ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<rj232_send>:
  endbr64
  test   %esi,%esi
  jle    L12
  push   %rbp
  movslq %esi,%rsi
  push   %rbx
  lea    (%rdi,%rsi,1),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L6:
  movzbl (%rbx),%esi
  add    $0x1,%rbx
  mov    $0x10,%edi
  call   <rj_writebyte@plt>
  cmp    %rbx,%rbp
  jne    L6
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L12:
  xor    %eax,%eax
  ret
```

assistant: ```c
int rj232_send(unsigned char *buf, int len) {
  int i, rval;
  for (i = 0; i < len; i++)
    rval = rj_writebyte(0x10, *buf++);
  return rval;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<rj232_send>:
  endbr64
  test   %esi,%esi
  jle    L16
  push   %rbp
  movslq %esi,%rsi
  push   %rbx
  lea    (%rdi,%rsi,1),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax,%rax,1)
L11:
  movzbl (%rbx),%esi
  add    $0x1,%rbx
  mov    $0x10,%edi
  call   <rj_writebyte@plt>
  cmp    %rbx,%rbp
  jne    L11
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L16:
  xor    %eax,%eax
  ret
```

assistant: ```c
int rj232_send(unsigned char *buf, int len) {
  int i, rval;
  for (i = 0; i < len; i++)
    rval = rj_writebyte(0x10, *buf++);
  return rval;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<get_digits_count>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  movl   $0x1,-0x4(%rbp)
  cmpl   $0x1,-0x18(%rbp)
  seta   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  cmpl   $0x1,-0x18(%rbp)
  ja     L0
  mov    $0x0,%eax
  jmp    L20
L22:
  mov    -0x14(%rbp),%eax
  mov    $0x0,%edx
  divl   -0x18(%rbp)
  mov    %eax,-0x14(%rbp)
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x14(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jae    L22
  mov    -0x4(%rbp),%eax
L20:
  leave
  ret
```

assistant: ```c
unsigned get_digits_count(unsigned num, unsigned base) {
  unsigned digits_count = 1;
  assert(base > 1);
  if (base <= 1)
    return 0;
  while (num >= base) {
    num /= base;
    digits_count += 1;
  }
  return digits_count;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<get_digits_count>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  mov    %esi,%ebx
  cmp    $0x1,%esi
  seta   %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%ecx
  cmp    $0x1,%ebx
  jbe    L30
  cmp    %ebp,%ebx
  ja     L18
  mov    $0x1,%ecx
L9:
  mov    %ebp,%eax
  mov    $0x0,%edx
  div    %ebx
  mov    %eax,%ebp
  add    $0x1,%ecx
  cmp    %eax,%ebx
  jbe    L9
L30:
  mov    %ecx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L18:
  mov    $0x1,%ecx
  jmp    L30
```

assistant: ```c
unsigned get_digits_count(unsigned num, unsigned base) {
  unsigned digits_count = 1;
  assert(base > 1);
  if (base <= 1)
    return 0;
  while (num >= base) {
    num /= base;
    digits_count += 1;
  }
  return digits_count;
}
```
user: What is the c source code of the assembly code below?

