ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  je     L19
  lea    D6(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L19:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x6(%rsp),%rsi
  lea    0x4(%rsp),%rdi
  mov    %ax,0x4(%rsp)
  call   <RandomFunc@plt>
  movzwl 0x6(%rsp),%esi
  cmp    $0x2098,%si
  je     L12
L6:
  xor    %eax,%eax
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L12:
  lea    D19(%rip),%rdi
  call   <puts@plt>
  movzwl 0x6(%rsp),%esi
  jmp    L6
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D6
datatype: string
name: read_data
arguments: data_tag: D14
datatype: string
name: read_data
arguments: data_tag: D19
datatype: string

tool: D6: Call this program with %i arguments


tool: D14: %u


tool: D19: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned short input[1];
  unsigned short output[1];
  int randomFuns_i5;
  unsigned short randomFuns_value6;
  int randomFuns_main_i7;
  {
    megaInit();
    if (argc != 2) {
      printf("Call this program with %i arguments\n", 1);
      exit(-1);
    } else {
    }
    randomFuns_i5 = 0;
    while (randomFuns_i5 < 1) {
      randomFuns_value6 = (unsigned short)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 8344) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<get_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <getsym@plt>
  mov    $0x0,%eax
  call   <readfactor@plt>
  nop
  pop    %rbp
  ret
```

assistant: ```c
void get_() {
  getsym();
  readfactor();
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<get_>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <getsym@plt>
  mov    $0x0,%eax
  call   <readfactor@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void get_() {
  getsym();
  readfactor();
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <getsym@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <readfactor@plt>
```

assistant: ```c
void get_() {
  getsym();
  readfactor();
}
```
user: Translate the below assembly code into C programming language:

```asm
<get_>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <getsym@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1070 <readfactor@plt>
```

assistant: ```c
void get_() {
  getsym();
  readfactor();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x10(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  lea    D28(%rip),%rax
  mov    %rax,-0x18(%rbp)
  lea    D7(%rip),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  lea    -0x20(%rbp),%rcx
  mov    %rcx,%r10
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <ech2.0>
  mov    -0x10(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L0
  call   <__stack_chk_fail@plt>
L0:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D28
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D7
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D23
type_infered: string
name: extract_symbolic_data
arguments: data_tag: D19
type_infered: string

tool: D28: coucou

tool: D7: hello

tool: D23: prems0,*deuz : %s,%s


tool: D19: ech coucou,hello : %s,%s


assistant: ```c
int main() {
  void ech1(char *a, char *b) {
    char *t;
    printf("aff a : %s\n", a);
    t = a;
    printf("t : %s\n", t);
    a = b;
    b = t;
    printf("aff a : %s\n", a);
  }
  void ech2(char **a, char **b) {
    char **t;
    printf("aff a : %s\n", a);
    *t = *a;
    printf("t : %s\n", t);
    *a = *b;
    *b = *t;
    printf("aff a : %s\n", a);
  }
  void echint(int *a, int *b) {
    int *t;
    t = a;
    a = b;
    b = t;
  }
  char *prems[1];
  char *deuz[1];
  prems[0] = "coucou";
  *deuz = "hello";
  printf("prems0,*deuz : %s,%s\n", prems[0], *deuz);
  ech2(prems, deuz);
  printf("ech coucou,hello : %s,%s\n", *prems, *deuz);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    D14(%rip),%rsi
  mov    %rsi,(%rsp)
  lea    D16(%rip),%r13
  mov    %r13,%rdx
