x28,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %r15d,%edi
  mov    %r9,0x18(%rsp)
  mov    %r8,0x10(%rsp)
  mov    %ecx,0x8(%rsp)
  call   <malloc@plt>
  mov    %r15d,%edi
  mov    %rax,(%rsp)
  call   <malloc@plt>
  lea    0x0(,%r12,8),%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    %r15d,%edi
  mov    %rax,%r14
  call   <malloc@plt>
  mov    0x18(%rsp),%r9
  mov    %r12d,%edx
  mov    %ebp,%esi
  mov    %rbx,%rdi
  mov    %rax,%r15
  push   %r9
  push   %r13
  push   %r14
  push   %rax
  mov    0x20(%rsp),%r9
  mov    0x30(%rsp),%r8
  mov    0x28(%rsp),%ecx
  call   <iterate_conjgrad@plt>
  mov    0x20(%rsp),%r10
  add    $0x20,%rsp
  mov    %eax,%r12d
  test   %r10,%r10
  je     L29
  mov    %r10,%rdi
  call   <free@plt>
  jmp    L29
```

assistant: ```c
int conjugate_gradient(double *A, int m, int n, int stride, double *b, double *x0) {
  double *y, *res, *dir, *tmp;
  int err;
  err = 0;
  y = NULL;
  if (m >= n) {
    tmp = malloc(m * sizeof tmp[0]);
    dir = malloc(m * sizeof dir[0]);
    res = malloc(m * sizeof res[0]);
    if ((err = iterate_conjgrad(A, m, n, stride, b, x0, res, dir, tmp, NULL)) == -1) {
      err = -1;
      goto err_out;
    }
  } else {
    y = malloc(m * sizeof y[0]);
    tmp = malloc(m * sizeof tmp[0]);
    dir = malloc(n * sizeof dir[0]);
    res = malloc(m * sizeof res[0]);
    if ((err = iterate_conjgrad(A, m, n, stride, b, y, res, dir, tmp, x0)) == -1) {
      err = -1;
      goto err_out;
    }
  }
err_out:
  if (y != NULL)
    free(y);
  free(res);
  free(dir);
  free(tmp);
  return err;
}
```
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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <say_hello@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  say_hello();
  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
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <say_hello@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  say_hello();
  return 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
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <say_hello@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  say_hello();
  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    $0x8,%rsp
  xor    %eax,%eax
  call   <say_hello@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  say_hello();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<predict_classification>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    %ecx,-0x30(%rbp)
  mov    %r8d,-0x3c(%rbp)
  mov    %r9,-0x48(%rbp)
  movl   $0x0,-0x20(%rbp)
  jmp    L25
L44:
  mov    -0x20(%rbp),%eax
  imul   -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  imul   -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rcx
  mov    -0x3c(%rbp),%edx
  mov    -0x38(%rbp),%rsi
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <predict_classification_@plt>
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L51
L8:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x18(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  addl   $0x1,-0x1c(%rbp)
L51:
  mov    -0x1c(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L8
  movl   $0x0,-0x1c(%rbp)
  jmp    L58
L9:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  divsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L58:
  mov    -0x1c(%rbp),%eax
  cmp    -0x3c(%rbp),%eax
  jl     L9
  addl   $0x1,-0x20(%rbp)
L25:
  mov    -0x20(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L44
  nop
  nop
  leave
  ret
```

assistant: ```c
void predict_classification(double *x, int size, struct SimpleTreeNode *node, int num_attrs, int cls_vals, double *p) {
  int i, j;
  double *xx, *pp;
  for (i = 0; i < size; i++) {
    xx = x + i * num_attrs;
    pp = p + i * cls_vals;
    predict_classification_(xx, node, cls_vals, pp);
    double sum = 0;
    for (j = 0; j < cls_vals; j++) {
      sum += pp[j];
    }
    for (j = 0; j < cls_vals; j++) {
      pp[j] /= sum;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<predict_classification>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdx,0x8(%rsp)
  test   %esi,%esi
  jle    L20
  mov    %rdi,%r12
  mov    %esi,%r15d
  mov    %r8d,%r14d
  movslq %ecx,%rcx
  lea    0x0(,%rcx,8),%rax
  mov    %rax,0x10(%rsp)
  movslq %r8d,%rax
  shl    $0x3,%rax
  mov    %rax,0x18(%rsp)
  mov    %r9,%rbp
  lea    -0x1(%r8),%eax
  lea    0x8(%r9,%rax,8),%rbx
  mov    $0x0,%r13d
L28:
  mov    %rbp,%rcx
