ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<split>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,%eax
  mov    %edx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    %al,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L64
L39:
  movl   $0x0,-0x4(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0x100,%esi
  mov    %rax,%rdi
  call   <bzero@plt>
  jmp    L40
L26:
  addl   $0x1,-0x8(%rbp)
L40:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  je     L26
  jmp    L9
L59:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cmp    %eax,-0x8(%rbp)
  jge    L3
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    %al,-0x1c(%rbp)
  jne    L59
L3:
  addl   $0x1,-0xc(%rbp)
L64:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jle    L39
  nop
  nop
  leave
  ret
```

assistant: ```c
void split(char *cad, char token, int pos, char devolver[]) {
  int i, j, k;
  i = j = k = 0;
  while (i <= pos) {
    k = 0;
    bzero(devolver, 256);
    for (; cad[j] == token; j++) {
    }
    for (; j < strlen(cad) && cad[j] != token; j++, k++) {
      devolver[k] = cad[j];
    }
    i++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<split>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0xc(%rsp)
  mov    %rcx,(%rsp)
  test   %edx,%edx
  js     L42
  mov    %rdi,%rbp
  mov    %esi,%r13d
  mov    $0x0,%r12d
  mov    $0x0,%r15d
  jmp    L32
L15:
  mov    %al,(%r14,%rbx,1)
  add    $0x1,%rbx
L14:
  mov    %ebx,%r12d
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jle    L27
  movzbl 0x0(%rbp,%rbx,1),%eax
  cmp    %r13b,%al
  jne    L15
L27:
  add    $0x1,%r15d
  cmp    %r15d,0xc(%rsp)
  jl     L42
L32:
  mov    $0x100,%esi
  mov    (%rsp),%rdi
  call   <bzero@plt>
  movslq %r12d,%rax
  cmp    0x0(%rbp,%rax,1),%r13b
  jne    L73
  lea    0x1(%r12),%eax
  cltq
L6:
  mov    %eax,%r12d
  add    $0x1,%rax
  cmp    %r13b,-0x1(%rbp,%rax,1)
  je     L6
L73:
  movslq %r12d,%r12
  mov    %r12,%rbx
  mov    (%rsp),%r14
  sub    %r12,%r14
  jmp    L14
L42:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void split(char *cad, char token, int pos, char devolver[]) {
  int i, j, k;
  i = j = k = 0;
  while (i <= pos) {
    k = 0;
    bzero(devolver, 256);
    for (; cad[j] == token; j++) {
    }
    for (; j < strlen(cad) && cad[j] != token; j++, k++) {
      devolver[k] = cad[j];
    }
    i++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<split>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0xc(%rsp)
  test   %edx,%edx
  js     L0
  movl   $0x0,0x8(%rsp)
  mov    %rdi,%r15
  mov    %rcx,%r14
  mov    %esi,%ebp
  xor    %ebx,%ebx
L35:
  mov    $0x100,%esi
  mov    %r14,%rdi
  call   <bzero@plt>
  movslq %ebx,%rdx
  cmp    (%r15,%rdx,1),%bpl
  jne    L43
  lea    0x1(%rbx),%eax
  cltq
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %eax,%ebx
  add    $0x1,%rax
  cmp    %bpl,-0x1(%r15,%rax,1)
  je     L11
  movslq %ebx,%rdx
L43:
  mov    %r14,%r12
  mov    %rdx,%r13
  sub    %rdx,%r12
  jmp    L16
  nopl   0x0(%rax,%rax,1)
L46:
  movzbl (%r15,%r13,1),%eax
  cmp    %bpl,%al
  je     L21
  mov    %al,(%r12,%r13,1)
  add    $0x1,%r13
L16:
  mov    %r15,%rdi
  mov    %r13d,%ebx
  call   <strlen@plt>
  cmp    %r13d,%eax
  jg     L46
L21:
  addl   $0x1,0x8(%rsp)
  mov    0x8(%rsp),%eax
  cmp    %eax,0xc(%rsp)
  jge    L35
L0:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void split(char *cad, char token, int pos, char devolver[]) {
  int i, j, k;
  i = j = k = 0;
  while (i <= pos) {
    k = 0;
    bzero(devolver, 256);
    for (; cad[j] == token; j++) {
    }
    for (; j < strlen(cad) && cad[j] != token; j++, k++) {
      devolver[k] = cad[j];
