ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<WSSIMPatch>:
  endbr64
  lea    0x1(%rcx,%rcx,1),%eax
  mov    %rdx,%r8
  movaps %xmm0,%xmm7
  mov    %eax,%edx
  imul   %eax,%edx
  imul   %eax,%edx
  test   %edx,%edx
  jle    L43
  pxor   %xmm9,%xmm9
  movslq %edx,%rdx
  xor    %eax,%eax
  movaps %xmm9,%xmm5
  movaps %xmm9,%xmm8
L4:
  movss  (%rdi,%rax,4),%xmm0
  movss  (%rsi,%rax,4),%xmm4
  movss  (%r8,%rax,4),%xmm6
  add    $0x1,%rax
  subss  %xmm7,%xmm0
  subss  %xmm1,%xmm4
  addss  %xmm6,%xmm8
  mulss  %xmm4,%xmm0
  mulss  %xmm6,%xmm0
  addss  %xmm0,%xmm5
  cmp    %rax,%rdx
  jne    L4
L18:
  divss  %xmm8,%xmm5
  comiss %xmm5,%xmm9
  ja     L25
  addss  %xmm5,%xmm5
  movss  D10(%rip),%xmm4
  addss  %xmm4,%xmm5
L6:
  movaps %xmm7,%xmm0
  addss  %xmm3,%xmm2
  addss  %xmm7,%xmm0
  mulss  %xmm7,%xmm7
  addss  %xmm4,%xmm2
  mulss  %xmm1,%xmm0
  mulss  %xmm1,%xmm1
  addss  %xmm4,%xmm0
  addss  %xmm1,%xmm7
  mulss  %xmm5,%xmm0
  addss  %xmm4,%xmm7
  mulss  %xmm2,%xmm7
  divss  %xmm7,%xmm0
  ret
  nopl   0x0(%rax)
L25:
  movss  D10(%rip),%xmm4
  movaps %xmm4,%xmm5
  jmp    L6
  nopl   (%rax)
L43:
  pxor   %xmm9,%xmm9
  movaps %xmm9,%xmm5
  movaps %xmm9,%xmm8
  jmp    L18
```

tool_calls: name: disassemble_data
arguments: data_label: D10
datatype: f32

tool: D10: 1e-06

assistant: ```c
float WSSIMPatch(float *PatchImg, float *PatchTemplate, float *kernel, float Mean, float TMean, float Var, float TVar, int f) {
  float d, weight, CoC;
  int i;
  weight = 0.0;
  d = 0.0;
  CoC = 0.0;
  float eps = 0.000001;
  for (i = 0; i < (2 * f + 1) * (2 * f + 1) * (2 * f + 1); i++) {
    CoC = CoC + (PatchImg[i] - Mean) * (PatchTemplate[i] - TMean) * kernel[i];
    weight = weight + kernel[i];
  }
  CoC = CoC / weight;
  if (CoC < 0)
    CoC = 0;
  d = (2 * Mean * TMean + eps) * (2 * CoC + eps);
  d = d / ((Mean * Mean + TMean * TMean + eps) * (Var + TVar + eps));
  return d;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ease8InOutQuad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  and    $0x80,%eax
  test   %eax,%eax
  je     L14
  mov    $0xff,%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L14:
  mov    -0xc(%rbp),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <scale8@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  and    $0x80,%eax
  test   %eax,%eax
  je     L8
  mov    $0xff,%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L8:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
uint8_t ease8InOutQuad(uint8_t i) {
  uint8_t j = i;
  if (j & 0x80) {
    j = 255 - j;
  }
  uint8_t jj = scale8(j, j);
  uint8_t jj2 = jj << 1;
  if (i & 0x80) {
    jj2 = 255 - jj2;
  }
  return jj2;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ease8InOutQuad>:
  endbr64
  push   %rbx
  test   $0x80,%dil
  je     L16
  mov    $0xff,%ebx
  mov    %ebx,%eax
  sub    %edi,%eax
  mov    %eax,%esi
  mov    %eax,%edi
  call   <scale8@plt>
  lea    (%rax,%rax,1),%edx
  mov    %ebx,%eax
  sub    %edx,%eax
L3:
  pop    %rbx
  ret
L16:
  mov    %edi,%esi
  call   <scale8@plt>
  add    %eax,%eax
  jmp    L3
```

