ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xe90,%rsp
  mov    $0x3e80,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x3e88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x2710,%edi
  movl   $0x1,0x3e7c(%rsp)
  mov    $0x1,%edx
  lea    0x3e7c(%rsp),%r8
L9:
  mov    %r8,%rcx
  xor    %eax,%eax
  jmp    L17
  nopl   0x0(%rax)
L6:
  mov    -0x4(%rcx),%edx
  mov    %rsi,%rcx
L17:
  lea    (%rax,%rdx,2),%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  lea    -0x4(%rcx),%rsi
  mov    %edx,(%rcx)
  cmp    %rbx,%rcx
  jne    L6
  sub    $0x1,%edi
  je     L85
  mov    0x3e7c(%rsp),%edx
  jmp    L9
L85:
  xor    %edx,%edx
L77:
  mov    (%rbx,%rdx,4),%esi
  mov    %edx,%eax
  test   %esi,%esi
  je     L87
  add    $0x1,%eax
  lea    D5(%rip),%r12
  movslq %eax,%rbp
  jmp    L51
L74:
  mov    (%rbx,%rbp,4),%esi
  add    $0x1,%rbp
L51:
  xor    %eax,%eax
  mov    %r12,%rdi
  call   <printf@plt>
  cmp    $0xfa0,%ebp
  jne    L74
L15:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x3e88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L69
  add    $0x3e90,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L87:
  add    $0x1,%rdx
  cmp    $0xfa0,%rdx
  jne    L77
  jmp    L15
L69:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D5
data_type: string

tool: D5: %d

assistant: ```c
int main() {
  int array[4000] = {0};
  int i, j, jinw, print_flag = 0;
  array[4000 - 1] = 1;
  for (j = 0; j < 10000; j++) {
    jinw = 0;
    for (i = 4000 - 1; i >= 0; i--) {
      int tmp = array[i] * 2 + jinw;
      array[i] = tmp % 10;
      jinw = tmp / 10;
    }
  }
  for (j = 0; j < 4000; j++) {
    if (array[j] != 0 && print_flag == 0)
      print_flag = 1;
    if (print_flag == 1)
      printf("%d", array[j]);
  }
  printf("\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0xe90,%rsp
  mov    $0x3e80,%edx
  xor    %esi,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x3e88(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  mov    $0x2710,%r8d
  movl   $0x1,0x3e7c(%rsp)
  lea    0x3e7c(%rsp),%r9
L37:
  mov    %r9,%rcx
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L25:
  mov    (%rcx),%edx
  mov    %rcx,%rdi
  sub    $0x4,%rcx
  lea    (%rax,%rdx,2),%edx
  movslq %edx,%rax
  mov    %edx,%esi
  imul   $0x66666667,%rax,%rax
  sar    $0x1f,%esi
  sar    $0x22,%rax
  sub    %esi,%eax
  lea    (%rax,%rax,4),%esi
  add    %esi,%esi
  sub    %esi,%edx
  mov    %edx,0x4(%rcx)
  cmp    %rdi,%rbx
  jne    L25
  sub    $0x1,%r8d
  jne    L37
  xor    %edx,%edx
L47:
  mov    (%rbx,%rdx,4),%esi
  mov    %edx,%eax
  test   %esi,%esi
  je     L73
  add    $0x1,%eax
  lea    D5(%rip),%r12
  movslq %eax,%rbp
  jmp    L64
L51:
  mov    (%rbx,%rbp,4),%esi
  add    $0x1,%rbp
L64:
  xor    %eax,%eax
  mov    %r12,%rdi
  call   <printf@plt>
  cmp    $0xfa0,%ebp
  jne    L51
L15:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x3e88(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L55
  add    $0x3e90,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L73:
  add    $0x1,%rdx
  cmp    $0xfa0,%rdx
  jne    L47
  jmp    L15
L55:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
datatype: string

