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   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    $0x4,%edi
  call   <epoll_create@plt>
  mov    %eax,%ebx
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  mov    D21(%rip),%rdx
  or     (%rdx),%eax
  mov    %eax,(%rsp)
  movq   $0x0,0x8(%rsp)
  mov    %rsp,%rbp
  mov    %rbp,%rcx
  mov    $0x0,%edx
  mov    D3(%rip),%rax
  mov    (%rax),%esi
  mov    %ebx,%edi
  call   <epoll_ctl@plt>
  mov    $0x14,%ecx
  mov    $0x4,%edx
  mov    %rbp,%rsi
  mov    %ebx,%edi
  call   <epoll_wait@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  mov    $0x0,%eax
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  struct epoll_event ev[4];
  int nfds;
  int efd;
  efd = epoll_create(4);
  ev[0].events = EPOLLIN | EPOLLOUT;
  ev[0].data.fd = 0;
  epoll_ctl(efd, EPOLL_CTL_ADD, 0, &(ev[0]));
  if ((nfds = epoll_wait(efd, ev, 4, 20))) {
  }
  return (0);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x4,%edi
  push   %rbp
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  call   <epoll_create@plt>
  mov    D16(%rip),%rdx
  mov    %r12,%rcx
  movq   $0x0,0x8(%rsp)
  mov    %eax,%ebp
  mov    D8(%rip),%rax
  mov    %ebp,%edi
  mov    (%rax),%eax
  or     (%rdx),%eax
  xor    %edx,%edx
  mov    %eax,(%rsp)
  mov    D19(%rip),%rax
  mov    (%rax),%esi
  call   <epoll_ctl@plt>
  mov    $0x14,%ecx
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    $0x4,%edx
  call   <epoll_wait@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  struct epoll_event ev[4];
  int nfds;
  int efd;
  efd = epoll_create(4);
  ev[0].events = EPOLLIN | EPOLLOUT;
  ev[0].data.fd = 0;
  epoll_ctl(efd, EPOLL_CTL_ADD, 0, &(ev[0]));
  if ((nfds = epoll_wait(efd, ev, 4, 20))) {
  }
  return (0);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r12
  mov    $0x4,%edi
  push   %rbp
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  call   <epoll_create@plt>
  mov    D5(%rip),%rdx
  mov    %r12,%rcx
  movq   $0x0,0x8(%rsp)
  mov    %eax,%ebp
  mov    D26(%rip),%rax
  mov    %ebp,%edi
  mov    (%rax),%eax
  or     (%rdx),%eax
  xor    %edx,%edx
  mov    %eax,(%rsp)
  mov    D0(%rip),%rax
  mov    (%rax),%esi
  call   <epoll_ctl@plt>
  mov    $0x14,%ecx
  mov    %r12,%rsi
  mov    %ebp,%edi
  mov    $0x4,%edx
  call   <epoll_wait@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x58,%rsp
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(int argc, char **argv) {
  struct epoll_event ev[4];
  int nfds;
  int efd;
  efd = epoll_create(4);
  ev[0].events = EPOLLIN | EPOLLOUT;
  ev[0].data.fd = 0;
  epoll_ctl(efd, EPOLL_CTL_ADD, 0, &(ev[0]));
  if ((nfds = epoll_wait(efd, ev, 4, 20))) {
  }
  return (0);
}
```
user: What is the c source code of the assembly code below:

```asm
<machowmany>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x830,%rsp
  mov    %rdi,-0x2828(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x2810(%rbp),%rax
  mov    %rax,%rdi
  call   <get_string@plt>
  mov    -0x2828(%rbp),%rax
  mov    %rax,%rdi
  call   <get_string@plt>
  lea    -0x2810(%rbp),%rdx
  mov    -0x2828(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <s_count@plt>
  mov    %eax,-0x2814(%rbp)
  mov    -0x2814(%rbp),%edx
  mov    -0x2828(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <long2str@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
char *machowmany(char *arg) {
  int i;
  char arg1[10240];
  get_string(arg1);
  get_string(arg);
  i = s_count(arg, arg1);
  return (long2str(arg, i));
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<machowmany>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x818,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x2808(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <get_string@plt>
  mov    %rbx,%rdi
  call   <get_string@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <s_count@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <long2str@plt>
  mov    0x2808(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x2818,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *machowmany(char *arg) {
  int i;
  char arg1[10240];
  get_string(arg1);
  get_string(arg);
  i = s_count(arg, arg1);
  return (long2str(arg, i));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<machowmany>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x2808(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %rdi,%rbp
  mov    %r12,%rdi
  call   <get_string@plt>
  mov    %rbp,%rdi
  call   <get_string@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <s_count@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <long2str@plt>
  mov    0x2808(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L7
  add    $0x2818,%rsp
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *machowmany(char *arg) {
  int i;
  char arg1[10240];
  get_string(arg1);
  get_string(arg);
  i = s_count(arg, arg1);
  return (long2str(arg, i));
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<machowmany>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x818,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x2808(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %rdi,%rbp
  mov    %r12,%rdi
  call   <get_string@plt>
  mov    %rbp,%rdi
  call   <get_string@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <s_count@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <long2str@plt>
  mov    0x2808(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x2818,%rsp
  pop    %rbp
  pop    %r12
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *machowmany(char *arg) {
  int i;
  char arg1[10240];
  get_string(arg1);
  get_string(arg);
  i = s_count(arg, arg1);
  return (long2str(arg, i));
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<s_wlf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  movss  %xmm0,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <s_wl32@plt>
  nop
  leave
  ret
```

assistant: ```c
void s_wlf(struct serializer *s, float f) {
  s_wl32(s, *(uint32_t *)&f);
  return;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<s_wlf>:
  endbr64
  sub    $0x8,%rsp
  movd   %xmm0,%esi
  call   <s_wl32@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void s_wlf(struct serializer *s, float f) {
  s_wl32(s, *(uint32_t *)&f);
  return;
}
```
user: What is the c source code of the assembly code below?

```asm
<s_wlf>:
  endbr64
  movd   %xmm0,%esi
  jmp    1050 <s_wl32@plt>
```

assistant: ```c
void s_wlf(struct serializer *s, float f) {
  s_wl32(s, *(uint32_t *)&f);
  return;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<s_wlf>:
  endbr64
  movd   %xmm0,%esi
  jmp    1050 <s_wl32@plt>
```

assistant: ```c
void s_wlf(struct serializer *s, float f) {
  s_wl32(s, *(uint32_t *)&f);
  return;
}
```
user: What is the c source code of the assembly code below:

```asm
<__assert_double_equal>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  movsd  %xmm0,-0x20(%rbp)
  movsd  %xmm1,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    %rdx,-0x38(%rbp)
  movsd  -0x20(%rbp),%xmm0
  ucomisd -0x28(%rbp),%xmm0
  setnp  %al
  mov    $0x0,%edx
  movsd  -0x20(%rbp),%xmm0
  ucomisd -0x28(%rbp),%xmm0
  cmovne %edx,%eax
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  je     L13
  mov    $0x0,%eax
  call   <test_case_assert_pass@plt>
  jmp    L9
L13:
  mov    $0x0,%eax
  call   <test_case_assert_fail@plt>
L9:
  nop
  leave
  ret
```

assistant: ```c
void __assert_double_equal(const char *spec, double expected, double actual, int line, const char *file) {
  int pass = expected == actual;
  if (pass) {
    test_case_assert_pass();
  } else {
    test_case_assert_fail();
  };
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<__assert_double_equal>:
  endbr64
  sub    $0x8,%rsp
  ucomisd %xmm1,%xmm0
  jp     L0
  jne    L0
  mov    $0x0,%eax
  call   <test_case_assert_pass@plt>
L17:
  add    $0x8,%rsp
  ret
L0:
  mov    $0x0,%eax
  call   <test_case_assert_fail@plt>
  jmp    L17
```

assistant: ```c
void __assert_double_equal(const char *spec, double expected, double actual, int line, const char *file) {
  int pass = expected == actual;
  if (pass) {
    test_case_assert_pass();
  } else {
    test_case_assert_fail();
  };
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<__assert_double_equal>:
  endbr64
  xor    %eax,%eax
  ucomisd %xmm1,%xmm0
  jp     L2
  jne    L2
  jmp    1070 <test_case_assert_pass@plt>
  nopl   0x0(%rax,%rax,1)
L2:
  jmp    1060 <test_case_assert_fail@plt>
```

