ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<__readString>:
  endbr64
  push   %rbx
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  call   <open@plt>
  cmp    $0xffffffff,%eax
  je     L6
  mov    %eax,%edi
  call   <__bufFromFd@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L14
  mov    %rax,%rdi
  call   <trim@plt>
L14:
  mov    %rbx,%rax
  pop    %rbx
  ret
L6:
  mov    $0x0,%ebx
  jmp    L14
```

assistant: ```c
char *__readString(char *name) {
  int fd;
  char *buf;
  fd = open(name, O_RDONLY);
  if (fd == -1)
    return NULL;
  buf = __bufFromFd(fd);
  if (buf)
    trim(buf);
  return buf;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<__readString>:
  endbr64
  mov    D0(%rip),%rax
  push   %r12
  mov    (%rax),%esi
  call   <open@plt>
  cmp    $0xffffffff,%eax
  je     L13
  mov    %eax,%edi
  call   <__bufFromFd@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L16
  mov    %rax,%rdi
  call   <trim@plt>
L16:
  mov    %r12,%rax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L13:
  xor    %r12d,%r12d
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *__readString(char *name) {
  int fd;
  char *buf;
  fd = open(name, O_RDONLY);
  if (fd == -1)
    return NULL;
  buf = __bufFromFd(fd);
  if (buf)
    trim(buf);
  return buf;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<__readString>:
  endbr64
  mov    D5(%rip),%rax
  push   %r12
  mov    (%rax),%esi
  call   <open@plt>
  cmp    $0xffffffff,%eax
  je     L9
  mov    %eax,%edi
  call   <__bufFromFd@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L1
  mov    %rax,%rdi
  call   <trim@plt>
L1:
  mov    %r12,%rax
  pop    %r12
  ret
  cs nopw 0x0(%rax,%rax,1)
L9:
  xor    %r12d,%r12d
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *__readString(char *name) {
  int fd;
  char *buf;
  fd = open(name, O_RDONLY);
  if (fd == -1)
    return NULL;
  buf = __bufFromFd(fd);
  if (buf)
    trim(buf);
  return buf;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<mmap_error>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  js     L15
  mov    $0x0,%eax
  jmp    L10
L15:
  cmpq   $0xfffffffffffff800,-0x8(%rbp)
  jge    L2
  mov    $0x0,%eax
  jmp    L10
L2:
  mov    -0x8(%rbp),%rax
L10:
  pop    %rbp
  ret
```

assistant: ```c
int mmap_error(void *ptr) {
  long ret = (long)ptr;
  if (ret >= 0)
    return 0;
  if (ret < -2048)
    return 0;
  return (int)ret;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<mmap_error>:
  endbr64
  lea    0x800(%rdi),%rax
  cmp    $0x7ff,%rax
  mov    $0x0,%eax
  cmovbe %rdi,%rax
  ret
```

assistant: ```c
int mmap_error(void *ptr) {
  long ret = (long)ptr;
  if (ret >= 0)
    return 0;
  if (ret < -2048)
    return 0;
  return (int)ret;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<mmap_error>:
  endbr64
  lea    0x800(%rdi),%rax
  cmp    $0x7ff,%rax
  mov    $0x0,%eax
  cmovbe %edi,%eax
  ret
```

