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

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    $0x0,%r12d
  mov    $0x0,%ebx
  lea    D26(%rip),%r14
  lea    D27(%rip),%r13
L14:
  add    $0x1,%ebx
  mov    %ebx,%esi
  mov    %r14,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  add    %eax,%r12d
  add    $0x33,%rbp
  cmp    $0xa,%ebx
  jne    L14
  mov    %r12d,%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L13
  mov    $0x0,%eax
  add    $0x210,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L13:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D26
datatype: string
name: read_data
arguments: tag: D27
datatype: string
name: read_data
arguments: tag: D13
datatype: string

tool: D26: Please enter string #%d: 

tool: D27: %s

tool: D13: The total length of all 10 strings is: %d


assistant: ```c
int main(void) {
  char inputStrings[10][51];
  int i = 0, totalStrLen = 0;
  for (i = 0; i < 10; i++) {
    printf("Please enter string #%d: ", i + 1);
    scanf("%s", inputStrings[i]);
    totalStrLen += strlen(inputStrings[i]);
  }
  printf("The total length of all 10 strings is: %d\n", totalStrLen);
  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   %r14
  lea    D3(%rip),%r14
  push   %r13
  lea    D9(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  nopl   0x0(%rax)
L1:
  add    $0x1,%ebx
  mov    %r14,%rdi
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  add    $0x33,%rbp
  call   <strlen@plt>
  add    %eax,%r12d
  cmp    $0xa,%ebx
  jne    L1
  mov    %r12d,%esi
  lea    D13(%rip),%rdi
  call   <printf@plt>
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x210,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D3
type_guessed: string
name: retrieve_data
arguments: tag: D9
type_guessed: string
name: retrieve_data
arguments: tag: D13
type_guessed: string

tool: D3: Please enter string #%d: 

tool: D9: %s

tool: D13: The total length of all 10 strings is: %d


assistant: ```c
int main(void) {
  char inputStrings[10][51];
  int i = 0, totalStrLen = 0;
  for (i = 0; i < 10; i++) {
    printf("Please enter string #%d: ", i + 1);
    scanf("%s", inputStrings[i]);
    totalStrLen += strlen(inputStrings[i]);
  }
  printf("The total length of all 10 strings is: %d\n", totalStrLen);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  lea    D22(%rip),%r14
  push   %r13
  lea    D27(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x210,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x208(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  nopl   0x0(%rax)
L18:
  add    $0x1,%ebx
  mov    %r14,%rdi
  mov    %ebx,%esi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <scanf@plt>
  mov    %rbp,%rdi
  add    $0x33,%rbp
  call   <strlen@plt>
  add    %eax,%r12d
  cmp    $0xa,%ebx
  jne    L18
  mov    %r12d,%esi
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    0x208(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x210,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D22
type_infered: string
name: extract_symbolic_data
arguments: label: D27
type_infered: string
name: extract_symbolic_data
arguments: label: D3
type_infered: string

tool: D22: Please enter string #%d: 