    }
    i++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<split>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edx,0xc(%rsp)
  test   %edx,%edx
  js     L38
  movl   $0x0,0x8(%rsp)
  mov    %rdi,%r15
  mov    %rcx,%r14
  mov    %esi,%ebp
  xor    %ebx,%ebx
L51:
  mov    $0x100,%esi
  mov    %r14,%rdi
  call   <bzero@plt>
  movslq %ebx,%rdx
  cmp    (%r15,%rdx,1),%bpl
  jne    L25
  lea    0x1(%rbx),%eax
  cltq
  nopl   0x0(%rax,%rax,1)
L11:
  mov    %eax,%ebx
  add    $0x1,%rax
  cmp    %bpl,-0x1(%r15,%rax,1)
  je     L11
  movslq %ebx,%rdx
L25:
  mov    %r14,%r12
  mov    %rdx,%r13
  sub    %rdx,%r12
  jmp    L0
  nopl   0x0(%rax,%rax,1)
L1:
  movzbl (%r15,%r13,1),%eax
  cmp    %bpl,%al
  je     L29
  mov    %al,(%r12,%r13,1)
  add    $0x1,%r13
L0:
  mov    %r15,%rdi
  mov    %r13d,%ebx
  call   <strlen@plt>
  cmp    %r13d,%eax
  jg     L1
L29:
  addl   $0x1,0x8(%rsp)
  mov    0x8(%rsp),%eax
  cmp    %eax,0xc(%rsp)
  jge    L51
L38:
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void split(char *cad, char token, int pos, char devolver[]) {
  int i, j, k;
  i = j = k = 0;
  while (i <= pos) {
    k = 0;
    bzero(devolver, 256);
    for (; cad[j] == token; j++) {
    }
    for (; j < strlen(cad) && cad[j] != token; j++, k++) {
      devolver[k] = cad[j];
    }
    i++;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<updateOverflow>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %rsi,-0x10(%rbp)
  mov    %al,-0x4(%rbp)
  subq   $0x8,-0x10(%rbp)
  cmpb   $0x0,-0x4(%rbp)
  je     L4
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  or     $0x10000000,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
  jmp    L9
L4:
  mov    -0x10(%rbp),%rax
  mov    (%rax),%eax
  and    $0xefffffff,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,(%rax)
L9:
  nop
  pop    %rbp
  ret
```

assistant: ```c
void updateOverflow(unsigned char ovf, unsigned int *reg) {
  reg -= 2;
  if (ovf) {
    *reg |= (1 << 28);
  } else {
    *reg &= ~(1 << 28);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<updateOverflow>:
  endbr64
  test   %dil,%dil
  je     L2
  mov    -0x8(%rsi),%eax
  or     $0x10000000,%eax
L11:
  mov    %eax,-0x8(%rsi)
  ret
L2:
  mov    -0x8(%rsi),%eax
  and    $0xefffffff,%eax
  jmp    L11
```

assistant: ```c
void updateOverflow(unsigned char ovf, unsigned int *reg) {
  reg -= 2;
  if (ovf) {
    *reg |= (1 << 28);
  } else {
    *reg &= ~(1 << 28);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<updateOverflow>:
  endbr64
  mov    -0x8(%rsi),%eax
  mov    %eax,%edx
  and    $0xefffffff,%eax
  or     $0x10000000,%edx
  test   %dil,%dil
  cmovne %edx,%eax
  mov    %eax,-0x8(%rsi)
  ret
```

assistant: ```c
void updateOverflow(unsigned char ovf, unsigned int *reg) {
  reg -= 2;
  if (ovf) {
    *reg |= (1 << 28);
  } else {
    *reg &= ~(1 << 28);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<updateOverflow>:
  endbr64
  mov    -0x8(%rsi),%eax
  mov    %eax,%edx
  and    $0xefffffff,%eax
  or     $0x10000000,%edx
  test   %dil,%dil
  cmovne %edx,%eax
  mov    %eax,-0x8(%rsi)
  ret
```

assistant: ```c
void updateOverflow(unsigned char ovf, unsigned int *reg) {
  reg -= 2;
  if (ovf) {
    *reg |= (1 << 28);
  } else {
    *reg &= ~(1 << 28);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%eax
  mov    $0x43,%ecx
  mov    $0x42,%edx
  mov    $0x41,%esi
  mov    %eax,%edi
  call   <towers@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D6
type_infered: string

tool: D6: %d

assistant: ```c
int main() {
  int x;
  scanf("%d", &x);
  towers(x, 'A', 'B', 'C');
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  mov    $0x43,%ecx
