ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strtoull>:
  endbr64
  push   %r14
  mov    %rdi,%rax
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  sub    $0x10,%rsp
  mov    %rdi,0x8(%rsp)
  jmp    L21
  nop
L5:
  add    $0x1,%rax
  mov    %rax,0x8(%rsp)
L21:
  movsbl (%rax),%edi
  call   <isspace@plt>
  test   %rax,%rax
  mov    0x8(%rsp),%rax
  jne    L5
  xor    %r13d,%r13d
  cmpb   $0x2d,(%rax)
  jne    L28
  add    $0x1,%rax
  mov    $0x1,%r13d
  mov    %rax,0x8(%rsp)
L28:
  mov    %ebp,%esi
  lea    0x8(%rsp),%rdi
  xor    %r14d,%r14d
  call   <strtoul_base@plt>
  movslq %eax,%rbp
  mov    0x8(%rsp),%rax
  mov    %rbp,%rbx
  jmp    L48
  xchg   %ax,%ax
L52:
  imul   %rbp,%r14
  mov    %eax,%eax
  add    %rax,%r14
  mov    0x8(%rsp),%rax
  add    $0x1,%rax
  mov    %rax,0x8(%rsp)
L48:
  movsbl (%rax),%edi
  call   <strtoul_charval@plt>
  cmp    %ebx,%eax
  jb     L52
  mov    %r14,%rax
  neg    %rax
  test   %r13d,%r13d
  cmovne %rax,%r14
  test   %r12,%r12
  je     L37
  mov    0x8(%rsp),%rax
  mov    %rax,(%r12)
L37:
  add    $0x10,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
unsigned long long strtoull(const char *p, char **endp, int base) {
  unsigned long long ret = 0;
  int negative = 0;
  unsigned int charval;
  while (isspace(*p))
    p++;
  if (*p == '-') {
    negative = 1;
    p++;
  }
  base = strtoul_base(&p, base);
  while (1) {
    charval = strtoul_charval(*p);
    if (charval >= (unsigned int)base)
      break;
    ret = ((ret * base) + charval);
    p++;
  }
  if (negative)
    ret = -ret;
  if (endp)
    *endp = (char *)p;
  return (ret);
}
```
user: Translate the below assembly code into C programming language:

```asm
<max_dvec>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  cmpl   $0x1,-0x14(%rbp)
  jne    L2
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm0
  jmp    L14
L2:
  movl   $0x1,-0xc(%rbp)
  jmp    L38
L6:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm1
  comisd %xmm1,%xmm0
  jbe    L25
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L25:
  addl   $0x1,-0xc(%rbp)
L38:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L6
  mov    -0x8(%rbp),%rax
  movsd  (%rax),%xmm0
L14:
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double max_dvec(int n, double *x) {
  double *ptr = x;
  if (n == 1)
    return *ptr;
  else {
    for (int i = 1; i < n; ++i) {
      if (x[i] > *ptr)
        ptr = &x[i];
    }
  }
  return *ptr;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<max_dvec>:
  endbr64
  cmp    $0x1,%edi
  je     L21
  jle    L29
  lea    0x8(%rsi),%rax
  lea    -0x2(%rdi),%edx
  lea    0x10(%rsi,%rdx,8),%rdx
L28:
  movsd  (%rax),%xmm0
  ucomisd (%rsi),%xmm0
  cmova  %rax,%rsi
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L28
L29:
  movsd  (%rsi),%xmm0
  ret
L21:
  movsd  (%rsi),%xmm0
  ret
```

assistant: ```c
double max_dvec(int n, double *x) {
  double *ptr = x;
  if (n == 1)
    return *ptr;
  else {
    for (int i = 1; i < n; ++i) {
      if (x[i] > *ptr)
        ptr = &x[i];
    }
  }
  return *ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<max_dvec>:
  endbr64
  movsd  (%rsi),%xmm0
  cmp    $0x1,%edi
  je     L14
  jle    L14
  lea    -0x2(%rdi),%edx
  lea    0x8(%rsi),%rax
  lea    0x10(%rsi,%rdx,8),%rdx
  nopl   0x0(%rax,%rax,1)
L6:
  movsd  (%rax),%xmm1
  add    $0x8,%rax
  maxsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L6
L14:
  ret
```

assistant: ```c
double max_dvec(int n, double *x) {
  double *ptr = x;
  if (n == 1)
    return *ptr;
  else {
    for (int i = 1; i < n; ++i) {
      if (x[i] > *ptr)
        ptr = &x[i];
    }
