ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<strncmp>:
  endbr64
  test   %edx,%edx
  je     L4
  mov    %edx,%edx
  xor    %ecx,%ecx
  jmp    L34
  xchg   %ax,%ax
L36:
  test   %r8b,%r8b
  je     L4
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  je     L4
L34:
  movzbl (%rdi,%rcx,1),%r8d
  mov    %r8d,%eax
  sub    (%rsi,%rcx,1),%al
  je     L36
  movsbl %al,%eax
  ret
  nop
L4:
  xor    %eax,%eax
  ret
```

assistant: ```c
int strncmp(const char *cs, const char *ct, int count) {
  signed char __res = 0;
  while (count) {
    if ((__res = *cs - *ct++) != 0 || !*cs++)
      break;
    count--;
  }
  return __res;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x118,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x108(%rsp)
  xor    %eax,%eax
  mov    $0x1f40,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    $0x0,%r12d
  mov    $0x3e8,%r15d
  lea    D4(%rip),%r14
  jmp    L33
L3:
  mov    %rdi,%rbx
L24:
  movb   $0x0,(%rbx)
  call   <atol@plt>
  add    $0x1,%rbx
  cmp    %rax,%rbp
  je     L31
  mov    0x0(%r13,%rbp,8),%rsi
  mov    %r12,%rdx
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L20:
  mov    %rbx,%rdi
  call   <puts@plt>
  add    $0x1,%r12
L33:
  mov    %rsp,%rdi
  call   <gets@plt>
  test   %rax,%rax
  je     L111
  movzbl (%rsp),%eax
  test   %al,%al
  je     L76
  cmp    $0x3a,%al
  setne  %al
  mov    %rsp,%rbx
  test   %al,%al
  je     L76
L54:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L138
  cmp    $0x3a,%al
  jne    L54
L138:
  movb   $0x0,(%rbx)
  mov    %rsp,%rdi
  call   <atol@plt>
  mov    %rax,%rbp
  lea    0x1(%rbx),%rdi
  movzbl 0x1(%rbx),%eax
  test   %al,%al
  je     L3
  cmp    $0x3a,%al
  je     L3
  mov    %rdi,%rbx
L136:
  add    $0x1,%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L24
  cmp    $0x3a,%al
  jne    L136
  jmp    L24
L31:
  mov    %r12,%rdx
  mov    %r12,%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x1(%r15),%rax
  cmp    %rbp,%rax
  jb     L96
L127:
  mov    %r12,0x0(%r13,%rbp,8)
  jmp    L20
L96:
  lea    0x1(%rbp),%r15
  lea    0x0(,%r15,8),%rsi
  mov    %r13,%rdi
  call   <realloc@plt>
  mov    %rax,%r13
  jmp    L127
L76:
  mov    %rsp,%rbx
  jmp    L138
L111:
  mov    0x108(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L84
  mov    $0x0,%eax
  add    $0x118,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L84:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
type: string

tool: D4: %d:%d:

assistant: ```c
main() {
  unsigned long *clusterID;
  unsigned long nClusters = 1000;
  unsigned long ruid = 0;
  unsigned long clusterIdVal, recordIdVal;
  char line[255];
  register char *ptr, *ptr2;
  clusterID = (unsigned long *)malloc(nClusters * sizeof(unsigned long));
  while (gets(line)) {
    ptr = line;
    while (*ptr && *ptr != ':')
      ptr++;
    *ptr = '\0';
    clusterIdVal = atol(line);
    ptr2 = ++ptr;
    while (*ptr2 && *ptr2 != ':')
      ptr2++;
    *ptr2 = '\0';
    recordIdVal = atol(ptr);
    ptr2++;
    if (clusterIdVal == recordIdVal) {
      printf("%d:%d:", ruid, ruid);
      if (clusterIdVal > nClusters - 1) {
        nClusters = clusterIdVal + 1;
        clusterID = (unsigned long *)realloc(clusterID, nClusters * sizeof(unsigned long));
      }
      clusterID[clusterIdVal] = ruid;
    } else
      printf("%d:%d:", clusterID[clusterIdVal], ruid);
    printf("%s\n", ptr2);
    ruid++;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<selection_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L15
L1:
  mov    -0x10(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  jmp    L11
L21:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L18
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L18:
  addl   $0x1,-0xc(%rbp)
L11:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L21
  mov    -0x10(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  je     L8
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
L8:
  addl   $0x1,-0x10(%rbp)
L15:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x10(%rbp)
  jl     L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void selection_sort(int num[], int tam) {
  int i, j, min, swap;
  for (i = 0; i < (tam - 1); i++) {
    min = i;
    for (j = (i + 1); j < tam; j++) {
      if (num[j] < num[min]) {
        min = j;
      }
    }
    if (i != min) {
      swap = num[i];
      num[i] = num[min];
      num[min] = swap;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<selection_sort>:
  endbr64
  cmp    $0x1,%esi
  jle    L19
  lea    -0x2(%rsi),%r10d
  add    $0x2,%r10
  mov    $0x1,%r9d
  mov    $0x0,%r8d
  jmp    L65
L36:
  add    $0x1,%r8
  add    $0x1,%r9
  cmp    %r10,%r9
  je     L19
L65:
  mov    %r8d,%edx
  cmp    %r9d,%esi
  jle    L36
  mov    %r9,%rax
L68:
  movslq %edx,%rcx
  mov    (%rdi,%rcx,4),%ecx
  cmp    %ecx,(%rdi,%rax,4)
  cmovl  %eax,%edx
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L68
  cmp    %r8d,%edx
  je     L36
  mov    (%rdi,%r8,4),%ecx
  movslq %edx,%rdx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%rdi,%r8,4)
  mov    %ecx,(%rax)
  jmp    L36
L19:
  ret
```