  mov    $0x42,%edx
  mov    $0x41,%esi
  mov    0x4(%rsp),%edi
  call   <towers@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D7
data_type: string

tool: D7: %d

assistant: ```c
int main() {
  int x;
  scanf("%d", &x);
  towers(x, 'A', 'B', 'C');
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D7(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  mov    $0x43,%ecx
  mov    $0x42,%edx
  mov    $0x41,%esi
  call   <towers@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D7
data_type: string

tool: D7: %d

assistant: ```c
int main() {
  int x;
  scanf("%d", &x);
  towers(x, 'A', 'B', 'C');
  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
  sub    $0x18,%rsp
  lea    D2(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edi
  mov    $0x43,%ecx
  mov    $0x42,%edx
  mov    $0x41,%esi
  call   <towers@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

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

tool: D2: %d

assistant: ```c
int main() {
  int x;
  scanf("%d", &x);
  towers(x, 'A', 'B', 'C');
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<memset28>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xa(%rbp),%rax
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movzwl -0xa(%rbp),%eax
  cmp    $0x101,%ax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <clang_analyzer_eval@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
void memset28() {
  short x;
  memset(&x, 1, sizeof(short));
  clang_analyzer_eval(x == 0x101);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<memset28>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rdi
  mov    $0x2,%edx
  mov    $0x1,%esi
  call   <memset@plt>
  cmpw   $0x101,0x6(%rsp)
  sete   %dil
  movzbl %dil,%edi
  call   <clang_analyzer_eval@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void memset28() {
  short x;
  memset(&x, 1, sizeof(short));
  clang_analyzer_eval(x == 0x101);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<memset28>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rdi
  call   <memset@plt>
  xor    %edi,%edi
  cmpw   $0x101,0x6(%rsp)
  sete   %dil
  call   <clang_analyzer_eval@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void memset28() {
  short x;
  memset(&x, 1, sizeof(short));
  clang_analyzer_eval(x == 0x101);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memset28>:
  endbr64
  sub    $0x18,%rsp
  mov    $0x2,%edx
  mov    $0x1,%esi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x6(%rsp),%rdi
  call   <memset@plt>
  xor    %edi,%edi
  cmpw   $0x101,0x6(%rsp)
  sete   %dil
  call   <clang_analyzer_eval@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x18,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void memset28() {
  short x;
  memset(&x, 1, sizeof(short));
  clang_analyzer_eval(x == 0x101);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ext2_bit_clear>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  sar    $0x5,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  and    $0x1f,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  mov    $0x1,%esi
  mov    %eax,%ecx
  shl    %cl,%esi
  mov    %esi,%eax
  not    %eax
  mov    %eax,%ecx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rsi
  mov    -0x18(%rbp),%rax
  add    %rsi,%rax
  and    %ecx,%edx
  mov    %edx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void ext2_bit_clear(uint32_t *btbl, uint32_t bit_num) {
  int byte = (bit_num >> 5);
  int bit = (bit_num & 0x1F);
  btbl[byte] &= ~(1 << bit);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ext2_bit_clear>:
  endbr64
  mov    %esi,%ecx
  mov    %esi,%eax