  mov    %r14d,%edx
  mov    0x8(%rsp),%rsi
  mov    %r12,%rdi
  call   <predict_classification_@plt>
  test   %r14d,%r14d
  jle    L5
  mov    %rbp,%rax
  pxor   %xmm1,%xmm1
L27:
  addsd  (%rax),%xmm1
  add    $0x8,%rax
  cmp    %rbx,%rax
  jne    L27
  mov    %rbp,%rax
L36:
  movsd  (%rax),%xmm0
  divsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  add    $0x8,%rax
  cmp    %rbx,%rax
  jne    L36
L5:
  add    $0x1,%r13d
  add    0x10(%rsp),%r12
  mov    0x18(%rsp),%rax
  add    %rax,%rbp
  add    %rax,%rbx
  cmp    %r13d,%r15d
  jne    L28
L20:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void predict_classification(double *x, int size, struct SimpleTreeNode *node, int num_attrs, int cls_vals, double *p) {
  int i, j;
  double *xx, *pp;
  for (i = 0; i < size; i++) {
    xx = x + i * num_attrs;
    pp = p + i * cls_vals;
    predict_classification_(xx, node, cls_vals, pp);
    double sum = 0;
    for (j = 0; j < cls_vals; j++) {
      sum += pp[j];
    }
    for (j = 0; j < cls_vals; j++) {
      pp[j] /= sum;
    }
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<predict_classification>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,0xc(%rsp)
  mov    %rdx,0x10(%rsp)
  test   %esi,%esi
  jle    L39
  movslq %ecx,%rcx
  movslq %r8d,%r10
  mov    %r8d,%r12d
  mov    %rdi,%r13
  lea    0x0(,%rcx,8),%rax
  lea    0x0(,%r10,8),%rbp
  mov    %r9,%r14
  xor    %r15d,%r15d
  mov    %rax,0x18(%rsp)
  lea    -0x1(%r8),%eax
  lea    0x8(%r9,%rax,8),%rbx
  nopl   0x0(%rax,%rax,1)
L41:
  mov    0x10(%rsp),%rsi
  mov    %r14,%rcx
  mov    %r12d,%edx
  mov    %r13,%rdi
  call   <predict_classification_@plt>
  test   %r12d,%r12d
  jle    L2
  mov    %r14,%rax
  pxor   %xmm1,%xmm1
  nopw   0x0(%rax,%rax,1)
L46:
  addsd  (%rax),%xmm1
  add    $0x8,%rax
  cmp    %rbx,%rax
  jne    L46
  mov    %r14,%rax
L35:
  movsd  (%rax),%xmm0
  add    $0x8,%rax
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rax)
  cmp    %rbx,%rax
  jne    L35
L2:
  add    $0x1,%r15d
  add    0x18(%rsp),%r13
  add    %rbp,%r14
  add    %rbp,%rbx
  cmp    %r15d,0xc(%rsp)
  jne    L41
L39:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void predict_classification(double *x, int size, struct SimpleTreeNode *node, int num_attrs, int cls_vals, double *p) {
  int i, j;
  double *xx, *pp;
  for (i = 0; i < size; i++) {
    xx = x + i * num_attrs;
    pp = p + i * cls_vals;
    predict_classification_(xx, node, cls_vals, pp);
    double sum = 0;
    for (j = 0; j < cls_vals; j++) {
      sum += pp[j];
    }
    for (j = 0; j < cls_vals; j++) {
      pp[j] /= sum;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<predict_classification>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %esi,0x4(%rsp)
  mov    %rdx,0x8(%rsp)
  test   %esi,%esi
  jle    L16
  movslq %ecx,%rcx
  movslq %r8d,%r14
  mov    %r8d,%r15d
  mov    %rdi,%r13
  lea    0x0(,%rcx,8),%rax
  mov    %r15d,%ebp
  mov    %r9,%rbx
  xor    %r12d,%r12d
  mov    %rax,0x10(%rsp)
  lea    0x0(,%r14,8),%rax
  and    $0xfffffffe,%ebp
  mov    %rax,0x18(%rsp)
  mov    %r8d,%eax
  shr    %eax
  lea    -0x1(%rax),%r8d
  shl    $0x4,%r8
  mov    %r8,%r14
  nopl   (%rax)
L105:
  mov    0x8(%rsp),%rsi
  mov    %rbx,%rcx
  mov    %r15d,%edx
  mov    %r13,%rdi
  call   <predict_classification_@plt>
  test   %r15d,%r15d
  jle    L32
  cmp    $0x1,%r15d
  je     L104
  lea    0x10(%r14),%rdx
  mov    %rbx,%rax
  pxor   %xmm0,%xmm0
  lea    (%rdx,%rbx,1),%rcx
  nopl   (%rax)
L92:
  movsd  (%rax),%xmm1
  add    $0x10,%rax
  addsd  %xmm1,%xmm0
  movsd  -0x8(%rax),%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rax,%rcx
  jne    L92
  movslq %ebp,%rax
  cmp    %ebp,%r15d
  je     L74
L55:
  addsd  (%rbx,%rax,8),%xmm0
  cmp    $0x1,%r15d
  je     L10
  lea    0x10(%r14),%rdx
L74:
  movapd %xmm0,%xmm2
  add    %rbx,%rdx
  mov    %rbx,%rax
  unpcklpd %xmm2,%xmm2
  nop
L3:
  movupd (%rax),%xmm1
  add    $0x10,%rax
  divpd  %xmm2,%xmm1
  movups %xmm1,-0x10(%rax)
  cmp    %rax,%rdx
  jne    L3
  movslq %ebp,%rax
  cmp    %ebp,%r15d
  je     L32
L64:
  lea    (%rbx,%rax,8),%rax
  movsd  (%rax),%xmm1
  divsd  %xmm0,%xmm1
  movsd  %xmm1,(%rax)
L32:
  add    $0x1,%r12d
  add    0x10(%rsp),%r13
  add    0x18(%rsp),%rbx
  cmp    %r12d,0x4(%rsp)
  jne    L105
L16:
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L10:
  xor    %eax,%eax
  jmp    L64
L104:
  pxor   %xmm0,%xmm0
  xor    %eax,%eax
  jmp    L55
```

