ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  movabs $0xa00000014,%rax
  mov    %rsp,%rdi
  lea    0x28(%rsp),%rsi
  mov    %rax,(%rsp)
  movabs $0x2700000005,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x1000000004,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x1a00000013,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x140000001f,%rax
  mov    %rax,0x20(%rsp)
  call   <sump@plt>
  lea    D4(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

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

tool: D4: The total number of marbles is %ld.


assistant: ```c
int main() {
  int marbles[10] = {20, 10, 5, 39, 4, 16, 19, 26, 31, 20};
  long answer;
  answer = sump(marbles, marbles + 10);
  printf("The total number of marbles is %ld.\n", answer);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  sub    $0x38,%rsp
  movdqa D7(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    D24(%rip),%rax
  mov    %rsp,%rdi
  lea    0x28(%rsp),%rsi
  movaps %xmm0,(%rsp)
  movdqa D22(%rip),%xmm0
  mov    %rax,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  call   <sump@plt>
  lea    D36(%rip),%rdi
  mov    %rax,%rsi
  call   <printf@plt>
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D7
datatype: byte[16]
name: disassemble_data
arguments: data_tag: D24
datatype: byte[8]
name: disassemble_data
arguments: data_tag: D22
datatype: byte[16]
name: disassemble_data
arguments: data_tag: D36
datatype: string

tool: {"D7": [0x14,0x00,0x00,0x00,0x0a,0x00,0x00,0x00,0x05,0x00,0x00,0x00,0x27,0x00,0x00,0x00]}

tool: {"D24": [0x1f,0x00,0x00,0x00,0x14,0x00,0x00,0x00]}

tool: {"D22": [0x04,0x00,0x00,0x00,0x10,0x00,0x00,0x00,0x13,0x00,0x00,0x00,0x1a,0x00,0x00,0x00]}

tool: D36: The total number of marbles is %ld.


