ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x3f8,%rsp
  lea    D26(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D21(%rip),%rdi
  call   <scanf@plt>
  cmpb   $0x0,(%rsp)
  je     L38
  xor    %eax,%eax
  nopl   0x0(%rax)
L32:
  add    $0x1,%eax
  movsbq %al,%rdx
  movsbl %al,%esi
  cmpb   $0x0,(%rsp,%rdx,1)
  jne    L32
L31:
  xor    %eax,%eax
  lea    D22(%rip),%rdi
  call   <printf@plt>
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  xor    %eax,%eax
  add    $0x3f8,%rsp
  ret
L38:
  xor    %esi,%esi
  jmp    L31
L37:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D26
type_guessed: string
name: read_data
arguments: data_label: D21
type_guessed: string
name: read_data
arguments: data_label: D22
type_guessed: string

tool: D26: Enter a string: 

tool: D21: %s

tool: D22: Length of string: %d

assistant: ```c
int main() {
  char s[1000], i;
  printf("Enter a string: ");
  scanf("%s", &s);
  for (i = 0; s[i] != '\0'; i++)
    ;
  printf("Length of string: %d", i);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x3f8,%rsp
  lea    D24(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x3e8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rsp,%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  cmpb   $0x0,(%rsp)
  je     L35
  xor    %eax,%eax
  nopl   0x0(%rax)
L22:
  add    $0x1,%eax
  movsbq %al,%rdx
  movsbl %al,%esi
  cmpb   $0x0,(%rsp,%rdx,1)
  jne    L22
L32:
  xor    %eax,%eax
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x3e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  xor    %eax,%eax
  add    $0x3f8,%rsp
  ret
L35:
  xor    %esi,%esi
  jmp    L32
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D24
datatype: string
name: disassemble_data
arguments: tag: D7
datatype: string
name: disassemble_data
arguments: tag: D13
datatype: string

tool: D24: Enter a string: 

tool: D7: %s

tool: D13: Length of string: %d

assistant: ```c
int main() {
  char s[1000], i;
  printf("Enter a string: ");
  scanf("%s", &s);
  for (i = 0; s[i] != '\0'; i++)
    ;
  printf("Length of string: %d", i);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<idx_msb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  je     L8
  bsr    -0x14(%rbp),%eax
  xor    $0x1f,%eax
  mov    %eax,%edx
  mov    $0x20,%eax
  sub    %edx,%eax
  mov    %eax,-0x4(%rbp)
L8:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int idx_msb(unsigned int x) {
  int ret = 0;
  if (x > 0) {
    ret = ((sizeof(x) << 3) - __builtin_clz(x));
  }
  return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<idx_msb>:
  endbr64
  mov    $0x0,%eax
  test   %edi,%edi
  je     L9
  bsr    %edi,%edi
  xor    $0x1f,%edi
  mov    $0x20,%eax
  sub    %edi,%eax
L9:
  ret
```

assistant: ```c
int idx_msb(unsigned int x) {
  int ret = 0;
  if (x > 0) {
    ret = ((sizeof(x) << 3) - __builtin_clz(x));
  }
  return ret;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<idx_msb>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L7
  bsr    %edi,%edi
  mov    $0x20,%eax
  xor    $0x1f,%edi
  sub    %edi,%eax
L7:
  ret
```