assistant: ```c
int mmap_error(void *ptr) {
  long ret = (long)ptr;
  if (ret >= 0)
    return 0;
  if (ret < -2048)
    return 0;
  return (int)ret;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<mmap_error>:
  endbr64
  lea    0x800(%rdi),%rax
  cmp    $0x7ff,%rax
  mov    $0x0,%eax
  cmovbe %edi,%eax
  ret
```

assistant: ```c
int mmap_error(void *ptr) {
  long ret = (long)ptr;
  if (ret >= 0)
    return 0;
  if (ret < -2048)
    return 0;
  return (int)ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inv>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x20(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L2
L7:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x10(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x10(%rbp),%rax
  mov    -0x1c(%rbp),%edx
  mov    %edx,(%rax)
  addq   $0x4,-0x18(%rbp)
  subq   $0x4,-0x10(%rbp)
L2:
  mov    -0x18(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jbe    L7
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type: string

tool: D2: The first value is:%d


assistant: ```c
void inv(int *x, int n) {
  int *p, m, temp, *i, *j;
  m = (n - 1) / 2;
  i = x;
  j = x + n - 1;
  p = x + m;
  printf("The first value is:%d\n", *x);
  for (; i <= p; i++, j--) {
    temp = *i;
    *i = *j;
    *j = temp;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<inv>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  movslq %esi,%rax
  lea    -0x4(%rdi,%rax,4),%rbp
  sub    $0x1,%esi
  mov    %esi,%eax
  shr    $0x1f,%eax
  add    %esi,%eax
  sar    %eax
  cltq
  lea    (%rdi,%rax,4),%r12
  mov    (%rdi),%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  cmp    %r12,%rbx
  ja     L14
L16:
  mov    (%rbx),%eax
  mov    0x0(%rbp),%edx
  mov    %edx,(%rbx)
  mov    %eax,0x0(%rbp)
  add    $0x4,%rbx
  sub    $0x4,%rbp
  cmp    %rbx,%r12
  jae    L16
L14:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D4
type: string

tool: D4: The first value is:%d


assistant: ```c
void inv(int *x, int n) {
  int *p, m, temp, *i, *j;
  m = (n - 1) / 2;
  i = x;
  j = x + n - 1;
  p = x + m;
  printf("The first value is:%d\n", *x);
  for (; i <= p; i++, j--) {
    temp = *i;
    *i = *j;
    *j = temp;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<inv>:
  endbr64
  movslq %esi,%rax
  sub    $0x1,%esi
  push   %r12
  push   %rbp
  lea    -0x4(%rdi,%rax,4),%rbp
  mov    %esi,%eax
  shr    $0x1f,%eax
  push   %rbx
  mov    %rdi,%rbx
  add    %esi,%eax
  mov    (%rdi),%esi
  sar    %eax
  cltq
  lea    (%rdi,%rax,4),%r12
  lea    D6(%rip),%rdi
  call   <printf@plt>
  cmp    %r12,%rbx
  ja     L5
  nopl   0x0(%rax,%rax,1)
L3:
  mov    0x0(%rbp),%edx
  mov    (%rbx),%eax
  add    $0x4,%rbx
  sub    $0x4,%rbp
  mov    %edx,-0x4(%rbx)
  mov    %eax,0x4(%rbp)
  cmp    %rbx,%r12
  jae    L3
L5:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: tag: D6
type_infered: string

tool: D6: The first value is:%d


assistant: ```c
void inv(int *x, int n) {
  int *p, m, temp, *i, *j;
  m = (n - 1) / 2;
  i = x;
  j = x + n - 1;
  p = x + m;
  printf("The first value is:%d\n", *x);
  for (; i <= p; i++, j--) {
    temp = *i;
    *i = *j;
    *j = temp;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<inv>:
  endbr64
  movslq %esi,%rax
  sub    $0x1,%esi
  push   %r15
  push   %r14
  push   %r13
  lea    0x0(,%rax,4),%r13
  push   %r12
  mov    %esi,%r12d
  lea    -0x4(%r13),%r14
  shr    $0x1f,%r12d
  push   %rbp
  lea    (%rdi,%r14,1),%rbp
  push   %rbx
  add    %esi,%r12d
  mov    %rdi,%rbx
  sar    %r12d
  movslq %r12d,%r12
  sub    $0x8,%rsp
  shl    $0x2,%r12
  mov    (%rdi),%esi
  lea    (%rdi,%r12,1),%r15
  lea    D6(%rip),%rdi
  call   <printf@plt>
  cmp    %r15,%rbx
  ja     L5
  mov    %r12,%rdi
  mov    %rbx,%rax
  shr    $0x2,%rdi
  mov    %rdi,%rdx
  lea    0x1(%rdi),%rsi
  neg    %rdx
  lea    0x0(,%rsi,4),%rcx
  lea    (%r14,%rdx,4),%rdx
  cmp    %rdx,%rcx
  setle  %dl
  test   %r13,%r13
  setle  %cl
  or     %cl,%dl
  je     L50
  cmp    $0x3,%r12
  jbe    L50
  cmp    $0xb,%r12
  jbe    L75
  mov    %rsi,%rcx
  lea    -0x10(%rbx,%r13,1),%rdx
  shr    $0x2,%rcx
  shl    $0x4,%rcx
  add    %rbx,%rcx
  nopl   (%rax)
L59:
  movdqu (%rdx),%xmm2
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  sub    $0x10,%rdx
  pshufd $0x1b,%xmm2,%xmm1
  pshufd $0x1b,%xmm0,%xmm0
  movups %xmm1,-0x10(%rax)
  movups %xmm0,0x10(%rdx)
  cmp    %rax,%rcx
  jne    L59
  mov    %rsi,%rax
  and    $0xfffffffffffffffc,%rax
  lea    0x0(,%rax,4),%rcx
  lea    (%rbx,%rcx,1),%rdx
  sub    %rcx,%rbp
  cmp    %rax,%rsi
  je     L5
L46:
  sub    %rax,%rsi
  cmp    %rax,%rdi
  je     L76
  lea    0x0(,%rax,4),%rcx
  lea    -0x8(%r13),%rax
  sub    %rcx,%rax
  lea    (%rbx,%rcx,1),%rdi
  add    %rax,%rbx
  movq   (%rdi),%xmm0
  mov    %rsi,%rax
  movq   (%rbx),%xmm1
  and    $0xfffffffffffffffe,%rax
  lea    0x0(,%rax,4),%rcx
  pshufd $0xe1,%xmm0,%xmm0
  pshufd $0xe1,%xmm1,%xmm1
  add    %rcx,%rdx
  sub    %rcx,%rbp
  movq   %xmm1,(%rdi)
  movq   %xmm0,(%rbx)
  cmp    %rax,%rsi
  je     L5
L76:
  mov    (%rdx),%eax
  mov    0x0(%rbp),%ecx
  mov    %ecx,(%rdx)
  mov    %eax,0x0(%rbp)
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L50:
  mov    0x0(%rbp),%edx
  mov    (%rbx),%eax
  add    $0x4,%rbx
  sub    $0x4,%rbp
  mov    %edx,-0x4(%rbx)
  mov    %eax,0x4(%rbp)
  cmp    %rbx,%r15
  jae    L50
  jmp    L5
L75:
  mov    %rbx,%rdx
  xor    %eax,%eax
  jmp    L46
```

