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

```asm
<strrev>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  add    %rbx,%rax
  cmp    %rax,%rbx
  jae    L19
L4:
  sub    $0x1,%rax
  movzbl (%rax),%edx
  movzbl (%rbx),%ecx
  mov    %cl,(%rax)
  add    $0x1,%rbx
  mov    %dl,-0x1(%rbx)
  cmp    %rbx,%rax
  ja     L4
L19:
  pop    %rbx
  ret
```

assistant: ```c
void strrev(char *str) {
  char *tail, tmp;
  tail = str + strlen(str);
  while (tail > str) {
    tmp = *--tail;
    *tail = *str;
    *str++ = tmp;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strrev>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  add    %rbx,%rax
  cmp    %rax,%rbx
  jae    L12
  nopw   0x0(%rax,%rax,1)
L2:
  movzbl (%rbx),%ecx
  movzbl -0x1(%rax),%edx
  sub    $0x1,%rax
  add    $0x1,%rbx
  mov    %cl,(%rax)
  mov    %dl,-0x1(%rbx)
  cmp    %rbx,%rax
  ja     L2
L12:
  pop    %rbx
  ret
```

assistant: ```c
void strrev(char *str) {
  char *tail, tmp;
  tail = str + strlen(str);
  while (tail > str) {
    tmp = *--tail;
    *tail = *str;
    *str++ = tmp;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<strrev>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cltq
  add    %rbx,%rax
  cmp    %rax,%rbx
  jae    L11
  nopw   0x0(%rax,%rax,1)
L18:
  movzbl (%rbx),%ecx
  movzbl -0x1(%rax),%edx
  sub    $0x1,%rax
  add    $0x1,%rbx
  mov    %cl,(%rax)
  mov    %dl,-0x1(%rbx)
  cmp    %rbx,%rax
  ja     L18
L11:
  pop    %rbx
  ret
```

assistant: ```c
void strrev(char *str) {
  char *tail, tmp;
  tail = str + strlen(str);
  while (tail > str) {
    tmp = *--tail;
    *tail = *str;
    *str++ = tmp;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<_print_char>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L14
  mov    $0x0,%edi
  call   <putchar@plt>
  jmp    L15
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L15:
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
datatype: string

tool: D1: %c

assistant: ```c
void _print_char(void *d) {
  if (d == NULL)
    putchar('\0');
  else
    printf("%c", *(char *)d);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<_print_char>:
  endbr64
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L5
  movsbl (%rdi),%esi
  lea    D2(%rip),%rdi
  call   <printf@plt>
L10:
  add    $0x8,%rsp
  ret
L5:
  call   <putchar@plt>
  jmp    L10
```

tool_calls: name: unpack_data
arguments: data_label: D2
datatype: string

tool: D2: %c

assistant: ```c
void _print_char(void *d) {
  if (d == NULL)
    putchar('\0');
  else
    printf("%c", *(char *)d);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<_print_char>:
  endbr64
  test   %rdi,%rdi
  je     L4
  movsbl (%rdi),%esi
  lea    D8(%rip),%rdi
  jmp    1070 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L4:
  jmp    1060 <putchar@plt>
```

tool_calls: name: unpack_data
arguments: tag: D8
type: string

tool: D8: %c

assistant: ```c
void _print_char(void *d) {
  if (d == NULL)
    putchar('\0');
  else
    printf("%c", *(char *)d);
}
```
user: Translate the below assembly code into C programming language:

```asm
<_print_char>:
  endbr64
  test   %rdi,%rdi
  je     L5
  movsbl (%rdi),%esi
  lea    D0(%rip),%rdi
  jmp    1070 <printf@plt>
  nopl   0x0(%rax,%rax,1)
L5:
  jmp    1060 <putchar@plt>
```

tool_calls: name: read_data
arguments: data_label: D0
type: string

tool: D0: %c

assistant: ```c
void _print_char(void *d) {
  if (d == NULL)
    putchar('\0');
  else
    printf("%c", *(char *)d);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<__fnv_1a>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  movb   $0x49,-0x9(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L10
L9:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  xor    %al,-0x9(%rbp)
  movzbl -0x9(%rbp),%eax
  mov    $0xffffffb3,%edx
  imul   %edx,%eax
  mov    %al,-0x9(%rbp)
  addl   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L9
  movzbl -0x9(%rbp),%eax
  leave
  ret
```

