ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<extend_plane_borders>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,0x28(%rsp)
  mov    %esi,0x1c(%rsp)
  mov    %edx,%ebx
  mov    %ecx,0x30(%rsp)
  mov    %r8d,0x34(%rsp)
  mov    %r9d,0x8(%rsp)
  movslq %ecx,%rbp
  lea    -0x1(%rdi,%rbp,1),%r12
  movslq %r9d,%rax
  mov    %rax,0x20(%rsp)
  mov    %rdi,%r15
  sub    %rax,%r15
  mov    %r15,0x38(%rsp)
  add    %rdi,%rbp
  test   %edx,%edx
  js     L58
  mov    %rdi,%r14
  mov    $0x0,%r13d
  movslq %esi,%rax
  mov    %rax,0x10(%rsp)
L34:
  movzbl (%r14),%esi
  mov    0x8(%rsp),%edx
  mov    %r15,%rdi
  call   <memset@plt>
  movzbl (%r12),%esi
  mov    0x88(%rsp),%edx
  mov    %rbp,%rdi
  call   <memset@plt>
  mov    0x10(%rsp),%rax
  add    %rax,%r14
  add    %rax,%r12
  add    %rax,%r15
  add    %rax,%rbp
  add    $0x1,%r13d
  cmp    %r13d,%ebx
  jge    L34
L58:
  sub    $0x1,%ebx
  mov    0x1c(%rsp),%ecx
  imul   %ecx,%ebx
  movslq %ebx,%r14
  mov    0x20(%rsp),%rdi
  sub    %rdi,%r14
  mov    0x28(%rsp),%rax
  add    %rax,%r14
  mov    0x34(%rsp),%edx
  mov    %edx,%ebp
  neg    %ebp
  imul   %ecx,%ebp
  movslq %ebp,%rbp
  sub    %rdi,%rbp
  add    %rax,%rbp
  add    %ecx,%ebx
  movslq %ebx,%rbx
  sub    %rdi,%rbx
  add    %rax,%rbx
  mov    0x8(%rsp),%eax
  add    0x88(%rsp),%eax
  mov    0x30(%rsp),%edi
  lea    0x1(%rdi,%rax,1),%r13d
  test   %edx,%edx
  jle    L10
  mov    $0x0,%r12d
  movslq %ecx,%r15
  mov    %r14,0x8(%rsp)
  mov    0x38(%rsp),%r14
  mov    %rbx,0x10(%rsp)
  mov    %r15,%rbx
  mov    %edx,%r15d
L36:
  mov    %r13d,%edx
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <memcpy@plt>
  add    %rbx,%rbp
  add    $0x1,%r12d
  cmp    %r12d,%r15d
  jne    L36
  mov    0x8(%rsp),%r14
  mov    0x10(%rsp),%rbx
L10:
  cmpl   $0x0,0x80(%rsp)
  jle    L30
  mov    $0x0,%ebp
  movslq 0x1c(%rsp),%r12
  mov    0x80(%rsp),%r15d
L12:
  mov    %r13d,%edx
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <memcpy@plt>
  add    %r12,%rbx
  add    $0x1,%ebp
  cmp    %ebp,%r15d
  jne    L12
L30:
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void extend_plane_borders(unsigned char *s, int sp, int h, int w, int et, int el, int eb, int er) {
  int i;
  unsigned char *src_ptr1, *src_ptr2;
  unsigned char *dest_ptr1, *dest_ptr2;
  int linesize;
  src_ptr1 = s;
  src_ptr2 = s + w - 1;
  dest_ptr1 = s - el;
  dest_ptr2 = s + w;
  for (i = 0; i < h - 0 + 1; i++) {
    memset(dest_ptr1, src_ptr1[0], el);
    memset(dest_ptr2, src_ptr2[0], er);
    src_ptr1 += sp;
    src_ptr2 += sp;
    dest_ptr1 += sp;
    dest_ptr2 += sp;
  }
  src_ptr1 = s - el;
  src_ptr2 = s + sp * (h - 1) - el;
  dest_ptr1 = s + sp * (-et) - el;
  dest_ptr2 = s + sp * (h)-el;
  linesize = el + er + w + 1;
  for (i = 0; i < (int)et; i++) {
    memcpy(dest_ptr1, src_ptr1, linesize);
    dest_ptr1 += sp;
  }
  for (i = 0; i < (int)eb; i++) {
    memcpy(dest_ptr2, src_ptr2, linesize);
    dest_ptr2 += sp;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x64,-0x40(%rbp)
  movl   $0xc8,-0x3c(%rbp)
  movl   $0x12c,-0x38(%rbp)
  movl   $0x190,-0x34(%rbp)
  movl   $0x1f4,-0x30(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x44(%rbp)
L43:
  lea    -0x48(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%edx
  mov    -0x48(%rbp),%eax
  add    $0x1,%edx
  cltq
  mov    %edx,-0x20(%rbp,%rax,4)
  mov    -0x48(%rbp),%eax
  cmp    $0xffffffff,%eax
  je     L21
  jmp    L43
L21:
  nop
  movl   $0x0,-0x48(%rbp)
  jmp    L34
L35:
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%edx
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%eax
  mov    %edx,%esi
  imul   %eax,%esi
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%ecx
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%edx
  mov    -0x48(%rbp),%eax
  mov    %esi,%r8d
  mov    %eax,%esi
  lea    D11(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x20(%rbp,%rax,4),%edx
  mov    -0x48(%rbp),%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%eax
  imul   %edx,%eax
  add    %eax,-0x44(%rbp)
  mov    -0x48(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x48(%rbp)
L34:
  mov    -0x48(%rbp),%eax
  cmp    $0x4,%eax
  jle    L35
  mov    -0x44(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L39
  call   <__stack_chk_fail@plt>
L39:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D23
type_infered: string
name: read_data
arguments: label: D11
type_infered: string
name: read_data
arguments: label: D4
type_infered: string

