ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_fun>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  mov    %edi,%r14d
  mov    $0x4,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r12
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%rbx
  call   <alloca@plt>
  mov    %r14d,(%r12)
  mov    %r13d,(%rbx)
  movl   $0x0,(%rax)
  mov    (%r12),%edx
  mov    (%rbx),%ecx
  cmp    %edx,%ecx
  je     L53
  mov    %rax,%r8
  jmp    L56
  nopl   0x0(%rax,%rax,1)
L10:
  add    $0x1,%ecx
  mov    %ecx,(%rbx)
L36:
  mov    (%r8),%eax
  add    $0x1,%eax
  mov    %eax,(%r8)
  mov    (%r12),%edx
  mov    (%rbx),%ecx
  cmp    %ecx,%edx
  je     L29
L56:
  cmp    %ecx,%edx
  jg     L10
  add    $0x1,%edx
  mov    %edx,(%r12)
  jmp    L36
  nopl   (%rax)
L53:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L29:
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L51
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L51:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_fun(int x, int y) {
  int *x_ref = alloca(sizeof(int));
  int *y_ref = alloca(sizeof(int));
  int *c = alloca(sizeof(int));
  *x_ref = x;
  *y_ref = y;
  *c = 0;
  while (!(*x_ref == *y_ref)) {
    if (*x_ref > *y_ref) {
      *y_ref = *y_ref + 1;
    } else {
      *x_ref = *x_ref + 1;
    }
    *c = *c + 1;
  }
  return *c;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_fun>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r14
  mov    %edi,%r14d
  mov    $0x4,%edi
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x28(%rbp)
  xor    %eax,%eax
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%r12
  call   <alloca@plt>
  mov    $0x4,%edi
  mov    %rax,%rbx
  call   <alloca@plt>
  mov    %r14d,(%r12)
  mov    %r13d,(%rbx)
  mov    %rax,%r8
  movl   $0x0,(%rax)
  mov    (%r12),%edx
  mov    (%rbx),%ecx
  cmp    %edx,%ecx
  jne    L32
  jmp    L50
  nopl   0x0(%rax,%rax,1)
L16:
  add    $0x1,%ecx
  mov    %ecx,(%rbx)
L21:
  mov    (%r8),%eax
  add    $0x1,%eax
  mov    %eax,(%r8)
  mov    (%r12),%edx
  mov    (%rbx),%ecx
  cmp    %ecx,%edx
  je     L20
L32:
  cmp    %ecx,%edx
  jg     L16
  add    $0x1,%edx
  mov    %edx,(%r12)
  jmp    L21
  nopl   (%rax)
L50:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L20:
  mov    -0x28(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L47
  lea    -0x20(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %rbp
  ret
L47:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int test_fun(int x, int y) {
  int *x_ref = alloca(sizeof(int));
  int *y_ref = alloca(sizeof(int));
  int *c = alloca(sizeof(int));
  *x_ref = x;
  *y_ref = y;
  *c = 0;
  while (!(*x_ref == *y_ref)) {
    if (*x_ref > *y_ref) {
      *y_ref = *y_ref + 1;
    } else {
      *x_ref = *x_ref + 1;
    }
    *c = *c + 1;
  }
  return *c;
}
```
user: What is the c source code of the assembly code below:

```asm
<data_haswonderstage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <data_getwonder@plt>
  cmp    %eax,-0x18(%rbp)
  je     L21
  mov    $0x0,%eax
  jmp    L13
L21:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <data_getwonderside@plt>
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  cmp    %eax,-0x1c(%rbp)
  jg     L59
  mov    $0x0,%eax
  jmp    L13
L59:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <data_getwonderstages@plt>
  mov    %eax,%ebx
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <data_getwonderside@plt>
  mov    %eax,%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %ebx,%eax
  cmp    %eax,-0x1c(%rbp)
  jg     L32
  mov    $0x1,%eax
  jmp    L13
L32:
  mov    $0x0,%eax
L13:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int data_haswonderstage(int p, int wonder, int stage) {
  if (data_getwonder(p) != wonder)
    return 0;
  if (data_getwonderside(p) * 3 + 1 > stage)
    return 0;
  if (data_getwonderstages(p) + data_getwonderside(p) * 3 + 1 > stage)
    return 1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<data_haswonderstage>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%ebp
  mov    %edx,%r12d
  call   <data_getwonder@plt>
  mov    %eax,%edx
  mov    $0x0,%eax
  cmp    %ebp,%edx
  je     L2
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  mov    %ebx,%edi
  call   <data_getwonderside@plt>
  lea    (%rax,%rax,2),%edx
  mov    $0x0,%eax
  cmp    %r12d,%edx
  jge    L5
  mov    %ebx,%edi
  call   <data_getwonderstages@plt>
  mov    %eax,%ebp
  mov    %ebx,%edi
  call   <data_getwonderside@plt>
  lea    (%rax,%rax,2),%eax
  add    %ebp,%eax
  cmp    %r12d,%eax
  setge  %al
  movzbl %al,%eax
  jmp    L5
```

assistant: ```c
int data_haswonderstage(int p, int wonder, int stage) {
  if (data_getwonder(p) != wonder)
    return 0;
  if (data_getwonderside(p) * 3 + 1 > stage)
    return 0;
  if (data_getwonderstages(p) + data_getwonderside(p) * 3 + 1 > stage)
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<data_haswonderstage>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  call   <data_getwonder@plt>
  cmp    %ebx,%eax
  je     L9
L6:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nop
L9:
  mov    %ebp,%edi
  call   <data_getwonderside@plt>
  lea    (%rax,%rax,2),%eax
  cmp    %r12d,%eax
  jge    L6
  mov    %ebp,%edi
  call   <data_getwonderstages@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <data_getwonderside@plt>
  lea    (%rax,%rax,2),%eax
  add    %ebx,%eax
  pop    %rbx
  pop    %rbp
  cmp    %r12d,%eax
  pop    %r12
  setge  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int data_haswonderstage(int p, int wonder, int stage) {
  if (data_getwonder(p) != wonder)
    return 0;
  if (data_getwonderside(p) * 3 + 1 > stage)
    return 0;
  if (data_getwonderstages(p) + data_getwonderside(p) * 3 + 1 > stage)
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<data_haswonderstage>:
  endbr64
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %esi,%ebx
  call   <data_getwonder@plt>
  cmp    %ebx,%eax
  je     L12
L9:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
  nop
L12:
  mov    %ebp,%edi
  call   <data_getwonderside@plt>
  lea    (%rax,%rax,2),%eax
  cmp    %r12d,%eax
  jge    L9
  mov    %ebp,%edi
  call   <data_getwonderstages@plt>
  mov    %ebp,%edi
  mov    %eax,%ebx
  call   <data_getwonderside@plt>
  lea    (%rax,%rax,2),%eax
  add    %ebx,%eax
  pop    %rbx
  pop    %rbp
  cmp    %r12d,%eax
  pop    %r12
  setge  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
int data_haswonderstage(int p, int wonder, int stage) {
  if (data_getwonder(p) != wonder)
    return 0;
  if (data_getwonderside(p) * 3 + 1 > stage)
    return 0;
  if (data_getwonderstages(p) + data_getwonderside(p) * 3 + 1 > stage)
    return 1;
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rand_serial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L23
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L72
L23:
  mov    $0x0,%eax
  call   <BN_new@plt>
  mov    %rax,-0x8(%rbp)
L72:
  cmpq   $0x0,-0x8(%rbp)
  jne    L51
  mov    $0x0,%eax
  jmp    L3
L51:
  mov    -0x8(%rbp),%rax
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x40,%esi
  mov    %rax,%rdi
  call   <BN_pseudo_rand@plt>
  test   %eax,%eax
  je     L77
