ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  movabs $0x2073696874206968,%rax
  movabs $0x6e69727473207369,%rdx
  mov    %rax,(%rsp)
  mov    %rdx,0x8(%rsp)
  movq   $0x67,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  movq   $0x0,0x28(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movq   $0x0,0x40(%rsp)
  movq   $0x0,0x48(%rsp)
  movq   $0x0,0x50(%rsp)
  movq   $0x0,0x58(%rsp)
  movl   $0x0,0x60(%rsp)
  mov    %rsp,%rdi
  call   <reverse1@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x78,%rsp
  ret
L1:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void main() {
  char str[100] = "hi this is string";
  reverse1(str);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x78,%rsp
  movdqa D9(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movl   $0x0,0x60(%rsp)
  movaps %xmm0,(%rsp)
  pxor   %xmm0,%xmm0
  movq   $0x67,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  call   <reverse1@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x78,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
type: byte[16]

tool: {"D9": [0x68,0x69,0x20,0x74,0x68,0x69,0x73,0x20,0x69,0x73,0x20,0x73,0x74,0x72,0x69,0x6e]}

assistant: ```c
void main() {
  char str[100] = "hi this is string";
  reverse1(str);
}
```
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    $0x78,%rsp
  movdqa D4(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  movl   $0x0,0x60(%rsp)
  movaps %xmm0,(%rsp)
  pxor   %xmm0,%xmm0
  movq   $0x67,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movaps %xmm0,0x20(%rsp)
  movaps %xmm0,0x30(%rsp)
  movaps %xmm0,0x40(%rsp)
  movaps %xmm0,0x50(%rsp)
  call   <reverse1@plt>
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x78,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D4
datatype: byte[16]

tool: {"D4": [0x68,0x69,0x20,0x74,0x68,0x69,0x73,0x20,0x69,0x73,0x20,0x73,0x74,0x72,0x69,0x6e]}

assistant: ```c
void main() {
  char str[100] = "hi this is string";
  reverse1(str);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x80(%rbp),%rax
  mov    $0x32,%esi
  mov    %rax,%rdi
  call   <getLine@plt>
  lea    -0x40(%rbp),%rdx
  lea    -0x80(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <convertToPostfix@plt>
  lea    -0x40(%rbp),%rax
  mov    %rax,%rsi
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type_infered: string
name: parse_data
arguments: tag: D18
type_infered: string

tool: D0: Please enter infix expression: 

tool: D18: Postfix expression: %s


assistant: ```c
int main() {
  char infix[50];
  char postfix[50];
  printf("Please enter infix expression: ");
  getLine(infix, 50);
  convertToPostfix(infix, postfix);
  printf("Postfix expression: %s\n", postfix);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    D0(%rip),%rdi
  call   <printf@plt>
  mov    %rsp,%rbp
  mov    $0x32,%esi
  mov    %rbp,%rdi
  call   <getLine@plt>
  lea    0x40(%rsp),%rbx
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  call   <convertToPostfix@plt>
  mov    %rbx,%rsi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x88,%rsp
  pop    %rbx
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D0
datatype: string
name: read_data
arguments: data_label: D1
datatype: string

tool: D0: Please enter infix expression: 

tool: D1: Postfix expression: %s


assistant: ```c
int main() {
  char infix[50];
  char postfix[50];
  printf("Please enter infix expression: ");
  getLine(infix, 50);
  convertToPostfix(infix, postfix);
  printf("Postfix expression: %s\n", postfix);
  return 0;
}
```
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
  push   %r12
  lea    D17(%rip),%rdi
  push   %rbp
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x40(%rsp),%rbp
  call   <printf@plt>
  mov    $0x32,%esi
  mov    %r12,%rdi
  call   <getLine@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <convertToPostfix@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D17
type: string
name: retrieve_data
arguments: tag: D5
type: string

tool: D17: Please enter infix expression: 

tool: D5: Postfix expression: %s


assistant: ```c
int main() {
  char infix[50];
  char postfix[50];
  printf("Please enter infix expression: ");
  getLine(infix, 50);
  convertToPostfix(infix, postfix);
  printf("Postfix expression: %s\n", postfix);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  lea    D16(%rip),%rdi
  push   %rbp
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x40(%rsp),%rbp
  call   <printf@plt>
  mov    $0x32,%esi
  mov    %r12,%rdi
  call   <getLine@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <convertToPostfix@plt>
  xor    %eax,%eax
  mov    %rbp,%rsi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D16
type_infered: string
name: extract_symbolic_data
arguments: data_label: D6
type_infered: string