  sar    $0x5,%eax
  cltq
  mov    $0xfffffffe,%edx
  rol    %cl,%edx
  and    %edx,(%rdi,%rax,4)
  ret
```

assistant: ```c
void ext2_bit_clear(uint32_t *btbl, uint32_t bit_num) {
  int byte = (bit_num >> 5);
  int bit = (bit_num & 0x1F);
  btbl[byte] &= ~(1 << bit);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ext2_bit_clear>:
  endbr64
  mov    %esi,%eax
  mov    %esi,%ecx
  mov    $0xfffffffe,%edx
  sar    $0x5,%eax
  rol    %cl,%edx
  cltq
  and    %edx,(%rdi,%rax,4)
  ret
```

assistant: ```c
void ext2_bit_clear(uint32_t *btbl, uint32_t bit_num) {
  int byte = (bit_num >> 5);
  int bit = (bit_num & 0x1F);
  btbl[byte] &= ~(1 << bit);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<ext2_bit_clear>:
  endbr64
  mov    %esi,%eax
  mov    %esi,%ecx
  mov    $0xfffffffe,%edx
  sar    $0x5,%eax
  rol    %cl,%edx
  cltq
  and    %edx,(%rdi,%rax,4)
  ret
```

assistant: ```c
void ext2_bit_clear(uint32_t *btbl, uint32_t bit_num) {
  int byte = (bit_num >> 5);
  int bit = (bit_num & 0x1F);
  btbl[byte] &= ~(1 << bit);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<get_new>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L47
L21:
  cmpl   $0x0,-0x10(%rbp)
  je     L67
  cmpl   $0x1,-0x10(%rbp)
  jne    L14
L67:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  sub    $0x1,%edx
  mov    %edx,(%rax)
  jmp    L69
L14:
  mov    -0x10(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  movslq %eax,%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L39
  mov    -0x10(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  add    $0x1,%edx
  mov    %edx,(%rax)
  jmp    L69
L39:
  mov    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
L69:
  addl   $0x1,-0x10(%rbp)
L47:
  mov    -0x10(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L21
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
int *get_new(char *model) {
  int len, i;
  int *new;
  len = strlen(model);
  new = malloc(sizeof(int) * len);
  for (i = 0; i < len; i++) {
    if (i == 0 || i == 1) {
      new[i] = i - 1;
    } else {
      if (model[i - 1] == model[new[i - 1]]) {
        new[i] = new[i - 1] + 1;
      } else {
        new[i] = 0;
      }
    }
  }
  return new;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<get_new>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <strlen@plt>
  mov    %eax,%ebp
  lea    0x0(,%rax,4),%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  jle    L6
  mov    %ebp,%esi
  mov    $0x0,%edx
  jmp    L51
L8:
  lea    -0x1(%rdx),%ecx
  mov    %ecx,(%rax,%rdx,4)
L40:
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  je     L6
L51:
  cmp    $0x1,%edx
  jbe    L8
  mov    -0x4(%rax,%rdx,4),%ecx
  movslq %ecx,%rdi
  movzbl (%rbx,%rdi,1),%edi
  cmp    %dil,-0x1(%rbx,%rdx,1)
  je     L39
  movl   $0x0,(%rax,%rdx,4)
  jmp    L40
L39:
  add    $0x1,%ecx
  mov    %ecx,(%rax,%rdx,4)
  jmp    L40
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int *get_new(char *model) {
  int len, i;
  int *new;
  len = strlen(model);
  new = malloc(sizeof(int) * len);
  for (i = 0; i < len; i++) {
    if (i == 0 || i == 1) {
      new[i] = i - 1;
    } else {
      if (model[i - 1] == model[new[i - 1]]) {
        new[i] = new[i - 1] + 1;
      } else {
        new[i] = 0;
      }
    }
  }
  return new;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<get_new>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  movslq %eax,%rbp
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  jle    L2
  xor    %edx,%edx
  nopl   (%rax)
L14:
  lea    -0x1(%rdx),%ecx
  cmp    $0x1,%rdx
  jbe    L6
  movslq -0x4(%rax,%rdx,4),%rdi
  xor    %ecx,%ecx