  cmpq   $0x0,-0x20(%rbp)
  je     L26
  mov    -0x20(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <BN_to_ASN1_INTEGER@plt>
  test   %eax,%eax
  je     L67
L26:
  movl   $0x1,-0xc(%rbp)
  jmp    L94
L77:
  nop
  jmp    L94
L67:
  nop
L94:
  cmpq   $0x0,-0x18(%rbp)
  jne    L37
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <BN_free@plt>
L37:
  mov    -0xc(%rbp),%eax
L3:
  leave
  ret
```

assistant: ```c
int rand_serial(BIGNUM *b, ASN1_INTEGER *ai) {
  BIGNUM *btmp;
  int ret = 0;
  if (b)
    btmp = b;
  else
    btmp = BN_new();
  if (!btmp)
    return 0;
  if (!BN_pseudo_rand(btmp, 64, 0, 0))
    goto error;
  if (ai && !BN_to_ASN1_INTEGER(btmp, ai))
    goto error;
  ret = 1;
error:
  if (!b)
    BN_free(btmp);
  return ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rand_serial>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%r12
  test   %rdi,%rdi
  je     L66
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x40,%esi
  call   <BN_pseudo_rand@plt>
  mov    %eax,%ebp
  mov    %rbx,%r13
  test   %eax,%eax
  je     L75
L31:
  mov    $0x1,%ebp
  test   %r12,%r12
  je     L43
  mov    %r12,%rsi
  mov    %r13,%rdi
  call   <BN_to_ASN1_INTEGER@plt>
  test   %eax,%eax
  setne  %bpl
  movzbl %bpl,%ebp
L43:
  test   %rbx,%rbx
  je     L46
L75:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L66:
  mov    $0x0,%eax
  call   <BN_new@plt>
  mov    %rax,%r13
  test   %rax,%rax
  jne    L60
  mov    $0x0,%ebp
  jmp    L75
L60:
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x40,%esi
  mov    %rax,%rdi
  call   <BN_pseudo_rand@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  jne    L31
L46:
  mov    %r13,%rdi
  call   <BN_free@plt>
  jmp    L75
```

assistant: ```c
int rand_serial(BIGNUM *b, ASN1_INTEGER *ai) {
  BIGNUM *btmp;
  int ret = 0;
  if (b)
    btmp = b;
  else
    btmp = BN_new();
  if (!btmp)
    return 0;
  if (!BN_pseudo_rand(btmp, 64, 0, 0))
    goto error;
  if (ai && !BN_to_ASN1_INTEGER(btmp, ai))
    goto error;
  ret = 1;
error:
  if (!b)
    BN_free(btmp);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rand_serial>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  sub    $0x10,%rsp
  test   %rdi,%rdi
  je     L54
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    $0x40,%esi
  mov    %rdi,%rbp
  call   <BN_pseudo_rand@plt>
  test   %eax,%eax
  je     L48
  mov    $0x1,%r12d
  test   %r13,%r13
  je     L31
  mov    %r13,%rsi
  mov    %rbp,%rdi
  xor    %r12d,%r12d
  call   <BN_to_ASN1_INTEGER@plt>
  test   %eax,%eax
  setne  %r12b
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L54:
  xor    %eax,%eax
  call   <BN_new@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L7
L48:
  xor    %r12d,%r12d
L31:
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    $0x40,%esi
  mov    %rax,0x8(%rsp)
  call   <BN_pseudo_rand@plt>
  mov    0x8(%rsp),%rdi
  test   %eax,%eax
  mov    %eax,%r12d
  je     L23
  mov    $0x1,%r12d
  test   %r13,%r13
  je     L23
  mov    %r13,%rsi
  mov    %rdi,0x8(%rsp)
  xor    %r12d,%r12d
  call   <BN_to_ASN1_INTEGER@plt>
  mov    0x8(%rsp),%rdi
  test   %eax,%eax
  setne  %r12b
L23:
  call   <BN_free@plt>
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int rand_serial(BIGNUM *b, ASN1_INTEGER *ai) {
  BIGNUM *btmp;
  int ret = 0;
  if (b)
    btmp = b;
  else
    btmp = BN_new();
  if (!btmp)
    return 0;
  if (!BN_pseudo_rand(btmp, 64, 0, 0))
    goto error;
  if (ai && !BN_to_ASN1_INTEGER(btmp, ai))
    goto error;
  ret = 1;
error:
  if (!b)
    BN_free(btmp);