tool: D27: %s

tool: D3: The total length of all 10 strings is: %d


assistant: ```c
int main(void) {
  char inputStrings[10][51];
  int i = 0, totalStrLen = 0;
  for (i = 0; i < 10; i++) {
    printf("Please enter string #%d: ", i + 1);
    scanf("%s", inputStrings[i]);
    totalStrLen += strlen(inputStrings[i]);
  }
  printf("The total length of all 10 strings is: %d\n", totalStrLen);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<outb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,%eax
  mov    %rsi,-0x10(%rbp)
  mov    %al,-0x4(%rbp)
  movzbl -0x4(%rbp),%eax
  imul   $0x1010101,%eax,%eax
  mov    -0x10(%rbp),%rdx
  mov    %rdx,%rcx
  shl    $0x5,%rcx
  movabs $0xfffffc01c0000000,%rdx
  add    %rcx,%rdx
  mov    %eax,(%rdx)
  mov    $0x0,%eax
  call   <mb@plt>
  nop
  leave
  ret
```

assistant: ```c
extern void outb(unsigned char b, unsigned long addr) {
  *(volatile unsigned int *)((addr << 5) + ((0xfffffc0000000000UL) + 0x1c0000000UL) + 0x00) = b * 0x01010101;
  mb();
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<outb>:
  endbr64
  sub    $0x8,%rsp
  shl    $0x5,%rsi
  movabs $0xfffffc01c0000000,%rax
  add    %rax,%rsi
  movzbl %dil,%edi
  imul   $0x1010101,%edi,%edi
  mov    %edi,(%rsi)
  mov    $0x0,%eax
  call   <mb@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
extern void outb(unsigned char b, unsigned long addr) {
  *(volatile unsigned int *)((addr << 5) + ((0xfffffc0000000000UL) + 0x1c0000000UL) + 0x00) = b * 0x01010101;
  mb();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<outb>:
  endbr64
  movzbl %dil,%edi
  shl    $0x5,%rsi
  movabs $0xfffffc01c0000000,%rax
  imul   $0x1010101,%edi,%edi
  add    %rax,%rsi
  xor    %eax,%eax
  mov    %edi,(%rsi)
  jmp    1050 <mb@plt>
```

assistant: ```c
extern void outb(unsigned char b, unsigned long addr) {
  *(volatile unsigned int *)((addr << 5) + ((0xfffffc0000000000UL) + 0x1c0000000UL) + 0x00) = b * 0x01010101;
  mb();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<outb>:
  endbr64
  movzbl %dil,%edi
  shl    $0x5,%rsi
  movabs $0xfffffc01c0000000,%rax
  imul   $0x1010101,%edi,%edi
  add    %rax,%rsi
  xor    %eax,%eax
  mov    %edi,(%rsi)
  jmp    1050 <mb@plt>
```

assistant: ```c
extern void outb(unsigned char b, unsigned long addr) {
  *(volatile unsigned int *)((addr << 5) + ((0xfffffc0000000000UL) + 0x1c0000000UL) + 0x00) = b * 0x01010101;
  mb();
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<get_A>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L8
L14:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x14(%rbp),%edx
  mov    %edx,%eax
  shl    $0x3,%eax
  lea    (%rax,%rdx,1),%ecx
  mov    -0xc(%rbp),%edx
  mov    %edx,%eax
  add    %eax,%eax
  add    %edx,%eax
  lea    (%rcx,%rax,1),%edx
  mov    -0x18(%rbp),%eax
  add    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L8:
  cmpl   $0x2,-0xc(%rbp)
  jle    L14
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double get_A(const int atom_i, const int cart_i, const double q[3], const double *born) {
  int i;
  double sum;
  sum = 0;
  for (i = 0; i < 3; i++) {
    sum += q[i] * born[atom_i * 9 + i * 3 + cart_i];
  }
  return sum;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<get_A>:
  endbr64
  mov    %rcx,%r8
  mov    %rdx,%rax
  lea    (%rdi,%rdi,8),%ecx
  movslq %ecx,%rcx
  movslq %esi,%rsi
  add    %rsi,%rcx
  lea    (%r8,%rcx,8),%rcx
  add    $0x18,%rdx
  pxor   %xmm1,%xmm1
L2:
  movsd  (%rax),%xmm0
  mulsd  (%rcx),%xmm0
  addsd  %xmm0,%xmm1
  add    $0x8,%rax
  add    $0x18,%rcx
  cmp    %rdx,%rax
  jne    L2
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double get_A(const int atom_i, const int cart_i, const double q[3], const double *born) {
  int i;
  double sum;
  sum = 0;
  for (i = 0; i < 3; i++) {
    sum += q[i] * born[atom_i * 9 + i * 3 + cart_i];
  }
  return sum;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<get_A>:
  endbr64
  lea    (%rdi,%rdi,8),%eax
  movslq %esi,%rsi
  pxor   %xmm1,%xmm1
  cltq
  add    %rsi,%rax
  lea    (%rcx,%rax,8),%rax
  lea    0x18(%rdx),%rcx
L8:
  movsd  (%rdx),%xmm0
  mulsd  (%rax),%xmm0
  add    $0x8,%rdx
  add    $0x18,%rax
  addsd  %xmm0,%xmm1
  cmp    %rdx,%rcx
  jne    L8
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double get_A(const int atom_i, const int cart_i, const double q[3], const double *born) {
  int i;
  double sum;
  sum = 0;
  for (i = 0; i < 3; i++) {
    sum += q[i] * born[atom_i * 9 + i * 3 + cart_i];
  }
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<get_A>:
  endbr64
  movslq %edi,%rax
  movslq %esi,%r8
  movupd (%rdx),%xmm2
  pxor   %xmm1,%xmm1
  lea    (%rax,%rax,8),%rax
  add    %r8,%rax
  movupd 0x10(%rcx,%rax,8),%xmm0
  movlpd (%rcx,%rax,8),%xmm0
  lea    0x6(%rdi,%rdi,8),%eax
  mulpd  %xmm2,%xmm0
  add    %esi,%eax
  cltq
  addsd  %xmm0,%xmm1
  unpckhpd %xmm0,%xmm0
  addsd  %xmm0,%xmm1
  movsd  (%rcx,%rax,8),%xmm0
  mulsd  0x10(%rdx),%xmm0
  addsd  %xmm1,%xmm0
  ret
```