assistant: ```c
int main() {
  int marbles[10] = {20, 10, 5, 39, 4, 16, 19, 26, 31, 20};
  long answer;
  answer = sump(marbles, marbles + 10);
  printf("The total number of marbles is %ld.\n", answer);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strCopy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L9
L18:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x8(%rbp)
  addq   $0x1,-0x10(%rbp)
L9:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L18
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void strCopy(char target[], char source[]) {
  for (; *source != '\0'; target++, source++) {
    *target = *source;
  }
  *target = *source;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strCopy>:
  endbr64
  movzbl (%rsi),%eax
  test   %al,%al
  je     L12
L18:
  mov    %al,(%rdi)
  add    $0x1,%rdi
  add    $0x1,%rsi
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L18
L12:
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void strCopy(char target[], char source[]) {
  for (; *source != '\0'; target++, source++) {
    *target = *source;
  }
  *target = *source;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strCopy>:
  endbr64
  jmp    L12
  cs nopw 0x0(%rax,%rax,1)
L10:
  mov    %al,(%rdi)
  add    $0x1,%rsi
  add    $0x1,%rdi
L12:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L10
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void strCopy(char target[], char source[]) {
  for (; *source != '\0'; target++, source++) {
    *target = *source;
  }
  *target = *source;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strCopy>:
  endbr64
  jmp    L14
  cs nopw 0x0(%rax,%rax,1)
L15:
  mov    %al,(%rdi)
  add    $0x1,%rsi
  add    $0x1,%rdi
L14:
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L15
  movb   $0x0,(%rdi)
  ret
```

assistant: ```c
void strCopy(char target[], char source[]) {
  for (; *source != '\0'; target++, source++) {
    *target = *source;
  }
  *target = *source;
}
```
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    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x0,%eax
  call   <testLibmemcachedCompat@plt>
  mov    $0x0,%eax
  call   <testAddMultipleTimes@plt>
  mov    $0x0,%eax
  call   <testRemoveNode@plt>
  mov    $0x0,%eax
  call   <testRingSorted@plt>
  mov    $0x0,%eax
  call   <testKnownNextHighestItemOnRing@plt>
  mov    $0x0,%eax
  call   <testEmptyRingItemSearchReturnsNull@plt>
  mov    $0x0,%eax
  call   <testEmptyRingSearchReturnsNull@plt>
  mov    $0x0,%eax
  call   <testKnownSlotsOnRing@plt>
  mov    $0x0,%eax
  call   <testKnownMultipleSlotsOnRing@plt>
  mov    $0x0,%eax
  call   <runBenchmark@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  testLibmemcachedCompat();
  testAddMultipleTimes();
  testRemoveNode();
  testRingSorted();
  testKnownNextHighestItemOnRing();
  testEmptyRingItemSearchReturnsNull();
  testEmptyRingSearchReturnsNull();
  testKnownSlotsOnRing();
  testKnownMultipleSlotsOnRing();
  runBenchmark();
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <testLibmemcachedCompat@plt>
  mov    $0x0,%eax
  call   <testAddMultipleTimes@plt>
  mov    $0x0,%eax
  call   <testRemoveNode@plt>
  mov    $0x0,%eax
  call   <testRingSorted@plt>
  mov    $0x0,%eax
  call   <testKnownNextHighestItemOnRing@plt>
  mov    $0x0,%eax
  call   <testEmptyRingItemSearchReturnsNull@plt>
  mov    $0x0,%eax
  call   <testEmptyRingSearchReturnsNull@plt>
  mov    $0x0,%eax
  call   <testKnownSlotsOnRing@plt>
  mov    $0x0,%eax
  call   <testKnownMultipleSlotsOnRing@plt>
  mov    $0x0,%eax
  call   <runBenchmark@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  testLibmemcachedCompat();
  testAddMultipleTimes();
  testRemoveNode();
  testRingSorted();
  testKnownNextHighestItemOnRing();
  testEmptyRingItemSearchReturnsNull();
  testEmptyRingSearchReturnsNull();
  testKnownSlotsOnRing();
  testKnownMultipleSlotsOnRing();
  runBenchmark();
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <testLibmemcachedCompat@plt>
  xor    %eax,%eax
  call   <testAddMultipleTimes@plt>
  xor    %eax,%eax
  call   <testRemoveNode@plt>
  xor    %eax,%eax
  call   <testRingSorted@plt>
  xor    %eax,%eax
  call   <testKnownNextHighestItemOnRing@plt>
  xor    %eax,%eax
  call   <testEmptyRingItemSearchReturnsNull@plt>
  xor    %eax,%eax
  call   <testEmptyRingSearchReturnsNull@plt>
  xor    %eax,%eax
  call   <testKnownSlotsOnRing@plt>
  xor    %eax,%eax
  call   <testKnownMultipleSlotsOnRing@plt>
  xor    %eax,%eax
  call   <runBenchmark@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  testLibmemcachedCompat();
  testAddMultipleTimes();
  testRemoveNode();
  testRingSorted();
  testKnownNextHighestItemOnRing();
  testEmptyRingItemSearchReturnsNull();
  testEmptyRingSearchReturnsNull();
  testKnownSlotsOnRing();
  testKnownMultipleSlotsOnRing();
  runBenchmark();
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <testLibmemcachedCompat@plt>
  xor    %eax,%eax
  call   <testAddMultipleTimes@plt>
  xor    %eax,%eax
  call   <testRemoveNode@plt>
  xor    %eax,%eax
  call   <testRingSorted@plt>
  xor    %eax,%eax
  call   <testKnownNextHighestItemOnRing@plt>
  xor    %eax,%eax
  call   <testEmptyRingItemSearchReturnsNull@plt>
  xor    %eax,%eax
  call   <testEmptyRingSearchReturnsNull@plt>
  xor    %eax,%eax
  call   <testKnownSlotsOnRing@plt>
  xor    %eax,%eax
  call   <testKnownMultipleSlotsOnRing@plt>
  xor    %eax,%eax
  call   <runBenchmark@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char **argv) {
  testLibmemcachedCompat();
  testAddMultipleTimes();
  testRemoveNode();
  testRingSorted();
  testKnownNextHighestItemOnRing();
  testEmptyRingItemSearchReturnsNull();
  testEmptyRingSearchReturnsNull();
  testKnownSlotsOnRing();
  testKnownMultipleSlotsOnRing();
  runBenchmark();
  return 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    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L30
L59:
  mov    $0x0,%eax
  call   <dice@plt>
  cmp    $0x5,%eax
  ja     L19
  mov    %eax,%eax
  lea    0x0(,%rax,4),%rdx
  lea    D51(%rip),%rax
  mov    (%rdx,%rax,1),%eax
  cltq
  lea    D51(%rip),%rdx
  add    %rdx,%rax
  notrack jmp *%rax
  addl   $0x1,-0x18(%rbp)
  jmp    L76
  addl   $0x1,-0x14(%rbp)
  jmp    L76
  addl   $0x1,-0x10(%rbp)
  jmp    L76
  addl   $0x1,-0xc(%rbp)
  jmp    L76
  addl   $0x1,-0x8(%rbp)
  jmp    L76
L19:
  addl   $0x1,-0x4(%rbp)
L76:
  addl   $0x1,-0x1c(%rbp)
L30:
  cmpl   $0x13,-0x1c(%rbp)
  jle    L59
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D76(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%eax
  mov    %eax,%esi
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%esi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D74
type_infered: string
name: parse_data
arguments: tag: D76
type_infered: string
name: parse_data
arguments: tag: D26
type_infered: string
name: parse_data
arguments: tag: D46
type_infered: string
name: parse_data
arguments: tag: D33
type_infered: string
name: parse_data
arguments: tag: D53
type_infered: string