tool_calls: name: unpack_data
arguments: data_label: D6
type_infered: string

tool: D6: The first value is:%d


assistant: ```c
void inv(int *x, int n) {
  int *p, m, temp, *i, *j;
  m = (n - 1) / 2;
  i = x;
  j = x + n - 1;
  p = x + m;
  printf("The first value is:%d\n", *x);
  for (; i <= p; i++, j--) {
    temp = *i;
    *i = *j;
    *j = temp;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x34(%rbp)
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x40(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x38(%rbp)
  jmp    L10
L1:
  lea    -0x30(%rbp),%rdx
  mov    -0x38(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x38(%rbp)
L10:
  mov    -0x40(%rbp),%eax
  cmp    %eax,-0x38(%rbp)
  jle    L1
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x3c(%rbp),%rax
  mov    %rax,%rsi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x30(%rbp),%eax
  mov    %eax,-0x34(%rbp)
  movl   $0x1,-0x38(%rbp)
  jmp    L39
L26:
  mov    -0x3c(%rbp),%eax
  imul   -0x34(%rbp),%eax
  mov    %eax,%edx
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%eax
  add    %edx,%eax
  mov    %eax,-0x34(%rbp)
  addl   $0x1,-0x38(%rbp)
L39:
  mov    -0x40(%rbp),%eax
  cmp    %eax,-0x38(%rbp)
  jle    L26
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D31(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L35
  call   <__stack_chk_fail@plt>
L35:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D40
data_type: string
name: unpack_data
arguments: tag: D29
data_type: string
name: unpack_data
arguments: tag: D3
data_type: string
name: unpack_data
arguments: tag: D54
data_type: string
name: unpack_data
arguments: tag: D31
data_type: string

tool: D40: 
 Enter the degree of the polynomial : 

tool: D29: %d

tool: D3: 
 Enter the %d coefficients


tool: D54: 
 Enter the value of x :

tool: D31: 
 The sum of polynomial = %d

assistant: ```c
int main() {
  int n, i, x, a[10], poly = 0;
  printf("\n Enter the degree of the polynomial : ");
  scanf("%d", &n);
  printf("\n Enter the %d coefficients\n", n + 1);
  for (i = 0; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  printf("\n Enter the value of x :");
  scanf("%d", &x);
  poly = a[0];
  for (i = 1; i <= n; i++) {
    poly = poly * x + a[i];
  }
  printf("\n The sum of polynomial = %d", poly);
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    D11(%rip),%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D49(%rip),%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  lea    0x1(%rax),%esi
  lea    D43(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmpl   $0x0,0x8(%rsp)
  js     L48
  lea    0x10(%rsp),%rbp
  mov    $0x0,%ebx
  lea    D49(%rip),%r12
L15:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0x8(%rsp)
  jge    L15
L48:
  lea    D39(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    D49(%rip),%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%esi
  mov    0x8(%rsp),%edx
  test   %edx,%edx
  jle    L21
  mov    0xc(%rsp),%ecx
  lea    0x14(%rsp),%rax
  lea    -0x1(%rdx),%edx
  lea    0x18(%rsp,%rdx,4),%rdx
L43:
  imul   %ecx,%esi
  add    (%rax),%esi
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L43
L21:
  lea    D54(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D11
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D49
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D43
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D39
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D54
type_guessed: string

