ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<strcat>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r12
  call   <__get_array_size@plt>
  mov    %eax,%ebp
  mov    %r12,%rdi
  call   <__get_array_size@plt>
  cmp    %eax,%ebp
  jge    L16
L7:
  jmp    L7
L16:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *strcat(char *s1, const char *s2) {
  int size1;
  int size2;
  size1 = __get_array_size(s1);
  size2 = __get_array_size(s2);
  if (size2 > size1)
    while (1)
      ;
  return s1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<strcat>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <__get_array_size@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <__get_array_size@plt>
  cmp    %eax,%ebx
  jl     L8
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L8:
  jmp    L8
```

assistant: ```c
char *strcat(char *s1, const char *s2) {
  int size1;
  int size2;
  size1 = __get_array_size(s1);
  size2 = __get_array_size(s2);
  if (size2 > size1)
    while (1)
      ;
  return s1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<strcat>:
  endbr64
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  call   <__get_array_size@plt>
  mov    %rbp,%rdi
  mov    %eax,%ebx
  call   <__get_array_size@plt>
  cmp    %eax,%ebx
  jl     L9
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  jmp    L9
```

assistant: ```c
char *strcat(char *s1, const char *s2) {
  int size1;
  int size2;
  size1 = __get_array_size(s1);
  size2 = __get_array_size(s2);
  if (size2 > size1)
    while (1)
      ;
  return s1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<_getline>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L10
L35:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x8(%rbp)
L10:
  mov    -0x8(%rbp),%eax
  cltq
  cmp    %rax,-0x20(%rbp)
  jbe    L68
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <getc@plt>
  mov    %eax,-0x4(%rbp)
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L68
  cmpl   $0xa,-0x4(%rbp)
  jne    L35
L68:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  jmp    L71
L11:
  mov    D5(%rip),%rax
  mov    (%rax),%eax
  mov    %eax,%edi
  call   <getc@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0xa,-0x4(%rbp)
  jne    L20
  subl   $0x1,-0x8(%rbp)
L20:
  addl   $0x1,-0x8(%rbp)
L71:
  mov    D18(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  je     L62
  cmpl   $0xa,-0x4(%rbp)
  jne    L11
L62:
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int _getline(char *buf, size_t buf_size) {
  int i;
  int c = 0;
  for (i = 0; i < buf_size && (c = getc(stdin)) != EOF && c != '\n'; ++i)
    buf[i] = c;
  buf[i] = '\0';
  while (c != EOF && c != '\n') {
    c = getc(stdin);
    if (c == '\n')
      --i;
    ++i;
  }
  return i;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<_getline>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %rsi,%r13
  test   %rsi,%rsi
  je     L43
  mov    $0x0,%ebx
  mov    D25(%rip),%r15
L82:
  mov    %ebx,%r12d
  mov    %ebx,%ebp
  mov    (%r15),%edi
  call   <getc@plt>
  mov    D22(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L38
  cmp    $0xa,%eax
  je     L38
  mov    %al,(%r14,%rbx,1)
  lea    0x1(%r12),%ebp
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L82
L38:
  movb   $0x0,(%r14,%rbx,1)
  mov    D22(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L91
  cmp    $0xa,%eax
  je     L91
  mov    D25(%rip),%rbx
  jmp    L22
L43:
  mov    %rsi,%rbx
  mov    $0x0,%ebp
  mov    $0x0,%eax
  jmp    L38
L31:
  add    $0x1,%ebp
  mov    D22(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L91
L22:
  mov    (%rbx),%edi
  call   <getc@plt>
  cmp    $0xa,%eax
  jne    L31
L91:
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int _getline(char *buf, size_t buf_size) {
  int i;