assistant: ```c
void selection_sort(int num[], int tam) {
  int i, j, min, swap;
  for (i = 0; i < (tam - 1); i++) {
    min = i;
    for (j = (i + 1); j < tam; j++) {
      if (num[j] < num[min]) {
        min = j;
      }
    }
    if (i != min) {
      swap = num[i];
      num[i] = num[min];
      num[min] = swap;
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<selection_sort>:
  endbr64
  cmp    $0x1,%esi
  jle    L4
  lea    -0x2(%rsi),%r10d
  mov    $0x1,%r9d
  xor    %r8d,%r8d
  add    $0x2,%r10
  nopw   0x0(%rax,%rax,1)
L36:
  mov    %r8d,%edx
  cmp    %r9d,%esi
  jle    L31
  mov    %r9,%rax
  nopl   0x0(%rax,%rax,1)
L45:
  movslq %edx,%rcx
  mov    (%rdi,%rcx,4),%ecx
  cmp    %ecx,(%rdi,%rax,4)
  cmovl  %eax,%edx
  add    $0x1,%rax
  cmp    %eax,%esi
  jg     L45
  cmp    %r8d,%edx
  je     L31
  movslq %edx,%rdx
  mov    (%rdi,%r8,4),%ecx
  lea    (%rdi,%rdx,4),%rax
  mov    (%rax),%edx
  mov    %edx,(%rdi,%r8,4)
  mov    %ecx,(%rax)
L31:
  add    $0x1,%r9
  add    $0x1,%r8
  cmp    %r10,%r9
  jne    L36
L4:
  ret
```

assistant: ```c
void selection_sort(int num[], int tam) {
  int i, j, min, swap;
  for (i = 0; i < (tam - 1); i++) {
    min = i;
    for (j = (i + 1); j < tam; j++) {
      if (num[j] < num[min]) {
        min = j;
      }
    }
    if (i != min) {
      swap = num[i];
      num[i] = num[min];
      num[min] = swap;
    }
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<selection_sort>:
  endbr64
  push   %r13
  lea    -0x1(%rsi),%r13d
  push   %r12
  push   %rbp
  push   %rbx
  test   %r13d,%r13d
  jle    L64
  mov    %rdi,%rbx
  lea    0x4(%rdi),%r11
  xor    %r10d,%r10d
  nopl   (%rax)
L27:
  mov    %r10d,%ebp
  add    $0x1,%r10d
  cmp    %r10d,%esi
  jle    L24
  mov    -0x4(%r11),%r12d
  mov    %r11,%rax
  mov    %ebp,%r9d
  mov    %r10d,%edx
  mov    %r12d,%edi
  jmp    L18
  xchg   %ax,%ax
L31:
  movslq %r9d,%rcx
  add    $0x1,%edx
  add    $0x4,%rax
  lea    (%rbx,%rcx,4),%r8
  cmp    %esi,%edx
  je     L59
L18:
  mov    (%rax),%ecx
  mov    %rax,%r8
  cmp    %edi,%ecx
  jge    L31
  mov    %edx,%r9d
  add    $0x1,%edx
  mov    %ecx,%edi
  add    $0x4,%rax
  cmp    %esi,%edx
  jne    L18
L59:
  cmp    %ebp,%r9d
  je     L24
  mov    %edi,-0x4(%r11)
  mov    %r12d,(%r8)
L24:
  add    $0x4,%r11
  cmp    %r13d,%r10d
  jne    L27
L64:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void selection_sort(int num[], int tam) {
  int i, j, min, swap;
  for (i = 0; i < (tam - 1); i++) {
    min = i;
    for (j = (i + 1); j < tam; j++) {
      if (num[j] < num[min]) {
        min = j;
      }
    }
    if (i != min) {
      swap = num[i];
      num[i] = num[min];
      num[min] = swap;
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <f@plt>
  mov    $0x0,%eax
  call   <foo@plt>
  test   %rax,%rax
  jne    L3
  mov    $0x0,%eax
  call   <g@plt>
  mov    $0x0,%eax
  jmp    L18
L3:
  nop
  nop
L18:
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  f();
  if (foo())
    return;
  g();
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <f@plt>
  mov    $0x0,%eax
  call   <foo@plt>
  test   %rax,%rax
  je     L13
L1:
  add    $0x8,%rsp
  ret
L13:
  mov    $0x0,%eax
  call   <g@plt>
  mov    $0x0,%eax
  jmp    L1
```