assistant: ```c
double get_A(const int atom_i, const int cart_i, const double q[3], const double *born) {
  int i;
  double sum;
  sum = 0;
  for (i = 0; i < 3; i++) {
    sum += q[i] * born[atom_i * 9 + i * 3 + cart_i];
  }
  return sum;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cmd_insert>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L34
L6:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <do_insert@plt>
  mov    %eax,-0x4(%rbp)
L34:
  cmpl   $0x0,-0x4(%rbp)
  je     L6
  cmpl   $0x0,-0x4(%rbp)
  jle    L39
  mov    -0x4(%rbp),%eax
  jmp    L8
L39:
  mov    $0x0,%eax
L8:
  leave
  ret
```

assistant: ```c
int cmd_insert(struct edna *a) {
  int err = 0;
  while (!err)
    err = do_insert(a);
  if (err > 0)
    return err;
  else
    return 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
<cmd_insert>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
L7:
  mov    %rbx,%rdi
  call   <do_insert@plt>
  test   %eax,%eax
  je     L7
  mov    $0x0,%edx
  cmovs  %edx,%eax
  pop    %rbx
  ret
```

assistant: ```c
int cmd_insert(struct edna *a) {
  int err = 0;
  while (!err)
    err = do_insert(a);
  if (err > 0)
    return err;
  else
    return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<cmd_insert>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %rbx,%rdi
  call   <do_insert@plt>
  test   %eax,%eax
  je     L1
  mov    $0x0,%edx
  pop    %rbx
  cmovs  %edx,%eax
  ret
```

assistant: ```c
int cmd_insert(struct edna *a) {
  int err = 0;
  while (!err)
    err = do_insert(a);
  if (err > 0)
    return err;
  else
    return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<cmd_insert>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  nopl   0x0(%rax,%rax,1)
L1:
  mov    %rbx,%rdi
  call   <do_insert@plt>
  test   %eax,%eax
  je     L1
  mov    $0x0,%edx
  pop    %rbx
  cmovs  %edx,%eax
  ret
```