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%r12
  mov    %r12,%rsi
  lea    D59(%rip),%rbp
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%ebx
  mov    (%rsp),%rax
  mov    %rax,(%rbx)
  mov    %rbx,%rsi
  lea    D39(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r13,(%rsp)
  mov    (%rbx),%rbx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rdx
  mov    (%rsp),%rsi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D14
type_guessed: string
name: parse_data
arguments: data_label: D16
type_guessed: string
name: parse_data
arguments: data_label: D5
type_guessed: string
name: parse_data
arguments: data_label: D59
type_guessed: string
name: parse_data
arguments: data_label: D39
type_guessed: string
name: parse_data
arguments: data_label: D2
type_guessed: string

tool: D14: coucou

tool: D16: hello

tool: D5: prems0,*deuz : %s,%s


tool: D59: aff a : %s


tool: D39: t : %s


tool: D2: ech coucou,hello : %s,%s


assistant: ```c
int main() {
  void ech1(char *a, char *b) {
    char *t;
    printf("aff a : %s\n", a);
    t = a;
    printf("t : %s\n", t);
    a = b;
    b = t;
    printf("aff a : %s\n", a);
  }
  void ech2(char **a, char **b) {
    char **t;
    printf("aff a : %s\n", a);
    *t = *a;
    printf("t : %s\n", t);
    *a = *b;
    *b = *t;
    printf("aff a : %s\n", a);
  }
  void echint(int *a, int *b) {
    int *t;
    t = a;
    a = b;
    b = t;
  }
  char *prems[1];
  char *deuz[1];
  prems[0] = "coucou";
  *deuz = "hello";
  printf("prems0,*deuz : %s,%s\n", prems[0], *deuz);
  ech2(prems, deuz);
  printf("ech coucou,hello : %s,%s\n", *prems, *deuz);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %r13
  lea    D48(%rip),%r13
  lea    D4(%rip),%rsi
  push   %r12
  mov    %r13,%rdx
  lea    D56(%rip),%rdi
  push   %rbp
  lea    D46(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %rsi,(%rsp)
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%rax
  mov    %rbx,%rsi
  lea    D57(%rip),%rdi
  mov    %rax,(%rbx)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,(%rsp)
  mov    %r12,%rsi
  mov    (%rbx),%r13
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%rsi
  xor    %eax,%eax
  mov    %r13,%rdx
  lea    D31(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D48
type: string
name: unpack_data
arguments: label: D4
type: string
name: unpack_data
arguments: label: D56
type: string
name: unpack_data
arguments: label: D46
type: string
name: unpack_data
arguments: label: D57
type: string
name: unpack_data
arguments: label: D31
type: string

tool: D48: hello

tool: D4: coucou

tool: D56: prems0,*deuz : %s,%s


tool: D46: aff a : %s


tool: D57: t : %s


tool: D31: ech coucou,hello : %s,%s


assistant: ```c
int main() {
  void ech1(char *a, char *b) {
    char *t;
    printf("aff a : %s\n", a);
    t = a;
    printf("t : %s\n", t);
    a = b;
    b = t;
    printf("aff a : %s\n", a);
  }
  void ech2(char **a, char **b) {
    char **t;
    printf("aff a : %s\n", a);
    *t = *a;
    printf("t : %s\n", t);
    *a = *b;
    *b = *t;
    printf("aff a : %s\n", a);
  }
  void echint(int *a, int *b) {
    int *t;
    t = a;
    a = b;
    b = t;
  }
  char *prems[1];
  char *deuz[1];
  prems[0] = "coucou";
  *deuz = "hello";
  printf("prems0,*deuz : %s,%s\n", prems[0], *deuz);
  ech2(prems, deuz);
  printf("ech coucou,hello : %s,%s\n", *prems, *deuz);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r13
  lea    D9(%rip),%r13
  lea    D17(%rip),%rsi
  push   %r12
  mov    %r13,%rdx
  lea    D51(%rip),%rdi
  push   %rbp
  lea    D27(%rip),%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %rsi,(%rsp)
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%rax
  mov    %rbx,%rsi
  lea    D4(%rip),%rdi
  mov    %rax,(%rbx)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,(%rsp)
  mov    %r12,%rsi
  mov    (%rbx),%r13