tool: D23: %d

tool: D11: menu:%d %d %d %6d


tool: D4: Total: %d


assistant: ```c
main() {
  int i;
  int menu[5] = {100, 200, 300, 400, 500};
  int count[5] = {};
  int total = 0;
  while (1) {
    scanf("%d", &i);
    count[i] += 1;
    if (i == -1) {
      break;
    }
  }
  for (i = 0; i < 5; i++) {
    printf("menu:%d %d %d %6d\n", i, menu[i], count[i], count[i] * menu[i]);
    total += count[i] * menu[i];
  }
  printf("Total: %d\n", total);
}
```
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    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movl   $0x64,0x10(%rsp)
  movl   $0xc8,0x14(%rsp)
  movl   $0x12c,0x18(%rsp)
  movl   $0x190,0x1c(%rsp)
  movl   $0x1f4,0x20(%rsp)
  movq   $0x0,0x30(%rsp)
  movq   $0x0,0x38(%rsp)
  movl   $0x0,0x40(%rsp)
  lea    D10(%rip),%rbx
L6:
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  movslq %eax,%rdx
  addl   $0x1,0x30(%rsp,%rdx,4)
  cmp    $0xffffffff,%eax
  jne    L6
  movl   $0x0,0xc(%rsp)
  mov    $0x0,%ebx
  mov    $0x0,%esi
  lea    D28(%rip),%rbp
L27:
  movslq %esi,%rax
  mov    0x30(%rsp,%rax,4),%ecx
  mov    0x10(%rsp,%rax,4),%edx
  mov    %ecx,%r8d
  imul   %edx,%r8d
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  movslq %esi,%rdx
  mov    0x30(%rsp,%rdx,4),%eax
  imul   0x10(%rsp,%rdx,4),%eax
  add    %eax,%ebx
  add    $0x1,%esi
  mov    %esi,0xc(%rsp)
  cmp    $0x4,%esi
  jle    L27
  mov    %ebx,%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    $0x0,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D10
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D28
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D9
type_guessed: string

tool: D10: %d

tool: D28: menu:%d %d %d %6d


tool: D9: Total: %d


assistant: ```c
main() {
  int i;
  int menu[5] = {100, 200, 300, 400, 500};
  int count[5] = {};
  int total = 0;
  while (1) {
    scanf("%d", &i);
    count[i] += 1;
    if (i == -1) {
      break;
    }
  }
  for (i = 0; i < 5; i++) {
    printf("menu:%d %d %d %6d\n", i, menu[i], count[i], count[i] * menu[i]);
    total += count[i] * menu[i];
  }
  printf("Total: %d\n", total);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  pxor   %xmm0,%xmm0
  push   %rbx
  lea    D1(%rip),%rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  movabs $0xc800000064,%rax
  movl   $0x1f4,0x20(%rsp)
  lea    0xc(%rsp),%rbp
  movl   $0x0,0x40(%rsp)
  movaps %xmm0,0x30(%rsp)
  mov    %rax,0x10(%rsp)
  movabs $0x1900000012c,%rax
  mov    %rax,0x18(%rsp)
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movslq 0xc(%rsp),%rdx
  addl   $0x1,0x30(%rsp,%rdx,4)
  cmp    $0xffffffff,%edx
  jne    L6
  movl   $0x0,0xc(%rsp)
  xor    %ebp,%ebp
  xor    %esi,%esi
  xor    %eax,%eax
  mov    $0x64,%edx
  lea    D6(%rip),%rbx
  jmp    L28
  cs nopw 0x0(%rax,%rax,1)
L18:
  movslq %esi,%rax
  mov    0x10(%rsp,%rax,4),%edx
L28:
  mov    0x30(%rsp,%rax,4),%ecx
  mov    %rbx,%rdi
  xor    %eax,%eax
  mov    %ecx,%r8d
  imul   %edx,%r8d
  call   <printf@plt>
  movslq 0xc(%rsp),%rdx
  mov    0x30(%rsp,%rdx,4),%eax
  imul   0x10(%rsp,%rdx,4),%eax
  mov    %rdx,%rsi
  add    $0x1,%esi
  mov    %esi,0xc(%rsp)
  add    %eax,%ebp
  cmp    $0x4,%esi
  jle    L18
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L21:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D1
data_type: string
name: unpack_data
arguments: data_tag: D6
data_type: string
name: unpack_data
arguments: data_tag: D13
data_type: string

tool: D1: %d

tool: D6: menu:%d %d %d %6d


tool: D13: Total: %d


assistant: ```c
main() {
  int i;
  int menu[5] = {100, 200, 300, 400, 500};
  int count[5] = {};
  int total = 0;
  while (1) {
    scanf("%d", &i);
    count[i] += 1;
    if (i == -1) {
      break;
    }
  }
  for (i = 0; i < 5; i++) {
    printf("menu:%d %d %d %6d\n", i, menu[i], count[i], count[i] * menu[i]);
    total += count[i] * menu[i];
  }
  printf("Total: %d\n", total);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  lea    D7(%rip),%rbx
  sub    $0x58,%rsp
  movdqa D23(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  movl   $0x1f4,0x20(%rsp)
  lea    0xc(%rsp),%rbp
  movl   $0x0,0x40(%rsp)
