ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<eembed_byte_to_hex_chars>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %edx,%edi
  push   %rbp
  shr    $0x4,%dil
  mov    %rsi,%rbp
  push   %rbx
  movzbl %dil,%edi
  mov    %edx,%ebx
  and    $0xf,%ebx
  call   <eembed_nibble_to_hex@plt>
  mov    %ebx,%edi
  mov    %al,(%r12)
  call   <eembed_nibble_to_hex@plt>
  pop    %rbx
  mov    %al,0x0(%rbp)
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void eembed_byte_to_hex_chars(char *hi, char *lo, unsigned char byte) {
  *hi = eembed_nibble_to_hex((byte & 0xF0) >> 4);
  *lo = eembed_nibble_to_hex((byte & 0x0F));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<eembed_byte_to_hex_chars>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  mov    %edx,%edi
  push   %rbp
  shr    $0x4,%dil
  mov    %rsi,%rbp
  push   %rbx
  movzbl %dil,%edi
  mov    %edx,%ebx
  and    $0xf,%ebx
  call   <eembed_nibble_to_hex@plt>
  mov    %ebx,%edi
  mov    %al,(%r12)
  call   <eembed_nibble_to_hex@plt>
  pop    %rbx
  mov    %al,0x0(%rbp)
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
void eembed_byte_to_hex_chars(char *hi, char *lo, unsigned char byte) {
  *hi = eembed_nibble_to_hex((byte & 0xF0) >> 4);
  *lo = eembed_nibble_to_hex((byte & 0x0F));
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<CalcAllTass17Elem>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  movsd  %xmm0,-0x58(%rbp)
  mov    %rdi,-0x60(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x40(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  mov    %rdx,%rdi
  movq   %rax,%xmm0
  call   <CalcLon@plt>
  movl   $0x0,-0x44(%rbp)
  jmp    L23
L1:
  mov    -0x44(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  add    %eax,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x60(%rbp),%rax
  add    %rax,%rdx
  mov    -0x44(%rbp),%esi
  lea    -0x40(%rbp),%rcx
  mov    -0x58(%rbp),%rax
  mov    %rcx,%rdi
  movq   %rax,%xmm0
  call   <CalcTass17Elem@plt>
  addl   $0x1,-0x44(%rbp)
L23:
  cmpl   $0x7,-0x44(%rbp)
  jle    L1
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
void CalcAllTass17Elem(const double t, double elem[(8 * 6)]) {
  int body;
  double lon[7];
  CalcLon(t, lon);
  for (body = 0; body < 8; body++)
    CalcTass17Elem(t, lon, body, elem + (body * 6));
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<CalcAllTass17Elem>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  movq   %xmm0,%r12
  mov    %rdi,%rbp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  call   <CalcLon@plt>
  mov    $0x0,%ebx
  mov    %rsp,%r13
L9:
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r13,%rdi
  movq   %r12,%xmm0
  call   <CalcTass17Elem@plt>
  add    $0x1,%ebx
  add    $0x30,%rbp
  cmp    $0x8,%ebx
  jne    L9
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L14:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void CalcAllTass17Elem(const double t, double elem[(8 * 6)]) {
  int body;
  double lon[7];
  CalcLon(t, lon);
  for (body = 0; body < 8; body++)
    CalcTass17Elem(t, lon, body, elem + (body * 6));
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<CalcAllTass17Elem>:
  endbr64
  push   %r14
  movq   %xmm0,%r14
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <CalcLon@plt>
  nopl   0x0(%rax,%rax,1)
L12:
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  movq   %r14,%xmm0
  add    $0x30,%rbp
  call   <CalcTass17Elem@plt>
  cmp    $0x8,%ebx
  jne    L12
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L19
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void CalcAllTass17Elem(const double t, double elem[(8 * 6)]) {
  int body;