assistant: ```c
int main() {
  f();
  if (foo())
    return;
  g();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <f@plt>
  xor    %eax,%eax
  call   <foo@plt>
  test   %rax,%rax
  jne    L7
  xor    %eax,%eax
  call   <g@plt>
  xor    %eax,%eax
L7:
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  f();
  if (foo())
    return;
  g();
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <f@plt>
  xor    %eax,%eax
  call   <foo@plt>
  test   %rax,%rax
  jne    L4
  xor    %eax,%eax
  call   <g@plt>
  xor    %eax,%eax
L4:
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  f();
  if (foo())
    return;
  g();
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<time>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
time_t time(time_t *tloc) {
  assert(0);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<time>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
time_t time(time_t *tloc) {
  assert(0);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<time>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
time_t time(time_t *tloc) {
  assert(0);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<time>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
time_t time(time_t *tloc) {
  assert(0);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<double_sort_up>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movsd  -0x8(%rbp),%xmm0
  comisd -0x10(%rbp),%xmm0
  jbe    L2
  mov    $0xffffffff,%eax
  jmp    L45
L2:
  movsd  -0x10(%rbp),%xmm0
  ucomisd -0x8(%rbp),%xmm0
  jp     L0
  movsd  -0x10(%rbp),%xmm0
  ucomisd -0x8(%rbp),%xmm0
  jne    L0
  mov    $0x0,%eax
  jmp    L45
L0:
  mov    $0x1,%eax
L45:
  pop    %rbp
  ret
```

assistant: ```c
int double_sort_up(const void *_a, const void *_b) {
  double a = *((double *)_a);
  double b = *((double *)_b);
  if (a < b)
    return -1;
  if (a == b)
    return 0;
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<double_sort_up>:
  endbr64
  movsd  (%rdi),%xmm1
  movsd  (%rsi),%xmm0
  mov    $0xffffffff,%eax
  comisd %xmm1,%xmm0
  ja     L9
  ucomisd %xmm0,%xmm1
  setp   %al
  movzbl %al,%eax
  mov    $0x1,%edx
  cmovne %edx,%eax
L9:
  ret
```

assistant: ```c
int double_sort_up(const void *_a, const void *_b) {
  double a = *((double *)_a);
  double b = *((double *)_b);
  if (a < b)
    return -1;
  if (a == b)
    return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<double_sort_up>:
  endbr64
  movsd  (%rdi),%xmm1
  movsd  (%rsi),%xmm0
  mov    $0xffffffff,%eax
  comisd %xmm1,%xmm0
  ja     L7
  xor    %eax,%eax
  ucomisd %xmm0,%xmm1
  mov    $0x1,%edx
  setp   %al
  cmovne %edx,%eax
L7:
  ret
```

assistant: ```c
int double_sort_up(const void *_a, const void *_b) {
  double a = *((double *)_a);
  double b = *((double *)_b);
  if (a < b)
    return -1;
  if (a == b)
    return 0;
  return 1;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<double_sort_up>:
  endbr64
  movsd  (%rdi),%xmm1
  movsd  (%rsi),%xmm0
  mov    $0xffffffff,%eax
  comisd %xmm1,%xmm0
  ja     L1
  xor    %eax,%eax
  ucomisd %xmm0,%xmm1
  mov    $0x1,%edx
  setp   %al
  cmovne %edx,%eax
L1:
  ret
```