tool: D74: Quantidade de dados com o lado 1 = %d


tool: D76: Quantidade de dados com o lado 2 = %d


tool: D26: Quantidade de dados com o lado 3 = %d


tool: D46: Quantidade de dados com o lado 4 = %d


tool: D33: Quantidade de dados com o lado 5 = %d


tool: D53: Quantidade de dados com o lado 6 = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i, num1 = 0, num2 = 0, num3 = 0, num4 = 0, num5 = 0, num6 = 0, numero;
  for (i = 0; i < 20; i += 1) {
    switch (dice()) {
    case 1:
      num1 += 1;
      break;
    case 2:
      num2 += 1;
      break;
    case 3:
      num3 += 1;
      break;
    case 4:
      num4 += 1;
      break;
    case 5:
      num5 += 1;
      break;
    default:
      num6 += 1;
    }
  }
  printf("Quantidade de dados com o lado 1 = %d\n", num1);
  printf("Quantidade de dados com o lado 2 = %d\n", num2);
  printf("Quantidade de dados com o lado 3 = %d\n", num3);
  printf("Quantidade de dados com o lado 4 = %d\n", num4);
  printf("Quantidade de dados com o lado 5 = %d\n", num5);
  printf("Quantidade de dados com o lado 6 = %d\n", num6);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    $0x14,%ebx
  movl   $0x0,0xc(%rsp)
  movl   $0x0,0x8(%rsp)
  mov    $0x0,%r12d
  mov    $0x0,%r13d
  mov    $0x0,%r14d
  mov    $0x0,%r15d
  lea    D10(%rip),%rbp
  jmp    L8
  add    $0x1,%r15d
L77:
  sub    $0x1,%ebx
  je     L2
L8:
  mov    $0x0,%eax
  call   <dice@plt>
  cmp    $0x5,%eax
  ja     L55
  mov    %eax,%eax
  movslq 0x0(%rbp,%rax,4),%rax
  add    %rbp,%rax
  notrack jmp *%rax
  add    $0x1,%r14d
  jmp    L77
  add    $0x1,%r13d
  jmp    L77
  add    $0x1,%r12d
  jmp    L77
  addl   $0x1,0x8(%rsp)
  jmp    L77
L55:
  addl   $0x1,0xc(%rsp)
  jmp    L77
L2:
  mov    %r15d,%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    %r14d,%esi
  lea    D44(%rip),%rdi
  call   <printf@plt>
  mov    %r13d,%esi
  lea    D54(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D38(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%esi
  lea    D58(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    D66(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: label: D13
type_guessed: string
name: parse_data
arguments: label: D44
type_guessed: string
name: parse_data
arguments: label: D54
type_guessed: string
name: parse_data
arguments: label: D38
type_guessed: string
name: parse_data
arguments: label: D58
type_guessed: string
name: parse_data
arguments: label: D66
type_guessed: string