assistant: ```c
int idx_msb(unsigned int x) {
  int ret = 0;
  if (x > 0) {
    ret = ((sizeof(x) << 3) - __builtin_clz(x));
  }
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<idx_msb>:
  endbr64
  xor    %eax,%eax
  test   %edi,%edi
  je     L8
  bsr    %edi,%edi
  mov    $0x20,%eax
  xor    $0x1f,%edi
  sub    %edi,%eax
L8:
  ret
```

assistant: ```c
int idx_msb(unsigned int x) {
  int ret = 0;
  if (x > 0) {
    ret = ((sizeof(x) << 3) - __builtin_clz(x));
  }
  return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D51(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D57(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D44
type: string
name: disassemble_data
arguments: label: D5
type: string
name: disassemble_data
arguments: label: D33
type: string
name: disassemble_data
arguments: label: D51
type: string
name: disassemble_data
arguments: label: D57
type: string
name: disassemble_data
arguments: label: D28
type: string

tool: D44: Clbin41 : convert lst files into bin files for emu41


tool: D5: Syntax: clbin41 [options] sourcefile destfile 


tool: D33:         if sourcefile='null', create empty bin file


tool: D51: Option: /r convert unpacked rom file into packed bin file


tool: D57:         /b convert packed bin file into unpacked rom file


tool: D28: V1.9 JFG 2005


assistant: ```c
void usage(void) {
  printf("Clbin41 : convert lst files into bin files for emu41\n");
  printf("Syntax: clbin41 [options] sourcefile destfile \n");
  printf("        if sourcefile='null', create empty bin file\n");
  printf("Option: /r convert unpacked rom file into packed bin file\n");
  printf("        /b convert packed bin file into unpacked rom file\n");
  printf("V1.9 JFG 2005\n");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D29(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D42(%rip),%rdi
  call   <printf@plt>
  lea    D23(%rip),%rdi
  call   <printf@plt>
  lea    D1(%rip),%rdi
  call   <printf@plt>
  lea    D56(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D29
type: string
name: extract_symbolic_data
arguments: label: D15
type: string
name: extract_symbolic_data
arguments: label: D42
type: string
name: extract_symbolic_data
arguments: label: D23
type: string
name: extract_symbolic_data
arguments: label: D1
type: string
name: extract_symbolic_data
arguments: label: D56
type: string

tool: D29: Clbin41 : convert lst files into bin files for emu41


tool: D15: Syntax: clbin41 [options] sourcefile destfile 


tool: D42:         if sourcefile='null', create empty bin file


tool: D23: Option: /r convert unpacked rom file into packed bin file


tool: D1:         /b convert packed bin file into unpacked rom file


tool: D56: V1.9 JFG 2005


assistant: ```c
void usage(void) {
  printf("Clbin41 : convert lst files into bin files for emu41\n");
  printf("Syntax: clbin41 [options] sourcefile destfile \n");
  printf("        if sourcefile='null', create empty bin file\n");
  printf("Option: /r convert unpacked rom file into packed bin file\n");
  printf("        /b convert packed bin file into unpacked rom file\n");
  printf("V1.9 JFG 2005\n");
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D54(%rip),%rdi
  call   <printf@plt>
  lea    D21(%rip),%rdi
  call   <printf@plt>
  lea    D39(%rip),%rdi
  call   <printf@plt>
  lea    D8(%rip),%rdi
  call   <printf@plt>
  lea    D5(%rip),%rdi
  call   <printf@plt>
  lea    D47(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: tag: D54
type: string
name: unpack_data
arguments: tag: D21
type: string
name: unpack_data
arguments: tag: D39
type: string
name: unpack_data
arguments: tag: D8
type: string
name: unpack_data
arguments: tag: D5
type: string
name: unpack_data
arguments: tag: D47
type: string

tool: D54: Clbin41 : convert lst files into bin files for emu41


tool: D21: Syntax: clbin41 [options] sourcefile destfile 


tool: D39:         if sourcefile='null', create empty bin file


tool: D8: Option: /r convert unpacked rom file into packed bin file


tool: D5:         /b convert packed bin file into unpacked rom file


tool: D47: V1.9 JFG 2005


assistant: ```c
void usage(void) {
  printf("Clbin41 : convert lst files into bin files for emu41\n");
  printf("Syntax: clbin41 [options] sourcefile destfile \n");
  printf("        if sourcefile='null', create empty bin file\n");
  printf("Option: /r convert unpacked rom file into packed bin file\n");
  printf("        /b convert packed bin file into unpacked rom file\n");
  printf("V1.9 JFG 2005\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D38(%rip),%rdi
  call   <printf@plt>
  lea    D57(%rip),%rdi
  call   <printf@plt>
  lea    D2(%rip),%rdi
  call   <printf@plt>
  lea    D34(%rip),%rdi
  call   <printf@plt>
  lea    D51(%rip),%rdi
  call   <printf@plt>
  lea    D58(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D38
type_guessed: string
name: disassemble_data
arguments: tag: D57
type_guessed: string
name: disassemble_data
arguments: tag: D2
type_guessed: string
name: disassemble_data
arguments: tag: D34
type_guessed: string
name: disassemble_data
arguments: tag: D51
type_guessed: string
name: disassemble_data
arguments: tag: D58
type_guessed: string

tool: D38: Clbin41 : convert lst files into bin files for emu41


tool: D57: Syntax: clbin41 [options] sourcefile destfile 


tool: D2:         if sourcefile='null', create empty bin file


tool: D34: Option: /r convert unpacked rom file into packed bin file


tool: D51:         /b convert packed bin file into unpacked rom file


tool: D58: V1.9 JFG 2005


assistant: ```c
void usage(void) {
  printf("Clbin41 : convert lst files into bin files for emu41\n");
  printf("Syntax: clbin41 [options] sourcefile destfile \n");
  printf("        if sourcefile='null', create empty bin file\n");
  printf("Option: /r convert unpacked rom file into packed bin file\n");
  printf("        /b convert packed bin file into unpacked rom file\n");
  printf("V1.9 JFG 2005\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<r_list_fini>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void r_list_fini(struct r_list *list) {
  do {
    ;
  } while (0);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<r_list_fini>:
  endbr64
  ret
```