assistant: ```c
int cmd_insert(struct edna *a) {
  int err = 0;
  while (!err)
    err = do_insert(a);
  if (err > 0)
    return err;
  else
    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
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    D4(%rip),%eax
  test   %eax,%eax
  jne    L29
  movl   $0x1,D4(%rip)
  mov    D14(%rip),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L26
L24:
  subq   $0x8,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rdx
  mov    $0x0,%eax
  call   *%rdx
L26:
  mov    D1(%rip),%rax
  cmp    %rax,-0x8(%rbp)
  ja     L24
L29:
  nop
  leave
  ret
```

assistant: ```c
void __main() {
  static int initialized;
  if (!initialized) {
    typedef void (*pfunc)();
    extern pfunc __ctors[];
    extern pfunc __ctors_end[];
    pfunc *p;
    initialized = 1;
    for (p = __ctors_end; p > __ctors;)
      (*--p)();
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<__main>:
  endbr64
  cmpl   $0x0,D48(%rip)
  jne    L17
  movl   $0x1,D48(%rip)
  mov    D28(%rip),%rax
  cmp    %rax,D50(%rip)
  jbe    L17
  push   %rbx
  mov    D50(%rip),%rbx
L18:
  sub    $0x8,%rbx
  mov    $0x0,%eax
  call   *(%rbx)
  cmp    D28(%rip),%rbx
  ja     L18
  pop    %rbx
  ret
L17:
  ret
```

assistant: ```c
void __main() {
  static int initialized;
  if (!initialized) {
    typedef void (*pfunc)();
    extern pfunc __ctors[];
    extern pfunc __ctors_end[];
    pfunc *p;
    initialized = 1;
    for (p = __ctors_end; p > __ctors;)
      (*--p)();
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<__main>:
  endbr64
  mov    D4(%rip),%eax
  test   %eax,%eax
  jne    L18
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D5(%rip),%rbx
  mov    D25(%rip),%rbp
  movl   $0x1,D4(%rip)
  cmp    %rbp,%rbx
  jbe    L12
  nopl   0x0(%rax)
L22:
  sub    $0x8,%rbx
  xor    %eax,%eax
  call   *(%rbx)
  cmp    %rbp,%rbx
  ja     L22
L12:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L18:
  ret
```

assistant: ```c
void __main() {
  static int initialized;
  if (!initialized) {
    typedef void (*pfunc)();
    extern pfunc __ctors[];
    extern pfunc __ctors_end[];
    pfunc *p;
    initialized = 1;
    for (p = __ctors_end; p > __ctors;)
      (*--p)();
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<__main>:
  endbr64
  mov    D13(%rip),%eax
  test   %eax,%eax
  jne    L0
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    D24(%rip),%rbx
  mov    D4(%rip),%rbp
  movl   $0x1,D13(%rip)
  cmp    %rbp,%rbx
  jbe    L1
  nopl   0x0(%rax)
L5:
  sub    $0x8,%rbx
  xor    %eax,%eax
  call   *(%rbx)
  cmp    %rbp,%rbx
  ja     L5
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L0:
  ret
```

assistant: ```c
void __main() {
  static int initialized;
  if (!initialized) {
    typedef void (*pfunc)();
    extern pfunc __ctors[];
    extern pfunc __ctors_end[];
    pfunc *p;
    initialized = 1;
    for (p = __ctors_end; p > __ctors;)
      (*--p)();
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<multiplyStr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %edx,-0x30(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  je     L12
  cmpl   $0x0,-0x2c(%rbp)
  js     L12
  cmpl   $0x0,-0x30(%rbp)
  jg     L36
L12:
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  jmp    L47
L36:
  movl   $0x1,-0x10(%rbp)
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L0
L8:
  mov    -0xc(%rbp),%eax
  cltd
  idivl  -0x2c(%rbp)
  mov    %edx,%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x11(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x11(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x10(%rbp)
  mov    -0x10(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L0:
  mov    -0x2c(%rbp),%eax
  imul   -0x30(%rbp),%eax
  cmp    %eax,-0xc(%rbp)
  jl     L8
  mov    -0x10(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x8(%rbp),%rax
L47:
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D5
datatype: string

