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
<var_y>:
  endbr64
  movslq %edx,%rdx
  lea    0x8(%rdi),%r9
  mov    $0x3,%r10d
  mov    $0x0,%eax
L10:
  mov    %rdi,%r8
L12:
  movzbl (%r8),%ecx
  movzbl (%r8,%rdx,1),%esi
  sub    %esi,%ecx
  mov    %ecx,%esi
  sar    $0x1f,%esi
  xor    %ecx,%esi
  shr    $0x1f,%ecx
  add    %esi,%ecx
  add    %ecx,%eax
  add    $0x1,%r8
  cmp    %r9,%r8
  jne    L12
  add    %rdx,%rdi
  add    %rdx,%r9
  sub    $0x1,%r10d
  jne    L10
  shl    $0x2,%eax
  ret
```

assistant: ```c
int var_y(unsigned char *a, unsigned char *b, int s) {
  int i, j, var = 0;
  for (i = 3; i; i--) {
    for (j = 0; j < 8; j++) {
      var += (((a[j] - a[j + s]) ^ ((a[j] - a[j + s]) >> 31)) - ((a[j] - a[j + s]) >> 31));
    }
    a += s;
    b += s;
  }
  return 4 * var;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<var_y>:
  endbr64
  movslq %edx,%rdx
  lea    0x8(%rdi),%r9
  mov    $0x3,%r10d
  xor    %eax,%eax
L19:
  mov    %rdi,%r8
  cs nopw 0x0(%rax,%rax,1)
L7:
  movzbl (%r8,%rdx,1),%esi
  movzbl (%r8),%ecx
  add    $0x1,%r8
  sub    %esi,%ecx
  mov    %ecx,%esi
  sar    $0x1f,%esi
  xor    %ecx,%esi
  shr    $0x1f,%ecx
  add    %esi,%ecx
  add    %ecx,%eax
  cmp    %r8,%r9
  jne    L7
  add    %rdx,%rdi
  add    %rdx,%r9
  sub    $0x1,%r10d
  jne    L19
  shl    $0x2,%eax
  ret
```

assistant: ```c
int var_y(unsigned char *a, unsigned char *b, int s) {
  int i, j, var = 0;
  for (i = 3; i; i--) {
    for (j = 0; j < 8; j++) {
      var += (((a[j] - a[j + s]) ^ ((a[j] - a[j + s]) >> 31)) - ((a[j] - a[j + s]) >> 31));
    }
    a += s;
    b += s;
  }
  return 4 * var;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<touch_stack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x10000,-0x2c(%rbp)
  mov    -0x2c(%rbp),%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,%rsi
  mov    $0x0,%edi
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L51:
  cmp    %rdx,%rsp
  je     L39
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L51
L39:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L24
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L24:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rdx
  mov    -0x2c(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  mov    $0x0,%eax
  call   <fill@plt>
  movl   $0x0,-0x34(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x30(%rbp)
  jmp    L28
L59:
  mov    -0x20(%rbp),%rdx
  mov    -0x30(%rbp),%eax
  cltq
  movzbl (%rdx,%rax,1),%eax
  movsbl %al,%eax
  add    %eax,-0x34(%rbp)
  subl   $0x1,-0x30(%rbp)
L28:
  cmpl   $0x0,-0x30(%rbp)
  jns    L59
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbx,%rsp
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L37
  call   <__stack_chk_fail@plt>
L37:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
datatype: string

tool: D1: stack is hopefully touched enough: %d


assistant: ```c
void touch_stack(void) {
  int stack_size = 1024 * 64;
  char hepp[stack_size];
  fill(stack_size, hepp);
  int ret = 0;
  for (int i = stack_size - 1; i >= 0; i--) {
    ret += hepp[i];
    __asm__("");
  }
  printf("stack is hopefully touched enough: %d\n", ret);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<touch_stack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10000(%rsp),%rax
L12:
  cmp    %rax,%rsp
  je     L26
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L12
L26:
  mov    %rsp,%rsi
  mov    $0x10000,%edi
  mov    $0x0,%eax
  call   <fill@plt>
  lea    0xffff(%rsp),%rax
  mov    %rsp,%rcx
  mov    $0x0,%esi
L38:
  movsbl (%rax),%edx
  add    %edx,%esi
  mov    %rax,%rdx
  sub    $0x1,%rax
  cmp    %rcx,%rdx
  jne    L38
  lea    D0(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  leave
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D0
type: string