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    (%rsp),%rsi
  xor    %eax,%eax
  mov    %r13,%rdx
  lea    D55(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D9
type_infered: string
name: read_data
arguments: data_tag: D17
type_infered: string
name: read_data
arguments: data_tag: D51
type_infered: string
name: read_data
arguments: data_tag: D27
type_infered: string
name: read_data
arguments: data_tag: D4
type_infered: string
name: read_data
arguments: data_tag: D55
type_infered: string

tool: D9: hello

tool: D17: coucou

tool: D51: prems0,*deuz : %s,%s


tool: D27: aff a : %s


tool: D4: t : %s


tool: D55: ech coucou,hello : %s,%s


assistant: ```c
int main() {
  void ech1(char *a, char *b) {
    char *t;
    printf("aff a : %s\n", a);
    t = a;
    printf("t : %s\n", t);
    a = b;
    b = t;
    printf("aff a : %s\n", a);
  }
  void ech2(char **a, char **b) {
    char **t;
    printf("aff a : %s\n", a);
    *t = *a;
    printf("t : %s\n", t);
    *a = *b;
    *b = *t;
    printf("aff a : %s\n", a);
  }
  void echint(int *a, int *b) {
    int *t;
    t = a;
    a = b;
    b = t;
  }
  char *prems[1];
  char *deuz[1];
  prems[0] = "coucou";
  *deuz = "hello";
  printf("prems0,*deuz : %s,%s\n", prems[0], *deuz);
  ech2(prems, deuz);
  printf("ech coucou,hello : %s,%s\n", *prems, *deuz);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D18(%rip),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x28(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L16
  mov    $0xff,%eax
  jmp    L53
L16:
  movl   $0x0,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
  movl   $0x0,-0x34(%rbp)
  mov    $0x640,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  jmp    L28
L56:
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x500,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  jmp    L61
L15:
  lea    -0x34(%rbp),%rdx
  mov    -0x28(%rbp),%rax
  lea    D10(%rip),%rcx
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <fscanf@plt>
  mov    -0x2c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rax,%rdx
  mov    -0x34(%rbp),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x30(%rbp)
L61:
  cmpl   $0x13f,-0x30(%rbp)
  jle    L15
  movl   $0x0,-0x30(%rbp)
  addl   $0x1,-0x2c(%rbp)
L28:
  cmpl   $0xc7,-0x2c(%rbp)
  jle    L56
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <TaskA@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <TaskB@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <TaskC@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <TaskD@plt>
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L53:
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L60
  call   <__stack_chk_fail@plt>
L60:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D13
data_type: string
name: unpack_data
arguments: label: D18
data_type: string
name: unpack_data
arguments: label: D7
data_type: string
name: unpack_data
arguments: label: D10
data_type: string

tool: D13: Matura 2017 Informatyka - Zadanie 6


tool: D18: r

tool: D7: dane.txt

tool: D10: %d

assistant: ```c
int main() {
  printf("Matura 2017 Informatyka - Zadanie 6\n");
  FILE *InputFile = fopen("dane.txt", "r");
  if (InputFile == NULL) {
    return 255;
  }
  int PixelIndex = 0;
  int CurrentIndex = 0;
  int PixelBuffer = 0;
  int **PixelStorage = (int **)malloc(200 * sizeof(*PixelStorage));
  while (CurrentIndex < 200) {
    PixelStorage[CurrentIndex] = malloc(320 * sizeof(**PixelStorage));
    while (PixelIndex < 320) {
      fscanf(InputFile, "%d", &PixelBuffer);
      PixelStorage[CurrentIndex][PixelIndex] = PixelBuffer;
      PixelIndex++;
    }
    PixelIndex = 0;
    CurrentIndex++;
  }
  TaskA(PixelStorage);
  TaskB(PixelStorage);
  TaskC(PixelStorage);
  TaskD(PixelStorage);
  fclose(InputFile);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D11(%rip),%rsi
  lea    D6(%rip),%rdi
  call   <fopen@plt>