assistant: ```c
void r_list_fini(struct r_list *list) {
  do {
    ;
  } while (0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<r_list_fini>:
  endbr64
  ret
```

assistant: ```c
void r_list_fini(struct r_list *list) {
  do {
    ;
  } while (0);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<r_list_fini>:
  endbr64
  ret
```

assistant: ```c
void r_list_fini(struct r_list *list) {
  do {
    ;
  } while (0);
}
```
user: What is the c source code of the assembly code below:

```asm
<combinar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    %r9,-0x30(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x28(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jne    L5
  movl   $0x0,-0x10(%rbp)
  jmp    L80
L44:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    0x10(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x10(%rbp)
L80:
  mov    -0x10(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L44
  mov    0x28(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x8(%rbp)
  jge    L31
  mov    -0xc(%rbp),%eax
  cmp    0x18(%rbp),%eax
  jl     L31
  mov    0x20(%rbp),%rcx
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%esi
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <retiraRepetidos@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  sub    -0x4(%rbp),%eax
  cmp    %eax,0x18(%rbp)
  jg     L31
  mov    0x28(%rbp),%rax
  mov    -0x8(%rbp),%edx
  mov    %edx,(%rax)
  jmp    L31
L5:
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L75
L6:
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    -0x10(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x28(%rbp),%eax
  lea    0x1(%rax),%r10d
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x30(%rbp),%r8
  mov    -0x20(%rbp),%rdx
  mov    -0x18(%rbp),%esi
  mov    -0x14(%rbp),%eax
  push   0x28(%rbp)
  push   0x20(%rbp)
  mov    0x18(%rbp),%edi
  push   %rdi
  push   0x10(%rbp)
  mov    %r8,%r9
  mov    %r10d,%r8d
  mov    %eax,%edi
  call   <combinar>
  add    $0x20,%rsp
  addl   $0x1,-0x10(%rbp)
L75:
  mov    -0x10(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L6
L31:
  nop
  leave
  ret
```

assistant: ```c
void combinar(int n, int nParls, int *comb, int prox, int tam, int *somaInlf, int *salario, int k, int **infl, int *menor) {
  int i, p = 0, s = 0, r;
  if (tam == nParls) {
    for (i = 0; i < nParls; i++) {
      p += somaInlf[comb[i]];
      s += salario[comb[i]];
    }
    if (s < *menor && p >= k) {
      r = retiraRepetidos(n, nParls, comb, infl);
      if (p - r >= k)
        *menor = s;
    }
  } else
    for (i = prox; i < n; i++) {
      comb[tam] = i;
      combinar(n, nParls, comb, i + 1, tam + 1, somaInlf, salario, k, infl, menor);
    }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<combinar>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%ebp
  mov    %esi,%r13d
  mov    %rdx,(%rsp)
  mov    %r9,%r12
  mov    0x50(%rsp),%r14
  cmp    %esi,%r8d
  je     L15
  mov    %ecx,%ebx
  cmp    %edi,%ecx
  jge    L4
  movslq %r8d,%rax
  mov    (%rsp),%rsi
  lea    (%rsi,%rax,4),%r15
  lea    0x1(%r8),%eax
  mov    %eax,0xc(%rsp)
  mov    %r13d,%r12d
  mov    %r9,%r13
L19:
  mov    %ebx,(%r15)
  add    $0x1,%ebx
  push   0x68(%rsp)
  push   0x68(%rsp)
  mov    0x68(%rsp),%eax
  push   %rax
  push   %r14
  mov    %r13,%r9
  mov    0x2c(%rsp),%r8d
  mov    %ebx,%ecx
  mov    0x20(%rsp),%rdx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <combinar>
  add    $0x20,%rsp
  cmp    %ebp,%ebx
  jne    L19
L4:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L15:
  test   %esi,%esi
  jle    L93
  mov    (%rsp),%rdi
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rcx
  mov    $0x0,%ebx
  mov    $0x0,%r15d
L72:
  movslq (%rax),%rdx
  add    (%r12,%rdx,4),%r15d
  add    (%r14,%rdx,4),%ebx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L72
L55:
  mov    0x68(%rsp),%rax
  cmp    %ebx,(%rax)
  jle    L4
  cmp    %r15d,0x58(%rsp)
  jg     L4
  mov    0x60(%rsp),%rcx
  mov    (%rsp),%rdx
  mov    %r13d,%esi
  mov    %ebp,%edi
  call   <retiraRepetidos@plt>
  sub    %eax,%r15d
  cmp    0x58(%rsp),%r15d
  jl     L4
  mov    0x68(%rsp),%rax
  mov    %ebx,(%rax)
  jmp    L4
L93:
  mov    $0x0,%ebx
  mov    $0x0,%r15d
  jmp    L55
```