  return ret;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<rand_serial>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  sub    $0x10,%rsp
  test   %rdi,%rdi
  je     L23
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    $0x40,%esi
  mov    %rdi,%rbp
  call   <BN_pseudo_rand@plt>
  test   %eax,%eax
  je     L41
  mov    $0x1,%r12d
  test   %r13,%r13
  je     L6
  mov    %r13,%rsi
  mov    %rbp,%rdi
  xor    %r12d,%r12d
  call   <BN_to_ASN1_INTEGER@plt>
  test   %eax,%eax
  setne  %r12b
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  cs nopw 0x0(%rax,%rax,1)
L23:
  xor    %eax,%eax
  call   <BN_new@plt>
  mov    %rax,%rdi
  test   %rax,%rax
  jne    L35
L41:
  xor    %r12d,%r12d
L6:
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L35:
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    $0x40,%esi
  mov    %rax,0x8(%rsp)
  call   <BN_pseudo_rand@plt>
  mov    0x8(%rsp),%rdi
  test   %eax,%eax
  mov    %eax,%r12d
  je     L0
  mov    $0x1,%r12d
  test   %r13,%r13
  je     L0
  mov    %r13,%rsi
  mov    %rdi,0x8(%rsp)
  xor    %r12d,%r12d
  call   <BN_to_ASN1_INTEGER@plt>
  mov    0x8(%rsp),%rdi
  test   %eax,%eax
  setne  %r12b
L0:
  call   <BN_free@plt>
  add    $0x10,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int rand_serial(BIGNUM *b, ASN1_INTEGER *ai) {
  BIGNUM *btmp;
  int ret = 0;
  if (b)
    btmp = b;
  else
    btmp = BN_new();
  if (!btmp)
    return 0;
  if (!BN_pseudo_rand(btmp, 64, 0, 0))
    goto error;
  if (ai && !BN_to_ASN1_INTEGER(btmp, ai))
    goto error;
  ret = 1;
error:
  if (!b)
    BN_free(btmp);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<mallocInts>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L4
L7:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L4:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L7
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *mallocInts(int size) {
  int *ints = malloc(sizeof(int) * size);
  int i;
  for (i = 0; i < size; i++) {
    ints[i] = i;
  }
  return ints;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mallocInts>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  lea    0x0(,%rdi,4),%edi
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L4
  mov    %ebx,%edi
  mov    $0x0,%edx
L2:
  mov    %edx,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdi,%rdx
  jne    L2
L4:
  pop    %rbx
  ret
```

assistant: ```c
int *mallocInts(int size) {
  int *ints = malloc(sizeof(int) * size);
  int i;
  for (i = 0; i < size; i++) {
    ints[i] = i;
  }
  return ints;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<mallocInts>:
  endbr64
  push   %rbp
  push   %rbx
  movslq %edi,%rbx
  lea    0x0(,%rbx,4),%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L4
  xor    %edx,%edx
  nop
L9:
  mov    %edx,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rbx,%rdx
  jne    L9
L4:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *mallocInts(int size) {
  int *ints = malloc(sizeof(int) * size);
  int i;
  for (i = 0; i < size; i++) {
    ints[i] = i;
  }
  return ints;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<mallocInts>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  shl    $0x2,%edi
  call   <malloc@plt>
  test   %ebx,%ebx
  jle    L44
  lea    -0x1(%rbx),%edx
  cmp    $0x2,%edx
  jbe    L49
  mov    %ebx,%ecx
  movdqa D8(%rip),%xmm0
  movdqa D10(%rip),%xmm2
  mov    %rax,%rdx
  shr    $0x2,%ecx
  shl    $0x4,%rcx
  add    %rax,%rcx
  nopw   0x0(%rax,%rax,1)
L24:
  movdqa %xmm0,%xmm1
  add    $0x10,%rdx
  paddd  %xmm2,%xmm0
  movups %xmm1,-0x10(%rdx)
  cmp    %rcx,%rdx
  jne    L24
  mov    %ebx,%edx