  test   %rax,%rax
  je     L2
  mov    %rax,%r12
  movl   $0x0,0x14(%rsp)
  mov    $0x640,%edi
  call   <malloc@plt>
  mov    %rax,0x8(%rsp)
  mov    %rax,%r14
  lea    0x640(%rax),%r15
  lea    D4(%rip),%r13
L29:
  mov    $0x500,%edi
  call   <malloc@plt>
  mov    %r14,%rbp
  mov    %rax,(%r14)
  mov    $0x0,%ebx
L26:
  lea    0x14(%rsp),%rdx
  mov    %r13,%rsi
  mov    %r12,%rdi
  call   <fscanf@plt>
  mov    0x0(%rbp),%rax
  mov    0x14(%rsp),%edx
  mov    %edx,(%rax,%rbx,1)
  add    $0x4,%rbx
  cmp    $0x500,%rbx
  jne    L26
  add    $0x8,%r14
  cmp    %r15,%r14
  jne    L29
  mov    0x8(%rsp),%rbx
  mov    %rbx,%rdi
  call   <TaskA@plt>
  mov    %rbx,%rdi
  call   <TaskB@plt>
  mov    %rbx,%rdi
  call   <TaskC@plt>
  mov    %rbx,%rdi
  call   <TaskD@plt>
  mov    %r12,%rdi
  call   <fclose@plt>
  mov    $0x0,%eax
L7:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L44
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L2:
  mov    $0xff,%eax
  jmp    L7
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D11
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D6
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D4
type_guessed: string

tool: D2: Matura 2017 Informatyka - Zadanie 6


tool: D11: r

tool: D6: dane.txt

tool: D4: %d

assistant: ```c
int main() {
  printf("Matura 2017 Informatyka - Zadanie 6\n");
  FILE *InputFile = fopen("dane.txt", "r");
  if (InputFile == NULL) {
    return 255;
  }
  int PixelIndex = 0;
  int CurrentIndex = 0;
  int PixelBuffer = 0;
  int **PixelStorage = (int **)malloc(200 * sizeof(*PixelStorage));
  while (CurrentIndex < 200) {
    PixelStorage[CurrentIndex] = malloc(320 * sizeof(**PixelStorage));
    while (PixelIndex < 320) {
      fscanf(InputFile, "%d", &PixelBuffer);
      PixelStorage[CurrentIndex][PixelIndex] = PixelBuffer;
      PixelIndex++;
    }
    PixelIndex = 0;
    CurrentIndex++;
  }
  TaskA(PixelStorage);
  TaskB(PixelStorage);
  TaskC(PixelStorage);
  TaskD(PixelStorage);
  fclose(InputFile);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  lea    D14(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D21(%rip),%rsi
  lea    D35(%rip),%rdi
  call   <fopen@plt>
  test   %rax,%rax
  je     L32
  mov    $0x640,%edi
  movl   $0x0,0x14(%rsp)
  mov    %rax,%rbp
  lea    0x14(%rsp),%r13
  call   <malloc@plt>
  lea    D39(%rip),%r12
  mov    %rax,%r15
  mov    %rax,%r14
  lea    0x640(%rax),%rax
  mov    %rax,0x8(%rsp)
L18:
  mov    $0x500,%edi
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,(%r14)
  nopl   0x0(%rax)
L27:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <fscanf@plt>
  mov    (%r14),%rax
  mov    0x14(%rsp),%edx
  mov    %edx,(%rax,%rbx,1)
  add    $0x4,%rbx
  cmp    $0x500,%rbx
  jne    L27
  add    $0x8,%r14
  cmp    0x8(%rsp),%r14
  jne    L18
  mov    %r15,%rdi
  call   <TaskA@plt>
  mov    %r15,%rdi
  call   <TaskB@plt>
  mov    %r15,%rdi
  call   <TaskC@plt>
  mov    %r15,%rdi
  call   <TaskD@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L3:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L29
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L32:
  mov    $0xff,%eax
  jmp    L3
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D14
data_type: string
name: parse_data
arguments: data_tag: D21
data_type: string
name: parse_data
arguments: data_tag: D35
data_type: string
name: parse_data
arguments: data_tag: D39
data_type: string

tool: D14: Matura 2017 Informatyka - Zadanie 6


tool: D21: r

tool: D35: dane.txt

tool: D39: %d

assistant: ```c
int main() {
  printf("Matura 2017 Informatyka - Zadanie 6\n");
  FILE *InputFile = fopen("dane.txt", "r");
  if (InputFile == NULL) {
    return 255;
  }
  int PixelIndex = 0;
  int CurrentIndex = 0;
  int PixelBuffer = 0;
  int **PixelStorage = (int **)malloc(200 * sizeof(*PixelStorage));
  while (CurrentIndex < 200) {