  int c = 0;
  for (i = 0; i < buf_size && (c = getc(stdin)) != EOF && c != '\n'; ++i)
    buf[i] = c;
  buf[i] = '\0';
  while (c != EOF && c != '\n') {
    c = getc(stdin);
    if (c == '\n')
      --i;
    ++i;
  }
  return i;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<_getline>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L81
  mov    D34(%rip),%rbp
  mov    D27(%rip),%r12
  mov    %rsi,%r13
  xor    %ebx,%ebx
  jmp    L82
  nopl   0x0(%rax,%rax,1)
L68:
  cmp    $0xa,%eax
  je     L65
  mov    %al,(%r15,%rbx,1)
  add    $0x1,%rbx
  add    $0x1,%r14d
  cmp    %r13,%rbx
  je     L91
L82:
  mov    0x0(%rbp),%edi
  mov    %ebx,%r14d
  call   <getc@plt>
  lea    (%r15,%rbx,1),%rdx
  cmp    %eax,(%r12)
  jne    L68
  cmp    $0xa,%eax
  setne  %cl
L70:
  movb   $0x0,(%rdx)
  cmp    %eax,(%r12)
  je     L28
  test   %cl,%cl
  je     L28
  mov    D34(%rip),%rbp
  jmp    L111
  nopl   0x0(%rax,%rax,1)
L44:
  add    $0x1,%r14d
  cmp    %eax,(%r12)
  je     L28
L111:
  mov    0x0(%rbp),%edi
  call   <getc@plt>
  cmp    $0xa,%eax
  jne    L44
L28:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L91:
  lea    (%r15,%rbx,1),%rdx
  mov    $0x1,%ecx
  jmp    L70
  nopl   0x0(%rax,%rax,1)
L65:
  movb   $0x0,(%rdx)
  jmp    L28
L81:
  mov    D27(%rip),%r12
  mov    %rdi,%rdx
  xor    %r14d,%r14d
  xor    %eax,%eax
  mov    $0x1,%ecx
  jmp    L70
```

assistant: ```c
int _getline(char *buf, size_t buf_size) {
  int i;
  int c = 0;
  for (i = 0; i < buf_size && (c = getc(stdin)) != EOF && c != '\n'; ++i)
    buf[i] = c;
  buf[i] = '\0';
  while (c != EOF && c != '\n') {
    c = getc(stdin);
    if (c == '\n')
      --i;
    ++i;
  }
  return i;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<_getline>:
  endbr64
  push   %r15
  mov    %rdi,%r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  je     L28
  mov    D29(%rip),%rbp
  mov    D8(%rip),%r12
  mov    %rsi,%r13
  xor    %ebx,%ebx
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L42:
  cmp    $0xa,%eax
  je     L2
  mov    %al,(%r15,%rbx,1)
  add    $0x1,%rbx
  add    $0x1,%r14d
  cmp    %rbx,%r13
  je     L76
L18:
  mov    0x0(%rbp),%edi
  mov    %ebx,%r14d
  call   <getc@plt>
  lea    (%r15,%rbx,1),%rdx
  cmp    %eax,(%r12)
  jne    L42
  cmp    $0xa,%eax
  setne  %cl
L109:
  movb   $0x0,(%rdx)
  cmp    %eax,(%r12)
  je     L49
  test   %cl,%cl
  je     L49
  mov    D29(%rip),%rbp
  jmp    L118
  nopl   0x0(%rax,%rax,1)
L37:
  add    $0x1,%r14d
  cmp    %eax,(%r12)
  je     L49
L118:
  mov    0x0(%rbp),%edi
  call   <getc@plt>
  cmp    $0xa,%eax
  jne    L37
L49:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L76:
  lea    (%r15,%r13,1),%rdx
  mov    $0x1,%ecx
  jmp    L109
  nopl   0x0(%rax,%rax,1)
L2:
  movb   $0x0,(%rdx)
  jmp    L49
L28:
  mov    D8(%rip),%r12
  mov    %rdi,%rdx
  xor    %r14d,%r14d
  xor    %eax,%eax
  mov    $0x1,%ecx
  jmp    L109
```

assistant: ```c
int _getline(char *buf, size_t buf_size) {
  int i;
  int c = 0;
  for (i = 0; i < buf_size && (c = getc(stdin)) != EOF && c != '\n'; ++i)
    buf[i] = c;
  buf[i] = '\0';
  while (c != EOF && c != '\n') {
    c = getc(stdin);
    if (c == '\n')
      --i;
    ++i;
  }
  return i;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<vector_subtract>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L14
L1:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm1
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  subsd  %xmm1,%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x4(%rbp)
L14:
  cmpl   $0x2,-0x4(%rbp)
  jle    L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vector_subtract(double u[], double v[], double w[]) {
  for (int i = 0; i < 3; i++)
    w[i] = u[i] - v[i];
}
```
user: What is the c source code of the assembly code below?