  }
  return *ptr;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<max_dvec>:
  endbr64
  movsd  (%rsi),%xmm0
  cmp    $0x1,%edi
  je     L4
  jle    L4
  lea    -0x2(%rdi),%edx
  lea    0x8(%rsi),%rax
  lea    0x10(%rsi,%rdx,8),%rdx
  nopl   0x0(%rax,%rax,1)
L14:
  movsd  (%rax),%xmm1
  add    $0x8,%rax
  maxsd  %xmm0,%xmm1
  movapd %xmm1,%xmm0
  cmp    %rax,%rdx
  jne    L14
L4:
  ret
```

assistant: ```c
double max_dvec(int n, double *x) {
  double *ptr = x;
  if (n == 1)
    return *ptr;
  else {
    for (int i = 1; i < n; ++i) {
      if (x[i] > *ptr)
        ptr = &x[i];
    }
  }
  return *ptr;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x78(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x74(%rbp),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x74(%rbp),%edx
  mov    -0x78(%rbp),%eax
  lea    -0x70(%rbp),%rcx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <itooa@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D18
type_guessed: string
name: unpack_data
arguments: data_label: D13
type_guessed: string
name: unpack_data
arguments: data_label: D19
type_guessed: string

tool: D18: Enter the number to be converted : 


tool: D13: %d

tool: D19: Enter the base: 


assistant: ```c
int main() {
  int n, base;
  char str[100];
  printf("Enter the number to be converted : \n");
  scanf("%d", &n);
  printf("Enter the base: \n");
  scanf("%d", &base);
  itooa(n, str, base);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbx
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  lea    D28(%rip),%rdi
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  lea    D13(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D10(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    0x10(%rsp),%rsi
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%edi
  call   <itooa@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  mov    $0x0,%eax
  sub    $0xffffffffffffff80,%rsp
  pop    %rbx
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D28
type: string
name: unpack_data
arguments: data_tag: D13
type: string
name: unpack_data
arguments: data_tag: D10
type: string

tool: D28: Enter the number to be converted : 


tool: D13: %d

tool: D10: Enter the base: 


assistant: ```c
int main() {
  int n, base;
  char str[100];
  printf("Enter the number to be converted : \n");
  scanf("%d", &n);
  printf("Enter the base: \n");
  scanf("%d", &base);
  itooa(n, str, base);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D22(%rip),%rdi
  lea    D23(%rip),%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D25(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%edi
  lea    0x10(%rsp),%rsi
  call   <itooa@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  sub    $0xffffffffffffff80,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D22
data_type: string
name: read_data
arguments: data_label: D23
data_type: string
name: read_data
arguments: data_label: D25
data_type: string

tool: D22: Enter the number to be converted : 


tool: D23: %d

tool: D25: Enter the base: 


assistant: ```c
int main() {
  int n, base;
  char str[100];
  printf("Enter the number to be converted : \n");
  scanf("%d", &n);
  printf("Enter the base: \n");
  scanf("%d", &base);
  itooa(n, str, base);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D6(%rip),%rdi
  lea    D15(%rip),%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x8(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  lea    D20(%rip),%rdi
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  mov    0x8(%rsp),%edi
  lea    0x10(%rsp),%rsi