tool: D13: Quantidade de dados com o lado 1 = %d


tool: D44: Quantidade de dados com o lado 2 = %d


tool: D54: Quantidade de dados com o lado 3 = %d


tool: D38: Quantidade de dados com o lado 4 = %d


tool: D58: Quantidade de dados com o lado 5 = %d


tool: D66: Quantidade de dados com o lado 6 = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i, num1 = 0, num2 = 0, num3 = 0, num4 = 0, num5 = 0, num6 = 0, numero;
  for (i = 0; i < 20; i += 1) {
    switch (dice()) {
    case 1:
      num1 += 1;
      break;
    case 2:
      num2 += 1;
      break;
    case 3:
      num3 += 1;
      break;
    case 4:
      num4 += 1;
      break;
    case 5:
      num5 += 1;
      break;
    default:
      num6 += 1;
    }
  }
  printf("Quantidade de dados com o lado 1 = %d\n", num1);
  printf("Quantidade de dados com o lado 2 = %d\n", num2);
  printf("Quantidade de dados com o lado 3 = %d\n", num3);
  printf("Quantidade de dados com o lado 4 = %d\n", num4);
  printf("Quantidade de dados com o lado 5 = %d\n", num5);
  printf("Quantidade de dados com o lado 6 = %d\n", num6);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %r15
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D31(%rip),%rbp
  push   %rbx
  mov    $0x14,%ebx
  sub    $0x18,%rsp
  movl   $0x0,0xc(%rsp)
  movl   $0x0,0x8(%rsp)
  nopw   0x0(%rax,%rax,1)
L10:
  xor    %eax,%eax
  call   <dice@plt>
  cmp    $0x5,%eax
  ja     L28
  mov    %eax,%eax
  movslq 0x0(%rbp,%rax,4),%rax
  add    %rbp,%rax
  notrack jmp *%rax
  nopl   (%rax)
  addl   $0x1,0x8(%rsp)
  nopl   (%rax)
L4:
  sub    $0x1,%ebx
  jne    L10
  mov    %r15d,%esi
  lea    D69(%rip),%rdi
  call   <printf@plt>
  mov    %r14d,%esi
  lea    D24(%rip),%rdi
  call   <printf@plt>
  mov    %r13d,%esi
  lea    D50(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D36(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%esi
  lea    D14(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
  add    $0x1,%r12d
  jmp    L4
  cs nopw 0x0(%rax,%rax,1)
  add    $0x1,%r13d
  jmp    L4
  nopl   0x0(%rax)
  add    $0x1,%r14d
  jmp    L4
  nopl   0x0(%rax)
  add    $0x1,%r15d
  jmp    L4
L28:
  addl   $0x1,0xc(%rsp)
  jmp    L4
```

tool_calls: name: retrieve_data
arguments: tag: D69
data_type: string
name: retrieve_data
arguments: tag: D24
data_type: string
name: retrieve_data
arguments: tag: D50
data_type: string
name: retrieve_data
arguments: tag: D36
data_type: string
name: retrieve_data
arguments: tag: D14
data_type: string
name: retrieve_data
arguments: tag: D23
data_type: string

tool: D69: Quantidade de dados com o lado 1 = %d


tool: D24: Quantidade de dados com o lado 2 = %d


tool: D50: Quantidade de dados com o lado 3 = %d


tool: D36: Quantidade de dados com o lado 4 = %d


tool: D14: Quantidade de dados com o lado 5 = %d


tool: D23: Quantidade de dados com o lado 6 = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i, num1 = 0, num2 = 0, num3 = 0, num4 = 0, num5 = 0, num6 = 0, numero;
  for (i = 0; i < 20; i += 1) {
    switch (dice()) {
    case 1:
      num1 += 1;
      break;
    case 2:
      num2 += 1;
      break;
    case 3:
      num3 += 1;
      break;
    case 4:
      num4 += 1;
      break;
    case 5:
      num5 += 1;
      break;
    default:
      num6 += 1;
    }
  }
  printf("Quantidade de dados com o lado 1 = %d\n", num1);
  printf("Quantidade de dados com o lado 2 = %d\n", num2);
  printf("Quantidade de dados com o lado 3 = %d\n", num3);
  printf("Quantidade de dados com o lado 4 = %d\n", num4);
  printf("Quantidade de dados com o lado 5 = %d\n", num5);
  printf("Quantidade de dados com o lado 6 = %d\n", num6);
  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   %r15
  xor    %r15d,%r15d
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  lea    D48(%rip),%rbp
  push   %rbx
  mov    $0x14,%ebx