assistant: ```c
uint8_t ease8InOutQuad(uint8_t i) {
  uint8_t j = i;
  if (j & 0x80) {
    j = 255 - j;
  }
  uint8_t jj = scale8(j, j);
  uint8_t jj2 = jj << 1;
  if (i & 0x80) {
    jj2 = 255 - jj2;
  }
  return jj2;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ease8InOutQuad>:
  endbr64
  push   %rbx
  test   $0x80,%dil
  je     L6
  mov    $0xff,%ebx
  mov    %ebx,%r8d
  sub    %edi,%r8d
  mov    %r8d,%esi
  mov    %r8d,%edi
  call   <scale8@plt>
  lea    (%rax,%rax,1),%edx
  mov    %ebx,%eax
  pop    %rbx
  sub    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %edi,%esi
  call   <scale8@plt>
  pop    %rbx
  add    %eax,%eax
  ret
```

assistant: ```c
uint8_t ease8InOutQuad(uint8_t i) {
  uint8_t j = i;
  if (j & 0x80) {
    j = 255 - j;
  }
  uint8_t jj = scale8(j, j);
  uint8_t jj2 = jj << 1;
  if (i & 0x80) {
    jj2 = 255 - jj2;
  }
  return jj2;
}
```
user: What is the c source code of the assembly code below?

```asm
<ease8InOutQuad>:
  endbr64
  push   %rbx
  test   $0x80,%dil
  je     L2
  mov    $0xff,%ebx
  mov    %ebx,%r8d
  sub    %edi,%r8d
  mov    %r8d,%esi
  mov    %r8d,%edi
  call   <scale8@plt>
  lea    (%rax,%rax,1),%edx
  mov    %ebx,%eax
  pop    %rbx
  sub    %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  mov    %edi,%esi
  call   <scale8@plt>
  pop    %rbx
  add    %eax,%eax
  ret
```

assistant: ```c
uint8_t ease8InOutQuad(uint8_t i) {
  uint8_t j = i;
  if (j & 0x80) {
    j = 255 - j;
  }
  uint8_t jj = scale8(j, j);
  uint8_t jj2 = jj << 1;
  if (i & 0x80) {
    jj2 = 255 - jj2;
  }
  return jj2;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<int_code>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  movsbl -0x4(%rbp),%eax
  sub    $0x41,%eax
  cmp    $0x34,%eax
  ja     L46
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D4(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D4(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x1,%eax
  jmp    L21
  mov    $0x2,%eax
  jmp    L21
  mov    $0x3,%eax
  jmp    L21
  mov    $0x4,%eax
  jmp    L21
L46:
  mov    $0x0,%eax
L21:
  pop    %rbp
  ret
```

assistant: ```c
uint int_code(char c) {
  switch (c) {
  case 'C':
  case 'c':
    return 1;
  case 'G':
  case 'g':
    return 2;
  case 'A':
  case 'a':
    return 3;
  case 'T':
  case 't':
  case 'U':
  case 'u':
    return 4;
  default:
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<int_code>:
  endbr64
  sub    $0x41,%edi
  cmp    $0x34,%dil
  ja     L6
  movzbl %dil,%edi
  lea    D7(%rip),%rdx
  movslq (%rdx,%rdi,4),%rax
  add    %rdx,%rax
  notrack jmp *%rax
  mov    $0x2,%eax
  ret
  mov    $0x3,%eax
  ret
  mov    $0x4,%eax
  ret
L6:
  mov    $0x0,%eax
  ret
  mov    $0x1,%eax
  ret
```