  movaps %xmm0,0x10(%rsp)
  pxor   %xmm0,%xmm0
  movaps %xmm0,0x30(%rsp)
  nopl   0x0(%rax)
L7:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  movslq 0xc(%rsp),%rdx
  addl   $0x1,0x30(%rsp,%rdx,4)
  cmp    $0xffffffff,%edx
  jne    L7
  movl   $0x0,0xc(%rsp)
  xor    %ebp,%ebp
  xor    %esi,%esi
  xor    %eax,%eax
  mov    $0x64,%edx
  lea    D10(%rip),%rbx
  jmp    L38
  cs nopw 0x0(%rax,%rax,1)
L35:
  movslq %esi,%rax
  mov    0x10(%rsp,%rax,4),%edx
L38:
  mov    0x30(%rsp,%rax,4),%ecx
  mov    %rbx,%rdi
  xor    %eax,%eax
  mov    %ecx,%r8d
  imul   %edx,%r8d
  call   <printf@plt>
  movslq 0xc(%rsp),%rdx
  mov    0x30(%rsp,%rdx,4),%eax
  imul   0x10(%rsp,%rdx,4),%eax
  mov    %rdx,%rsi
  add    $0x1,%esi
  mov    %esi,0xc(%rsp)
  add    %eax,%ebp
  cmp    $0x4,%esi
  jle    L35
  xor    %eax,%eax
  mov    %ebp,%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D7
type: string
name: read_data
arguments: label: D23
type: byte[16]
name: read_data
arguments: label: D10
type: string
name: read_data
arguments: label: D13
type: string

tool: D7: %d

tool: {"D23": [0x64,0x00,0x00,0x00,0xc8,0x00,0x00,0x00,0x2c,0x01,0x00,0x00,0x90,0x01,0x00,0x00]}

tool: D10: menu:%d %d %d %6d


tool: D13: Total: %d


assistant: ```c
main() {
  int i;
  int menu[5] = {100, 200, 300, 400, 500};
  int count[5] = {};
  int total = 0;
  while (1) {
    scanf("%d", &i);
    count[i] += 1;
    if (i == -1) {
      break;
    }
  }
  for (i = 0; i < 5; i++) {
    printf("menu:%d %d %d %6d\n", i, menu[i], count[i], count[i] * menu[i]);
    total += count[i] * menu[i];
  }
  printf("Total: %d\n", total);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<tem_repeticao>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L29
L32:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x20(%rbp)
  jne    L4
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%eax
  jmp    L14
L4:
  mov    -0x20(%rbp),%eax
  mov    %eax,%esi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L29:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L32
  mov    $0x0,%eax
L14:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D5
type_infered: string
name: extract_symbolic_data
arguments: label: D2
type_infered: string

tool: D5: Ja existe %d 


tool: D2: Nao existe %d 


assistant: ```c
int tem_repeticao(int *v, int n, int numero_gerado) {
  int i = 0;
  for (i = 0; i < n; ++i) {
    if (v[i] == numero_gerado) {
      printf("Ja existe %d \n", numero_gerado);
      return 1;
    } else {
      printf("Nao existe %d \n", numero_gerado);
    }
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<tem_repeticao>:
  endbr64
  test   %esi,%esi
  jle    L9
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edx,%ebp
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%r13
  lea    D6(%rip),%r12
L16:
  cmp    %ebp,(%rbx)
  je     L39
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r13,%rbx
  jne    L16
  mov    $0x0,%eax
  jmp    L30
L39:
  mov    %ebp,%esi
  lea    D11(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%eax
L30:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L9:
  mov    $0x0,%eax
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
datatype: string
name: unpack_data
arguments: data_tag: D11
datatype: string

tool: D6: Nao existe %d 


tool: D11: Ja existe %d 


assistant: ```c
int tem_repeticao(int *v, int n, int numero_gerado) {
  int i = 0;
  for (i = 0; i < n; ++i) {
    if (v[i] == numero_gerado) {
      printf("Ja existe %d \n", numero_gerado);
      return 1;
    } else {
      printf("Nao existe %d \n", numero_gerado);
    }
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<tem_repeticao>:
  endbr64
  test   %esi,%esi
  jle    L29
  push   %r13
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r13
  lea    D4(%rip),%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L33:
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r13,%rbx
  je     L8
L6:
  mov    %ebp,%esi
  cmp    %ebp,(%rbx)
  jne    L33
  lea    D13(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L8:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  xor    %eax,%eax
  ret
```