tool: D11: 
 Enter the degree of the polynomial : 

tool: D49: %d

tool: D43: 
 Enter the %d coefficients


tool: D39: 
 Enter the value of x :

tool: D54: 
 The sum of polynomial = %d

assistant: ```c
int main() {
  int n, i, x, a[10], poly = 0;
  printf("\n Enter the degree of the polynomial : ");
  scanf("%d", &n);
  printf("\n Enter the %d coefficients\n", n + 1);
  for (i = 0; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  printf("\n Enter the value of x :");
  scanf("%d", &x);
  poly = a[0];
  for (i = 1; i <= n; i++) {
    poly = poly * x + a[i];
  }
  printf("\n The sum of polynomial = %d", poly);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D23(%rip),%rdi
  lea    D11(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  lea    D49(%rip),%rdi
  lea    0x1(%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  js     L42
  lea    0x10(%rsp),%rbp
  xor    %ebx,%ebx
L45:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    %ebx,0x8(%rsp)
  jge    L45
L42:
  lea    D47(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%edx
  mov    0x10(%rsp),%esi
  test   %edx,%edx
  jle    L44
  sub    $0x1,%edx
  mov    0xc(%rsp),%ecx
  lea    0x14(%rsp),%rax
  lea    0x18(%rsp,%rdx,4),%rdx
L8:
  imul   %ecx,%esi
  add    $0x4,%rax
  add    -0x4(%rax),%esi
  cmp    %rdx,%rax
  jne    L8
L44:
  xor    %eax,%eax
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D23
type_guessed: string
name: unpack_data
arguments: tag: D11
type_guessed: string
name: unpack_data
arguments: tag: D49
type_guessed: string
name: unpack_data
arguments: tag: D47
type_guessed: string
name: unpack_data
arguments: tag: D38
type_guessed: string

tool: D23: 
 Enter the degree of the polynomial : 

tool: D11: %d

tool: D49: 
 Enter the %d coefficients


tool: D47: 
 Enter the value of x :

tool: D38: 
 The sum of polynomial = %d

assistant: ```c
int main() {
  int n, i, x, a[10], poly = 0;
  printf("\n Enter the degree of the polynomial : ");
  scanf("%d", &n);
  printf("\n Enter the %d coefficients\n", n + 1);
  for (i = 0; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  printf("\n Enter the value of x :");
  scanf("%d", &x);
  poly = a[0];
  for (i = 1; i <= n; i++) {
    poly = poly * x + a[i];
  }
  printf("\n The sum of polynomial = %d", poly);
  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
  push   %r12
  lea    D46(%rip),%rdi
  lea    D7(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  lea    D24(%rip),%rdi
  lea    0x1(%rax),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%eax
  test   %eax,%eax
  js     L45
  lea    0x10(%rsp),%rbp
  xor    %ebx,%ebx
L68:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  cmp    %ebx,0x8(%rsp)
  jge    L68
L45:
  lea    D23(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x8(%rsp),%eax
  mov    0x10(%rsp),%esi
  test   %eax,%eax
  jle    L56
  mov    0xc(%rsp),%edx
  imul   %edx,%esi
  add    0x14(%rsp),%esi
  cmp    $0x1,%eax
  je     L56
  imul   %edx,%esi
  add    0x18(%rsp),%esi
  cmp    $0x2,%eax
  je     L56
  imul   %edx,%esi
  add    0x1c(%rsp),%esi
  cmp    $0x3,%eax
  je     L56
  imul   %edx,%esi
  add    0x20(%rsp),%esi
  cmp    $0x4,%eax
  je     L56
  imul   %edx,%esi
  add    0x24(%rsp),%esi
  cmp    $0x5,%eax
  je     L56
  imul   %edx,%esi
  add    0x28(%rsp),%esi
  cmp    $0x6,%eax
  je     L56
  imul   %edx,%esi
  add    0x2c(%rsp),%esi
  cmp    $0x7,%eax
  je     L56
  imul   %edx,%esi
  add    0x30(%rsp),%esi
  cmp    $0x8,%eax
  je     L56
  imul   %edx,%esi
  add    0x34(%rsp),%esi
L56:
  xor    %eax,%eax
  lea    D2(%rip),%rdi
  call   <printf@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L48
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D46
type: string
name: extract_symbolic_data
arguments: tag: D7
type: string
name: extract_symbolic_data
arguments: tag: D24
type: string
name: extract_symbolic_data
arguments: tag: D23
type: string
name: extract_symbolic_data
arguments: tag: D2
type: string