tool: D0: stack is hopefully touched enough: %d


assistant: ```c
void touch_stack(void) {
  int stack_size = 1024 * 64;
  char hepp[stack_size];
  fill(stack_size, hepp);
  int ret = 0;
  for (int i = stack_size - 1; i >= 0; i--) {
    ret += hepp[i];
    __asm__("");
  }
  printf("stack is hopefully touched enough: %d\n", ret);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<touch_stack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10000(%rsp),%rax
  cmp    %rax,%rsp
  je     L18
L29:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rax,%rsp
  jne    L29
L18:
  mov    $0x10000,%edi
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <fill@plt>
  lea    0xffff(%rsp),%rax
  mov    %rsp,%rcx
  xor    %esi,%esi
  nopl   0x0(%rax)
L36:
  movsbl (%rax),%edx
  add    %edx,%esi
  mov    %rax,%rdx
  sub    $0x1,%rax
  cmp    %rcx,%rdx
  jne    L36
  xor    %eax,%eax
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  leave
  ret
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D3
type_guessed: string

tool: D3: stack is hopefully touched enough: %d


assistant: ```c
void touch_stack(void) {
  int stack_size = 1024 * 64;
  char hepp[stack_size];
  fill(stack_size, hepp);
  int ret = 0;
  for (int i = stack_size - 1; i >= 0; i--) {
    ret += hepp[i];
    __asm__("");
  }
  printf("stack is hopefully touched enough: %d\n", ret);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<touch_stack>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x10000(%rsp),%rax
  cmp    %rax,%rsp
  je     L1
L8:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rax,%rsp
  jne    L8
L1:
  mov    $0x10000,%edi
  xor    %eax,%eax
  mov    %rsp,%rsi
  call   <fill@plt>
  lea    0xffff(%rsp),%rax
  mov    %rsp,%rcx
  xor    %esi,%esi
  nopl   0x0(%rax)
L37:
  movsbl (%rax),%edx
  add    %edx,%esi
  mov    %rax,%rdx
  sub    $0x1,%rax
  cmp    %rcx,%rdx
  jne    L37
  xor    %eax,%eax
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  leave
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D6
type_guessed: string