assistant: ```c
void predict_classification(double *x, int size, struct SimpleTreeNode *node, int num_attrs, int cls_vals, double *p) {
  int i, j;
  double *xx, *pp;
  for (i = 0; i < size; i++) {
    xx = x + i * num_attrs;
    pp = p + i * cls_vals;
    predict_classification_(xx, node, cls_vals, pp);
    double sum = 0;
    for (j = 0; j < cls_vals; j++) {
      sum += pp[j];
    }
    for (j = 0; j < cls_vals; j++) {
      pp[j] /= sum;
    }
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<BN_reciprocal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0xffffffff,-0xc(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_start@plt>
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_get@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L38
  mov    -0x24(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <BN_set_bit@plt>
  test   %eax,%eax
  je     L45
  mov    -0x30(%rbp),%rsi
  mov    -0x20(%rbp),%rcx
  mov    -0x8(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rsi,%r8
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <BN_div@plt>
  test   %eax,%eax
  je     L56
  mov    -0x24(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L59
L38:
  nop
  jmp    L59
L45:
  nop
  jmp    L59
L56:
  nop
L59:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <bn_check_top@plt>
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_CTX_end@plt>
  mov    -0xc(%rbp),%eax
  leave
  ret
```

assistant: ```c
int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx) {
  int ret = -1;
  BIGNUM *t;
  BN_CTX_start(ctx);
  if ((t = BN_CTX_get(ctx)) == NULL)
    goto err;
  if (!BN_set_bit(t, len))
    goto err;
  if (!BN_div(r, NULL, t, m, ctx))
    goto err;
  ret = len;
err:
  bn_check_top(r);
  BN_CTX_end(ctx);
  return (ret);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<BN_reciprocal>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    %rsi,%r14
  mov    %edx,%ebp
  mov    %rcx,%rbx
  mov    %rcx,%rdi
  call   <BN_CTX_start@plt>
  mov    %rbx,%rdi
  call   <BN_CTX_get@plt>
  test   %rax,%rax
  je     L30
  mov    %rax,%r12
  mov    %ebp,%esi
  mov    %rax,%rdi
  call   <BN_set_bit@plt>
  test   %eax,%eax
  je     L12
  mov    %rbx,%r8
  mov    %r14,%rcx
  mov    %r12,%rdx
  mov    $0x0,%esi
  mov    %r13,%rdi
  call   <BN_div@plt>
  test   %eax,%eax
  mov    $0xffffffff,%eax
  cmove  %eax,%ebp
L6:
  mov    %r13,%rdi
  call   <bn_check_top@plt>
  mov    %rbx,%rdi
  call   <BN_CTX_end@plt>
  mov    %ebp,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L30:
  mov    $0xffffffff,%ebp
  jmp    L6
L12:
  mov    $0xffffffff,%ebp
  jmp    L6
```