  sub    $0x18,%rsp
  movl   $0x0,0xc(%rsp)
  movl   $0x0,0x8(%rsp)
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  call   <dice@plt>
  cmp    $0x5,%eax
  ja     L68
  mov    %eax,%eax
  movslq 0x0(%rbp,%rax,4),%rax
  add    %rbp,%rax
  notrack jmp *%rax
  nopl   (%rax)
  addl   $0x1,0x8(%rsp)
  nopl   (%rax)
L8:
  sub    $0x1,%ebx
  jne    L4
  mov    %r15d,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    %r14d,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    %r13d,%esi
  lea    D28(%rip),%rdi
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D61(%rip),%rdi
  call   <printf@plt>
  mov    0x8(%rsp),%esi
  lea    D23(%rip),%rdi
  call   <printf@plt>
  mov    0xc(%rsp),%esi
  lea    D40(%rip),%rdi
  call   <printf@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
  add    $0x1,%r12d
  jmp    L8
  cs nopw 0x0(%rax,%rax,1)
  add    $0x1,%r13d
  jmp    L8
  nopl   0x0(%rax)
  add    $0x1,%r14d
  jmp    L8
  nopl   0x0(%rax)
  add    $0x1,%r15d
  jmp    L8
L68:
  addl   $0x1,0xc(%rsp)
  jmp    L8
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D3
datatype: string
name: extract_symbolic_data
arguments: data_label: D8
datatype: string
name: extract_symbolic_data
arguments: data_label: D28
datatype: string
name: extract_symbolic_data
arguments: data_label: D61
datatype: string
name: extract_symbolic_data
arguments: data_label: D23
datatype: string
name: extract_symbolic_data
arguments: data_label: D40
datatype: string

tool: D3: Quantidade de dados com o lado 1 = %d


tool: D8: Quantidade de dados com o lado 2 = %d


tool: D28: Quantidade de dados com o lado 3 = %d


tool: D61: Quantidade de dados com o lado 4 = %d


tool: D23: Quantidade de dados com o lado 5 = %d


tool: D40: Quantidade de dados com o lado 6 = %d


assistant: ```c
int main(int argc, char *argv[]) {
  int i, num1 = 0, num2 = 0, num3 = 0, num4 = 0, num5 = 0, num6 = 0, numero;
  for (i = 0; i < 20; i += 1) {
    switch (dice()) {
    case 1:
      num1 += 1;
      break;
    case 2:
      num2 += 1;
      break;
    case 3:
      num3 += 1;
      break;
    case 4:
      num4 += 1;
      break;
    case 5:
      num5 += 1;
      break;
    default:
      num6 += 1;
    }
  }
  printf("Quantidade de dados com o lado 1 = %d\n", num1);
  printf("Quantidade de dados com o lado 2 = %d\n", num2);
  printf("Quantidade de dados com o lado 3 = %d\n", num3);
  printf("Quantidade de dados com o lado 4 = %d\n", num4);
  printf("Quantidade de dados com o lado 5 = %d\n", num5);
  printf("Quantidade de dados com o lado 6 = %d\n", num6);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L17
L5:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <ePrimo@plt>
  test   %eax,%eax
  je     L15
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <ePrimo@plt>
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L15:
  addl   $0x1,-0x10(%rbp)
L17:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L5
  mov    -0xc(%rbp),%eax
  mov    %eax,%esi
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D46
type_infered: string
name: disassemble_data
arguments: label: D9
type_infered: string
name: disassemble_data
arguments: label: D24
type_infered: string
name: disassemble_data
arguments: label: D4
type_infered: string
name: disassemble_data
arguments: label: D16
type_infered: string

tool: D46: Informe inicio do intervalo

tool: D9: %d

tool: D24: Informe fim do intervalo

tool: D4: %d e primo


tool: D16: no total foram %d primos

assistant: ```c
main() {
  int first, last, i, count = 0;
  puts("Informe inicio do intervalo");
  scanf("%d", &first);
  puts("Informe fim do intervalo");
  scanf("%d", &last);
  for (i = first; i < last + 1; i++) {
    if (ePrimo(i)) {
      printf("%d e primo\n", ePrimo(i));
      count++;
    }
  }
  printf("no total foram %d primos", count);
}
```