assistant: ```c
uint64_t __fnv_1a(const char *key) {
  int i, len = strlen(key);
  uint64_t h = 14695981039346656073ULL;
  for (i = 0; i < len; ++i) {
    h = h ^ (unsigned char)key[i];
    h = h * 1099511628211ULL;
  }
  return h;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<__fnv_1a>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L16
  mov    %rbx,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rsi
  mov    $0x49,%eax
  mov    $0xffffffb3,%ecx
L9:
  xor    (%rdx),%al
  imul   %ecx,%eax
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L9
L4:
  pop    %rbx
  ret
L16:
  mov    $0x49,%eax
  jmp    L4
```

assistant: ```c
uint64_t __fnv_1a(const char *key) {
  int i, len = strlen(key);
  uint64_t h = 14695981039346656073ULL;
  for (i = 0; i < len; ++i) {
    h = h ^ (unsigned char)key[i];
    h = h * 1099511628211ULL;
  }
  return h;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<__fnv_1a>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L4
  sub    $0x1,%eax
  mov    %rbx,%rdi
  mov    $0xffffffb3,%edx
  lea    0x1(%rbx,%rax,1),%rcx
  mov    $0x49,%eax
  cs nopw 0x0(%rax,%rax,1)
L13:
  xor    (%rdi),%al
  add    $0x1,%rdi
  imul   %edx,%eax
  cmp    %rcx,%rdi
  jne    L13
  pop    %rbx
  ret
L4:
  mov    $0x49,%eax
  pop    %rbx
  ret
```

assistant: ```c
uint64_t __fnv_1a(const char *key) {
  int i, len = strlen(key);
  uint64_t h = 14695981039346656073ULL;
  for (i = 0; i < len; ++i) {
    h = h ^ (unsigned char)key[i];
    h = h * 1099511628211ULL;
  }
  return h;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<__fnv_1a>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  test   %eax,%eax
  jle    L0
  sub    $0x1,%eax
  mov    %rbx,%rdi
  mov    $0xffffffb3,%edx
  lea    0x1(%rbx,%rax,1),%rcx
  mov    $0x49,%eax
  cs nopw 0x0(%rax,%rax,1)
L8:
  xor    (%rdi),%al
  add    $0x1,%rdi
  imul   %edx,%eax
  cmp    %rdi,%rcx
  jne    L8
  pop    %rbx
  ret
L0:
  mov    $0x49,%eax
  pop    %rbx
  ret
```

assistant: ```c
uint64_t __fnv_1a(const char *key) {
  int i, len = strlen(key);
  uint64_t h = 14695981039346656073ULL;
  for (i = 0; i < len; ++i) {
    h = h ^ (unsigned char)key[i];
    h = h * 1099511628211ULL;
  }
  return h;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test_lastprivate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L176:
  addl   $0x1,-0x4(%rbp)
L0:
  cmpl   $0xf,-0x4(%rbp)
  jle    L176
  movl   $0x0,-0x4(%rbp)
  jmp    L69
L11:
  addl   $0x1,-0x4(%rbp)
L69:
  cmpl   $0xf,-0x4(%rbp)
  jle    L11
  movl   $0x0,-0x4(%rbp)
  jmp    L6
L139:
  addl   $0x1,-0x4(%rbp)
L6:
  cmpl   $0xf,-0x4(%rbp)
  jle    L139
  movl   $0x0,-0x4(%rbp)
  jmp    L101
L59:
  addl   $0x1,-0x4(%rbp)
L101:
  cmpl   $0xf,-0x4(%rbp)
  jle    L59
  movl   $0x0,-0x4(%rbp)
  jmp    L150
L56:
  addl   $0x1,-0x4(%rbp)
L150:
  cmpl   $0xf,-0x4(%rbp)
  jle    L56
  movl   $0x0,-0x4(%rbp)
  jmp    L113
L38:
  addl   $0x1,-0x4(%rbp)
L113:
  cmpl   $0xf,-0x4(%rbp)
  jle    L38
  movl   $0x0,-0x4(%rbp)
  jmp    L90
L173:
  addl   $0x1,-0x4(%rbp)
L90:
  cmpl   $0xf,-0x4(%rbp)
  jle    L173
  movl   $0x0,-0x4(%rbp)
  jmp    L73
L116:
  addl   $0x1,-0x4(%rbp)
L73:
  cmpl   $0xf,-0x4(%rbp)
  jle    L116
  movl   $0x0,-0x4(%rbp)
  jmp    L172
L88:
  addl   $0x1,-0x4(%rbp)
L172:
  cmpl   $0xf,-0x4(%rbp)
  jle    L88
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void test_lastprivate() {
  int i;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(int)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(0)
  for (i = 0; i < 16; ++i)
    ;
  int x, y, z;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x, y)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x, y, z)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<test_lastprivate>:
  endbr64
  ret
```