assistant: ```c
int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx) {
  int ret = -1;
  BIGNUM *t;
  BN_CTX_start(ctx);
  if ((t = BN_CTX_get(ctx)) == NULL)
    goto err;
  if (!BN_set_bit(t, len))
    goto err;
  if (!BN_div(r, NULL, t, m, ctx))
    goto err;
  ret = len;
err:
  bn_check_top(r);
  BN_CTX_end(ctx);
  return (ret);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<BN_reciprocal>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %rdi,%r14
  mov    %rcx,%rdi
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  call   <BN_CTX_start@plt>
  mov    %rbp,%rdi
  call   <BN_CTX_get@plt>
  test   %rax,%rax
  je     L28
  mov    %r13d,%esi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <BN_set_bit@plt>
  test   %eax,%eax
  je     L28
  xor    %esi,%esi
  mov    %rbp,%r8
  mov    %r15,%rcx
  mov    %r12,%rdx
  mov    %r14,%rdi
  call   <BN_div@plt>
  test   %eax,%eax
  je     L28
L8:
  mov    %r14,%rdi
  call   <bn_check_top@plt>
  mov    %rbp,%rdi
  call   <BN_CTX_end@plt>
  pop    %rbp
  mov    %r13d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L28:
  mov    $0xffffffff,%r13d
  jmp    L8
```

assistant: ```c
int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx) {
  int ret = -1;
  BIGNUM *t;
  BN_CTX_start(ctx);
  if ((t = BN_CTX_get(ctx)) == NULL)
    goto err;
  if (!BN_set_bit(t, len))
    goto err;
  if (!BN_div(r, NULL, t, m, ctx))
    goto err;
  ret = len;
err:
  bn_check_top(r);
  BN_CTX_end(ctx);
  return (ret);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<BN_reciprocal>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %rdi,%r14
  mov    %rcx,%rdi
  push   %r13
  mov    %edx,%r13d
  push   %r12
  push   %rbp
  mov    %rcx,%rbp
  call   <BN_CTX_start@plt>
  mov    %rbp,%rdi
  call   <BN_CTX_get@plt>
  test   %rax,%rax
  je     L30
  mov    %r13d,%esi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <BN_set_bit@plt>
  test   %eax,%eax
  je     L30
  xor    %esi,%esi
  mov    %rbp,%r8
  mov    %r15,%rcx
  mov    %r12,%rdx
  mov    %r14,%rdi
  call   <BN_div@plt>
  test   %eax,%eax
  je     L30
L16:
  mov    %r14,%rdi
  call   <bn_check_top@plt>
  mov    %rbp,%rdi
  call   <BN_CTX_end@plt>
  pop    %rbp
  mov    %r13d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L30:
  mov    $0xffffffff,%r13d
  jmp    L16
```