assistant: ```c
uint int_code(char c) {
  switch (c) {
  case 'C':
  case 'c':
    return 1;
  case 'G':
  case 'g':
    return 2;
  case 'A':
  case 'a':
    return 3;
  case 'T':
  case 't':
  case 'U':
  case 'u':
    return 4;
  default:
    return 0;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<int_code>:
  endbr64
  sub    $0x41,%edi
  xor    %eax,%eax
  cmp    $0x34,%dil
  ja     L3
  movzbl %dil,%edi
  lea    D0(%rip),%rax
  movsbl (%rax,%rdi,1),%eax
L3:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D0
type_infered: string

tool: D0: 

assistant: ```c
uint int_code(char c) {
  switch (c) {
  case 'C':
  case 'c':
    return 1;
  case 'G':
  case 'g':
    return 2;
  case 'A':
  case 'a':
    return 3;
  case 'T':
  case 't':
  case 'U':
  case 'u':
    return 4;
  default:
    return 0;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<int_code>:
  endbr64
  sub    $0x41,%edi
  xor    %eax,%eax
  cmp    $0x34,%dil
  ja     L1
  movzbl %dil,%edi
  lea    D2(%rip),%rax
  movsbl (%rax,%rdi,1),%eax
L1:
  ret
```

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

tool: D2: 

assistant: ```c
uint int_code(char c) {
  switch (c) {
  case 'C':
  case 'c':
    return 1;
  case 'G':
  case 'g':
    return 2;
  case 'A':
  case 'a':
    return 3;
  case 'T':
  case 't':
  case 'U':
  case 'u':
    return 4;
  default:
    return 0;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  lea    0x1ddf826b(%rax,%rax,1),%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned int copy11;
  unsigned int copy12;
  char copy13;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615UL;
    if (state[0UL] & 1UL) {
      if (state[0UL] & 1UL) {
        if (state[0UL] & 1UL) {
          state[0UL] += state[0UL];
        } else {
          copy11 = *((unsigned int *)(&state[0UL]) + 1);
          *((unsigned int *)(&state[0UL]) + 1) = *((unsigned int *)(&state[0UL]) + 0);
          *((unsigned int *)(&state[0UL]) + 0) = copy11;
          copy11 = *((unsigned int *)(&state[0UL]) + 0);
          *((unsigned int *)(&state[0UL]) + 0) = *((unsigned int *)(&state[0UL]) + 1);
          *((unsigned int *)(&state[0UL]) + 1) = copy11;
        }
      } else if ((state[0UL] >> 2UL) & 1UL) {
        copy12 = *((unsigned int *)(&state[0UL]) + 0);
        *((unsigned int *)(&state[0UL]) + 0) = *((unsigned int *)(&state[0UL]) + 1);
        *((unsigned int *)(&state[0UL]) + 1) = copy12;
      } else {
        copy13 = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = *((char *)(&state[0UL]) + 7);
        *((char *)(&state[0UL]) + 7) = copy13;
      }
    } else {
      state[0UL] += state[0UL];
    }
    output[0UL] = state[0UL] + 634099461UL;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<RandomFunc>:
  endbr64
  mov    (%rdi),%rax
  lea    0x1ddf826b(%rax,%rax,1),%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned long input[1], unsigned long output[1]) {
  unsigned long state[1];
  unsigned int copy11;
  unsigned int copy12;
  char copy13;
  {
    state[0UL] = (input[0UL] + 914778474UL) - 981234615UL;
    if (state[0UL] & 1UL) {
      if (state[0UL] & 1UL) {
        if (state[0UL] & 1UL) {
          state[0UL] += state[0UL];
        } else {
          copy11 = *((unsigned int *)(&state[0UL]) + 1);
          *((unsigned int *)(&state[0UL]) + 1) = *((unsigned int *)(&state[0UL]) + 0);
          *((unsigned int *)(&state[0UL]) + 0) = copy11;
          copy11 = *((unsigned int *)(&state[0UL]) + 0);
          *((unsigned int *)(&state[0UL]) + 0) = *((unsigned int *)(&state[0UL]) + 1);
          *((unsigned int *)(&state[0UL]) + 1) = copy11;
        }
      } else if ((state[0UL] >> 2UL) & 1UL) {
        copy12 = *((unsigned int *)(&state[0UL]) + 0);
        *((unsigned int *)(&state[0UL]) + 0) = *((unsigned int *)(&state[0UL]) + 1);
        *((unsigned int *)(&state[0UL]) + 1) = copy12;
      } else {
        copy13 = *((char *)(&state[0UL]) + 2);
        *((char *)(&state[0UL]) + 2) = *((char *)(&state[0UL]) + 7);
        *((char *)(&state[0UL]) + 7) = copy13;
      }
    } else {
      state[0UL] += state[0UL];
    }
    output[0UL] = state[0UL] + 634099461UL;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<process_ssdp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %rdx,-0x18(%rbp)
  mov    %ecx,-0x1c(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void process_ssdp(struct Seaper *seap, struct NetFrame *frame, const unsigned char *px, unsigned length) {
  seap;
  frame;
  px;
  length;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<process_ssdp>:
  endbr64
  ret
```