```asm
<get_digits_count>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x8,%rsp
  cmp    $0x1,%esi
  seta   %dil
  call   <assert@plt>
  xor    %r8d,%r8d
  cmp    $0x1,%ebp
  jbe    L18
  mov    $0x1,%r8d
  cmp    %ebx,%ebp
  ja     L18
  xchg   %ax,%ax
L13:
  mov    %ebx,%eax
  xor    %edx,%edx
  add    $0x1,%r8d
  div    %ebp
  mov    %eax,%ebx
  cmp    %eax,%ebp
  jbe    L13
L18:
  add    $0x8,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned get_digits_count(unsigned num, unsigned base) {
  unsigned digits_count = 1;
  assert(base > 1);
  if (base <= 1)
    return 0;
  while (num >= base) {
    num /= base;
    digits_count += 1;
  }
  return digits_count;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_digits_count>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %edi,%ebx
  xor    %edi,%edi
  sub    $0x8,%rsp
  cmp    $0x1,%esi
  seta   %dil
  call   <assert@plt>
  xor    %r8d,%r8d
  cmp    $0x1,%ebp
  jbe    L14
  mov    $0x1,%r8d
  cmp    %ebx,%ebp
  ja     L14
  xchg   %ax,%ax
L0:
  mov    %ebx,%eax
  xor    %edx,%edx
  add    $0x1,%r8d
  div    %ebp
  mov    %eax,%ebx
  cmp    %eax,%ebp
  jbe    L0
L14:
  add    $0x8,%rsp
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
unsigned get_digits_count(unsigned num, unsigned base) {
  unsigned digits_count = 1;
  assert(base > 1);
  if (base <= 1)
    return 0;
  while (num >= base) {
    num /= base;
    digits_count += 1;
  }
  return digits_count;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    D12(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,-0x8(%rbp)
  movq   $0x0,-0x8(%rbp)
  movq   $0x0,-0x8(%rbp)
  movq   $0x0,-0x8(%rbp)
  movq   $0x0,-0x8(%rbp)
  movq   $0x0,-0x8(%rbp)
  movq   $0x0,-0x8(%rbp)
  movq   $0x0,-0x8(%rbp)
  mov    D7(%rip),%rax
  mov    (%rax),%eax
  mov    -0x8(%rbp),%rdx
  mov    $0x8,%ecx
  mov    %eax,%esi
  mov    $0x141,%edi
  call   <syscall@plt>
  leave
  ret
```

assistant: ```c
int main(void) {
  union bpf_attr attr;
  attr.prog_type = BPF_PROG_TYPE_KPROBE;
  attr.insn_cnt = 0;
  attr.insns = 0;
  attr.license = 0;
  attr.log_buf = 0;
  attr.log_size = 0;
  attr.log_level = 0;
  attr.kern_version = 0;
  attr = attr;
  return syscall(321, BPF_PROG_LOAD, attr, sizeof(attr));
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x8,%ecx
  mov    $0x0,%edx
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  mov    $0x141,%edi
  call   <syscall@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  union bpf_attr attr;
  attr.prog_type = BPF_PROG_TYPE_KPROBE;
  attr.insn_cnt = 0;
  attr.insns = 0;
  attr.license = 0;
  attr.log_buf = 0;
  attr.log_size = 0;
  attr.log_level = 0;
  attr.kern_version = 0;
  attr = attr;
  return syscall(321, BPF_PROG_LOAD, attr, sizeof(attr));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  mov    D8(%rip),%rax
  mov    $0x8,%ecx
  xor    %edx,%edx
  mov    $0x141,%edi
  mov    (%rax),%esi
  jmp    1050 <syscall@plt>
```