assistant: ```c
void test_lastprivate() {
  int i;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(int)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(0)
  for (i = 0; i < 16; ++i)
    ;
  int x, y, z;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x, y)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x, y, z)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: Translate the below assembly code into C programming language:

```asm
<test_lastprivate>:
  endbr64
  ret
```

assistant: ```c
void test_lastprivate() {
  int i;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(int)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(0)
  for (i = 0; i < 16; ++i)
    ;
  int x, y, z;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x, y)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x, y, z)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<test_lastprivate>:
  endbr64
  ret
```

assistant: ```c
void test_lastprivate() {
  int i;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(,
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(, )
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate()
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(int)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(0)
  for (i = 0; i < 16; ++i)
    ;
  int x, y, z;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x, y)
  for (i = 0; i < 16; ++i)
    ;
#pragma omp target
#pragma omp teams
#pragma omp distribute simd lastprivate(x, y, z)
  for (i = 0; i < 16; ++i)
    ;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<calc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x200,%rsp
  mov    %edi,-0x1f4(%rbp)
  mov    %esi,-0x1f8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1f0(%rbp),%rax
  lea    D6(%rip),%rdx
  mov    $0x3c,%ecx
  mov    %rax,%rdi
  mov    %rdx,%rsi
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    %rsi,%rdx
  mov    %rdi,%rax
  mov    (%rdx),%ecx
  mov    %ecx,(%rax)
  mov    -0x1f8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x1f4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rcx,%rax
  mov    -0x1f0(%rbp,%rax,4),%eax
  cmp    $0xffffffff,%eax
  jne    L5
  mov    -0x1f4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x1f8(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rcx,%rax
  mov    -0x1f0(%rbp,%rax,4),%eax
  jmp    L26
L5:
  mov    -0x1f8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x1f4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  add    %rdx,%rax
  add    %rcx,%rax
  mov    -0x1f0(%rbp,%rax,4),%eax
L26:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D6
type: string

tool: 

assistant: ```c
int calc(int x, int y) {
  int distance[11][11] = {{0, 996, 723, 890, 1286, 305, 564, 245, 424, 1026, 884},
                          {-1, 0, 750, 104, 286, 711, 576, 747, 590, 505, 543},
                          {-1, -1, 0, 668, 979, 593, 224, 515, 194, 534, 905},
                          {-1, -1, -1, 0, 316, 607, 472, 645, 502, 434, 467},
                          {
                              -1,
                              -1,
                              -1,
                              -1,
                              0,
                              890,
                              769,
                              938,
                              757,
                              750,
                              400,
                          },
                          {-1, -1, -1, -1, -1, 0, 386, 106, 315, 832, 559},
                          {-1, -1, -1, -1, -1, -1, 0, 348, 112, 447, 681},
                          {-1, -1, -1, -1, -1, -1, -1, 0, 249, 799, 665},
                          {-1, -1, -1, -1, -1, -1, -1, -1, 0, 497, 665},
                          {-1, -1, -1, -1, -1, -1, -1, -1, -1, 0, 901},
                          {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 0}};
  if (distance[x][y] == -1) {
    return distance[y][x];
  } else
    return distance[x][y];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<calc>:
  endbr64
  sub    $0x1f8,%rsp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %fs:0x28,%rcx
  mov    %rcx,0x1e8(%rsp)
  xor    %ecx,%ecx
  mov    %rsp,%rdi
  lea    D8(%rip),%rsi
  mov    $0x3c,%ecx
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    (%rsi),%ecx
  mov    %ecx,(%rdi)
  movslq %eax,%rsi
  movslq %edx,%rcx
  lea    (%rcx,%rcx,4),%rdi
  lea    (%rcx,%rdi,2),%rcx
  add    %rsi,%rcx
  mov    (%rsp,%rcx,4),%ecx
  cmp    $0xffffffff,%ecx
  je     L6
L26:
  mov    0x1e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  mov    %ecx,%eax
  add    $0x1f8,%rsp
  ret
L6:
  movslq %edx,%rdx
  lea    (%rsi,%rsi,4),%rcx
  lea    (%rsi,%rcx,2),%rax
  add    %rdx,%rax
  mov    (%rsp,%rax,4),%ecx
  jmp    L26
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D8
datatype: string

tool: 

assistant: ```c
int calc(int x, int y) {
  int distance[11][11] = {{0, 996, 723, 890, 1286, 305, 564, 245, 424, 1026, 884},
                          {-1, 0, 750, 104, 286, 711, 576, 747, 590, 505, 543},
                          {-1, -1, 0, 668, 979, 593, 224, 515, 194, 534, 905},
                          {-1, -1, -1, 0, 316, 607, 472, 645, 502, 434, 467},
                          {
                              -1,
                              -1,
                              -1,
                              -1,
                              0,
                              890,
                              769,
                              938,
                              757,