  mov    %rdi,%rsi
  movzbl (%rbx,%rdi,1),%edi
  cmp    %dil,-0x1(%rbx,%rdx,1)
  jne    L6
  lea    0x1(%rsi),%ecx
L6:
  mov    %ecx,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%rbp
  jne    L14
L2:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *get_new(char *model) {
  int len, i;
  int *new;
  len = strlen(model);
  new = malloc(sizeof(int) * len);
  for (i = 0; i < len; i++) {
    if (i == 0 || i == 1) {
      new[i] = i - 1;
    } else {
      if (model[i - 1] == model[new[i - 1]]) {
        new[i] = new[i - 1] + 1;
      } else {
        new[i] = 0;
      }
    }
  }
  return new;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<get_new>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  movslq %eax,%rbp
  lea    0x0(,%rbp,4),%edi
  call   <malloc@plt>
  test   %ebp,%ebp
  jle    L36
  xor    %edx,%edx
  nopl   (%rax)
L28:
  lea    -0x1(%rdx),%ecx
  cmp    $0x1,%rdx
  jbe    L26
  movslq -0x4(%rax,%rdx,4),%rdi
  xor    %ecx,%ecx
  mov    %rdi,%rsi
  movzbl (%rbx,%rdi,1),%edi
  cmp    %dil,-0x1(%rbx,%rdx,1)
  jne    L26
  lea    0x1(%rsi),%ecx
L26:
  mov    %ecx,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%rbp
  jne    L28
L36:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int *get_new(char *model) {
  int len, i;
  int *new;
  len = strlen(model);
  new = malloc(sizeof(int) * len);
  for (i = 0; i < len; i++) {
    if (i == 0 || i == 1) {
      new[i] = i - 1;
    } else {
      if (model[i - 1] == model[new[i - 1]]) {
        new[i] = new[i - 1] + 1;
      } else {
        new[i] = 0;
      }
    }
  }
  return new;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<DoStrTest>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x130,%rsp
  mov    %rdi,-0x118(%rbp)
  mov    %rsi,-0x120(%rbp)
  mov    %rdx,-0x128(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x110(%rbp)
  movq   $0x0,-0x108(%rbp)
  movq   $0x0,-0x100(%rbp)
  movq   $0x0,-0xf8(%rbp)
  movq   $0x0,-0xf0(%rbp)
  movq   $0x0,-0xe8(%rbp)
  movq   $0x0,-0xe0(%rbp)
  movq   $0x0,-0xd8(%rbp)
  movq   $0x0,-0xd0(%rbp)
  movq   $0x0,-0xc8(%rbp)
  movq   $0x0,-0xc0(%rbp)
  movq   $0x0,-0xb8(%rbp)
  movq   $0x0,-0xb0(%rbp)
  movq   $0x0,-0xa8(%rbp)
  movq   $0x0,-0xa0(%rbp)
  movq   $0x0,-0x98(%rbp)
  movq   $0x0,-0x90(%rbp)
  movq   $0x0,-0x88(%rbp)
  movq   $0x0,-0x80(%rbp)
  movq   $0x0,-0x78(%rbp)
  movq   $0x0,-0x70(%rbp)
  movq   $0x0,-0x68(%rbp)
  movq   $0x0,-0x60(%rbp)
  movq   $0x0,-0x58(%rbp)
  movq   $0x0,-0x50(%rbp)
  movq   $0x0,-0x48(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movq   $0x0,-0x18(%rbp)
  mov    -0x120(%rbp),%rcx
  mov    -0x118(%rbp),%rdx
  lea    -0x110(%rbp),%rax
  mov    $0x100,%esi
  mov    %rax,%rdi
  call   <Testvsnprintf@plt>
  lea    -0x110(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  lea    0x1(%rax),%rdx
  mov    -0x128(%rbp),%rcx
  lea    -0x110(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  je     L2
  lea    -0x110(%rbp),%rsi
  mov    -0x128(%rbp),%rcx
  mov    -0x118(%rbp),%rdx
  mov    -0x120(%rbp),%rax
  mov    %rsi,%r8
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <Fail@plt>
L2:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
type_guessed: string

tool: D7: ERROR: failed to insert string "%s" into "%s"
Expected "%s" got "%s".


assistant: ```c
void DoStrTest(const char *formatstr, char *param, const char *checkstr) {
  char buf[256] = {0};
  Testvsnprintf(buf, 256, formatstr, param);
  if (memcmp(buf, checkstr, strlen(buf) + 1) != 0) {
    Fail("ERROR: failed to insert string \"%s\" into \"%s\"\n"
         "Expected \"%s\" got \"%s\".\n",
         param, formatstr, checkstr, buf);
  }
}
```