assistant: ```c
int main(void) {
  union bpf_attr attr;
  attr.prog_type = BPF_PROG_TYPE_KPROBE;
  attr.insn_cnt = 0;
  attr.insns = 0;
  attr.license = 0;
  attr.log_buf = 0;
  attr.log_size = 0;
  attr.log_level = 0;
  attr.kern_version = 0;
  attr = attr;
  return syscall(321, BPF_PROG_LOAD, attr, sizeof(attr));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  mov    D7(%rip),%rax
  mov    $0x8,%ecx
  xor    %edx,%edx
  mov    $0x141,%edi
  mov    (%rax),%esi
  jmp    1050 <syscall@plt>
```

assistant: ```c
int main(void) {
  union bpf_attr attr;
  attr.prog_type = BPF_PROG_TYPE_KPROBE;
  attr.insn_cnt = 0;
  attr.insns = 0;
  attr.license = 0;
  attr.log_buf = 0;
  attr.log_size = 0;
  attr.log_level = 0;
  attr.kern_version = 0;
  attr = attr;
  return syscall(321, BPF_PROG_LOAD, attr, sizeof(attr));
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dynamic_programming>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %esi,-0x38(%rbp)
  mov    %rdx,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rdi
  mov    -0x34(%rbp),%eax
  add    $0x1,%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rcx
  mov    $0x0,%ebx
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%ebx
  mov    $0x0,%edx
  div    %rbx
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L94:
  cmp    %rdx,%rsp
  je     L56
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L94
L56:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L54
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L54:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  movl   $0x1,(%rax)
  movl   $0x1,-0x30(%rbp)
  jmp    L52
L31:
  mov    -0x20(%rbp),%rax
  mov    -0x30(%rbp),%edx
  movslq %edx,%rdx
  movl   $0x0,(%rax,%rdx,4)
  addl   $0x1,-0x30(%rbp)
L52:
  mov    -0x30(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jle    L31
  movl   $0x0,-0x30(%rbp)
  jmp    L40
L70:
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x2c(%rbp)
  jmp    L73
L42:
  mov    -0x20(%rbp),%rax
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%ecx
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%esi
  mov    -0x2c(%rbp),%eax
  sub    %esi,%eax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  add    %eax,%ecx
  mov    -0x20(%rbp),%rax
  mov    -0x2c(%rbp),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x2c(%rbp)
L73:
  mov    -0x2c(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jle    L42
  addl   $0x1,-0x30(%rbp)
L40:
  mov    -0x30(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L70
  mov    -0x20(%rbp),%rax
  mov    -0x34(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  mov    %rdi,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L81
  call   <__stack_chk_fail@plt>
L81:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int dynamic_programming(int target, int size, int coins[size]) {
  int ways[target + 1];
  int i, j;
  ways[0] = 1;
  for (i = 1; i <= target; i++)
    ways[i] = 0;
  for (i = 0; i < size; i++) {
    for (j = coins[i]; j <= target; j++) {
      ways[j] += ways[j - coins[i]];
    }
  }
  return ways[target];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<dynamic_programming>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%r8d
  mov    %esi,%ecx
  mov    %rdx,%rsi
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x1(%rdi),%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdi
  and    $0xfffffffffffffff0,%rdi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L94:
  cmp    %rdx,%rsp
  je     L58
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L94
L58:
  mov    %rdi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L2
  orq    $0x0,-0x8(%rsp,%rax,1)
L2:
  lea    0x3(%rsp),%r9
  mov    %r9,%rax
  shr    $0x2,%rax
  and    $0xfffffffffffffffc,%r9
  movl   $0x1,0x0(,%rax,4)
  test   %r8d,%r8d
  jle    L85
  lea    0x4(%r9),%rax
  lea    -0x1(%r8),%edx
  lea    0x8(%r9,%rdx,4),%rdx
L8:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L8
L85:
  test   %ecx,%ecx
  jle    L28
  mov    %rsi,%rdi
  lea    -0x1(%rcx),%eax
  lea    0x4(%rsi,%rax,4),%r10
  lea    0x4(%r9),%r11
L55:
  mov    (%rdi),%edx
  cmp    %edx,%r8d
  jl     L49
  movslq %edx,%rcx
  lea    (%r9,%rcx,4),%rax
  mov    %r8d,%esi
  sub    %edx,%esi
  lea    (%rsi,%rcx,1),%rdx
  lea    (%r11,%rdx,4),%rsi
  neg    %rcx
  shl    $0x2,%rcx
L74:
  mov    (%rax,%rcx,1),%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L74
L49:
  add    $0x4,%rdi
  cmp    %r10,%rdi
  jne    L55
L28:
  movslq %r8d,%r8
  mov    (%r9,%r8,4),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L88
  leave
  ret
L88:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dynamic_programming(int target, int size, int coins[size]) {
  int ways[target + 1];
  int i, j;
  ways[0] = 1;
  for (i = 1; i <= target; i++)
    ways[i] = 0;
  for (i = 0; i < size; i++) {
    for (j = coins[i]; j <= target; j++) {
      ways[j] += ways[j - coins[i]];
    }
  }
  return ways[target];
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<dynamic_programming>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  mov    %esi,%r15d
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0x1(%rdi),%eax
  mov    %rsp,%rcx
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L26
L60:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L60
L26:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L45
L30:
  lea    0x3(%rsp),%r12
  movslq %ebx,%r13
  mov    %r12,%rax
  and    $0xfffffffffffffffc,%r12
  shr    $0x2,%rax
  movl   $0x1,0x0(,%rax,4)
  test   %ebx,%ebx
  jle    L46
  lea    0x0(,%r13,4),%rdx
  lea    0x4(%r12),%rdi
  xor    %esi,%esi
  call   <memset@plt>
L46:
  test   %r15d,%r15d
  jle    L57
  lea    -0x1(%r15),%eax
  mov    %r14,%rdi
  lea    0x4(%r12),%r9
  lea    0x4(%r14,%rax,4),%r8
  nopl   0x0(%rax)
L59:
  mov    (%rdi),%edx
  cmp    %edx,%ebx
  jl     L24
  mov    %ebx,%esi
  movslq %edx,%r10
  sub    %edx,%esi
  lea    0x0(,%r10,4),%rcx
  lea    (%rsi,%r10,1),%rdx
  lea    (%r12,%rcx,1),%rax
  neg    %rcx
  lea    (%r9,%rdx,4),%rsi
  nopl   0x0(%rax)
L62:
  mov    (%rax,%rcx,1),%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rsi,%rax
  jne    L62
L24:
  add    $0x4,%rdi
  cmp    %r8,%rdi
  jne    L59
L57:
  mov    (%r12,%r13,4),%eax
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L29
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L45:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L30
L29:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dynamic_programming(int target, int size, int coins[size]) {
  int ways[target + 1];
  int i, j;
  ways[0] = 1;
  for (i = 1; i <= target; i++)
    ways[i] = 0;
  for (i = 0; i < size; i++) {
    for (j = coins[i]; j <= target; j++) {
      ways[j] += ways[j - coins[i]];
    }
  }
  return ways[target];
}
```
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    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <megaInit@plt>
  cmpl   $0x2,-0x24(%rbp)
  je     L24
  mov    $0x1,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L24:
  movl   $0x0,-0x14(%rbp)
  jmp    L62
L15:
  mov    -0x14(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %al,-0x15(%rbp)
  mov    -0x14(%rbp),%eax
  cltq
  movzbl -0x15(%rbp),%edx
  mov    %dl,-0xa(%rbp,%rax,1)
  addl   $0x1,-0x14(%rbp)
L62:
  cmpl   $0x0,-0x14(%rbp)
  jle    L15
  lea    -0x9(%rbp),%rdx
  lea    -0xa(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <RandomFunc@plt>
  movzbl -0x9(%rbp),%eax
  cmp    $0x78,%al
  jne    L26
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L26:
  movl   $0x0,-0x10(%rbp)
  jmp    L55
L31:
  mov    -0x10(%rbp),%eax
  cltq
  movzbl -0x9(%rbp,%rax,1),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L55:
  cmpl   $0x0,-0x10(%rbp)
  jle    L31
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L60
  call   <__stack_chk_fail@plt>
L60:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D14
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D28
type_guessed: string