tool: D16: Please enter infix expression: 

tool: D6: Postfix expression: %s


assistant: ```c
int main() {
  char infix[50];
  char postfix[50];
  printf("Please enter infix expression: ");
  getLine(infix, 50);
  convertToPostfix(infix, postfix);
  printf("Postfix expression: %s\n", postfix);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mm>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    %ecx,%eax
  mov    %r8,-0x40(%rbp)
  mov    %r9d,-0x38(%rbp)
  mov    0x18(%rbp),%edx
  mov    %al,-0x34(%rbp)
  mov    %edx,%eax
  mov    %al,-0x44(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    0x10(%rbp),%esi
  mov    %eax,%edi
  call   <alloc_array@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L35
L13:
  movl   $0x0,-0x18(%rbp)
  jmp    L28
L58:
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L33
L27:
  movsbl -0x34(%rbp),%ecx
  mov    -0x30(%rbp),%edx
  mov    -0x14(%rbp),%esi
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <map@plt>
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm2
  movsd  %xmm2,-0x50(%rbp)
  movsbl -0x44(%rbp),%edx
  mov    -0x18(%rbp),%esi
  mov    -0x14(%rbp),%eax
  mov    %edx,%ecx
  mov    0x10(%rbp),%edx
  mov    %eax,%edi
  call   <map@plt>
  lea    0x0(,%rax,8),%rdx
  mov    -0x40(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  -0x50(%rbp),%xmm0
  movsd  -0x10(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  addl   $0x1,-0x14(%rbp)
L33:
  mov    -0x14(%rbp),%eax
  cmp    0x10(%rbp),%eax
  jl     L27
  mov    -0x18(%rbp),%esi
  mov    -0x1c(%rbp),%eax
  mov    $0x6e,%ecx
  mov    0x10(%rbp),%edx
  mov    %eax,%edi
  call   <map@plt>
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x18(%rbp)
L28:
  mov    -0x18(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L58
  addl   $0x1,-0x1c(%rbp)
L35:
  mov    -0x1c(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L13
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
double *mm(double *A, int arows, int acols, char transA, double *B, int brows, int bcols, char transB) {
  double *C = alloc_array(arows, bcols);
  double sum;
  int i;
  int j;
  int k;
  for (i = 0; i < arows; i++) {
    for (j = 0; j < brows; j++) {
      sum = 0;
      for (k = 0; k < bcols; k++) {
        sum = sum + A[map(i, k, acols, transA)] * B[map(k, j, bcols, transB)];
      }
      C[map(i, j, bcols, 'n')] = sum;
    }
  }
  return C;
}
```
user: Translate the below assembly code into C programming language:

```asm
<mm>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,%r15
  mov    %esi,%r13d
  mov    %esi,0x2c(%rsp)
  mov    %edx,0x18(%rsp)
  mov    %r8,%r14
  mov    %r9d,0x28(%rsp)
  mov    0x70(%rsp),%r12d
  mov    %ecx,%ebx
  movzbl 0x78(%rsp),%ebp
  mov    %r12d,%esi
  mov    %r13d,%edi
  call   <alloc_array@plt>
  mov    %rax,0x20(%rsp)
  movl   $0x0,0x14(%rsp)
  movsbl %bl,%eax
  mov    %eax,0x1c(%rsp)
  test   %r13d,%r13d
  jg     L45
L13:
  mov    0x20(%rsp),%rax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L59:
  mov    0x1c(%rsp),%ecx
  mov    0x18(%rsp),%edx
  mov    %ebx,%esi
  mov    0x14(%rsp),%edi
  call   <map@plt>
  movsd  (%r15,%rax,8),%xmm1
  movsd  %xmm1,0x8(%rsp)
  movsbl %bpl,%ecx
  mov    %r12d,%edx
  mov    %r13d,%esi
  mov    %ebx,%edi
  call   <map@plt>
  movsd  0x8(%rsp),%xmm0
  mulsd  (%r14,%rax,8),%xmm0
  addsd  (%rsp),%xmm0
  movsd  %xmm0,(%rsp)
  add    $0x1,%ebx
  cmp    %ebx,%r12d
  jne    L59
L22:
  mov    $0x6e,%ecx
  mov    %r12d,%edx
  mov    %r13d,%esi
  mov    0x14(%rsp),%edi
  call   <map@plt>
  mov    0x20(%rsp),%rdx
  movsd  (%rsp),%xmm3
  movsd  %xmm3,(%rdx,%rax,8)
  add    $0x1,%r13d
  cmp    %r13d,0x28(%rsp)
  je     L2
L75:
  mov    $0x0,%ebx
  pxor   %xmm4,%xmm4
  movsd  %xmm4,(%rsp)
  test   %r12d,%r12d
  jg     L59
  jmp    L22
L2:
  addl   $0x1,0x14(%rsp)
  mov    0x14(%rsp),%eax
  cmp    %eax,0x2c(%rsp)
  je     L13
L45:
  mov    $0x0,%r13d
  cmpl   $0x0,0x28(%rsp)
  jg     L75
  jmp    L2
```

assistant: ```c
double *mm(double *A, int arows, int acols, char transA, double *B, int brows, int bcols, char transB) {
  double *C = alloc_array(arows, bcols);
  double sum;
  int i;
  int j;
  int k;
  for (i = 0; i < arows; i++) {
    for (j = 0; j < brows; j++) {
      sum = 0;
      for (k = 0; k < bcols; k++) {
        sum = sum + A[map(i, k, acols, transA)] * B[map(k, j, bcols, transB)];
      }
      C[map(i, j, bcols, 'n')] = sum;
    }
  }
  return C;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<mm>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  movsbl %cl,%r14d
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    0x80(%rsp),%ebx
  mov    %rdi,0x18(%rsp)
  mov    %r15d,%edi
  mov    %esi,0x3c(%rsp)
  movsbl 0x88(%rsp),%r13d
  mov    %ebx,%esi
  mov    %edx,0x24(%rsp)
  mov    %r8,0x28(%rsp)
  mov    %r9d,0x38(%rsp)
  call   <alloc_array@plt>
  mov    %rax,0x30(%rsp)
  test   %r15d,%r15d
  jle    L18
  xor    %r12d,%r12d
  nopl   0x0(%rax,%rax,1)
L50:
  mov    0x38(%rsp),%eax
  xor    %ebp,%ebp
  test   %eax,%eax
  jle    L16
  xchg   %ax,%ax
L43:
  movq   $0x0,0x8(%rsp)
  xor    %r15d,%r15d
  test   %ebx,%ebx
  jle    L59
L32:
  mov    0x24(%rsp),%edx
  mov    %r15d,%esi
  mov    %r14d,%ecx
  mov    %r12d,%edi
  call   <map@plt>
  mov    0x18(%rsp),%rdx
  mov    %r13d,%ecx
  mov    %r15d,%edi
  mov    %ebp,%esi
  add    $0x1,%r15d
  movsd  (%rdx,%rax,8),%xmm0
  mov    %ebx,%edx
  movsd  %xmm0,0x10(%rsp)
  call   <map@plt>
  mov    0x28(%rsp),%rcx
  movsd  0x10(%rsp),%xmm0
  mulsd  (%rcx,%rax,8),%xmm0
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  cmp    %r15d,%ebx
  jne    L32
L59:
  mov    %ebp,%esi
  mov    $0x6e,%ecx
  mov    %ebx,%edx
  mov    %r12d,%edi
  call   <map@plt>
  mov    0x30(%rsp),%rsi
  movsd  0x8(%rsp),%xmm2
  add    $0x1,%ebp
  movsd  %xmm2,(%rsi,%rax,8)
  cmp    %ebp,0x38(%rsp)
  jne    L43
L16:
  add    $0x1,%r12d
  cmp    %r12d,0x3c(%rsp)