assistant: ```c
void process_ssdp(struct Seaper *seap, struct NetFrame *frame, const unsigned char *px, unsigned length) {
  seap;
  frame;
  px;
  length;
}
```
user: Translate the below assembly code into C programming language:

```asm
<process_ssdp>:
  endbr64
  ret
```

assistant: ```c
void process_ssdp(struct Seaper *seap, struct NetFrame *frame, const unsigned char *px, unsigned length) {
  seap;
  frame;
  px;
  length;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<process_ssdp>:
  endbr64
  ret
```

assistant: ```c
void process_ssdp(struct Seaper *seap, struct NetFrame *frame, const unsigned char *px, unsigned length) {
  seap;
  frame;
  px;
  length;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x3a0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1390(%rbp),%rax
  mov    %rax,%rsi
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x1398(%rbp)
  movl   $0x0,-0x13a0(%rbp)
  movl   $0x0,-0x139c(%rbp)
  jmp    L53
L15:
  mov    -0x1398(%rbp),%eax
  cltq
  movzbl -0x1390(%rbp,%rax,1),%eax
  cmp    $0x62,%al
  jne    L22
  movl   $0x1,-0x1394(%rbp)
L22:
  mov    -0x1398(%rbp),%eax
  cltq
  movzbl -0x1390(%rbp,%rax,1),%eax
  cmp    $0x63,%al
  jne    L24
  movl   $0x0,-0x1394(%rbp)
L24:
  mov    -0x1394(%rbp),%edx
  lea    -0x139c(%rbp),%rcx
  lea    -0x13a0(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <do_door@plt>
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L35
L8:
  addl   $0x1,-0x1398(%rbp)
L35:
  mov    -0x1398(%rbp),%eax
  cltq
  movzbl -0x1390(%rbp,%rax,1),%eax
  cmp    $0xa,%al
  jne    L8
  addl   $0x1,-0x1398(%rbp)
L53:
  lea    -0x1390(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x1398(%rbp)
  jl     L15
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D22
data_type: string
name: read_data
arguments: label: D9
data_type: string
name: read_data
arguments: label: D27
data_type: string