  call   <itooa@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  sub    $0xffffffffffffff80,%rsp
  xor    %eax,%eax
  pop    %rbp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D6
type_infered: string
name: read_data
arguments: label: D15
type_infered: string
name: read_data
arguments: label: D20
type_infered: string

tool: D6: Enter the number to be converted : 


tool: D15: %d

tool: D20: Enter the base: 


assistant: ```c
int main() {
  int n, base;
  char str[100];
  printf("Enter the number to be converted : \n");
  scanf("%d", &n);
  printf("Enter the base: \n");
  scanf("%d", &base);
  itooa(n, str, base);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<kclrscr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movq   $0xb8000,-0x8(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L17
L16:
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movq   $0xa200a20,(%rax)
  addq   $0x1,-0x10(%rbp)
L17:
  cmpq   $0x3e7,-0x10(%rbp)
  jbe    L16
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void kclrscr() {
  unsigned long *vidmem = (unsigned long *)0xb8000;
  unsigned long i = 0;
  while (i < (40 * 25)) {
    vidmem[i] = 0x0A200A20;
    ++i;
  };
}
```
user: Translate the below assembly code into C programming language:

```asm
<kclrscr>:
  endbr64
  mov    $0xb8000,%eax
L0:
  movq   $0xa200a20,(%rax)
  add    $0x8,%rax
  cmp    $0xb9f40,%rax
  jne    L0
  ret
```

assistant: ```c
void kclrscr() {
  unsigned long *vidmem = (unsigned long *)0xb8000;
  unsigned long i = 0;
  while (i < (40 * 25)) {
    vidmem[i] = 0x0A200A20;
    ++i;
  };
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<kclrscr>:
  endbr64
  mov    $0xb8000,%eax
  nopl   0x0(%rax)
L8:
  movq   $0xa200a20,(%rax)
  add    $0x8,%rax
  cmp    $0xb9f40,%rax
  jne    L8
  ret
```

assistant: ```c
void kclrscr() {
  unsigned long *vidmem = (unsigned long *)0xb8000;
  unsigned long i = 0;
  while (i < (40 * 25)) {
    vidmem[i] = 0x0A200A20;
    ++i;
  };
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<kclrscr>:
  endbr64
  mov    $0xb8000,%eax
  nopl   0x0(%rax)
L7:
  movq   $0xa200a20,(%rax)
  add    $0x8,%rax
  cmp    $0xb9f40,%rax
  jne    L7
  ret
```

assistant: ```c
void kclrscr() {
  unsigned long *vidmem = (unsigned long *)0xb8000;
  unsigned long i = 0;
  while (i < (40 * 25)) {
    vidmem[i] = 0x0A200A20;
    ++i;
  };
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<SplitByLine>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L199
  mov    %rdi,%r13
  mov    %rsi,%rbp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L216
  mov    $0x0,%r14d
  mov    $0x0,%edx
  jmp    L26
L3:
  mov    %rcx,%rdx
L26:
  cmp    $0xa,%al
  sete   %al
  movzbl %al,%eax
  add    %rax,%r14
  lea    0x1(%rdx),%rcx
  movzbl 0x1(%r13,%rdx,1),%eax
  test   %al,%al
  jne    L3
  cmpb   $0xa,0x0(%r13,%rdx,1)
  je     L23
L44:
  add    $0x1,%r14
L23:
  lea    0x0(,%r14,8),%rdi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  test   %r14,%r14
  jne    L207
  jmp    L180
L85:
  sub    $0x1,%rdi
  jmp    L76
L49:
  cmpb   $0xa,0x1(%rax)
  je     L89
L126:
  mov    0x0(%rbp),%rsi
  mov    (%rsi,%r8,1),%rsi
  mov    %dl,(%rsi,%rcx,1)
  add    $0x1,%rcx
  add    $0x1,%rax
L113:
  lea    (%rcx,%rbx,1),%rsi
  mov    %rax,%rdi
  movzbl (%rax),%edx
  cmp    $0xd,%dl
  je     L49
  test   %dl,%dl
  je     L89
  cmp    $0xa,%dl
  jne    L126
L89:
  mov    0x0(%rbp),%rax
  mov    (%rax,%r8,1),%rax
  movb   $0x0,(%rax,%rcx,1)
  cmpb   $0xd,(%rdi)
  sete   %bl
  movzbl %bl,%ebx
  lea    0x1(%rsi,%rbx,1),%rbx
  add    $0x1,%r12
  cmp    %r12,%r14
  je     L180
L207:
  lea    0x0(%r13,%rbx,1),%r15
  movzbl (%r15),%eax