tool_calls: name: disassemble_data
arguments: label: D4
type: string
name: disassemble_data
arguments: label: D13
type: string

tool: D4: Nao existe %d 


tool: D13: Ja existe %d 


assistant: ```c
int tem_repeticao(int *v, int n, int numero_gerado) {
  int i = 0;
  for (i = 0; i < n; ++i) {
    if (v[i] == numero_gerado) {
      printf("Ja existe %d \n", numero_gerado);
      return 1;
    } else {
      printf("Nao existe %d \n", numero_gerado);
    }
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<tem_repeticao>:
  endbr64
  test   %esi,%esi
  jle    L33
  push   %r13
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r13
  lea    D19(%rip),%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L36
  nopl   0x0(%rax,%rax,1)
L31:
  mov    %r12,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r13,%rbx
  je     L35
L36:
  mov    %ebp,%esi
  cmp    %ebp,(%rbx)
  jne    L31
  lea    D4(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L35:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L33:
  xor    %eax,%eax
  ret
```

tool_calls: name: parse_data
arguments: data_label: D19
datatype: string
name: parse_data
arguments: data_label: D4
datatype: string

tool: D19: Nao existe %d 


tool: D4: Ja existe %d 


assistant: ```c
int tem_repeticao(int *v, int n, int numero_gerado) {
  int i = 0;
  for (i = 0; i < n; ++i) {
    if (v[i] == numero_gerado) {
      printf("Ja existe %d \n", numero_gerado);
      return 1;
    } else {
      printf("Nao existe %d \n", numero_gerado);
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Mutation_int_Gibbs>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,%r15
  mov    %rsi,%r14
  mov    %edx,0x10(%rsp)
  mov    %rcx,0x20(%rsp)
  mov    %r8,0x8(%rsp)
  mov    %r9d,0x14(%rsp)
  movsd  %xmm0,0x28(%rsp)
  mov    0x88(%rsp),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  test   %edx,%edx
  jle    L17
  lea    0x1(%rdx),%ecx
  mov    $0x1,%eax
L21:
  mov    (%r15,%rax,4),%edx
  mov    %edx,(%r12,%rax,4)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L21
L17:
  cmpl   $0x0,0x80(%rsp)
  jle    L30
  mov    $0x1,%r13d
  jmp    L63
L72:
  mov    $0x1,%eax
  sub    0x0(%rbp),%eax
  mov    %eax,0x0(%rbp)
L59:
  add    $0x1,%r13d
  cmp    %r13d,0x80(%rsp)
  jl     L30
L63:
  mov    0x10(%rsp),%esi
  mov    $0x1,%edi
  call   <integerrng@plt>
  mov    %eax,%ebx
  lea    0x30(%rsp),%rdi
  mov    0x14(%rsp),%edx
  mov    0x8(%rsp),%rsi
  movsd  (%r14),%xmm0
  call   <self_adj_index_search@plt>
  movslq %ebx,%rbx
  lea    (%r12,%rbx,4),%rbp
  mov    $0x1,%eax
  sub    0x0(%rbp),%eax
  mov    %eax,0x0(%rbp)