```asm
<vector_subtract>:
  endbr64
  mov    $0x0,%eax
L5:
  movsd  (%rdi,%rax,1),%xmm0
  subsd  (%rsi,%rax,1),%xmm0
  movsd  %xmm0,(%rdx,%rax,1)
  add    $0x8,%rax
  cmp    $0x18,%rax
  jne    L5
  ret
```

assistant: ```c
void vector_subtract(double u[], double v[], double w[]) {
  for (int i = 0; i < 3; i++)
    w[i] = u[i] - v[i];
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<vector_subtract>:
  endbr64
  xor    %eax,%eax
L6:
  movsd  (%rdi,%rax,1),%xmm0
  subsd  (%rsi,%rax,1),%xmm0
  movsd  %xmm0,(%rdx,%rax,1)
  add    $0x8,%rax
  cmp    $0x18,%rax
  jne    L6
  ret
```

assistant: ```c
void vector_subtract(double u[], double v[], double w[]) {
  for (int i = 0; i < 3; i++)
    w[i] = u[i] - v[i];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vector_subtract>:
  endbr64
  lea    0x8(%rsi),%rax
  cmp    %rax,%rdx
  je     L7
  lea    0x8(%rdi),%rax
  cmp    %rax,%rdx
  je     L7
  movupd (%rdi),%xmm0
  movupd (%rsi),%xmm1
  subpd  %xmm1,%xmm0
  movups %xmm0,(%rdx)
  movsd  0x10(%rdi),%xmm0
  subsd  0x10(%rsi),%xmm0
  movsd  %xmm0,0x10(%rdx)
  ret
  nopl   (%rax)
L7:
  movsd  (%rdi),%xmm0
  subsd  (%rsi),%xmm0
  movsd  %xmm0,(%rdx)
  movsd  0x8(%rdi),%xmm0
  subsd  0x8(%rsi),%xmm0
  movsd  %xmm0,0x8(%rdx)
  movsd  0x10(%rdi),%xmm0
  subsd  0x10(%rsi),%xmm0
  movsd  %xmm0,0x10(%rdx)
  ret
```

assistant: ```c
void vector_subtract(double u[], double v[], double w[]) {
  for (int i = 0; i < 3; i++)
    w[i] = u[i] - v[i];
}
```
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    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x0,%eax
  call   <othello_INIT@plt>
  movl   $0x1,-0x4(%rbp)
  jmp    L9
L10:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <othello_ARGS@plt>
  addl   $0x1,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L10
  mov    $0x0,%eax
  call   <event_LOOP@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
main(int argc, char **argv) {
  int i;
  othello_INIT();
  for (i = 1; i < argc; i++)
    othello_ARGS(argv[i]);
  event_LOOP();
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %rsi,%r12
  mov    $0x0,%eax
  call   <othello_INIT@plt>
  cmp    $0x1,%ebp
  jle    L6
  lea    0x8(%r12),%rbx
  lea    -0x2(%rbp),%eax
  lea    0x10(%r12,%rax,8),%rbp
L1:
  mov    (%rbx),%rdi
  call   <othello_ARGS@plt>
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L1
L6:
  mov    $0x0,%eax
  call   <event_LOOP@plt>
  mov    $0x0,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
main(int argc, char **argv) {
  int i;
  othello_INIT();
  for (i = 1; i < argc; i++)
    othello_ARGS(argv[i]);
  event_LOOP();
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  call   <othello_INIT@plt>
  cmp    $0x1,%ebp
  jle    L3
  lea    -0x2(%rbp),%eax
  lea    0x8(%r12),%rbx
  lea    0x10(%r12,%rax,8),%rbp
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <othello_ARGS@plt>
  cmp    %rbp,%rbx
  jne    L14
L3:
  xor    %eax,%eax
  call   <event_LOOP@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
main(int argc, char **argv) {
  int i;
  othello_INIT();
  for (i = 1; i < argc; i++)
    othello_ARGS(argv[i]);
  event_LOOP();
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r12
  xor    %eax,%eax
  mov    %rsi,%r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  call   <othello_INIT@plt>
  cmp    $0x1,%ebp
  jle    L14
  lea    -0x2(%rbp),%eax
  lea    0x8(%r12),%rbx
  lea    0x10(%r12,%rax,8),%rbp
  cs nopw 0x0(%rax,%rax,1)
L6:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <othello_ARGS@plt>
  cmp    %rbp,%rbx
  jne    L6
L14:
  xor    %eax,%eax
  call   <event_LOOP@plt>
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
main(int argc, char **argv) {
  int i;
  othello_INIT();
  for (i = 1; i < argc; i++)
    othello_ARGS(argv[i]);
  event_LOOP();
}
```
user: Translate the below assembly code into C programming language:

```asm
<asprintf_portable>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xd0,%rsp
  mov    %rdi,-0xc8(%rbp)
  mov    %rsi,-0xd0(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L26
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L26:
  mov    -0xd0(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb8(%rbp),%edx
  mov    -0xd0(%rbp),%rax
  mov    %edx,%ecx
  mov    %rax,%rdx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <vsnprintf@plt>
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xd0(%rbp),%rdx