tool: D5: %d

assistant: ```c
int main() {
  int array[4000] = {0};
  int i, j, jinw, print_flag = 0;
  array[4000 - 1] = 1;
  for (j = 0; j < 10000; j++) {
    jinw = 0;
    for (i = 4000 - 1; i >= 0; i--) {
      int tmp = array[i] * 2 + jinw;
      array[i] = tmp % 10;
      jinw = tmp / 10;
    }
  }
  for (j = 0; j < 4000; j++) {
    if (array[j] != 0 && print_flag == 0)
      print_flag = 1;
    if (print_flag == 1)
      printf("%d", array[j]);
  }
  printf("\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Vec_IntSortCompareUnsigned>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jae    L0
  mov    $0xffffffff,%eax
  jmp    L9
L0:
  mov    -0x8(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jbe    L12
  mov    $0x1,%eax
  jmp    L9
L12:
  mov    $0x0,%eax
L9:
  pop    %rbp
  ret
```

assistant: ```c
int Vec_IntSortCompareUnsigned(unsigned *pp1, unsigned *pp2) {
  if (*pp1 < *pp2)
    return -1;
  if (*pp1 > *pp2)
    return 1;
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Vec_IntSortCompareUnsigned>:
  endbr64
  mov    (%rdi),%ecx
  mov    (%rsi),%edx
  mov    $0xffffffff,%eax
  cmp    %edx,%ecx
  jb     L2
  seta   %al
  movzbl %al,%eax
L2:
  ret
```

assistant: ```c
int Vec_IntSortCompareUnsigned(unsigned *pp1, unsigned *pp2) {
  if (*pp1 < *pp2)
    return -1;
  if (*pp1 > *pp2)
    return 1;
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Vec_IntSortCompareUnsigned>:
  endbr64
  mov    (%rsi),%eax
  cmp    %eax,(%rdi)
  mov    $0xffffffff,%edx
  seta   %al
  movzbl %al,%eax
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int Vec_IntSortCompareUnsigned(unsigned *pp1, unsigned *pp2) {
  if (*pp1 < *pp2)
    return -1;
  if (*pp1 > *pp2)
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<Vec_IntSortCompareUnsigned>:
  endbr64
  mov    (%rsi),%eax
  cmp    %eax,(%rdi)
  mov    $0xffffffff,%edx
  seta   %al
  movzbl %al,%eax
  cmovb  %edx,%eax
  ret
```

assistant: ```c
int Vec_IntSortCompareUnsigned(unsigned *pp1, unsigned *pp2) {
  if (*pp1 < *pp2)
    return -1;
  if (*pp1 > *pp2)
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<findChar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %al,-0x1c(%rbp)
  movw   $0x0,-0x2(%rbp)
  jmp    L6
L26:
  movswq -0x2(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L3
  movzwl -0x2(%rbp),%eax
  jmp    L35
L3:
  movzwl -0x2(%rbp),%eax
  add    $0x1,%eax
  mov    %ax,-0x2(%rbp)
L6:
  movswq -0x2(%rbp),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L26
  mov    $0xffffffff,%eax
L35:
  pop    %rbp
  ret
```