assistant: ```c
int BN_reciprocal(BIGNUM *r, const BIGNUM *m, int len, BN_CTX *ctx) {
  int ret = -1;
  BIGNUM *t;
  BN_CTX_start(ctx);
  if ((t = BN_CTX_get(ctx)) == NULL)
    goto err;
  if (!BN_set_bit(t, len))
    goto err;
  if (!BN_div(r, NULL, t, m, ctx))
    goto err;
  ret = len;
err:
  bn_check_top(r);
  BN_CTX_end(ctx);
  return (ret);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x8(%rax),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L34:
  cmp    %rdx,%rsp
  je     L14
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L34
L14:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L31
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L31:
  mov    %rsp,%rax
  add    $0xf,%rax
  shr    $0x4,%rax
  shl    $0x4,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D8(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type: string

tool: D8: alloca %p %d


assistant: ```c
void *alloc(int size) {
  char *buf = alloca(size);
  printf("alloca %p %d\n", buf, size);
  return buf;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movslq %edi,%rax
  add    $0x17,%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L4:
  cmp    %rcx,%rsp
  je     L31
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L4
L31:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L26
  orq    $0x0,-0x8(%rsp,%rax,1)
L26:
  lea    0xf(%rsp),%rbx
  and    $0xfffffffffffffff0,%rbx
  mov    %edi,%edx
  mov    %rbx,%rsi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  mov    %rbx,%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D3
type_guessed: string

tool: D3: alloca %p %d


assistant: ```c
void *alloc(int size) {
  char *buf = alloca(size);
  printf("alloca %p %d\n", buf, size);
  return buf;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movslq %edi,%rax
  mov    %rsp,%rsi
  add    $0x17,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L24
L16:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L16
L24:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L22
L40:
  lea    0xf(%rsp),%rsi
  mov    %edi,%edx
  lea    D1(%rip),%rdi
  xor    %eax,%eax
  and    $0xfffffffffffffff0,%rsi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  leave
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L22:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L40
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type_guessed: string

tool: D1: alloca %p %d


assistant: ```c
void *alloc(int size) {
  char *buf = alloca(size);
  printf("alloca %p %d\n", buf, size);
  return buf;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movslq %edi,%rax
  mov    %rsp,%rsi
  add    $0x17,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rcx
  cmp    %rsi,%rsp
  je     L6
L8:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L8
L6:
  and    $0xfff,%ecx
  sub    %rcx,%rsp
  test   %rcx,%rcx
  jne    L13
L2:
  lea    0xf(%rsp),%rsi
  mov    %edi,%edx
  lea    D7(%rip),%rdi
  xor    %eax,%eax
  and    $0xfffffffffffffff0,%rsi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  leave
  xor    %eax,%eax
  ret
  nopl   0x0(%rax)
L13:
  orq    $0x0,-0x8(%rsp,%rcx,1)
  jmp    L2
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D7
type: string

tool: D7: alloca %p %d


assistant: ```c
void *alloc(int size) {
  char *buf = alloca(size);
  printf("alloca %p %d\n", buf, size);
  return buf;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test_nontemporal>:
  endbr64
  ret
```

assistant: ```c
void test_nontemporal() {
  int i;
#pragma omp target teams distribute simd nontemporal(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(int)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(0)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(x, y)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(x, y, z)
  for (i = 0; i < 16; ++i)
    ;
  int x, y;
#pragma omp target teams distribute simd nontemporal(x:)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(x:, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(x) nontemporal(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd private(x) nontemporal(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(x) private(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(x, y : 0)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd nontemporal(x) lastprivate(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd lastprivate(x) nontemporal(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target teams distribute simd order
  for (int i = 0; i < 10; ++i)
    ;
#pragma omp target teams distribute simd order(
  for (int i = 0; i < 10; ++i)
    ;
#pragma omp target teams distribute simd order(none
  for (int i = 0; i < 10; ++i)
    ;
#pragma omp target teams distribute simd order(concurrent
  for (int i = 0; i < 10; ++i)
    ;
#pragma omp target teams distribute simd order(concurrent)
  for (int i = 0; i < 10; ++i)
    ;
}
```