    PixelStorage[CurrentIndex] = malloc(320 * sizeof(**PixelStorage));
    while (PixelIndex < 320) {
      fscanf(InputFile, "%d", &PixelBuffer);
      PixelStorage[CurrentIndex][PixelIndex] = PixelBuffer;
      PixelIndex++;
    }
    PixelIndex = 0;
    CurrentIndex++;
  }
  TaskA(PixelStorage);
  TaskB(PixelStorage);
  TaskC(PixelStorage);
  TaskD(PixelStorage);
  fclose(InputFile);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D23(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D17(%rip),%rsi
  lea    D20(%rip),%rdi
  call   <fopen@plt>
  test   %rax,%rax
  je     L36
  mov    $0x640,%edi
  movl   $0x0,0x14(%rsp)
  mov    %rax,%rbp
  lea    0x14(%rsp),%r13
  call   <malloc@plt>
  lea    D18(%rip),%r12
  mov    %rax,%r15
  mov    %rax,%r14
  lea    0x640(%rax),%rax
  mov    %rax,0x8(%rsp)
L48:
  mov    $0x500,%edi
  xor    %ebx,%ebx
  call   <malloc@plt>
  mov    %rax,(%r14)
  nopl   0x0(%rax)
L9:
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <fscanf@plt>
  mov    (%r14),%rax
  mov    0x14(%rsp),%edx
  mov    %edx,(%rax,%rbx,1)
  add    $0x4,%rbx
  cmp    $0x500,%rbx
  jne    L9
  add    $0x8,%r14
  cmp    0x8(%rsp),%r14
  jne    L48
  mov    %r15,%rdi
  call   <TaskA@plt>
  mov    %r15,%rdi
  call   <TaskB@plt>
  mov    %r15,%rdi
  call   <TaskC@plt>
  mov    %r15,%rdi
  call   <TaskD@plt>
  mov    %rbp,%rdi
  call   <fclose@plt>
  xor    %eax,%eax
L25:
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L36:
  mov    $0xff,%eax
  jmp    L25
L19:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D23
data_type: string
name: retrieve_data
arguments: tag: D17
data_type: string
name: retrieve_data
arguments: tag: D20
data_type: string
name: retrieve_data
arguments: tag: D18
data_type: string

tool: D23: Matura 2017 Informatyka - Zadanie 6


tool: D17: r

tool: D20: dane.txt

tool: D18: %d

assistant: ```c
int main() {
  printf("Matura 2017 Informatyka - Zadanie 6\n");
  FILE *InputFile = fopen("dane.txt", "r");
  if (InputFile == NULL) {
    return 255;
  }
  int PixelIndex = 0;
  int CurrentIndex = 0;
  int PixelBuffer = 0;
  int **PixelStorage = (int **)malloc(200 * sizeof(*PixelStorage));
  while (CurrentIndex < 200) {
    PixelStorage[CurrentIndex] = malloc(320 * sizeof(**PixelStorage));
    while (PixelIndex < 320) {
      fscanf(InputFile, "%d", &PixelBuffer);
      PixelStorage[CurrentIndex][PixelIndex] = PixelBuffer;
      PixelIndex++;
    }
    PixelIndex = 0;
    CurrentIndex++;
  }
  TaskA(PixelStorage);
  TaskB(PixelStorage);
  TaskC(PixelStorage);
  TaskD(PixelStorage);
  fclose(InputFile);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<my_find_prime_sup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L14:
  addl   $0x1,-0x4(%rbp)
L1:
  mov    -0x14(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edi
  call   <my_is_prime@plt>
  test   %rax,%rax
  je     L14
  mov    -0x14(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
int my_find_prime_sup(int nb) {
  int i;
  for (i = 0; my_is_prime(nb + i) == 0; i++)
    ;
  return (nb + i);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<my_find_prime_sup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
L3:
  mov    %ebx,%ebp
  mov    %ebx,%edi
  call   <my_is_prime@plt>
  add    $0x1,%ebx
  test   %rax,%rax
  je     L3
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int my_find_prime_sup(int nb) {
  int i;
  for (i = 0; my_is_prime(nb + i) == 0; i++)
    ;
  return (nb + i);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_find_prime_sup>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  nopl   (%rax)
L9:
  mov    %ebx,%edi
  mov    %ebx,%r12d
  add    $0x1,%ebx
  call   <my_is_prime@plt>
  test   %rax,%rax
  je     L9
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
int my_find_prime_sup(int nb) {
  int i;
  for (i = 0; my_is_prime(nb + i) == 0; i++)
    ;
  return (nb + i);
}
```