assistant: ```c
short findChar(char *alphabet, char ch) {
  for (short i = 0; 0 != alphabet[i]; i++) {
    if (alphabet[i] == ch) {
      return (i);
    }
  }
  return (-1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<findChar>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L37
  mov    $0x0,%edx
L7:
  cmp    %al,%sil
  je     L26
  add    $0x1,%edx
  movswq %dx,%rax
  movzbl (%rdi,%rax,1),%eax
  test   %al,%al
  jne    L7
  mov    $0xffffffff,%edx
L26:
  mov    %edx,%eax
  ret
L37:
  mov    $0xffffffff,%edx
  jmp    L26
```

assistant: ```c
short findChar(char *alphabet, char ch) {
  for (short i = 0; 0 != alphabet[i]; i++) {
    if (alphabet[i] == ch) {
      return (i);
    }
  }
  return (-1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<findChar>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L32
  xor    %r8d,%r8d
  jmp    L39
L3:
  add    $0x1,%r8d
  movswq %r8w,%rax
  movzbl (%rdi,%rax,1),%eax
  test   %al,%al
  je     L32
L39:
  cmp    %al,%sil
  jne    L3
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L32:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
short findChar(char *alphabet, char ch) {
  for (short i = 0; 0 != alphabet[i]; i++) {
    if (alphabet[i] == ch) {
      return (i);
    }
  }
  return (-1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<findChar>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L2
  xor    %r8d,%r8d
  jmp    L30
L37:
  add    $0x1,%r8d
  movswq %r8w,%rax
  movzbl (%rdi,%rax,1),%eax
  test   %al,%al
  je     L2
L30:
  cmp    %al,%sil
  jne    L37
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L2:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
short findChar(char *alphabet, char ch) {
  for (short i = 0; 0 != alphabet[i]; i++) {
    if (alphabet[i] == ch) {
      return (i);
    }
  }
  return (-1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Sys_get_stdin>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movq   $0x14,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x20(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
L17:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x1c(%rbp)
  cmpl   $0xa,-0x1c(%rbp)
  jne    L36
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L20
L36:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x20(%rbp),%eax
  cltq
  mov    -0x18(%rbp),%rdx
  sub    $0x1,%rdx
  cmp    %rdx,%rax
  jne    L28
  mov    -0x8(%rbp),%rax
  add    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
L28:
  addl   $0x1,-0x20(%rbp)
  jmp    L17
L20:
  mov    -0x10(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *Sys_get_stdin() {
  size_t const granularity = 20;
  size_t len = granularity;
  char *retVal = malloc(len * (sizeof *retVal));
  int i = 0;
  assert(retVal != NULL);
  while (true) {
    int const c = getchar();
    if (c == '\n') {
      retVal[i] = '\0';
      break;
    }
    retVal[i] = c;
    if (i == (len - 1)) {
      len = len + granularity;
      retVal = realloc(retVal, len * (sizeof *retVal));
      assert(retVal != NULL);
    }
    ++i;
  }
  return retVal;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<Sys_get_stdin>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x14,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  mov    $0x0,%ebx
  mov    $0x14,%r12d
  jmp    L3
L27:
  add    $0x1,%rbx
L3:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%eax
  je     L10
  mov    %al,0x0(%rbp,%rbx,1)
  lea    -0x1(%r12),%rax
  cmp    %rbx,%rax
  jne    L27
  add    $0x14,%r12
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  setne  %dil
  movzbl %dil,%edi
  call   <assert@plt>
  jmp    L27
L10:
  movslq %ebx,%r13
  movb   $0x0,0x0(%rbp,%r13,1)
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *Sys_get_stdin() {
  size_t const granularity = 20;
  size_t len = granularity;
  char *retVal = malloc(len * (sizeof *retVal));
  int i = 0;
  assert(retVal != NULL);
  while (true) {
    int const c = getchar();
    if (c == '\n') {
      retVal[i] = '\0';
      break;
    }
    retVal[i] = c;
    if (i == (len - 1)) {
      len = len + granularity;
      retVal = realloc(retVal, len * (sizeof *retVal));
      assert(retVal != NULL);
    }
    ++i;
  }
  return retVal;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Sys_get_stdin>:
  endbr64
  push   %r12
  mov    $0x14,%edi
  mov    $0x14,%r12d
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbp
  setne  %dil
  xor    %ebx,%ebx
  call   <assert@plt>
  jmp    L31
  nopl   (%rax)
L15:
  add    $0x1,%rbx
L31:
  xor    %eax,%eax
  call   <getchar@plt>
  lea    0x0(%rbp,%rbx,1),%rdx
  cmp    $0xa,%eax
  je     L25
  mov    %al,(%rdx)
  lea    -0x1(%r12),%rax
  cmp    %rax,%rbx
  jne    L15
  add    $0x14,%r12
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <realloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbp
  setne  %dil
  call   <assert@plt>
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L25:
  movb   $0x0,(%rdx)
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *Sys_get_stdin() {
  size_t const granularity = 20;
  size_t len = granularity;
  char *retVal = malloc(len * (sizeof *retVal));
  int i = 0;
  assert(retVal != NULL);
  while (true) {
    int const c = getchar();
    if (c == '\n') {
      retVal[i] = '\0';
      break;
    }
    retVal[i] = c;
    if (i == (len - 1)) {
      len = len + granularity;
      retVal = realloc(retVal, len * (sizeof *retVal));
      assert(retVal != NULL);
    }
    ++i;
  }
  return retVal;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Sys_get_stdin>:
  endbr64
  push   %r12
  mov    $0x14,%edi
  mov    $0x14,%r12d
  push   %rbp
  push   %rbx
  call   <malloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbp
  setne  %dil
  xor    %ebx,%ebx
  call   <assert@plt>
  jmp    L25
  nopl   (%rax)
L11:
  add    $0x1,%rbx
L25:
  xor    %eax,%eax
  call   <getchar@plt>
  lea    0x0(%rbp,%rbx,1),%rdx
  cmp    $0xa,%eax
  je     L33
  mov    %al,(%rdx)
  lea    -0x1(%r12),%rax
  cmp    %rax,%rbx
  jne    L11
  add    $0x14,%r12
  mov    %rbp,%rdi
  mov    %r12,%rsi
  call   <realloc@plt>
  xor    %edi,%edi
  test   %rax,%rax
  mov    %rax,%rbp
  setne  %dil
  call   <assert@plt>
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L33:
  movb   $0x0,(%rdx)
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *Sys_get_stdin() {
  size_t const granularity = 20;
  size_t len = granularity;
  char *retVal = malloc(len * (sizeof *retVal));
  int i = 0;
  assert(retVal != NULL);
  while (true) {
    int const c = getchar();
    if (c == '\n') {
      retVal[i] = '\0';
      break;
    }
    retVal[i] = c;
    if (i == (len - 1)) {
      len = len + granularity;
      retVal = realloc(retVal, len * (sizeof *retVal));
      assert(retVal != NULL);
    }
    ++i;
  }
  return retVal;
}
```
user: What is the c source code of the assembly code below?

```asm
<swap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x20(%rbp),%rax
  mov    -0x4(%rbp),%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void swap(int *a, int *b) {
  int m;
  m = *a;
  *a = *b;
  *b = m;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void swap(int *a, int *b) {
  int m;
  m = *a;
  *a = *b;
  *b = m;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void swap(int *a, int *b) {
  int m;
  m = *a;
  *a = *b;
  *b = m;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<swap>:
  endbr64
  mov    (%rdi),%eax
  mov    (%rsi),%edx
  mov    %edx,(%rdi)
  mov    %eax,(%rsi)
  ret
```