  double lon[7];
  CalcLon(t, lon);
  for (body = 0; body < 8; body++)
    CalcTass17Elem(t, lon, body, elem + (body * 6));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<CalcAllTass17Elem>:
  endbr64
  push   %r14
  movq   %xmm0,%r14
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <CalcLon@plt>
  nopl   0x0(%rax,%rax,1)
L3:
  mov    %rbp,%rdx
  mov    %ebx,%esi
  mov    %r12,%rdi
  add    $0x1,%ebx
  movq   %r14,%xmm0
  add    $0x30,%rbp
  call   <CalcTass17Elem@plt>
  cmp    $0x8,%ebx
  jne    L3
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r14
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void CalcAllTass17Elem(const double t, double elem[(8 * 6)]) {
  int body;
  double lon[7];
  CalcLon(t, lon);
  for (body = 0; body < 8; body++)
    CalcTass17Elem(t, lon, body, elem + (body * 6));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<fill_with_numbers>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  movl   $0x0,-0x10(%rbp)
  jmp    L24
L37:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  jns    L12
L75:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idivl  -0x1c(%rbp)
  mov    %edx,-0x4(%rbp)
  movl   $0x1,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L26
L34:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L50
  movl   $0x0,-0x8(%rbp)
L50:
  addl   $0x1,-0xc(%rbp)
L26:
  mov    -0xc(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jl     L34
  cmpl   $0x0,-0x8(%rbp)
  je     L75
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  jmp    L77
L12:
  nop
L77:
  addl   $0x1,-0x10(%rbp)
L24:
  mov    -0x10(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L37
  nop
  nop
  leave
  ret
```

assistant: ```c
void fill_with_numbers(int *nums, int modulo, int grenze) {
  srand(time(NULL));
  int i, j;
  for (i = 0; i < grenze; i++) {
    if (*(nums + i) >= 0)
      continue;
    while (1) {
      int tmp = rand() % modulo;
      int ok = 1;
      for (j = 0; j < i; j++) {
        if (*(nums + j) == tmp)
          ok = 0;
      }
      if (ok) {
        *(nums + i) = tmp;
        break;
      }
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fill_with_numbers>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r13
  mov    %esi,0xc(%rsp)
  mov    %edx,%r14d
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %r14d,%r14d
  jle    L34
  mov    %r14d,%r14d
  mov    $0x0,%ebx
  lea    0x4(%r13),%r15
  mov    $0x0,%r12d
  jmp    L1
L55:
  mov    %edx,0x0(%r13,%rbx,4)
L47:
  add    $0x1,%rbx
  cmp    %r14,%rbx
  je     L34
L1:
  cmpl   $0x0,0x0(%r13,%rbx,4)
  jns    L47
  lea    -0x1(%rbx),%eax
  lea    (%r15,%rax,4),%rbp
L76:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idivl  0xc(%rsp)
  test   %ebx,%ebx
  jle    L55
  mov    %r13,%rax
  mov    $0x1,%ecx
L42:
  cmp    %edx,(%rax)
  cmove  %r12d,%ecx
  add    $0x4,%rax
  cmp    %rbp,%rax
  jne    L42
  test   %ecx,%ecx
  jne    L55
  jmp    L76
L34:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void fill_with_numbers(int *nums, int modulo, int grenze) {
  srand(time(NULL));
  int i, j;
  for (i = 0; i < grenze; i++) {
    if (*(nums + i) >= 0)
      continue;
    while (1) {
      int tmp = rand() % modulo;
      int ok = 1;
      for (j = 0; j < i; j++) {
        if (*(nums + j) == tmp)
          ok = 0;
      }
      if (ok) {
        *(nums + i) = tmp;
        break;
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<fill_with_numbers>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  movslq %edx,%r12
  push   %rbp
  mov    %rdi,%rbp
  xor    %edi,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %r12d,%r12d
  jle    L18
  lea    0x4(%rbp),%rax