tool: D5: Call this program with %i arguments


tool: D14: You win!

tool: D28: %u


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned char input[1];
  unsigned char output[1];
  int randomFuns_i5;
  unsigned char 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 char)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 120) {
      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: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  jne    L46
L32:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  mov    %al,0x6(%rsp)
  lea    0x7(%rsp),%rsi
  lea    0x6(%rsp),%rdi
  call   <RandomFunc@plt>
  cmpb   $0x78,0x7(%rsp)
  je     L24
L23:
  movzbl 0x7(%rsp),%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L46:
  mov    $0x1,%esi
  lea    D4(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L32
L24:
  lea    D7(%rip),%rdi
  call   <puts@plt>
  jmp    L23
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
data_type: string
name: extract_symbolic_data
arguments: data_label: D4
data_type: string
name: extract_symbolic_data
arguments: data_label: D7
data_type: string

tool: D1: %u


tool: D4: Call this program with %i arguments


tool: D7: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned char input[1];
  unsigned char output[1];
  int randomFuns_i5;
  unsigned char 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 char)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 120) {
      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: Demonstrate the C code that generates the assembly output below:

```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     L7
  lea    D11(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L7:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x7(%rsp),%rsi
  lea    0x6(%rsp),%rdi
  mov    %al,0x6(%rsp)
  call   <RandomFunc@plt>
  movzbl 0x7(%rsp),%esi
  cmp    $0x78,%sil
  je     L25
L16:
  xor    %eax,%eax
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L25:
  lea    D10(%rip),%rdi
  call   <puts@plt>
  movzbl 0x7(%rsp),%esi
  jmp    L16
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D11
type: string
name: read_data
arguments: data_tag: D4
type: string
name: read_data
arguments: data_tag: D10
type: string

tool: D11: Call this program with %i arguments


tool: D4: %u


tool: D10: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned char input[1];
  unsigned char output[1];
  int randomFuns_i5;
  unsigned char 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 char)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 120) {
      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: What would the C source look like for the assembly code provided here?

```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     L25
  lea    D13(%rip),%rdi
  mov    $0x1,%esi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%edi
  call   <exit@plt>
L25:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  lea    0x7(%rsp),%rsi
  lea    0x6(%rsp),%rdi
  mov    %al,0x6(%rsp)
  call   <RandomFunc@plt>
  movzbl 0x7(%rsp),%esi
  cmp    $0x78,%sil
  je     L4
L16:
  xor    %eax,%eax
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L4:
  lea    D2(%rip),%rdi
  call   <puts@plt>
  movzbl 0x7(%rsp),%esi
  jmp    L16
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D13
datatype: string
name: parse_data
arguments: label: D23
datatype: string
name: parse_data
arguments: label: D2
datatype: string

tool: D13: Call this program with %i arguments


tool: D23: %u


tool: D2: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned char input[1];
  unsigned char output[1];
  int randomFuns_i5;
  unsigned char 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 char)strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 120) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%u\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```