                              750,
                              400,
                          },
                          {-1, -1, -1, -1, -1, 0, 386, 106, 315, 832, 559},
                          {-1, -1, -1, -1, -1, -1, 0, 348, 112, 447, 681},
                          {-1, -1, -1, -1, -1, -1, -1, 0, 249, 799, 665},
                          {-1, -1, -1, -1, -1, -1, -1, -1, 0, 497, 665},
                          {-1, -1, -1, -1, -1, -1, -1, -1, -1, 0, 901},
                          {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 0}};
  if (distance[x][y] == -1) {
    return distance[y][x];
  } else
    return distance[x][y];
}
```
user: What is the c source code of the assembly code below:

```asm
<calc>:
  endbr64
  sub    $0x1f8,%rsp
  mov    %edi,%r8d
  movslq %esi,%rdx
  mov    $0x3c,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x1e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    D3(%rip),%rsi
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    (%rsi),%eax
  mov    %eax,(%rdi)
  movslq %r8d,%rax
  lea    (%rax,%rax,4),%rcx
  lea    (%rax,%rcx,2),%rcx
  add    %rdx,%rcx
  mov    (%rsp,%rcx,4),%r8d
  cmp    $0xffffffff,%r8d
  jne    L11
  lea    (%rdx,%rdx,4),%rcx
  lea    (%rdx,%rcx,2),%rdx
  add    %rdx,%rax
  mov    (%rsp,%rax,4),%r8d
L11:
  mov    0x1e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    %r8d,%eax
  add    $0x1f8,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D3
type_infered: string

tool: 

assistant: ```c
int calc(int x, int y) {
  int distance[11][11] = {{0, 996, 723, 890, 1286, 305, 564, 245, 424, 1026, 884},
                          {-1, 0, 750, 104, 286, 711, 576, 747, 590, 505, 543},
                          {-1, -1, 0, 668, 979, 593, 224, 515, 194, 534, 905},
                          {-1, -1, -1, 0, 316, 607, 472, 645, 502, 434, 467},
                          {
                              -1,
                              -1,
                              -1,
                              -1,
                              0,
                              890,
                              769,
                              938,
                              757,
                              750,
                              400,
                          },
                          {-1, -1, -1, -1, -1, 0, 386, 106, 315, 832, 559},
                          {-1, -1, -1, -1, -1, -1, 0, 348, 112, 447, 681},
                          {-1, -1, -1, -1, -1, -1, -1, 0, 249, 799, 665},
                          {-1, -1, -1, -1, -1, -1, -1, -1, 0, 497, 665},
                          {-1, -1, -1, -1, -1, -1, -1, -1, -1, 0, 901},
                          {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 0}};
  if (distance[x][y] == -1) {
    return distance[y][x];
  } else
    return distance[x][y];
}
```
user: Translate the below assembly code into C programming language:

```asm
<calc>:
  endbr64
  sub    $0x1f8,%rsp
  mov    %edi,%r8d
  movslq %esi,%rdx
  mov    $0x3c,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,0x1e8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  lea    D7(%rip),%rsi
  rep movsq %ds:(%rsi),%es:(%rdi)
  mov    (%rsi),%eax
  mov    %eax,(%rdi)
  movslq %r8d,%rax
  lea    (%rax,%rax,4),%rcx
  lea    (%rax,%rcx,2),%rcx
  add    %rdx,%rcx
  mov    (%rsp,%rcx,4),%r8d
  cmp    $0xffffffff,%r8d
  jne    L7
  lea    (%rdx,%rdx,4),%rcx
  lea    (%rdx,%rcx,2),%rdx
  add    %rdx,%rax
  mov    (%rsp,%rax,4),%r8d
L7:
  mov    0x1e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    %r8d,%eax
  add    $0x1f8,%rsp
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D7
data_type: string

tool: 

assistant: ```c
int calc(int x, int y) {
  int distance[11][11] = {{0, 996, 723, 890, 1286, 305, 564, 245, 424, 1026, 884},
                          {-1, 0, 750, 104, 286, 711, 576, 747, 590, 505, 543},
                          {-1, -1, 0, 668, 979, 593, 224, 515, 194, 534, 905},
                          {-1, -1, -1, 0, 316, 607, 472, 645, 502, 434, 467},
                          {
                              -1,
                              -1,
                              -1,
                              -1,
                              0,
                              890,
                              769,
                              938,
                              757,
                              750,
                              400,
                          },
                          {-1, -1, -1, -1, -1, 0, 386, 106, 315, 832, 559},
                          {-1, -1, -1, -1, -1, -1, 0, 348, 112, 447, 681},
                          {-1, -1, -1, -1, -1, -1, -1, 0, 249, 799, 665},
                          {-1, -1, -1, -1, -1, -1, -1, -1, 0, 497, 665},
                          {-1, -1, -1, -1, -1, -1, -1, -1, -1, 0, 901},
                          {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 0}};
  if (distance[x][y] == -1) {
    return distance[y][x];
  } else
    return distance[x][y];
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<check_prime>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L28
  cmpl   $0x1,-0x14(%rbp)
  je     L28
  cmpl   $0x0,-0x14(%rbp)
  jns    L16
L28:
  mov    $0x0,%eax
  jmp    L65
L47:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0x4(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L0
  mov    $0x0,%eax
  jmp    L65
L0:
  subl   $0x1,-0x4(%rbp)
L16:
  cmpl   $0x1,-0x4(%rbp)
  jg     L47
  mov    $0x1,%eax
L65:
  pop    %rbp
  ret
```