  and    $0xfffffffc,%edx
  test   $0x3,%bl
  je     L44
L60:
  movslq %edx,%rcx
  mov    %edx,(%rax,%rcx,4)
  lea    0x0(,%rcx,4),%rsi
  lea    0x1(%rdx),%ecx
  cmp    %ecx,%ebx
  jle    L44
  add    $0x2,%edx
  mov    %ecx,0x4(%rax,%rsi,1)
  cmp    %edx,%ebx
  jle    L44
  mov    %edx,0x8(%rax,%rsi,1)
L44:
  pop    %rbx
  ret
L49:
  xor    %edx,%edx
  jmp    L60
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
data_type: byte[16]
name: extract_symbolic_data
arguments: tag: D10
data_type: byte[16]

tool: {"D8": [0x00,0x00,0x00,0x00,0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00]}

tool: {"D10": [0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

assistant: ```c
int *mallocInts(int size) {
  int *ints = malloc(sizeof(int) * size);
  int i;
  for (i = 0; i < size; i++) {
    ints[i] = i;
  }
  return ints;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<atol>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movq   $0x0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L44
L32:
  addq   $0x1,-0x18(%rbp)
L44:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %rax,%rax
  jne    L32
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L55
  movl   $0x1,-0xc(%rbp)
  addq   $0x1,-0x18(%rbp)
  jmp    L55
L25:
  mov    -0x8(%rbp),%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  mov    %rax,%rdx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  sub    $0x30,%eax
  cltq
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  addq   $0x1,-0x18(%rbp)
L55:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L25
  cmpl   $0x0,-0xc(%rbp)
  je     L13
  negq   -0x8(%rbp)
L13:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long atol(const char *str) {
  long num = 0;
  int neg = 0;
  while (isspace(*str))
    str++;
  if (*str == '-') {
    neg = 1;
    str++;
  }
  while (isdigit(*str)) {
    num = 10 * num + (*str - '0');
    str++;
  }
  if (neg)
    num = -num;
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<atol>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  jmp    L9
L18:
  add    $0x1,%rbp
L9:
  movsbl 0x0(%rbp),%edi
  call   <isspace@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L18
  mov    $0x0,%r12d
  cmpb   $0x2d,0x0(%rbp)
  jne    L0
  add    $0x1,%rbp
  mov    $0x1,%r12d
  jmp    L0
L16:
  lea    (%rbx,%rbx,4),%rdx
  movsbl 0x0(%rbp),%eax
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rdx,2),%rbx
  add    $0x1,%rbp
L0:
  movsbl 0x0(%rbp),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L16
  mov    %rbx,%rax
  neg    %rax
  test   %r12d,%r12d
  cmovne %rax,%rbx
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
long atol(const char *str) {
  long num = 0;
  int neg = 0;
  while (isspace(*str))
    str++;
  if (*str == '-') {
    neg = 1;
    str++;
  }
  while (isdigit(*str)) {
    num = 10 * num + (*str - '0');
    str++;
  }
  if (neg)
    num = -num;
  return num;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<atol>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L30
  nopl   (%rax)
L3:
  add    $0x1,%rbx
L30:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L3
  movsbl (%rbx),%edi
  xor    %ebp,%ebp
  cmp    $0x2d,%dil
  jne    L12
  movsbl 0x1(%rbx),%edi
  mov    $0x1,%ebp
  add    $0x1,%rbx
  call   <isdigit@plt>
  test   %rax,%rax
  je     L13
  cs nopw 0x0(%rax,%rax,1)
L45:
  movsbl (%rbx),%eax
  lea    (%r12,%r12,4),%rdx
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rdx,2),%r12
L12:
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L45
L13:
  mov    %r12,%rax
  pop    %rbx
  neg    %rax
  test   %ebp,%ebp
  pop    %rbp