  mov    -0xb8(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0xb4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    -0xc8(%rbp),%rdx
  mov    %rax,(%rdx)
  mov    -0xc8(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  je     L12
  mov    -0xb4(%rbp),%eax
  lea    0x1(%rax),%esi
  mov    -0xc8(%rbp),%rax
  mov    (%rax),%rax
  mov    -0xb8(%rbp),%ecx
  mov    -0xd0(%rbp),%rdx
  mov    %rax,%rdi
  call   <vsnprintf@plt>
  jmp    L11
L12:
  mov    $0xffffffff,%eax
L11:
  mov    %eax,-0xb4(%rbp)
  mov    -0xb8(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0xb4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int asprintf_portable(char **str, const char *fmt, ...) {
  int len;
  va_list ap;
  va_start(ap, fmt);
  len = vsnprintf(NULL, 0, fmt, ap);
  va_end(ap);
  va_start(ap, fmt);
  *str = malloc(len + 1);
  len = (*str == NULL) ? -1 : vsnprintf(*str, len + 1, fmt, ap);
  va_end(ap);
  return len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<asprintf_portable>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%r12
  mov    $0x0,%ebx
  mov    %ebx,%edi
  call   <va_start@plt>
  mov    %ebx,%ecx
  mov    %r12,%rdx
  mov    $0x0,%esi
  mov    $0x0,%edi
  call   <vsnprintf@plt>
  mov    %eax,%ebp
  mov    %ebx,%edi
  call   <va_end@plt>
  mov    %r12,%rsi
  mov    %ebx,%edi
  call   <va_start@plt>
  add    $0x1,%ebp
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,0x0(%r13)
  test   %rax,%rax
  je     L16
  mov    %rax,%rdi
  mov    %ebx,%ecx
  mov    %r12,%rdx
  mov    %ebp,%esi
  call   <vsnprintf@plt>
  mov    %eax,%ebp
L13:
  mov    $0x0,%ebx
  mov    %ebx,%edi
  call   <va_end@plt>
  mov    %ebp,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  mov    $0xffffffff,%ebp
  jmp    L13
```

assistant: ```c
int asprintf_portable(char **str, const char *fmt, ...) {
  int len;
  va_list ap;
  va_start(ap, fmt);
  len = vsnprintf(NULL, 0, fmt, ap);
  va_end(ap);
  va_start(ap, fmt);
  *str = malloc(len + 1);
  len = (*str == NULL) ? -1 : vsnprintf(*str, len + 1, fmt, ap);
  va_end(ap);
  return len;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<asprintf_portable>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r13,%rdx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  add    $0x1,%r12d
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L11
  mov    %r12d,%esi
  mov    %rax,%rdi
  mov    %ebp,%ecx
  mov    %r13,%rdx
  call   <vsnprintf@plt>
  mov    %eax,%r12d
L14:
  xor    %edi,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  mov    $0xffffffff,%r12d
  jmp    L14
```

assistant: ```c
int asprintf_portable(char **str, const char *fmt, ...) {
  int len;
  va_list ap;
  va_start(ap, fmt);
  len = vsnprintf(NULL, 0, fmt, ap);
  va_end(ap);
  va_start(ap, fmt);
  *str = malloc(len + 1);
  len = (*str == NULL) ? -1 : vsnprintf(*str, len + 1, fmt, ap);
  va_end(ap);
  return len;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<asprintf_portable>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  mov    %ebp,%edi
  sub    $0x8,%rsp
  call   <va_start@plt>
  mov    %ebp,%ecx
  mov    %r13,%rdx
  xor    %esi,%esi
  xor    %edi,%edi
  call   <vsnprintf@plt>
  mov    %ebp,%edi
  mov    %eax,%r12d
  call   <va_end@plt>
  add    $0x1,%r12d
  mov    %r13,%rsi
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %r12d,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  test   %rax,%rax
  je     L17
  mov    %r12d,%esi
  mov    %rax,%rdi
  mov    %ebp,%ecx
  mov    %r13,%rdx
  call   <vsnprintf@plt>
  mov    %eax,%r12d
L18:
  xor    %edi,%edi
  call   <va_end@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax,%rax,1)
L17:
  mov    $0xffffffff,%r12d
  jmp    L18
```

assistant: ```c
int asprintf_portable(char **str, const char *fmt, ...) {
  int len;
  va_list ap;
  va_start(ap, fmt);
  len = vsnprintf(NULL, 0, fmt, ap);
  va_end(ap);
  va_start(ap, fmt);
  *str = malloc(len + 1);
  len = (*str == NULL) ? -1 : vsnprintf(*str, len + 1, fmt, ap);
  va_end(ap);
  return len;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<hint_msg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D40(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: tag: D41
datatype: string
name: read_data
arguments: tag: D40
datatype: string
name: read_data
arguments: tag: D7
datatype: string
name: read_data
arguments: tag: D18
datatype: string
name: read_data
arguments: tag: D45
datatype: string

tool: D41: Welcome;


tool: D40: This is a basic program for adding, subtracting, and multiplying polynomials.