assistant: ```c
int double_sort_up(const void *_a, const void *_b) {
  double a = *((double *)_a);
  double b = *((double *)_b);
  if (a < b)
    return -1;
  if (a == b)
    return 0;
  return 1;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x3fffc,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L50
L98:
  mov    $0x3fffc,%edx
  mov    $0x0,%esi
  mov    %r13,%rdi
  call   <memset@plt>
  movq   $0x1,(%rsp)
  mov    $0x0,%r12d
  lea    D6(%rip),%r15
  lea    D22(%rip),%r14
  mov    %rsp,%rbp
L56:
  add    $0x1,%r12d
  mov    (%rsp),%rax
  movl   $0x1,0x0(%r13,%rax,4)
  cmp    $0x17,%r12d
  jg     L14
  mov    (%rsp),%rsi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L28:
  mov    $0x20,%ebx
L69:
  mov    %rbp,%rdi
  call   <lfsr_calculate@plt>
  sub    $0x1,%ebx
  jne    L69
  mov    (%rsp),%rax
  cmpl   $0x1,0x0(%r13,%rax,4)
  jne    L56
  mov    %r12d,%esi
  lea    D59(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  cmp    $0xffff,%r12d
  je     L17
  lea    D42(%rip),%rdi
  call   <puts@plt>
L9:
  mov    %r13,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L99
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L50:
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L98
L14:
  cmp    $0x18,%r12d
  jne    L28
  mov    %r15,%rdi
  call   <puts@plt>
  jmp    L28
L17:
  lea    D26(%rip),%rdi
  call   <puts@plt>
  jmp    L9
L99:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D6
type: string
name: retrieve_data
arguments: label: D22
type: string
name: retrieve_data
arguments: label: D59
type: string
name: retrieve_data
arguments: label: D42
type: string
name: retrieve_data
arguments: label: D2
type: string
name: retrieve_data
arguments: label: D26
type: string

tool: D6:  ... etc etc ... 

tool: D22: My number is: %u


tool: D59: Got %u numbers before cycling!


tool: D42: Did I miss something?

tool: D2: Memory allocation failed!

tool: D26: Congratulations! It works!