assistant: ```c
void combinar(int n, int nParls, int *comb, int prox, int tam, int *somaInlf, int *salario, int k, int **infl, int *menor) {
  int i, p = 0, s = 0, r;
  if (tam == nParls) {
    for (i = 0; i < nParls; i++) {
      p += somaInlf[comb[i]];
      s += salario[comb[i]];
    }
    if (s < *menor && p >= k) {
      r = retiraRepetidos(n, nParls, comb, infl);
      if (p - r >= k)
        *menor = s;
    }
  } else
    for (i = prox; i < n; i++) {
      comb[tam] = i;
      combinar(n, nParls, comb, i + 1, tam + 1, somaInlf, salario, k, infl, menor);
    }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<combinar>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x18,%rsp
  cmp    %esi,%r8d
  je     L2
  cmp    %edi,%ecx
  jge    L9
  movslq %r8d,%rax
  lea    0x1(%r8),%r14d
  lea    (%rdx,%rax,4),%r15
  mov    %r14d,%ebx
  mov    %ecx,%r14d
  mov    %r15,0x8(%rsp)
  mov    %r9,%r15
L83:
  mov    0x8(%rsp),%rax
  mov    %r15,%r9
  mov    %ebx,%r8d
  mov    %r13,%rdx
  mov    %r12d,%esi
  mov    %ebp,%edi
  mov    %r14d,(%rax)
  add    $0x1,%r14d
  push   0x68(%rsp)
  mov    %r14d,%ecx
  push   0x68(%rsp)
  mov    0x68(%rsp),%eax
  push   %rax
  push   0x68(%rsp)
  call   <combinar>
  add    $0x20,%rsp
  cmp    %ebp,%r14d
  jne    L83
L9:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L2:
  test   %esi,%esi
  jle    L6
  lea    -0x1(%rsi),%ecx
  mov    %rdx,%rax
  xor    %ebx,%ebx
  xor    %r15d,%r15d
  lea    0x4(%rdx,%rcx,4),%rdi
  mov    0x50(%rsp),%rcx
  nopl   0x0(%rax)
L31:
  movslq (%rax),%rdx
  add    $0x4,%rax
  add    (%r9,%rdx,4),%r15d
  add    (%rcx,%rdx,4),%ebx
  cmp    %rax,%rdi
  jne    L31
L49:
  mov    0x68(%rsp),%rax
  cmp    %ebx,(%rax)
  jle    L9
  cmp    %r15d,0x58(%rsp)
  jg     L9
  mov    0x60(%rsp),%rcx
  mov    %r13,%rdx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <retiraRepetidos@plt>
  sub    %eax,%r15d
  cmp    0x58(%rsp),%r15d
  jl     L9
  mov    0x68(%rsp),%rax
  mov    %ebx,(%rax)
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L6:
  xor    %ebx,%ebx
  xor    %r15d,%r15d
  jmp    L49
```

assistant: ```c
void combinar(int n, int nParls, int *comb, int prox, int tam, int *somaInlf, int *salario, int k, int **infl, int *menor) {
  int i, p = 0, s = 0, r;
  if (tam == nParls) {
    for (i = 0; i < nParls; i++) {
      p += somaInlf[comb[i]];
      s += salario[comb[i]];
    }
    if (s < *menor && p >= k) {
      r = retiraRepetidos(n, nParls, comb, infl);
      if (p - r >= k)
        *menor = s;
    }
  } else
    for (i = prox; i < n; i++) {
      comb[tam] = i;
      combinar(n, nParls, comb, i + 1, tam + 1, somaInlf, salario, k, infl, menor);
    }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<combinar>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  sub    $0x18,%rsp
  cmp    %esi,%r8d
  je     L11
  cmp    %edi,%ecx
  jge    L32
  movslq %r8d,%rax
  lea    0x1(%r8),%r14d
  lea    (%rdx,%rax,4),%r15
  mov    %r14d,%ebx
  mov    %ecx,%r14d
  mov    %r15,0x8(%rsp)
  mov    %r9,%r15
L17:
  mov    0x8(%rsp),%rax
  mov    %r15,%r9
  mov    %ebx,%r8d
  mov    %r13,%rdx
  mov    %r12d,%esi
  mov    %ebp,%edi
  mov    %r14d,(%rax)
  add    $0x1,%r14d
  push   0x68(%rsp)
  mov    %r14d,%ecx
  push   0x68(%rsp)
  mov    0x68(%rsp),%eax
  push   %rax
  push   0x68(%rsp)
  call   <combinar>
  add    $0x20,%rsp
  cmp    %ebp,%r14d
  jne    L17
L32:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  test   %esi,%esi
  jle    L97
  lea    -0x1(%rsi),%ecx
  mov    %rdx,%rax
  xor    %ebx,%ebx
  xor    %r15d,%r15d
  lea    0x4(%rdx,%rcx,4),%rdi
  mov    0x50(%rsp),%rcx
  nopl   0x0(%rax)
L42:
  movslq (%rax),%rdx
  add    $0x4,%rax
  add    (%r9,%rdx,4),%r15d
  add    (%rcx,%rdx,4),%ebx
  cmp    %rax,%rdi
  jne    L42
L77:
  mov    0x68(%rsp),%rax
  cmp    %ebx,(%rax)
  jle    L32
  cmp    %r15d,0x58(%rsp)
  jg     L32
  mov    0x60(%rsp),%rcx
  mov    %r13,%rdx
  mov    %r12d,%esi
  mov    %ebp,%edi
  call   <retiraRepetidos@plt>
  sub    %eax,%r15d
  cmp    0x58(%rsp),%r15d
  jl     L32
  mov    0x68(%rsp),%rax
  mov    %ebx,(%rax)
  jmp    L32
  nopl   0x0(%rax,%rax,1)
L97:
  xor    %ebx,%ebx
  xor    %r15d,%r15d
  jmp    L77
```