tool: D6: stack is hopefully touched enough: %d


assistant: ```c
void touch_stack(void) {
  int stack_size = 1024 * 64;
  char hepp[stack_size];
  fill(stack_size, hepp);
  int ret = 0;
  for (int i = stack_size - 1; i >= 0; i--) {
    ret += hepp[i];
    __asm__("");
  }
  printf("stack is hopefully touched enough: %d\n", ret);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<dividedString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0x8(%rbp),%rdx
  add    $0x8,%rdx
  mov    %rax,(%rdx)
  movl   $0x0,-0xc(%rbp)
  jmp    L27
L29:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jge    L30
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
  jmp    L18
L30:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x8(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rdx
  mov    -0xc(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  cltq
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
L18:
  addl   $0x1,-0xc(%rbp)
L27:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0xc(%rbp)
  jl     L29
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char **dividedString(char *string, int nbChar) {
  int i = 0;
  char **result = (char **)malloc(2 * sizeof(char *));
  result[0] = (char *)malloc(nbChar * sizeof(char));
  result[1] = (char *)malloc((strlen(string) - nbChar) * sizeof(char));
  for (i = 0; i < strlen(string); i++) {
    if (i < nbChar)
      result[0][i] = string[i];
    else
      result[1][i - nbChar] = string[i];
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<dividedString>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    $0x10,%edi
  call   <malloc@plt>
  mov    %rax,%r13
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  mov    %rbp,%rdi
  call   <strlen@plt>
  sub    %r12d,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,0x8(%r13)
  mov    $0x0,%ebx
  jmp    L28
L7:
  movzbl 0x0(%rbp,%rbx,1),%ecx
  mov    %ebx,%eax
  sub    %r12d,%eax
  cltq
  mov    0x8(%r13),%rdx
  mov    %cl,(%rdx,%rax,1)
L36:
  add    $0x1,%rbx
L28:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L6
  cmp    %ebx,%r12d
  jle    L7
  movzbl 0x0(%rbp,%rbx,1),%edx
  mov    0x0(%r13),%rax
  mov    %dl,(%rax,%rbx,1)
  jmp    L36
L6:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **dividedString(char *string, int nbChar) {
  int i = 0;
  char **result = (char **)malloc(2 * sizeof(char *));
  result[0] = (char *)malloc(nbChar * sizeof(char));
  result[1] = (char *)malloc((strlen(string) - nbChar) * sizeof(char));
  for (i = 0; i < strlen(string); i++) {
    if (i < nbChar)
      result[0][i] = string[i];
    else
      result[1][i - nbChar] = string[i];
  }
  return result;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<dividedString>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x10,%edi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r12d,%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,0x0(%r13)
  call   <strlen@plt>
  sub    %r12d,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,0x8(%r13)
  jmp    L27
  nopw   0x0(%rax,%rax,1)
L2:
  mov    0x0(%r13),%rax
  mov    %dl,(%rax,%rbx,1)
L31:
  add    $0x1,%rbx
L27:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L34
  movzbl 0x0(%rbp,%rbx,1),%edx
  cmp    %r12d,%ebx
  jl     L2
  mov    %ebx,%eax
  mov    0x8(%r13),%rcx
  sub    %r12d,%eax
  cltq
  mov    %dl,(%rcx,%rax,1)
  jmp    L31
  nopl   0x0(%rax)
L34:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **dividedString(char *string, int nbChar) {
  int i = 0;
  char **result = (char **)malloc(2 * sizeof(char *));
  result[0] = (char *)malloc(nbChar * sizeof(char));
  result[1] = (char *)malloc((strlen(string) - nbChar) * sizeof(char));
  for (i = 0; i < strlen(string); i++) {
    if (i < nbChar)
      result[0][i] = string[i];
    else
      result[1][i - nbChar] = string[i];
  }
  return result;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<dividedString>:
  endbr64
  push   %r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  mov    $0x10,%edi
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r12d,%edi
  mov    %rax,%r13
  call   <malloc@plt>
  mov    %rbp,%rdi
  mov    %rax,0x0(%r13)
  call   <strlen@plt>
  sub    %r12d,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,0x8(%r13)
  jmp    L4
  nopw   0x0(%rax,%rax,1)
L16:
  mov    0x0(%r13),%rax
  mov    %dl,(%rax,%rbx,1)
L0:
  add    $0x1,%rbx
L4:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L36
  movzbl 0x0(%rbp,%rbx,1),%edx
  cmp    %r12d,%ebx
  jl     L16
  mov    %ebx,%eax
  mov    0x8(%r13),%rcx
  sub    %r12d,%eax
  cltq
  mov    %dl,(%rcx,%rax,1)
  jmp    L0
  nopl   0x0(%rax)
L36:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char **dividedString(char *string, int nbChar) {
  int i = 0;
  char **result = (char **)malloc(2 * sizeof(char *));
  result[0] = (char *)malloc(nbChar * sizeof(char));
  result[1] = (char *)malloc((strlen(string) - nbChar) * sizeof(char));
  for (i = 0; i < strlen(string); i++) {
    if (i < nbChar)
      result[0][i] = string[i];
    else
      result[1][i - nbChar] = string[i];
  }
  return result;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<test>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x2,-0x8(%rbp)
  movl   $0x3,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jle    L11
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <f@plt>
  mov    %eax,-0xc(%rbp)
  jmp    L5
L11:
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <g@plt>
  mov    %eax,-0xc(%rbp)
L5:
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  leave
  ret
```

assistant: ```c
int test() {
  int a = 2;
  int b = 3;
  int res;
  if (a > b)
    res = f(a, b);
  else
    res = g(a, b);
  return res + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <g@plt>
  add    $0x1,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int test() {
  int a = 2;
  int b = 3;
  int res;
  if (a > b)
    res = f(a, b);
  else
    res = g(a, b);
  return res + 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <g@plt>
  add    $0x8,%rsp
  add    $0x1,%eax
  ret
```