tool: D7: A polynomial is a function, that takes the general form of:
	->	a(n)*x^(n) + a(n-1)*x^(n-1) + a(n-2)*x^(n-2) + ... + a(0)
Where:	a(i) is an arbitrary constant, positive or negative, with or without a fraction, representing the coefficient of the unknown X.
	n is a positive integer, representing the power of the unknown X.



tool: D18: 	*Notes on input*
-> Order of the terms does not need to be decending or ascending based on it's power.
-> Presence of the brackets { ( and ) } is not neccessary.
-> Presence of '*' (multiplied by) sign between the coefficient and the unknow X is not neccessary.
-> Presence of '^' (raised to the power of) sign is required. Otherwise the input would be concidered incorrect.



tool: D45: 		****


assistant: ```c
void hint_msg() {
  printf("Welcome;\n");
  printf("This is a basic program for adding, subtracting, and multiplying polynomials.\n");
  printf("A polynomial is a function, that takes the general form of:\n\t->\ta(n)*x^(n) + a(n-1)*x^(n-1) + a(n-2)*x^(n-2) + ... + a(0)\nWhere:\ta(i) is an arbitrary constant, positive or negative, with or without a fraction, representing the coefficient of the unknown X.\n\tn is a positive integer, representing the power of the unknown X.\n\n");
  printf("\t*Notes on input*\n-> Order of the terms does not need to be decending or ascending based on it's power.\n-> Presence of the brackets { ( and ) } is not neccessary.\n-> Presence of '*' (multiplied by) sign between the coefficient and the unknow X is not neccessary.\n-> Presence of '^' (raised to the power of) sign is required. Otherwise the input would be concidered incorrect.\n\n");
  printf("\t\t****\n");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hint_msg>:
  endbr64
  sub    $0x8,%rsp
  lea    D38(%rip),%rdi
  call   <printf@plt>
  lea    D30(%rip),%rdi
  call   <printf@plt>
  lea    D37(%rip),%rdi
  call   <printf@plt>
  lea    D15(%rip),%rdi
  call   <printf@plt>
  lea    D48(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D38
type_infered: string
name: parse_data
arguments: tag: D30
type_infered: string
name: parse_data
arguments: tag: D37
type_infered: string
name: parse_data
arguments: tag: D15
type_infered: string
name: parse_data
arguments: tag: D48
type_infered: string

tool: D38: Welcome;


tool: D30: This is a basic program for adding, subtracting, and multiplying polynomials.


tool: D37: A polynomial is a function, that takes the general form of:
	->	a(n)*x^(n) + a(n-1)*x^(n-1) + a(n-2)*x^(n-2) + ... + a(0)
Where:	a(i) is an arbitrary constant, positive or negative, with or without a fraction, representing the coefficient of the unknown X.
	n is a positive integer, representing the power of the unknown X.



tool: D15: 	*Notes on input*
-> Order of the terms does not need to be decending or ascending based on it's power.
-> Presence of the brackets { ( and ) } is not neccessary.
-> Presence of '*' (multiplied by) sign between the coefficient and the unknow X is not neccessary.
-> Presence of '^' (raised to the power of) sign is required. Otherwise the input would be concidered incorrect.



tool: D48: 		****


assistant: ```c
void hint_msg() {
  printf("Welcome;\n");
  printf("This is a basic program for adding, subtracting, and multiplying polynomials.\n");
  printf("A polynomial is a function, that takes the general form of:\n\t->\ta(n)*x^(n) + a(n-1)*x^(n-1) + a(n-2)*x^(n-2) + ... + a(0)\nWhere:\ta(i) is an arbitrary constant, positive or negative, with or without a fraction, representing the coefficient of the unknown X.\n\tn is a positive integer, representing the power of the unknown X.\n\n");
  printf("\t*Notes on input*\n-> Order of the terms does not need to be decending or ascending based on it's power.\n-> Presence of the brackets { ( and ) } is not neccessary.\n-> Presence of '*' (multiplied by) sign between the coefficient and the unknow X is not neccessary.\n-> Presence of '^' (raised to the power of) sign is required. Otherwise the input would be concidered incorrect.\n\n");
  printf("\t\t****\n");
}
```