assistant: ```c
void swap(int *a, int *b) {
  int m;
  m = *a;
  *a = *b;
  *b = m;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<kill>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  leave
  ret
```

assistant: ```c
void kill(char *msg) {
  perror(msg);
  exit(1);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<kill>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void kill(char *msg) {
  perror(msg);
  exit(1);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<kill>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void kill(char *msg) {
  perror(msg);
  exit(1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<kill>:
  endbr64
  sub    $0x8,%rsp
  call   <perror@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

assistant: ```c
void kill(char *msg) {
  perror(msg);
  exit(1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<transfrom_base_k>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L13
  mov    -0x18(%rbp),%rax
  lea    D4(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  jmp    L11
L13:
  mov    -0x1c(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %eax,%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <transfrom_base_k>
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  cltd
  idivl  -0x20(%rbp)
  mov    %edx,%ecx
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movslq %ecx,%rax
  lea    D14(%rip),%rcx
  movzbl (%rax,%rcx,1),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L11:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D4
datatype: string
name: unpack_data
arguments: data_label: D14
datatype: string

tool: 

tool: D14: 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ

assistant: ```c
void transfrom_base_k(char *s, int n, int k) {
  if (n == 0) {
    strcpy(s, "");
    return;
  }
  int len;
  transfrom_base_k(s, n / k, k);
  len = strlen(s);
  s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n % k];
  s[len + 1] = '\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
<transfrom_base_k>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %esi,%esi
  je     L19
  mov    %esi,%eax
  mov    %edx,%ecx
  cltd
  idiv   %ecx
  mov    %edx,%ebp
  mov    %ecx,%edx
  mov    %eax,%esi
  call   <transfrom_base_k>
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  movslq %ebp,%rbp
  lea    D2(%rip),%rdx
  movzbl (%rdx,%rbp,1),%edx
  mov    %dl,(%rbx,%rax,1)
  movb   $0x0,0x1(%rbx,%rax,1)
L12:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  lea    D5(%rip),%rsi
  call   <strcpy@plt>
  jmp    L12
```

tool_calls: name: read_data
arguments: data_tag: D2
type_infered: string
name: read_data
arguments: data_tag: D5
type_infered: string

tool: D2: 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ

tool: 

assistant: ```c
void transfrom_base_k(char *s, int n, int k) {
  if (n == 0) {
    strcpy(s, "");
    return;
  }
  int len;
  transfrom_base_k(s, n / k, k);
  len = strlen(s);
  s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n % k];
  s[len + 1] = '\0';
}
```
user: Translate the below assembly code into C programming language:

```asm
<transfrom_base_k>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %esi,%esi
  je     L1
  mov    %esi,%eax
  mov    %edx,%r8d
  cltd
  idiv   %r8d
  movslq %edx,%rbx
  mov    %eax,%esi
  mov    %r8d,%edx
  call   <transfrom_base_k>
  mov    %rbp,%rdi
  call   <strlen@plt>
  lea    D14(%rip),%rdx
  movzbl (%rdx,%rbx,1),%edx
  cltq
  movb   $0x0,0x1(%rbp,%rax,1)
  mov    %dl,0x0(%rbp,%rax,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L1:
  add    $0x8,%rsp
  lea    D0(%rip),%rsi
  pop    %rbx
  pop    %rbp
  jmp    1060 <strcpy@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D14
data_type: string
name: extract_symbolic_data
arguments: label: D0
data_type: string

tool: D14: 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ

tool: 