  xor    %r13d,%r13d
  xor    %r15d,%r15d
  mov    %rax,0x8(%rsp)
  jmp    L61
  xchg   %ax,%ax
L6:
  add    $0x1,%r13
  cmp    %r12,%r13
  je     L18
L61:
  mov    0x0(%rbp,%r13,4),%eax
  test   %eax,%eax
  jns    L6
  mov    0x8(%rsp),%rsi
  lea    -0x1(%r13),%eax
  lea    (%rsi,%rax,4),%r14
  nop
L37:
  xor    %eax,%eax
  call   <rand@plt>
  cltd
  idiv   %ebx
  test   %r13,%r13
  je     L15
  mov    %rbp,%rax
  mov    $0x1,%ecx
  nopw   0x0(%rax,%rax,1)
L32:
  cmp    %edx,(%rax)
  cmove  %r15d,%ecx
  add    $0x4,%rax
  cmp    %r14,%rax
  jne    L32
  test   %ecx,%ecx
  je     L37
  nopl   0x0(%rax,%rax,1)
L15:
  mov    %edx,0x0(%rbp,%r13,4)
  add    $0x1,%r13
  cmp    %r12,%r13
  jne    L61
L18:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void fill_with_numbers(int *nums, int modulo, int grenze) {
  srand(time(NULL));
  int i, j;
  for (i = 0; i < grenze; i++) {
    if (*(nums + i) >= 0)
      continue;
    while (1) {
      int tmp = rand() % modulo;
      int ok = 1;
      for (j = 0; j < i; j++) {
        if (*(nums + j) == tmp)
          ok = 0;
      }
      if (ok) {
        *(nums + i) = tmp;
        break;
      }
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  imul   $0xfb16,%ax,%ax
  sub    $0x2e4,%ax
  mov    %ax,-0xa(%rbp)
  movw   $0x0,-0xc(%rbp)
  jmp    L27
L11:
  movzwl -0xc(%rbp),%eax
  lea    -0xa(%rbp),%rdx
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  mov    %al,-0xe(%rbp)
  movzwl -0xc(%rbp),%eax
  lea    -0xa(%rbp),%rdx
  cltq
  add    %rax,%rax
  add    %rdx,%rax
  movzwl -0xc(%rbp),%edx
  lea    -0xa(%rbp),%rcx
  movslq %edx,%rdx
  add    %rdx,%rdx
  add    %rcx,%rdx
  add    $0x1,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  movzwl -0xc(%rbp),%eax
  lea    -0xa(%rbp),%rdx
  cltq
  add    %rax,%rax
  add    %rax,%rdx
  movzbl -0xe(%rbp),%eax
  mov    %al,(%rdx)
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  lea    -0xa(%rbp),%rdx
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  lea    -0xa(%rbp),%rax
  movzbl -0xd(%rbp),%edx
  mov    %dl,(%rax)
  lea    -0xa(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  lea    -0xa(%rbp),%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  lea    -0xa(%rbp),%rax
  add    $0x1,%rax
  movzbl -0xd(%rbp),%edx
  mov    %dl,(%rax)
  movzwl -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0xc(%rbp)
L27:
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzwl (%rax),%eax
  cmp    %ax,-0xc(%rbp)
  jb     L11
  movzwl -0xa(%rbp),%eax
  lea    0x13eb(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L21
  call   <__stack_chk_fail@plt>
L21:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  char copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) * (unsigned short)64278;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
      *((char *)(&state[local1]) + 0) = copy11;
      copy12 = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
      copy12 = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = copy12;
      local1++;
    }
    output[0UL] = state[0UL] + 1047729131UL;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  imul   $0xfb16,(%rdi),%ax
  sub    $0x2e4,%ax
  mov    %ax,0x6(%rsp)
  movzwl 0x2(%rdi),%edx
  test   %dx,%dx
  je     L26
  lea    0x6(%rsp),%rax
  sub    $0x1,%edx
  movzwl %dx,%edx
  lea    0x8(%rsp,%rdx,2),%rdi
L20:
  movzbl 0x1(%rax),%edx
  movzbl (%rax),%ecx
  mov    %cl,0x1(%rax)
  mov    %dl,(%rax)
  add    $0x2,%rax
  cmp    %rdi,%rax
  jne    L20
L26:
  movzwl 0x6(%rsp),%eax
  add    $0x13eb,%ax
  mov    %ax,(%rsi)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x18,%rsp
  ret