  mov    0x10(%rsp),%esi
  mov    %r12,%rdi
  call   <cost@plt>
  movsd  %xmm0,(%rsp)
  lea    0x34(%rsp),%rdi
  mov    0x14(%rsp),%edx
  mov    0x8(%rsp),%rsi
  call   <self_adj_index_search@plt>
  movslq 0x34(%rsp),%rax
  movsd  (%rsp),%xmm0
  movsd  0x28(%rsp),%xmm3
  divsd  %xmm3,%xmm0
  mov    0x20(%rsp),%rcx
  addsd  (%rcx,%rax,8),%xmm0
  movslq 0x30(%rsp),%rax
  subsd  (%rcx,%rax,8),%xmm0
  movsd  (%r14),%xmm1
  divsd  %xmm3,%xmm1
  subsd  %xmm1,%xmm0
  call   <exp@plt>
  movsd  %xmm0,0x18(%rsp)
  mov    $0x0,%eax
  call   <uniformrng@plt>
  movsd  D12(%rip),%xmm4
  addsd  0x18(%rsp),%xmm4
  movsd  D12(%rip),%xmm2
  divsd  %xmm4,%xmm2
  comisd %xmm0,%xmm2
  jb     L72
  movsd  (%rsp),%xmm0
  movsd  %xmm0,(%r14)
  mov    0x0(%rbp),%eax
  mov    %eax,(%r15,%rbx,4)
  jmp    L59
L30:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D12
datatype: f64

tool: D12: 1

assistant: ```c
void Mutation_int_Gibbs(int *z, double *fz, int N_dimension, double *theta, double *grid_points, int grid_size, double temp, int Krep, int *z_new) {
  int k_old;
  int k_new;
  int i;
  double Pr;
  int lab;
  int gg;
  double fz_new;
  double rat;
  double un;
  for (i = 1; i <= N_dimension; i++)
    z_new[i] = z[i];
  for (gg = 1; gg <= Krep; gg++) {
    lab = integerrng(1, N_dimension);
    self_adj_index_search(&k_old, *fz, grid_points, grid_size);
    z_new[lab] = 1 - z_new[lab];
    fz_new = cost(z_new, N_dimension);
    self_adj_index_search(&k_new, fz_new, grid_points, grid_size);
    rat = theta[k_new] + fz_new / temp - theta[k_old] - *fz / temp;
    Pr = 1.0 / (1.0 + exp(rat));
    un = uniformrng();
    if (Pr >= un) {
      *fz = fz_new;
      z[lab] = z_new[lab];
    } else {
      z_new[lab] = 1 - z_new[lab];
    }
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<w_str_hash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L14
L8:
  mov    -0xc(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  add    %edx,%eax
  mov    %eax,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L14:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L8
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int w_str_hash(const void *p) {
  const char *s = (const char *)p;
  uint32_t hash = 0;
  while (*s) {
    hash = (hash << 4) + *s;
    s++;
  }
  return hash;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<w_str_hash>:
  endbr64
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L13
  mov    $0x0,%eax
L6:
  shl    $0x4,%eax
  movsbl %dl,%edx
  add    %edx,%eax
  add    $0x1,%rdi
  movzbl (%rdi),%edx
  test   %dl,%dl
  jne    L6
  ret
L13:
  mov    $0x0,%eax
  ret
```