assistant: ```c
void transfrom_base_k(char *s, int n, int k) {
  if (n == 0) {
    strcpy(s, "");
    return;
  }
  int len;
  transfrom_base_k(s, n / k, k);
  len = strlen(s);
  s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n % k];
  s[len + 1] = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<transfrom_base_k>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  test   %esi,%esi
  je     L20
  mov    %esi,%eax
  mov    %edx,%r8d
  cltd
  idiv   %r8d
  movslq %edx,%rbx
  test   %eax,%eax
  je     L1
  cltd
  idiv   %r8d
  movslq %edx,%rbp
  test   %eax,%eax
  je     L101
  cltd
  idiv   %r8d
  movslq %edx,%r12
  test   %eax,%eax
  je     L7
  cltd
  idiv   %r8d
  movslq %edx,%r13
  test   %eax,%eax
  je     L66
  cltd
  idiv   %r8d
  mov    %edx,0x8(%rsp)
  test   %eax,%eax
  je     L39
  cltd
  lea    D27(%rip),%r14
  idiv   %r8d
  mov    %edx,0xc(%rsp)
  mov    %eax,%esi
  mov    %r8d,%edx
  call   <transfrom_base_k>
  mov    %r15,%rdi
  call   <strlen@plt>
  movslq 0xc(%rsp),%rdx
  cltq
  movzbl (%r14,%rdx,1),%ecx
  movb   $0x0,0x1(%r15,%rax,1)
  mov    %cl,(%r15,%rax,1)
L40:
  mov    %r15,%rdi
  call   <strlen@plt>
  movslq 0x8(%rsp),%rdx
  cltq
  movzbl (%r14,%rdx,1),%ecx
  movb   $0x0,0x1(%r15,%rax,1)
  mov    %cl,(%r15,%rax,1)
L25:
  mov    %r15,%rdi
  call   <strlen@plt>
  movzbl (%r14,%r13,1),%ecx
  cltq
  mov    %cl,(%r15,%rax,1)
  movb   $0x0,0x1(%r15,%rax,1)
L45:
  mov    %r15,%rdi
  call   <strlen@plt>
  movzbl (%r14,%r12,1),%ecx
  cltq
  mov    %cl,(%r15,%rax,1)
  movb   $0x0,0x1(%r15,%rax,1)
L17:
  mov    %r15,%rdi
  call   <strlen@plt>
  movzbl (%r14,%rbp,1),%ecx
  cltq
  mov    %cl,(%r15,%rax,1)
  movb   $0x0,0x1(%r15,%rax,1)
  jmp    L49
L1:
  lea    D56(%rip),%rsi
  lea    D27(%rip),%r14
  call   <strcpy@plt>
L49:
  mov    %r15,%rdi
  call   <strlen@plt>
  movzbl (%r14,%rbx,1),%edx
  cltq
  movb   $0x0,0x1(%r15,%rax,1)
  mov    %dl,(%r15,%rax,1)
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L20:
  add    $0x18,%rsp
  lea    D56(%rip),%rsi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1060 <strcpy@plt>
  nopw   0x0(%rax,%rax,1)
L7:
  lea    D56(%rip),%rsi
  lea    D27(%rip),%r14
  call   <strcpy@plt>
  jmp    L45
  nopl   0x0(%rax,%rax,1)
L101:
  lea    D56(%rip),%rsi
  lea    D27(%rip),%r14
  call   <strcpy@plt>
  jmp    L17
  nopl   0x0(%rax,%rax,1)
L66:
  lea    D56(%rip),%rsi
  lea    D27(%rip),%r14
  call   <strcpy@plt>
  jmp    L25
  nopl   0x0(%rax,%rax,1)
L39:
  lea    D56(%rip),%rsi
  lea    D27(%rip),%r14
  call   <strcpy@plt>
  jmp    L40
```

tool_calls: name: unpack_data
arguments: tag: D27
type_infered: string
name: unpack_data
arguments: tag: D56
type_infered: string

tool: D27: 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ

tool: 

assistant: ```c
void transfrom_base_k(char *s, int n, int k) {
  if (n == 0) {
    strcpy(s, "");
    return;
  }
  int len;
  transfrom_base_k(s, n / k, k);
  len = strlen(s);
  s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n % k];
  s[len + 1] = '\0';
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<strreplc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  jmp    L24
L15:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2c,%al
  jne    L27
  mov    -0x8(%rbp),%rax
  movb   $0x20,(%rax)
L27:
  addq   $0x1,-0x8(%rbp)
L24:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L15
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void strreplc(char *s) {
  for (; *s != '\0'; s++)
    if (*s == ',')
      *s = ' ';
}
```