  jne    L50
L18:
  mov    0x30(%rsp),%rax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
double *mm(double *A, int arows, int acols, char transA, double *B, int brows, int bcols, char transB) {
  double *C = alloc_array(arows, bcols);
  double sum;
  int i;
  int j;
  int k;
  for (i = 0; i < arows; i++) {
    for (j = 0; j < brows; j++) {
      sum = 0;
      for (k = 0; k < bcols; k++) {
        sum = sum + A[map(i, k, acols, transA)] * B[map(k, j, bcols, transB)];
      }
      C[map(i, j, bcols, 'n')] = sum;
    }
  }
  return C;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<mm>:
  endbr64
  push   %r15
  mov    %esi,%r15d
  push   %r14
  movsbl %cl,%r14d
  push   %r13
  push   %r12
  push   %rbp
  mov    %r9d,%ebp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,0x18(%rsp)
  mov    0x80(%rsp),%ebx
  movzbl 0x88(%rsp),%edi
  mov    %esi,0x3c(%rsp)
  mov    %ebx,%esi
  mov    %edx,0x30(%rsp)
  mov    %dil,0x3b(%rsp)
  mov    %r15d,%edi
  mov    %r8,0x20(%rsp)
  mov    %r9d,0x34(%rsp)
  call   <alloc_array@plt>
  mov    %rax,0x28(%rsp)
  test   %r15d,%r15d
  jle    L39
  xor    %r12d,%r12d
  test   %ebp,%ebp
  jle    L39
  nopw   0x0(%rax,%rax,1)
L16:
  xor    %ebp,%ebp
  nopw   0x0(%rax,%rax,1)
L34:
  movq   $0x0,0x8(%rsp)
  movsbl 0x3b(%rsp),%r13d
  xor    %r15d,%r15d
  test   %ebx,%ebx
  jle    L24
  xchg   %ax,%ax
L12:
  mov    0x30(%rsp),%edx
  mov    %r15d,%esi
  mov    %r14d,%ecx
  mov    %r12d,%edi
  call   <map@plt>
  mov    0x18(%rsp),%rdx
  mov    %r13d,%ecx
  mov    %r15d,%edi
  mov    %ebp,%esi
  add    $0x1,%r15d
  movsd  (%rdx,%rax,8),%xmm0
  mov    %ebx,%edx
  movsd  %xmm0,0x10(%rsp)
  call   <map@plt>
  mov    0x20(%rsp),%rcx
  movsd  0x10(%rsp),%xmm0
  mulsd  (%rcx,%rax,8),%xmm0
  addsd  0x8(%rsp),%xmm0
  movsd  %xmm0,0x8(%rsp)
  cmp    %r15d,%ebx
  jne    L12
L24:
  mov    %ebp,%esi
  mov    $0x6e,%ecx
  mov    %ebx,%edx
  mov    %r12d,%edi
  call   <map@plt>
  mov    0x28(%rsp),%rsi
  movsd  0x8(%rsp),%xmm2
  add    $0x1,%ebp
  movsd  %xmm2,(%rsi,%rax,8)
  cmp    %ebp,0x34(%rsp)
  jne    L34
  add    $0x1,%r12d
  cmp    %r12d,0x3c(%rsp)
  jne    L16
L39:
  mov    0x28(%rsp),%rax
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
double *mm(double *A, int arows, int acols, char transA, double *B, int brows, int bcols, char transB) {
  double *C = alloc_array(arows, bcols);
  double sum;
  int i;
  int j;
  int k;
  for (i = 0; i < arows; i++) {
    for (j = 0; j < brows; j++) {
      sum = 0;
      for (k = 0; k < bcols; k++) {
        sum = sum + A[map(i, k, acols, transA)] * B[map(k, j, bcols, transB)];
      }
      C[map(i, j, bcols, 'n')] = sum;
    }
  }
  return C;
}
```
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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x5,-0x2c(%rbp)
  movl   $0x3,-0x28(%rbp)
  movl   $0xf,-0x24(%rbp)
  lea    -0x2c(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  lea    -0x28(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  lea    -0x24(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    -0x20(%rbp),%rdx
  sub    %rdx,%rax
  sar    $0x2,%rax
  mov    %rax,%rsi
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  lea    -0x18(%rbp),%rdx
  sub    %rdx,%rax
  sar    $0x3,%rax
  mov    %rax,%rsi
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D53
type_infered: string
name: parse_data
arguments: tag: D62
type_infered: string