  cmp    $0xa,%al
  je     L212
  mov    $0x0,%edi
  test   %al,%al
  je     L212
L147:
  add    $0x1,%rdi
  lea    (%rdi,%rbx,1),%rdx
  movzbl (%r15,%rdi,1),%eax
  cmp    $0xa,%al
  je     L166
  test   %al,%al
  jne    L147
L166:
  cmpb   $0xd,-0x1(%r13,%rdx,1)
  je     L85
L76:
  add    $0x1,%rdi
  call   <malloc@plt>
  lea    0x0(,%r12,8),%r8
  mov    0x0(%rbp),%rdx
  mov    %rax,(%rdx,%r12,8)
  mov    %r15,%rax
  mov    $0x0,%ecx
  jmp    L113
L212:
  mov    %rbx,%rdx
  mov    $0x0,%edi
  jmp    L166
L199:
  mov    $0x0,%r14d
L180:
  mov    %r14,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L216:
  mov    $0x0,%r14d
  cmpb   $0xa,-0x1(%rdi)
  jne    L44
  mov    $0x0,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%rbp)
  mov    $0x0,%r14d
  jmp    L180
```

assistant: ```c
unsigned long SplitByLine(char *text, char ***list) {
  unsigned long x, y, pos = 0, lines = 0;
  if (text == NULL)
    return 0;
  for (x ^= x; text[x] != 0; x++) {
    if (text[x] == '\n')
      lines++;
  }
  if (text[x - 1] != '\n')
    lines++;
  *list = (char **)malloc(lines * sizeof(char **));
  for (y ^= y; y < lines; y++) {
    for (x ^= x; text[pos + x] != '\n' && text[pos + x] != 0; x++)
      ;
    if (text[pos + x - 1] == '\r')
      x--;
    (*list)[y] = (char *)malloc(x + 1);
    for (x ^= x; !(text[pos + x] == '\r' && text[pos + x + 1] == '\n') && text[pos + x] != '\n' && text[pos + x] != 0; x++)
      *((*list)[y] + x) = text[pos + x];
    *((*list)[y] + x) = 0;
    pos += x;
    if (text[pos++] == '\r')
      pos++;
  }
  return lines;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<SplitByLine>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L56
  movzbl (%rdi),%eax
  mov    %rdi,%r14
  mov    %rsi,%r12
  test   %al,%al
  je     L152
  xor    %r15d,%r15d
  xor    %edx,%edx
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L130:
  mov    %rcx,%rdx
L0:
  cmp    $0xa,%al
  lea    0x1(%rdx),%rcx
  sete   %al
  movzbl %al,%eax
  add    %rax,%r15
  movzbl 0x1(%r14,%rdx,1),%eax
  test   %al,%al
  jne    L130
  cmpb   $0xa,(%r14,%rdx,1)
  je     L20
  add    $0x1,%r15
L20:
  lea    0x0(,%r15,8),%rdi
  call   <malloc@plt>
  mov    %rax,(%r12)
  test   %r15,%r15
  je     L56
L87:
  xor    %ebp,%ebp
  xor    %r13d,%r13d
  xchg   %ax,%ax
L73:
  lea    (%r14,%rbp,1),%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L164
  xor    %edi,%edi
  cmp    $0xa,%al
  je     L164
  nopl   0x0(%rax)
L94:
  add    $0x1,%rdi
  movzbl (%rbx,%rdi,1),%eax
  lea    (%rdi,%rbp,1),%rdx
  cmp    $0xa,%al
  je     L165
  test   %al,%al
  jne    L94
L165:
  cmpb   $0xd,-0x1(%r14,%rdx,1)
  je     L45
  add    $0x1,%rdi
L45:
  call   <malloc@plt>
  mov    (%r12),%rdx
  mov    %rbx,%rsi
  lea    0x1(%rbp),%r9
  lea    0x0(,%r13,8),%r8
  mov    %rax,(%rdx,%r13,8)
  xor    %edx,%edx
  jmp    L92
L117:
  cmp    $0xa,%cl
  je     L204
  test   %cl,%cl
  je     L204
L136:
  mov    %cl,(%rax,%rdx,1)
  mov    (%r12),%rax
  add    $0x1,%rdx
  add    $0x1,%rsi
  mov    (%rax,%r8,1),%rax
L92:
  movzbl (%rsi),%ecx
  lea    (%rdx,%rbp,1),%rdi
  cmp    $0xd,%cl
  jne    L117
  cmpb   $0xa,0x1(%rsi)
  lea    (%r9,%rdx,1),%r10
  jne    L136
  mov    %r10,%rbp
  nopl   (%rax)
L118:
  add    $0x2,%rdi
  movb   $0x0,(%rax,%rdx,1)
  cmpb   $0xd,(%rsi)
  cmove  %rdi,%rbp
  add    $0x1,%r13
  cmp    %r15,%r13
  jne    L73
L51:
  add    $0x8,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L204:
  lea    0x1(%rdi),%rbp
  jmp    L118
L164:
  mov    %rbp,%rdx
  xor    %edi,%edi
  jmp    L165
L129:
  xor    %edi,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
L56:
  xor    %r15d,%r15d