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  char copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) * (unsigned short)64278;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
      *((char *)(&state[local1]) + 0) = copy11;
      copy12 = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
      copy12 = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = copy12;
      local1++;
    }
    output[0UL] = state[0UL] + 1047729131UL;
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<RandomFunc>:
  endbr64
  sub    $0x18,%rsp
  movzwl 0x2(%rdi),%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  imul   $0xfb16,(%rdi),%ax
  sub    $0x2e4,%ax
  mov    %ax,0x6(%rsp)
  test   %dx,%dx
  je     L6
  sub    $0x1,%edx
  lea    0x6(%rsp),%rax
  movzwl %dx,%edx
  lea    0x8(%rsp,%rdx,2),%rdi
  nop
L3:
  movzbl 0x1(%rax),%edx
  movzbl (%rax),%ecx
  add    $0x2,%rax
  mov    %cl,-0x1(%rax)
  mov    %dl,-0x2(%rax)
  cmp    %rdi,%rax
  jne    L3
  movzwl 0x6(%rsp),%eax
L6:
  add    $0x13eb,%ax
  mov    %ax,(%rsi)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x18,%rsp
  ret
L20:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local1;
  char copy11;
  char copy12;
  {
    state[0UL] = (input[0UL] + 914778474UL) * (unsigned short)64278;
    local1 = 0UL;
    while (local1 < input[1UL]) {
      copy11 = *((char *)(&state[local1]) + 1);
      *((char *)(&state[local1]) + 1) = *((char *)(&state[local1]) + 0);
      *((char *)(&state[local1]) + 0) = copy11;
      copy12 = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = copy12;
      copy12 = *((char *)(&state[0UL]) + 0);
      *((char *)(&state[0UL]) + 0) = *((char *)(&state[0UL]) + 1);
      *((char *)(&state[0UL]) + 1) = copy12;
      local1++;
    }
    output[0UL] = state[0UL] + 1047729131UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ask_for_file>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1c(%rbp)
  movl   $0x1,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L29
L14:
  lea    -0x18(%rbp),%rcx
  lea    -0x1c(%rbp),%rdx
  lea    -0x10(%rbp),%rsi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <read_by_char@plt>
L29:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x14(%rbp)
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  je     L8
  cmpl   $0xa,-0x14(%rbp)
  jne    L14
L8:
  mov    -0x10(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
char *ask_for_file(void) {
  int c, charCount = 0, bytes = 1;
  char *filename;
  filename = malloc(sizeof(char) * bytes);
  while ((c = getchar()) != EOF && c != '\n')
    read_by_char(c, &filename, &charCount, &bytes);
  return filename;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ask_for_file>:
  endbr64
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,0x10(%rsp)
  lea    0xc(%rsp),%rbx
L27:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  mov    D1(%rip),%rax
  cmp    %edi,(%rax)
  je     L2
  cmp    $0xa,%edi
  je     L2
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rcx
  call   <read_by_char@plt>
  jmp    L27
L2:
  mov    0x10(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L12
  add    $0x20,%rsp
  pop    %rbx
  ret
L12:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ask_for_file(void) {
  int c, charCount = 0, bytes = 1;
  char *filename;
  filename = malloc(sizeof(char) * bytes);
  while ((c = getchar()) != EOF && c != '\n')
    read_by_char(c, &filename, &charCount, &bytes);
  return filename;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<ask_for_file>:
  endbr64
  push   %rbp
  mov    $0x1,%edi
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x8(%rsp)
  lea    0xc(%rsp),%rbp
  movl   $0x1,0xc(%rsp)