assistant: ```c
unsigned int w_str_hash(const void *p) {
  const char *s = (const char *)p;
  uint32_t hash = 0;
  while (*s) {
    hash = (hash << 4) + *s;
    s++;
  }
  return hash;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<w_str_hash>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L12
  nopl   (%rax)
L4:
  shl    $0x4,%eax
  add    $0x1,%rdi
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L4
  ret
  nopl   0x0(%rax)
L12:
  ret
```

assistant: ```c
unsigned int w_str_hash(const void *p) {
  const char *s = (const char *)p;
  uint32_t hash = 0;
  while (*s) {
    hash = (hash << 4) + *s;
    s++;
  }
  return hash;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<w_str_hash>:
  endbr64
  movsbl (%rdi),%edx
  xor    %eax,%eax
  test   %dl,%dl
  je     L12
  nopl   (%rax)
L16:
  shl    $0x4,%eax
  add    $0x1,%rdi
  add    %edx,%eax
  movsbl (%rdi),%edx
  test   %dl,%dl
  jne    L16
  ret
  nopl   0x0(%rax)
L12:
  ret
```

assistant: ```c
unsigned int w_str_hash(const void *p) {
  const char *s = (const char *)p;
  uint32_t hash = 0;
  while (*s) {
    hash = (hash << 4) + *s;
    s++;
  }
  return hash;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_binary_search_recursive_with_one_element_returns_index>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x18(%rbp)
  movl   $0x1,-0xc(%rbp)
  movl   $0x1,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x18(%rbp),%ecx
  lea    -0xc(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <binary_search_recursive@plt>
  mov    %eax,-0x10(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  je     L11
  mov    $0x2e,%ecx
  lea    D16(%rip),%rax
  mov    %rax,%rdx
  lea    D14(%rip),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <__test_fail@plt>
L11:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L10
  call   <__stack_chk_fail@plt>
L10:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D16
type_infered: string
name: unpack_data
arguments: label: D14
type_infered: string
name: unpack_data
arguments: label: D25
type_infered: string

tool: D16: test_binary_search_recursive_with_one_element_returns_index

tool: D14: /scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/michaelreneer/Algorithms/refs/heads/master/c/test/test_binary_search_recursive.c

tool: D25: index == 0

assistant: ```c
void test_binary_search_recursive_with_one_element_returns_index(void) {
  int length = 1;
  int array[] = {1};
  int element = 1;
  int index = binary_search_recursive(array, length, element);
  ((void)((index == 0) ? ((void)0) : __test_fail("index == 0", "/scratch/repos/new/home/jordi_armengol_estape/c-scraper/outputs/2022-01-22/02-19-57/repos/michaelreneer/Algorithms/refs/heads/master/c/test/test_binary_search_recursive.c", __FUNCTION__, 46)));
}
```