tool: D22: %[^EOF]s

tool: D9: Door: CLOSED

tool: D27: Door: %s


assistant: ```c
int main() {
  char commands[5000];
  scanf("%[^EOF]s", commands);
  printf("Door: CLOSED\n");
  int f = 0, door = 0, movement = 0, operation;
  for (; f < strlen(commands); f++) {
    if (commands[f] == 'b')
      operation = 1;
    if (commands[f] == 'c')
      operation = 0;
    printf("Door: %s\n", do_door(&door, &movement, operation));
    while (commands[f] != '\n')
      f++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x3a0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1398(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  lea    D18(%rip),%rdi
  call   <puts@plt>
  movl   $0x0,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  mov    $0x0,%ebx
  lea    D13(%rip),%r12
  jmp    L5
L40:
  mov    $0x1,%ebp
L52:
  lea    0xc(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  mov    %ebp,%edx
  call   <do_door@plt>
  mov    %rax,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movslq %ebx,%rax
  cmpb   $0xa,0x10(%rsp,%rax,1)
  je     L45
  lea    0x1(%rbx),%eax
  cltq
  lea    0xf(%rsp),%rdx
L27:
  mov    %eax,%ebx
  add    $0x1,%rax
  cmpb   $0xa,(%rdx,%rax,1)
  jne    L27
L45:
  add    $0x1,%ebx
L5:
  lea    0x10(%rsp),%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L18
  movslq %ebx,%rax
  movzbl 0x10(%rsp,%rax,1),%eax
  cmp    $0x62,%al
  je     L40
  cmp    $0x63,%al
  mov    $0x0,%eax
  cmove  %eax,%ebp
  jmp    L52
L18:
  mov    0x1398(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  mov    $0x0,%eax
  add    $0x13a0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D9
type_infered: string
name: read_data
arguments: label: D18
type_infered: string
name: read_data
arguments: label: D13
type_infered: string

tool: D9: %[^EOF]s

tool: D18: Door: CLOSED

tool: D13: Door: %s


assistant: ```c
int main() {
  char commands[5000];
  scanf("%[^EOF]s", commands);
  printf("Door: CLOSED\n");
  int f = 0, door = 0, movement = 0, operation;
  for (; f < strlen(commands); f++) {
    if (commands[f] == 'b')
      operation = 1;
    if (commands[f] == 'c')
      operation = 0;
    printf("Door: %s\n", do_door(&door, &movement, operation));
    while (commands[f] != '\n')
      f++;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x3a0,%rsp
  lea    D19(%rip),%rdi
  xor    %ebx,%ebx
  lea    D4(%rip),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x1398(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  mov    %r12,%rsi
  call   <scanf@plt>
  lea    D5(%rip),%rdi
  call   <puts@plt>
  mov    %r12,%rdi
  movl   $0x0,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L13
  nopw   0x0(%rax,%rax,1)
L68:
  movslq %ebx,%r14
  movzbl 0x10(%rsp,%r14,1),%eax
  cmp    $0x62,%al
  je     L26
  cmp    $0x63,%al
  mov    $0x0,%eax
  cmove  %eax,%ebp
L52:
  lea    0xc(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  mov    %ebp,%edx
  call   <do_door@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  cmpb   $0xa,0x10(%rsp,%r14,1)
  je     L29
  lea    0x1(%rbx),%eax
  lea    0xf(%rsp),%rdx
  cltq
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %eax,%ebx
  add    $0x1,%rax
  cmpb   $0xa,(%rdx,%rax,1)
  jne    L5
L29:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L68
L13:
  mov    0x1398(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  add    $0x13a0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L26:
  mov    $0x1,%ebp
  jmp    L52
L30:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D19
type_guessed: string
name: unpack_data
arguments: tag: D4
type_guessed: string
name: unpack_data
arguments: tag: D5
type_guessed: string