  call   <malloc@plt>
  mov    D5(%rip),%rbx
  mov    %rax,0x10(%rsp)
  jmp    L18
  nopw   0x0(%rax,%rax,1)
L3:
  cmp    $0xa,%eax
  je     L5
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rcx
  call   <read_by_char@plt>
L18:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L3
L5:
  mov    0x10(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ask_for_file(void) {
  int c, charCount = 0, bytes = 1;
  char *filename;
  filename = malloc(sizeof(char) * bytes);
  while ((c = getchar()) != EOF && c != '\n')
    read_by_char(c, &filename, &charCount, &bytes);
  return filename;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<ask_for_file>:
  endbr64
  push   %rbp
  mov    $0x1,%edi
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x8(%rsp)
  lea    0xc(%rsp),%rbp
  movl   $0x1,0xc(%rsp)
  call   <malloc@plt>
  mov    D2(%rip),%rbx
  mov    %rax,0x10(%rsp)
  jmp    L33
  nopw   0x0(%rax,%rax,1)
L25:
  cmp    $0xa,%eax
  je     L4
  lea    0x8(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rcx
  call   <read_by_char@plt>
L33:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L25
L4:
  mov    0x10(%rsp),%rax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ask_for_file(void) {
  int c, charCount = 0, bytes = 1;
  char *filename;
  filename = malloc(sizeof(char) * bytes);
  while ((c = getchar()) != EOF && c != '\n')
    read_by_char(c, &filename, &charCount, &bytes);
  return filename;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xb0,%rsp
  mov    %edi,-0xa4(%rbp)
  mov    %rsi,-0xb0(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x2073692073696874,%rax
  movabs $0x6d20676e6f6c2061,%rdx
  mov    %rax,-0x90(%rbp)
  mov    %rdx,-0x88(%rbp)
  movabs $0x7420656761737365,%rax
  movabs $0x6c6c697720746168,%rdx
  mov    %rax,-0x80(%rbp)
  mov    %rdx,-0x78(%rbp)
  movabs $0x6572697571657220,%rax
  movabs $0x6620676e69656220,%rdx
  mov    %rax,-0x70(%rbp)
  mov    %rdx,-0x68(%rbp)
  movabs $0x65746e656d676172,%rax
  movabs $0x6d206f746e692064,%rdx
  mov    %rax,-0x60(%rbp)
  mov    %rdx,-0x58(%rbp)
  movabs $0x20656c7069746c75,%rax
  movabs $0x207374656b636170,%rdx
  mov    %rax,-0x50(%rbp)
  mov    %rdx,-0x48(%rbp)
  movabs $0x656874207265766f,%rax
  movabs $0x6f72702050544e20,%rdx
  mov    %rax,-0x40(%rbp)
  mov    %rdx,-0x38(%rbp)
  movabs $0x59202e6c6f636f74,%rax
  movabs $0x75732049202e7961,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x2065706f68206572,%rax
  movabs $0x736b726f77207469,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movw   $0x2e,-0x10(%rbp)
L27:
  mov    $0x2,%edi
  call   <accept@plt>
  mov    %eax,-0x98(%rbp)
  cmpl   $0xffffffff,-0x98(%rbp)
  je     L27
  lea    -0x90(%rbp),%rcx
  mov    -0x98(%rbp),%eax
  mov    $0x82,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
  mov    %eax,-0x94(%rbp)
  cmpl   $0xffffffff,-0x94(%rbp)
  jne    L31
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L38
L31:
  mov    $0x0,%eax
L38:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L24
  call   <__stack_chk_fail@plt>
L24:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
type_infered: string

tool: D1: Error : Expected to write all bytes, but wrote -1

assistant: ```c
int main(int argc, char *argv[]) {
  int connection, bytes;
  char msg[] = "this is a long message that will require being fragmented into multiple packets over the NTP protocol. Yay. I sure hope it works.";
  do {
    connection = accept(2);
  } while (connection == -1);
  bytes = write(connection, msg, sizeof(msg));
  if (bytes == -1) {
    printf("Error : Expected to write all bytes, but wrote -1");
    return 1;
  }
  return 0;
}
```