tool: D5: Invalid Input


assistant: ```c
char *multiplyStr(char *str, int strLen, int factor) {
  if ((str == NULL) || (strLen < 0) || (factor < 1)) {
    printf("Invalid Input\n");
    return NULL;
  }
  int len = 1;
  char inp;
  char *output = malloc(sizeof(char));
  int i;
  for (i = 0; i < strLen * factor; i++) {
    inp = *(str + (i % strLen));
    *(output + len - 1) = inp;
    len++;
    output = realloc(output, len);
  }
  *(output + len - 1) = '\0';
  return output;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<multiplyStr>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edx,%r12d
  mov    %esi,%edx
  shr    $0x1f,%edx
  test   %r12d,%r12d
  setle  %al
  or     %dl,%al
  jne    L28
  mov    %rdi,%r13
  mov    %esi,%ebp
  test   %rdi,%rdi
  je     L28
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  imul   %ebp,%r12d
  test   %r12d,%r12d
  jle    L38
  mov    %r12d,%r14d
  mov    $0x0,%ebx
L3:
  mov    %ebx,%eax
  cltd
  idiv   %ebp
  movslq %edx,%rdx
  movzbl 0x0(%r13,%rdx,1),%eax
  mov    %al,(%rdi,%rbx,1)
  lea    0x2(%rbx),%esi
  call   <realloc@plt>
  mov    %rax,%rdi
  add    $0x1,%rbx
  cmp    %r14,%rbx
  jne    L3
  add    $0x1,%r12d
L32:
  movslq %r12d,%r12
  movb   $0x0,-0x1(%rdi,%r12,1)
L10:
  mov    %rdi,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L28:
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  jmp    L10
L38:
  mov    $0x1,%r12d
  jmp    L32
```

tool_calls: name: retrieve_data
arguments: tag: D5
type_guessed: string

tool: D5: Invalid Input


assistant: ```c
char *multiplyStr(char *str, int strLen, int factor) {
  if ((str == NULL) || (strLen < 0) || (factor < 1)) {
    printf("Invalid Input\n");
    return NULL;
  }
  int len = 1;
  char inp;
  char *output = malloc(sizeof(char));
  int i;
  for (i = 0; i < strLen * factor; i++) {
    inp = *(str + (i % strLen));
    *(output + len - 1) = inp;
    len++;
    output = realloc(output, len);
  }
  *(output + len - 1) = '\0';
  return output;
}
```
user: Translate the below assembly code into C programming language:

```asm
<multiplyStr>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  mov    %esi,%edx
  shr    $0x1f,%edx
  test   %r12d,%r12d
  push   %rbp
  setle  %al
  push   %rbx
  or     %dl,%al
  jne    L8
  mov    %rdi,%r14
  test   %rdi,%rdi
  je     L8
  mov    %esi,%ebp
  mov    $0x1,%edi
  call   <malloc@plt>
  imul   %ebp,%r12d
  mov    %rax,%rdi
  test   %r12d,%r12d
  je     L2
  movslq %r12d,%r13
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L37:
  mov    %ebx,%eax
  lea    0x2(%rbx),%esi
  cltd
  idiv   %ebp
  movslq %edx,%rdx
  movzbl (%r14,%rdx,1),%eax
  mov    %al,(%rdi,%rbx,1)
  add    $0x1,%rbx
  call   <realloc@plt>
  mov    %rax,%rdi
  cmp    %r13,%rbx
  jne    L37
  add    $0x1,%r12d
  movslq %r12d,%r12
  lea    -0x1(%rax,%r12,1),%rax
L2:
  movb   $0x0,(%rax)
  mov    %rdi,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  lea    D2(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  pop    %rbx
  pop    %rbp
  mov    %rdi,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
type_infered: string

tool: D2: Invalid Input


assistant: ```c
char *multiplyStr(char *str, int strLen, int factor) {
  if ((str == NULL) || (strLen < 0) || (factor < 1)) {
    printf("Invalid Input\n");
    return NULL;
  }
  int len = 1;
  char inp;
  char *output = malloc(sizeof(char));
  int i;
  for (i = 0; i < strLen * factor; i++) {
    inp = *(str + (i % strLen));
    *(output + len - 1) = inp;
    len++;
    output = realloc(output, len);
  }
  *(output + len - 1) = '\0';
  return output;
}
```
user: What is the c source code of the assembly code below:

```asm
<multiplyStr>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %edx,%r12d
  mov    %esi,%edx
  shr    $0x1f,%edx
  test   %r12d,%r12d
  push   %rbp
  setle  %al
  push   %rbx
  or     %dl,%al
  jne    L21
  mov    %rdi,%r14
  test   %rdi,%rdi
  je     L21
  mov    %esi,%ebp
  mov    $0x1,%edi
  call   <malloc@plt>
  imul   %ebp,%r12d
  mov    %rax,%rdi
  test   %r12d,%r12d
  je     L16
  movslq %r12d,%r13
  xor    %ebx,%ebx
  nopl   0x0(%rax,%rax,1)
L25:
  mov    %ebx,%eax
  lea    0x2(%rbx),%esi
  cltd
  idiv   %ebp
  movslq %edx,%rdx
  movzbl (%r14,%rdx,1),%eax
  mov    %al,(%rdi,%rbx,1)
  add    $0x1,%rbx
  call   <realloc@plt>
  mov    %rax,%rdi
  cmp    %rbx,%r13
  jne    L25
  add    $0x1,%r12d
  movslq %r12d,%r12
  lea    -0x1(%rax,%r12,1),%rax
L16:
  movb   $0x0,(%rax)
  mov    %rdi,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L21:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  pop    %rbx
  pop    %rbp
  mov    %rdi,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D6
type_guessed: string

tool: D6: Invalid Input


assistant: ```c
char *multiplyStr(char *str, int strLen, int factor) {
  if ((str == NULL) || (strLen < 0) || (factor < 1)) {
    printf("Invalid Input\n");
    return NULL;
  }
  int len = 1;
  char inp;
  char *output = malloc(sizeof(char));
  int i;
  for (i = 0; i < strLen * factor; i++) {
    inp = *(str + (i % strLen));
    *(output + len - 1) = inp;
    len++;
    output = realloc(output, len);
  }
  *(output + len - 1) = '\0';
  return output;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<insertion_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L11
L53:
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L47
L14:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L32
  mov    -0x4(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
  subl   $0x1,-0x4(%rbp)
L47:
  cmpl   $0x0,-0x4(%rbp)
  jne    L14
  jmp    L31
L32:
  nop
L31:
  addl   $0x1,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L53
  nop
  leave
  ret
```