  jmp    L51
L152:
  cmpb   $0xa,-0x1(%rdi)
  je     L129
  mov    $0x8,%edi
  mov    $0x1,%r15d
  call   <malloc@plt>
  mov    %rax,(%r12)
  jmp    L87
```

assistant: ```c
unsigned long SplitByLine(char *text, char ***list) {
  unsigned long x, y, pos = 0, lines = 0;
  if (text == NULL)
    return 0;
  for (x ^= x; text[x] != 0; x++) {
    if (text[x] == '\n')
      lines++;
  }
  if (text[x - 1] != '\n')
    lines++;
  *list = (char **)malloc(lines * sizeof(char **));
  for (y ^= y; y < lines; y++) {
    for (x ^= x; text[pos + x] != '\n' && text[pos + x] != 0; x++)
      ;
    if (text[pos + x - 1] == '\r')
      x--;
    (*list)[y] = (char *)malloc(x + 1);
    for (x ^= x; !(text[pos + x] == '\r' && text[pos + x + 1] == '\n') && text[pos + x] != '\n' && text[pos + x] != 0; x++)
      *((*list)[y] + x) = text[pos + x];
    *((*list)[y] + x) = 0;
    pos += x;
    if (text[pos++] == '\r')
      pos++;
  }
  return lines;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<SplitByLine>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L9
  movzbl (%rdi),%ecx
  mov    %rdi,%r14
  mov    %rsi,%r12
  test   %cl,%cl
  je     L138
  lea    0x1(%rdi),%rax
  xor    %r15d,%r15d
  jmp    L197
  nopl   (%rax)
L54:
  mov    %edx,%ecx
L197:
  xor    %edx,%edx
  cmp    $0xa,%cl
  sete   %dl
  add    $0x1,%rax
  add    %rdx,%r15
  movzbl -0x1(%rax),%edx
  test   %dl,%dl
  jne    L54
  cmp    $0xa,%cl
  je     L44
  add    $0x1,%r15
L44:
  lea    0x0(,%r15,8),%rdi
  call   <malloc@plt>
  mov    %rax,(%r12)
  test   %r15,%r15
  je     L9
L141:
  xor    %ebp,%ebp
  xor    %r13d,%r13d
  nopl   0x0(%rax)
L140:
  lea    (%r14,%rbp,1),%rbx
  movzbl (%rbx),%eax
  test   %al,%al
  je     L25
  xor    %edi,%edi
  cmp    $0xa,%al
  je     L25
  nopl   0x0(%rax)
L81:
  add    $0x1,%rdi
  movzbl (%rbx,%rdi,1),%eax
  lea    (%rdi,%rbp,1),%rdx
  cmp    $0xa,%al
  je     L173
  test   %al,%al
  jne    L81
L173:
  cmpb   $0xd,-0x1(%r14,%rdx,1)
  je     L158
  add    $0x1,%rdi
L158:
  call   <malloc@plt>
  mov    (%r12),%rdx
  mov    %rbx,%rsi
  lea    0x1(%rbp),%r9
  lea    0x0(,%r13,8),%r8
  mov    %rax,(%rdx,%r13,8)
  xor    %edx,%edx
  jmp    L195
L171:
  cmp    $0xa,%cl
  je     L56
  test   %cl,%cl
  je     L56
L191:
  mov    %cl,(%rax,%rdx,1)
  mov    (%r12),%rax
  add    $0x1,%rdx
  add    $0x1,%rsi
  mov    (%rax,%r8,1),%rax
L195:
  movzbl (%rsi),%ecx
  lea    (%rdx,%rbp,1),%rdi
  cmp    $0xd,%cl
  jne    L171
  cmpb   $0xa,0x1(%rsi)
  lea    (%r9,%rdx,1),%r10
  jne    L191
  mov    %r10,%rbp
  nopl   (%rax)
L55:
  add    $0x2,%rdi
  movb   $0x0,(%rax,%rdx,1)
  cmpb   $0xd,(%rsi)
  cmove  %rdi,%rbp
  add    $0x1,%r13
  cmp    %r13,%r15
  jne    L140
L118:
  add    $0x8,%rsp
  mov    %r15,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L56:
  lea    0x1(%rdi),%rbp
  jmp    L55
L25:
  mov    %rbp,%rdx
  xor    %edi,%edi
  jmp    L173
L53:
  xor    %edi,%edi
  call   <malloc@plt>
  mov    %rax,(%r12)
L9:
  xor    %r15d,%r15d
  jmp    L118
L138:
  cmpb   $0xa,-0x1(%rdi)
  je     L53
  mov    $0x8,%edi
  mov    $0x1,%r15d
  call   <malloc@plt>
  mov    %rax,(%r12)
  jmp    L141
```

assistant: ```c
unsigned long SplitByLine(char *text, char ***list) {
  unsigned long x, y, pos = 0, lines = 0;
  if (text == NULL)
    return 0;
  for (x ^= x; text[x] != 0; x++) {
    if (text[x] == '\n')
      lines++;
  }
  if (text[x - 1] != '\n')
    lines++;
  *list = (char **)malloc(lines * sizeof(char **));
  for (y ^= y; y < lines; y++) {
    for (x ^= x; text[pos + x] != '\n' && text[pos + x] != 0; x++)
      ;
    if (text[pos + x - 1] == '\r')
      x--;
    (*list)[y] = (char *)malloc(x + 1);