tool: D19: %[^EOF]s

tool: D4: Door: %s


tool: D5: Door: CLOSED

assistant: ```c
int main() {
  char commands[5000];
  scanf("%[^EOF]s", commands);
  printf("Door: CLOSED\n");
  int f = 0, door = 0, movement = 0, operation;
  for (; f < strlen(commands); f++) {
    if (commands[f] == 'b')
      operation = 1;
    if (commands[f] == 'c')
      operation = 0;
    printf("Door: %s\n", do_door(&door, &movement, operation));
    while (commands[f] != '\n')
      f++;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x3a0,%rsp
  lea    D19(%rip),%rdi
  xor    %ebx,%ebx
  lea    D25(%rip),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x1398(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  mov    %r12,%rsi
  call   <scanf@plt>
  lea    D5(%rip),%rdi
  call   <puts@plt>
  mov    %r12,%rdi
  movl   $0x0,0x8(%rsp)
  movl   $0x0,0xc(%rsp)
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L26
  nopw   0x0(%rax,%rax,1)
L9:
  movslq %ebx,%r14
  movzbl 0x10(%rsp,%r14,1),%eax
  cmp    $0x62,%al
  je     L12
  cmp    $0x63,%al
  mov    $0x0,%eax
  cmove  %eax,%ebp
L38:
  lea    0xc(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  mov    %ebp,%edx
  call   <do_door@plt>
  mov    %r13,%rdi
  mov    %rax,%rsi
  xor    %eax,%eax
  call   <printf@plt>
  cmpb   $0xa,0x10(%rsp,%r14,1)
  je     L18
  lea    0x1(%rbx),%eax
  lea    0xf(%rsp),%rdx
  cltq
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %eax,%ebx
  add    $0x1,%rax
  cmpb   $0xa,(%rdx,%rax,1)
  jne    L6
L18:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L9
L26:
  mov    0x1398(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x13a0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L12:
  mov    $0x1,%ebp
  jmp    L38
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D19
type_guessed: string
name: disassemble_data
arguments: data_label: D25
type_guessed: string
name: disassemble_data
arguments: data_label: D5
type_guessed: string

tool: D19: %[^EOF]s

tool: D25: Door: %s


tool: D5: Door: CLOSED

assistant: ```c
int main() {
  char commands[5000];
  scanf("%[^EOF]s", commands);
  printf("Door: CLOSED\n");
  int f = 0, door = 0, movement = 0, operation;
  for (; f < strlen(commands); f++) {
    if (commands[f] == 'b')
      operation = 1;
    if (commands[f] == 'c')
      operation = 0;
    printf("Door: %s\n", do_door(&door, &movement, operation));
    while (commands[f] != '\n')
      f++;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<print_linear_hist>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  movl   $0x28,-0x8(%rbp)
  movl   $0xffffffff,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L35
L61:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L39
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
L39:
  mov    -0x4(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jbe    L13
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L13:
  addl   $0x1,-0x14(%rbp)
L35:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L61
  cmpl   $0x0,-0x10(%rbp)
  js     L65
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x14(%rbp)
  jmp    L3
L78:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <print_stars@plt>
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  addl   $0x1,-0x14(%rbp)
L3:
  mov    -0x14(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jle    L78
  jmp    L23
L65:
  nop
L23:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D13
datatype: string
name: retrieve_data
arguments: data_tag: D20
datatype: string
name: retrieve_data
arguments: data_tag: D2
datatype: string

tool: D13:      %-13s : count     distribution


tool: D20:         %-10d : %-8d |

tool: D2: |

assistant: ```c
void print_linear_hist(unsigned int *vals, int vals_size, const char *val_type) {
  int i, stars_max = 40, idx_max = -1;
  unsigned int val, val_max = 0;
  for (i = 0; i < vals_size; i++) {
    val = vals[i];
    if (val > 0)
      idx_max = i;
    if (val > val_max)
      val_max = val;
  }
  if (idx_max < 0)
    return;
  printf("     %-13s : count     distribution\n", val_type);
  for (i = 0; i <= idx_max; i++) {
    val = vals[i];
    printf("        %-10d : %-8d |", i, val);
    print_stars(val, val_max, stars_max);
    printf("|\n");
  }
}
```