  cmovne %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
long atol(const char *str) {
  long num = 0;
  int neg = 0;
  while (isspace(*str))
    str++;
  if (*str == '-') {
    neg = 1;
    str++;
  }
  while (isdigit(*str)) {
    num = 10 * num + (*str - '0');
    str++;
  }
  if (neg)
    num = -num;
  return num;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<atol>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  jmp    L47
  nopl   (%rax)
L0:
  add    $0x1,%rbx
L47:
  movsbl (%rbx),%edi
  call   <isspace@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L0
  xor    %ebp,%ebp
  cmpb   $0x2d,(%rbx)
  jne    L21
  movsbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  mov    $0x1,%ebp
  call   <isdigit@plt>
  test   %rax,%rax
  je     L46
  nopw   0x0(%rax,%rax,1)
L7:
  movsbl (%rbx),%eax
  lea    (%r12,%r12,4),%rdx
  add    $0x1,%rbx
  sub    $0x30,%eax
  cltq
  lea    (%rax,%rdx,2),%r12
L21:
  movsbl (%rbx),%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L7
L46:
  mov    %r12,%rax
  pop    %rbx
  neg    %rax
  test   %ebp,%ebp
  pop    %rbp
  cmovne %rax,%r12
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
long atol(const char *str) {
  long num = 0;
  int neg = 0;
  while (isspace(*str))
    str++;
  if (*str == '-') {
    neg = 1;
    str++;
  }
  while (isdigit(*str)) {
    num = 10 * num + (*str - '0');
    str++;
  }
  if (neg)
    num = -num;
  return num;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<Mergesort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x1,-0xc(%rbp)
  jmp    L0
L17:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0xc(%rbp)
L0:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L17
  mov    -0x8(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  mov    $0x1,%edi
  call   <Mergesortd@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *Mergesort(int n, const double *key) {
  int a;
  int *order = (int *)malloc((n + 1) * sizeof(int));
  for (a = 1; a <= n; a++)
    order[a] = a;
  Mergesortd(1, n, key, order);
  return order;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<Mergesort>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %rsi,%r12
  lea    0x4(,%rdi,4),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %ebx,%ebx
  jle    L17
  lea    0x4(%rax),%rdx
  lea    0x1(%rbx),%ecx
  mov    $0x1,%eax
L18:
  mov    %eax,(%rdx)
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %ecx,%eax
  jne    L18
L17:
  mov    %rbp,%rcx
  mov    %r12,%rdx
  mov    %ebx,%esi
  mov    $0x1,%edi
  call   <Mergesortd@plt>
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *Mergesort(int n, const double *key) {
  int a;
  int *order = (int *)malloc((n + 1) * sizeof(int));
  for (a = 1; a <= n; a++)
    order[a] = a;
  Mergesortd(1, n, key, order);
  return order;
}
```
user: What is the c source code of the assembly code below:

```asm
<Mergesort>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  lea    0x4(,%rdi,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebp,%ebp
  jle    L19
  lea    0x4(%rax),%rcx
  mov    $0x1,%eax
  nopw   0x0(%rax,%rax,1)
L10:
  mov    %eax,%edi
  mov    %eax,(%rcx)
  add    $0x1,%eax
  add    $0x4,%rcx
  cmp    %edi,%ebp
  jne    L10
L19:
  mov    %r12,%rcx
  mov    %r13,%rdx
  mov    %ebp,%esi
  mov    $0x1,%edi
  call   <Mergesortd@plt>
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int *Mergesort(int n, const double *key) {
  int a;
  int *order = (int *)malloc((n + 1) * sizeof(int));
  for (a = 1; a <= n; a++)
    order[a] = a;
  Mergesortd(1, n, key, order);
  return order;
}
```