    for (x ^= x; !(text[pos + x] == '\r' && text[pos + x + 1] == '\n') && text[pos + x] != '\n' && text[pos + x] != 0; x++)
      *((*list)[y] + x) = text[pos + x];
    *((*list)[y] + x) = 0;
    pos += x;
    if (text[pos++] == '\r')
      pos++;
  }
  return lines;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<div_m32>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    $0xf7,%edi
  call   <put8@plt>
  mov    $0x35,%edi
  call   <put8@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <put32@plt>
  nop
  leave
  ret
```

assistant: ```c
void div_m32(void *_m32) {
  unsigned long *m32 = (unsigned long *)_m32;
  put8(0xF7);
  put8(0x35);
  put32((unsigned long)(m32));
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<div_m32>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    $0xf7,%edi
  call   <put8@plt>
  mov    $0x35,%edi
  call   <put8@plt>
  mov    %rbx,%rdi
  call   <put32@plt>
  pop    %rbx
  ret
```

assistant: ```c
void div_m32(void *_m32) {
  unsigned long *m32 = (unsigned long *)_m32;
  put8(0xF7);
  put8(0x35);
  put32((unsigned long)(m32));
}
```
user: Translate the below assembly code into C programming language:

```asm
<div_m32>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  mov    $0xf7,%edi
  call   <put8@plt>
  mov    $0x35,%edi
  call   <put8@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <put32@plt>
```

assistant: ```c
void div_m32(void *_m32) {
  unsigned long *m32 = (unsigned long *)_m32;
  put8(0xF7);
  put8(0x35);
  put32((unsigned long)(m32));
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<div_m32>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  mov    $0xf7,%edi
  call   <put8@plt>
  mov    $0x35,%edi
  call   <put8@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1070 <put32@plt>
```

assistant: ```c
void div_m32(void *_m32) {
  unsigned long *m32 = (unsigned long *)_m32;
  put8(0xF7);
  put8(0x35);
  put32((unsigned long)(m32));
}
```
user: What is the c source code of the assembly code below:

```asm
<lib_chop_filename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x18(%rbp),%rax
  mov    $0x3a,%esi
  mov    %rax,%rdi
  call   <lib_strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L26
  mov    -0x18(%rbp),%rax
  mov    $0x2f,%esi
  mov    %rax,%rdi
  call   <lib_strchr@plt>
  mov    %rax,-0x8(%rbp)
L26:
  cmpq   $0x0,-0x8(%rbp)
  je     L7
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
  mov    -0x28(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  mov    %rdx,(%rax)
  jmp    L29
L7:
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
L29:
  nop
  leave
  ret
```

assistant: ```c
void lib_chop_filename(char *str, char **host, char **file) {
  char *p;
  *host = str;
  p = lib_strchr(str, ':');
  if (!p)
    p = lib_strchr(str, '/');
  if (p) {
    *p++ = '\0';
    *file = p;
  } else {
    *file = 0;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<lib_chop_filename>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rdx,%rbp
  mov    %rdi,(%rsi)
  mov    $0x3a,%esi
  call   <lib_strchr@plt>
  test   %rax,%rax
  je     L23
L24:
  lea    0x1(%rax),%rdx
  movb   $0x0,(%rax)
L14:
  mov    %rdx,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L23:
  mov    $0x2f,%esi
  mov    %rbx,%rdi
  call   <lib_strchr@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  je     L14
  jmp    L24
```

assistant: ```c
void lib_chop_filename(char *str, char **host, char **file) {
  char *p;
  *host = str;
  p = lib_strchr(str, ':');
  if (!p)
    p = lib_strchr(str, '/');
  if (p) {
    *p++ = '\0';
    *file = p;
  } else {
    *file = 0;
  }
}
```