assistant: ```c
int test() {
  int a = 2;
  int b = 3;
  int res;
  if (a > b)
    res = f(a, b);
  else
    res = g(a, b);
  return res + 1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x3,%esi
  mov    $0x2,%edi
  call   <g@plt>
  add    $0x8,%rsp
  add    $0x1,%eax
  ret
```

assistant: ```c
int test() {
  int a = 2;
  int b = 3;
  int res;
  if (a > b)
    res = f(a, b);
  else
    res = g(a, b);
  return res + 1;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<temp_F2C>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x3e7,-0x4(%rbp)
  jne    L8
  mov    $0x3e7,%eax
  jmp    L2
L8:
  mov    -0x4(%rbp),%eax
  lea    -0x20(%rax),%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  D5(%rip),%xmm1
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <round@plt>
L2:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
type_guessed: f64

tool: D5: 9

assistant: ```c
int temp_F2C(int temp_F) {
  if (temp_F == 999)
    return 999;
  return round((temp_F - 32) * 5 / 9.0);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<temp_F2C>:
  endbr64
  mov    %edi,%eax
  cmp    $0x3e7,%edi
  je     L4
  sub    $0x8,%rsp
  lea    -0xa0(%rdi,%rdi,4),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  D2(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <round@plt>
  add    $0x8,%rsp
  ret
L4:
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D2
type_guessed: f64

tool: D2: 9

assistant: ```c
int temp_F2C(int temp_F) {
  if (temp_F == 999)
    return 999;
  return round((temp_F - 32) * 5 / 9.0);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<temp_F2C>:
  endbr64
  cmp    $0x3e7,%edi
  je     L4
  lea    -0xa0(%rdi,%rdi,4),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  D0(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  jmp    1050 <round@plt>
  nopl   0x0(%rax)
L4:
  mov    $0x3e7,%eax
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
type: f64

tool: D0: 9

assistant: ```c
int temp_F2C(int temp_F) {
  if (temp_F == 999)
    return 999;
  return round((temp_F - 32) * 5 / 9.0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<temp_F2C>:
  endbr64
  cmp    $0x3e7,%edi
  je     L1
  lea    -0xa0(%rdi,%rdi,4),%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  divsd  D1(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  jmp    1050 <round@plt>
  nopl   0x0(%rax)
L1:
  mov    $0x3e7,%eax
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_guessed: f64

tool: D1: 9

assistant: ```c
int temp_F2C(int temp_F) {
  if (temp_F == 999)
    return 999;
  return round((temp_F - 32) * 5 / 9.0);
}
```
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
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <megaInit@plt>
  cmpl   $0x2,-0x34(%rbp)
  je     L27
  mov    $0x1,%esi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
L27:
  movl   $0x0,-0x28(%rbp)
  jmp    L55
L48:
  mov    -0x28(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    $0xa,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x28(%rbp),%eax
  cltq
  mov    -0x20(%rbp),%rdx
  mov    %rdx,-0x18(%rbp,%rax,8)
  addl   $0x1,-0x28(%rbp)
L55:
  cmpl   $0x0,-0x28(%rbp)
  jle    L48
  lea    -0x10(%rbp),%rdx
  lea    -0x18(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <RandomFunc@plt>
  mov    -0x10(%rbp),%rax
  movabs $0xff1635531adb70a6,%rdx
  cmp    %rdx,%rax
  jne    L44
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L44:
  movl   $0x0,-0x24(%rbp)
  jmp    L21
L23:
  mov    -0x24(%rbp),%eax
  cltq
  mov    -0x10(%rbp,%rax,8),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x24(%rbp)
L21:
  cmpl   $0x0,-0x24(%rbp)
  jle    L23
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D18
type_infered: string
name: extract_symbolic_data
arguments: label: D5
type_infered: string
name: extract_symbolic_data
arguments: label: D23
type_infered: string

tool: D18: Call this program with %i arguments


tool: D5: You win!

tool: D23: %lu


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long input[1];
  unsigned long output[1];
  int randomFuns_i5;
  unsigned long 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 = strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 18380937560208404646UL) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%lu\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%ebp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <megaInit@plt>
  cmp    $0x2,%ebp
  jne    L48
L0:
  mov    0x8(%rbx),%rdi
  mov    $0xa,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  mov    %rax,0x8(%rsp)
  lea    0x10(%rsp),%rsi
  lea    0x8(%rsp),%rdi
  call   <RandomFunc@plt>
  movabs $0xff1635531adb70a6,%rax
  cmp    %rax,0x10(%rsp)
  je     L30
L45:
  mov    0x10(%rsp),%rsi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L44
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L48:
  mov    $0x1,%esi
  lea    D22(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%edi
  call   <exit@plt>
  jmp    L0
L30:
  lea    D13(%rip),%rdi
  call   <puts@plt>
  jmp    L45
L44:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D11
type_guessed: string
name: read_data
arguments: label: D22
type_guessed: string
name: read_data
arguments: label: D13
type_guessed: string

tool: D11: %lu


tool: D22: Call this program with %i arguments


tool: D13: You win!

assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long input[1];
  unsigned long output[1];
  int randomFuns_i5;
  unsigned long 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 = strtoul(argv[randomFuns_i5 + 1], 0, 10);
      input[randomFuns_i5] = randomFuns_value6;
      randomFuns_i5++;
    }
    RandomFunc(input, output);
    if (output[0] == 18380937560208404646UL) {
      printf("You win!\n");
    } else {
    }
    randomFuns_main_i7 = 0;
    while (randomFuns_main_i7 < 1) {
      printf("%lu\n", output[randomFuns_main_i7]);
      randomFuns_main_i7++;
    }
  }
}
```