assistant: ```c
int check_prime(int nb) {
  int i = nb - 1;
  if (nb == 0 || nb == 1 || nb < 0)
    return (0);
  while (i > 1) {
    if (nb % i == 0)
      return (0);
    i--;
  }
  return (1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<check_prime>:
  endbr64
  mov    $0x0,%ecx
  cmp    $0x1,%edi
  jle    L42
  lea    -0x1(%rdi),%ecx
  cmp    $0x1,%ecx
  jle    L26
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L22
L68:
  sub    $0x1,%ecx
  cmp    $0x1,%ecx
  je     L42
  mov    %edi,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L68
  mov    %edx,%ecx
L42:
  mov    %ecx,%eax
  ret
L26:
  mov    $0x1,%ecx
  jmp    L42
L22:
  mov    %edx,%ecx
  jmp    L42
```

assistant: ```c
int check_prime(int nb) {
  int i = nb - 1;
  if (nb == 0 || nb == 1 || nb < 0)
    return (0);
  while (i > 1) {
    if (nb % i == 0)
      return (0);
    i--;
  }
  return (1);
}
```
user: What is the c source code of the assembly code below?

```asm
<check_prime>:
  endbr64
  xor    %r8d,%r8d
  cmp    $0x1,%edi
  jle    L1
  lea    -0x1(%rdi),%r8d
  cmp    $0x2,%edi
  jne    L33
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L16:
  sub    $0x1,%r8d
  cmp    $0x1,%r8d
  je     L1
L33:
  mov    %edi,%eax
  cltd
  idiv   %r8d
  test   %edx,%edx
  jne    L16
  xor    %r8d,%r8d
L1:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int check_prime(int nb) {
  int i = nb - 1;
  if (nb == 0 || nb == 1 || nb < 0)
    return (0);
  while (i > 1) {
    if (nb % i == 0)
      return (0);
    i--;
  }
  return (1);
}
```