tool: D46: 
 Enter the degree of the polynomial : 

tool: D7: %d

tool: D24: 
 Enter the %d coefficients


tool: D23: 
 Enter the value of x :

tool: D2: 
 The sum of polynomial = %d

assistant: ```c
int main() {
  int n, i, x, a[10], poly = 0;
  printf("\n Enter the degree of the polynomial : ");
  scanf("%d", &n);
  printf("\n Enter the %d coefficients\n", n + 1);
  for (i = 0; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  printf("\n Enter the value of x :");
  scanf("%d", &x);
  poly = a[0];
  for (i = 1; i <= n; i++) {
    poly = poly * x + a[i];
  }
  printf("\n The sum of polynomial = %d", poly);
  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
<test1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,D16(%rip)
  mov    D16(%rip),%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x24(%rbp),%eax
  lea    (%rdx,%rax,1),%ebx
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <test1>
  add    %eax,%ebx
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <test2@plt>
  add    %ebx,%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int test1(int x) {
  static int thestatic;
  int y;
  thestatic = x;
  y = thestatic;
  return y + x + test1(x - 1) + test2(x - 1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<test1>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  lea    -0x1(%rdi),%ebp
  mov    %ebp,%edi
  call   <test1>
  lea    (%rax,%rbx,2),%ebx
  mov    %ebp,%edi
  call   <test2@plt>
  add    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int test1(int x) {
  static int thestatic;
  int y;
  thestatic = x;
  y = thestatic;
  return y + x + test1(x - 1) + test2(x - 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<test1>:
  endbr64
  push   %r13
  lea    -0x9(%rdi),%r13d
  push   %r12
  push   %rbp
  lea    (%rdi,%rdi,1),%ebp
  push   %rbx
  mov    %edi,%ebx
  mov    %r13d,%edi
  sub    $0x8,%rsp
  call   <test1>
  mov    %r13d,%edi
  lea    -0x10(%rbp,%rax,1),%r12d
  call   <test2@plt>
  lea    -0x8(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0xe(%rbp,%r12,1),%r12d
  lea    -0x7(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0xc(%rbp,%r12,1),%r12d
  lea    -0x6(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0xa(%rbp,%r12,1),%r12d
  lea    -0x5(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0x8(%rbp,%r12,1),%r12d
  lea    -0x4(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0x6(%rbp,%r12,1),%r12d
  lea    -0x3(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0x4(%rbp,%r12,1),%r12d
  lea    -0x2(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0x2(%rbp,%r12,1),%r12d
  lea    -0x1(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  add    %r12d,%ebp
  add    $0x8,%rsp
  pop    %rbx
  add    %ebp,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int test1(int x) {
  static int thestatic;
  int y;
  thestatic = x;
  y = thestatic;
  return y + x + test1(x - 1) + test2(x - 1);
}
```
user: Translate the below assembly code into C programming language:

```asm
<test1>:
  endbr64
  push   %r13
  lea    -0x9(%rdi),%r13d
  push   %r12
  push   %rbp
  lea    (%rdi,%rdi,1),%ebp
  push   %rbx
  mov    %edi,%ebx
  mov    %r13d,%edi
  sub    $0x8,%rsp
  call   <test1>
  mov    %r13d,%edi
  lea    -0x10(%rbp,%rax,1),%r12d
  call   <test2@plt>
  lea    -0x8(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0xe(%rbp,%r12,1),%r12d
  lea    -0x7(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0xc(%rbp,%r12,1),%r12d
  lea    -0x6(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0xa(%rbp,%r12,1),%r12d
  lea    -0x5(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0x8(%rbp,%r12,1),%r12d
  lea    -0x4(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0x6(%rbp,%r12,1),%r12d
  lea    -0x3(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0x4(%rbp,%r12,1),%r12d
  lea    -0x2(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  lea    -0x2(%rbp,%r12,1),%r12d
  lea    -0x1(%rbx),%edi
  add    %eax,%r12d
  call   <test2@plt>
  add    %r12d,%ebp
  add    $0x8,%rsp
  pop    %rbx
  add    %ebp,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int test1(int x) {
  static int thestatic;
  int y;
  thestatic = x;
  y = thestatic;
  return y + x + test1(x - 1) + test2(x - 1);
}
```