assistant: ```c
int main() {
  int8_t *numbers = (int8_t *)malloc(sizeof(int8_t) * 65535);
  if (numbers == NULL) {
    printf("Memory allocation failed!");
    exit(1);
  }
  memset(numbers, 0, sizeof(int8_t) * 65535);
  uint16_t reg = 0x1;
  uint32_t count = 0;
  int i;
  do {
    count++;
    numbers[reg] = 1;
    if (count < 24) {
      printf("My number is: %u\n", reg);
    } else if (count == 24) {
      printf(" ... etc etc ... \n");
    }
    for (i = 0; i < 32; i++)
      lfsr_calculate(&reg);
  } while (numbers[reg] != 1);
  printf("Got %u numbers before cycling!\n", count);
  if (count == 65535) {
    printf("Congratulations! It works!\n");
  } else {
    printf("Did I miss something?\n");
  }
  free(numbers);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x3fffc,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L92
L88:
  xor    %esi,%esi
  mov    $0x3fffc,%edx
  mov    %r13,%rdi
  xor    %r12d,%r12d
  call   <memset@plt>
  mov    %rsp,%rbp
  movq   $0x1,(%rsp)
  mov    $0x1,%esi
  lea    D4(%rip),%r15
  lea    D41(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L18:
  movl   $0x1,0x0(%r13,%rsi,4)
  add    $0x1,%r12d
  cmp    $0x17,%r12d
  jg     L23
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L46:
  mov    $0x20,%ebx
  cs nopw 0x0(%rax,%rax,1)
L6:
  mov    %rbp,%rdi
  call   <lfsr_calculate@plt>
  sub    $0x1,%ebx
  jne    L6
  mov    (%rsp),%rsi
  cmpl   $0x1,0x0(%r13,%rsi,4)
  jne    L18
  xor    %eax,%eax
  mov    %r12d,%esi
  lea    D50(%rip),%rdi
  call   <printf@plt>
  cmp    $0xffff,%r12d
  je     L40
  lea    D6(%rip),%rdi
  call   <puts@plt>
L27:
  mov    %r13,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L99
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L92:
  lea    D38(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L88
L40:
  lea    D35(%rip),%rdi
  call   <puts@plt>
  jmp    L27
L99:
  call   <__stack_chk_fail@plt>
L23:
  cmp    $0x18,%r12d
  jne    L46
  mov    %r15,%rdi
  call   <puts@plt>
  jmp    L46
```

tool_calls: name: unpack_data
arguments: label: D4
datatype: string
name: unpack_data
arguments: label: D41
datatype: string
name: unpack_data
arguments: label: D50
datatype: string
name: unpack_data
arguments: label: D6
datatype: string
name: unpack_data
arguments: label: D38
datatype: string
name: unpack_data
arguments: label: D35
datatype: string

tool: D4:  ... etc etc ... 

tool: D41: My number is: %u


tool: D50: Got %u numbers before cycling!


tool: D6: Did I miss something?

tool: D38: Memory allocation failed!

tool: D35: Congratulations! It works!

assistant: ```c
int main() {
  int8_t *numbers = (int8_t *)malloc(sizeof(int8_t) * 65535);
  if (numbers == NULL) {
    printf("Memory allocation failed!");
    exit(1);
  }
  memset(numbers, 0, sizeof(int8_t) * 65535);
  uint16_t reg = 0x1;
  uint32_t count = 0;
  int i;
  do {
    count++;
    numbers[reg] = 1;
    if (count < 24) {
      printf("My number is: %u\n", reg);
    } else if (count == 24) {
      printf(" ... etc etc ... \n");
    }
    for (i = 0; i < 32; i++)
      lfsr_calculate(&reg);
  } while (numbers[reg] != 1);
  printf("Got %u numbers before cycling!\n", count);
  if (count == 65535) {
    printf("Congratulations! It works!\n");
  } else {
    printf("Did I miss something?\n");
  }
  free(numbers);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x3fffc,%edi
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L75
L78:
  xor    %esi,%esi
  mov    $0x3fffc,%edx
  mov    %r13,%rdi
  xor    %r12d,%r12d
  call   <memset@plt>
  mov    %rsp,%rbp
  movq   $0x1,(%rsp)
  mov    $0x1,%esi
  lea    D36(%rip),%r15
  lea    D57(%rip),%r14
  nopl   0x0(%rax,%rax,1)
L33:
  movl   $0x1,0x0(%r13,%rsi,4)
  add    $0x1,%r12d
  cmp    $0x17,%r12d
  jg     L31
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L51:
  mov    $0x20,%ebx
  cs nopw 0x0(%rax,%rax,1)
L95:
  mov    %rbp,%rdi
  call   <lfsr_calculate@plt>
  sub    $0x1,%ebx
  jne    L95
  mov    (%rsp),%rsi
  cmpl   $0x1,0x0(%r13,%rsi,4)
  jne    L33
  xor    %eax,%eax
  mov    %r12d,%esi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  cmp    $0xffff,%r12d
  je     L49
  lea    D8(%rip),%rdi
  call   <puts@plt>
L37:
  mov    %r13,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L58
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L75:
  lea    D55(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  jmp    L78
L49:
  lea    D4(%rip),%rdi
  call   <puts@plt>
  jmp    L37
L58:
  call   <__stack_chk_fail@plt>
L31:
  cmp    $0x18,%r12d
  jne    L51
  mov    %r15,%rdi
  call   <puts@plt>
  jmp    L51
```

tool_calls: name: extract_symbolic_data
arguments: tag: D36
datatype: string
name: extract_symbolic_data
arguments: tag: D57
datatype: string
name: extract_symbolic_data
arguments: tag: D12
datatype: string
name: extract_symbolic_data
arguments: tag: D8
datatype: string
name: extract_symbolic_data
arguments: tag: D55
datatype: string
name: extract_symbolic_data
arguments: tag: D4
datatype: string

tool: D36:  ... etc etc ... 

tool: D57: My number is: %u


tool: D12: Got %u numbers before cycling!


tool: D8: Did I miss something?

tool: D55: Memory allocation failed!

tool: D4: Congratulations! It works!

assistant: ```c
int main() {
  int8_t *numbers = (int8_t *)malloc(sizeof(int8_t) * 65535);
  if (numbers == NULL) {
    printf("Memory allocation failed!");
    exit(1);
  }
  memset(numbers, 0, sizeof(int8_t) * 65535);
  uint16_t reg = 0x1;
  uint32_t count = 0;
  int i;
  do {
    count++;
    numbers[reg] = 1;
    if (count < 24) {
      printf("My number is: %u\n", reg);
    } else if (count == 24) {
      printf(" ... etc etc ... \n");
    }
    for (i = 0; i < 32; i++)
      lfsr_calculate(&reg);
  } while (numbers[reg] != 1);
  printf("Got %u numbers before cycling!\n", count);
  if (count == 65535) {
    printf("Congratulations! It works!\n");
  } else {
    printf("Did I miss something?\n");
  }
  free(numbers);
  return 0;
}
```