assistant: ```c
void combinar(int n, int nParls, int *comb, int prox, int tam, int *somaInlf, int *salario, int k, int **infl, int *menor) {
  int i, p = 0, s = 0, r;
  if (tam == nParls) {
    for (i = 0; i < nParls; i++) {
      p += somaInlf[comb[i]];
      s += salario[comb[i]];
    }
    if (s < *menor && p >= k) {
      r = retiraRepetidos(n, nParls, comb, infl);
      if (p - r >= k)
        *menor = s;
    }
  } else
    for (i = prox; i < n; i++) {
      comb[tam] = i;
      combinar(n, nParls, comb, i + 1, tam + 1, somaInlf, salario, k, infl, menor);
    }
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x8(%rbp)
  jmp    L16
L6:
  mov    $0x0,%eax
  call   <rand@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jns    L1
  mov    $0x0,%eax
  call   <abort@plt>
L1:
  addl   $0x1,-0x8(%rbp)
L16:
  cmpl   $0x270f,-0x8(%rbp)
  jle    L6
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  for (int i = 0; i < 10000; i++) {
    int random = rand();
    if (random < 0) {
      abort();
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2710,%ebx
  jmp    L14
L1:
  sub    $0x1,%ebx
  je     L38
L14:
  mov    $0x0,%eax
  call   <rand@plt>
  test   %eax,%eax
  jns    L1
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L1
L38:
  mov    $0x0,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  for (int i = 0; i < 10000; i++) {
    int random = rand();
    if (random < 0) {
      abort();
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2710,%ebx
  jmp    L5
  nopl   0x0(%rax)
L36:
  sub    $0x1,%ebx
  je     L27
L5:
  xor    %eax,%eax
  call   <rand@plt>
  test   %eax,%eax
  jns    L36
  xor    %eax,%eax
  call   <abort@plt>
  sub    $0x1,%ebx
  jne    L5
L27:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  for (int i = 0; i < 10000; i++) {
    int random = rand();
    if (random < 0) {
      abort();
    }
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x2710,%ebx
  jmp    L24
  nopl   0x0(%rax)
L10:
  sub    $0x1,%ebx
  je     L5
L24:
  xor    %eax,%eax
  call   <rand@plt>
  test   %eax,%eax
  jns    L10
  xor    %eax,%eax
  call   <abort@plt>
  sub    $0x1,%ebx
  jne    L24
L5:
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main() {
  for (int i = 0; i < 10000; i++) {
    int random = rand();
    if (random < 0) {
      abort();
    }
  }
}
```