assistant: ```c
int insertion_sort(int *elements, int size) {
  int iter = 0;
  int prev = 0;
  for (iter = 0; iter < size; iter++) {
    prev = iter;
    while (prev) {
      if (elements[prev] < elements[prev - 1]) {
        swap(&elements[prev], &elements[prev - 1]);
        prev--;
      } else {
        break;
      }
    }
  }
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<insertion_sort>:
  endbr64
  test   %esi,%esi
  jle    L35
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%r14d
  lea    0x4(%rdi),%r13
  mov    $0x0,%r12d
  jmp    L31
L60:
  add    $0x4,%r13
L31:
  add    $0x1,%r12d
  cmp    %r12d,%r14d
  je     L32
  test   %r12d,%r12d
  je     L60
  mov    %r13,%rbx
L6:
  lea    -0x4(%rbx),%rsi
  mov    -0x4(%rbx),%eax
  cmp    %eax,(%rbx)
  jge    L60
  mov    %rbx,%rdi
  call   <swap@plt>
  sub    $0x4,%rbx
  cmp    %rbp,%rbx
  jne    L6
  jmp    L60
L32:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L35:
  ret
```

assistant: ```c
int insertion_sort(int *elements, int size) {
  int iter = 0;
  int prev = 0;
  for (iter = 0; iter < size; iter++) {
    prev = iter;
    while (prev) {
      if (elements[prev] < elements[prev - 1]) {
        swap(&elements[prev], &elements[prev - 1]);
        prev--;
      } else {
        break;
      }
    }
  }
}
```