tool: D53: %ld


tool: D62: %lu


assistant: ```c
void main(void) {
  int x = 5, y = 3, z = 15;
  int *a = &x;
  int *b = &y;
  int *c = &z;
  printf("%ld\n", c - a);
  printf("%lu\n", a);
  printf("%lu\n", b);
  printf("%lu\n", c);
  printf("%ld\n", &c - &b);
  printf("%lu\n", &a);
  printf("%lu\n", &b);
  printf("%lu\n", &c);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movl   $0x5,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0xf,0xc(%rsp)
  lea    0x4(%rsp),%rax
  mov    %rax,0x10(%rsp)
  lea    0x8(%rsp),%rdx
  mov    %rdx,0x18(%rsp)
  lea    0xc(%rsp),%rsi
  mov    %rsi,0x20(%rsp)
  sub    %rax,%rsi
  sar    $0x2,%rsi
  lea    D19(%rip),%r13
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x10(%rsp),%rsi
  lea    D16(%rip),%rbx
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x20(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x20(%rsp),%rbp
  lea    0x18(%rsp),%r12
  mov    %rbp,%rsi
  sub    %r12,%rsi
  sar    $0x3,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D19
data_type: string
name: parse_data
arguments: tag: D16
data_type: string

tool: D19: %ld


tool: D16: %lu


assistant: ```c
void main(void) {
  int x = 5, y = 3, z = 15;
  int *a = &x;
  int *b = &y;
  int *c = &z;
  printf("%ld\n", c - a);
  printf("%lu\n", a);
  printf("%lu\n", b);
  printf("%lu\n", c);
  printf("%ld\n", &c - &b);
  printf("%lu\n", &a);
  printf("%lu\n", &b);
  printf("%lu\n", &c);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r14
  lea    D11(%rip),%r14
  push   %r13
  mov    %r14,%rdi
  push   %r12
  push   %rbp
  lea    D15(%rip),%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x4(%rsp),%rax
  movl   $0x5,0x4(%rsp)
  mov    %rsi,0x20(%rsp)
  sub    %rax,%rsi
  lea    0x8(%rsp),%rdx
  lea    0x20(%rsp),%r12
  sar    $0x2,%rsi
  mov    %rax,0x10(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%r13
  mov    %rdx,0x18(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0xf,0xc(%rsp)
  call   <printf@plt>
  mov    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x20(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  sub    %r13,%rsi
  sar    $0x3,%rsi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D11
datatype: string
name: unpack_data
arguments: data_label: D15
datatype: string

tool: D11: %ld


tool: D15: %lu


assistant: ```c
void main(void) {
  int x = 5, y = 3, z = 15;
  int *a = &x;
  int *b = &y;
  int *c = &z;
  printf("%ld\n", c - a);
  printf("%lu\n", a);
  printf("%lu\n", b);
  printf("%lu\n", c);
  printf("%ld\n", &c - &b);
  printf("%lu\n", &a);
  printf("%lu\n", &b);
  printf("%lu\n", &c);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r14
  lea    D12(%rip),%r14
  push   %r13
  mov    %r14,%rdi
  push   %r12
  push   %rbp
  lea    D9(%rip),%rbp
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    0x4(%rsp),%rax
  movl   $0x5,0x4(%rsp)
  mov    %rsi,0x20(%rsp)
  sub    %rax,%rsi
  lea    0x8(%rsp),%rdx
  lea    0x20(%rsp),%r12
  sar    $0x2,%rsi
  mov    %rax,0x10(%rsp)
  xor    %eax,%eax
  lea    0x18(%rsp),%r13
  mov    %rdx,0x18(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0xf,0xc(%rsp)
  call   <printf@plt>
  mov    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x20(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  sub    %r13,%rsi
  sar    $0x3,%rsi
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x38,%rsp
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D12
type_infered: string
name: parse_data
arguments: label: D9
type_infered: string

tool: D12: %ld


tool: D9: %lu


assistant: ```c
void main(void) {
  int x = 5, y = 3, z = 15;
  int *a = &x;
  int *b = &y;
  int *c = &z;
  printf("%ld\n", c - a);
  printf("%lu\n", a);
  printf("%lu\n", b);
  printf("%lu\n", c);
  printf("%ld\n", &c - &b);
  printf("%lu\n", &a);
  printf("%lu\n", &b);
  printf("%lu\n", &c);
}
```