assistant: ```c
void __assert_double_equal(const char *spec, double expected, double actual, int line, const char *file) {
  int pass = expected == actual;
  if (pass) {
    test_case_assert_pass();
  } else {
    test_case_assert_fail();
  };
}
```
user: Translate the below assembly code into C programming language:

```asm
<__assert_double_equal>:
  endbr64
  xor    %eax,%eax
  ucomisd %xmm1,%xmm0
  jp     L16
  jne    L16
  jmp    1070 <test_case_assert_pass@plt>
  nopl   0x0(%rax,%rax,1)
L16:
  jmp    1060 <test_case_assert_fail@plt>
```

assistant: ```c
void __assert_double_equal(const char *spec, double expected, double actual, int line, const char *file) {
  int pass = expected == actual;
  if (pass) {
    test_case_assert_pass();
  } else {
    test_case_assert_fail();
  };
}
```
user: What is the c source code of the assembly code below?

```asm
<spaces>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  jmp    L14
L4:
  mov    $0x20,%edi
  call   <putchar@plt>
L14:
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  test   %eax,%eax
  jne    L4
  nop
  nop
  leave
  ret
```

assistant: ```c
void spaces(n) int n;
{
  while (n--)
    putchar(' ');
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<spaces>:
  endbr64
  push   %rbx
  mov    %edi,%ebx
  test   %edi,%edi
  je     L2
L4:
  mov    $0x20,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L4
L2:
  pop    %rbx
  ret
```

assistant: ```c
void spaces(n) int n;
{
  while (n--)
    putchar(' ');
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<spaces>:
  endbr64
  test   %edi,%edi
  je     L1
  push   %rbx
  mov    %edi,%ebx
  nopl   0x0(%rax,%rax,1)
L0:
  mov    $0x20,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L0
  pop    %rbx
  ret
  nopl   0x0(%rax)
L1:
  ret
```

assistant: ```c
void spaces(n) int n;
{
  while (n--)
    putchar(' ');
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<spaces>:
  endbr64
  test   %edi,%edi
  je     L11
  push   %rbx
  mov    %edi,%ebx
  nopl   0x0(%rax,%rax,1)
L13:
  mov    $0x20,%edi
  call   <putchar@plt>
  sub    $0x1,%ebx
  jne    L13
  pop    %rbx
  ret
  nopl   0x0(%rax)
L11:
  ret
```

assistant: ```c
void spaces(n) int n;
{
  while (n--)
    putchar(' ');
}
```
user: Translate the below assembly code into C programming language:

```asm
<str2intlist>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %rsi,-0x40(%rbp)
  mov    %rdx,-0x48(%rbp)
  movl   $0x0,-0x24(%rbp)
  movq   $0x0,-0x20(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x18(%rbp)
L10:
  mov    -0x38(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,-0x20(%rbp)
  mov    -0x38(%rbp),%rdx
  mov    -0x40(%rbp),%rcx
  mov    -0x20(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <strtok_r@plt>
  mov    %rax,-0x20(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L15
  mov    -0x24(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x48(%rbp),%rax
  mov    (%rax),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,%rdx
  mov    -0x48(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x48(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <atof@plt>
  mov    %eax,(%rbx)
  addl   $0x1,-0x24(%rbp)
  jmp    L10
L15:
  nop
  mov    -0x38(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  mov    %rdx,(%rax)
  mov    -0x24(%rbp),%eax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
unsigned int str2intlist(char **string, const char *sep, int **list) {
  int i = 0;
  char *str = (char *)NULL;
  char *orig_string = *string;
  ;
  while (1) {
    str = *string;
    str = strtok_r(str, sep, string);
    ;
    if (str == NULL)
      break;
    else {
      *list = (int *)realloc(*list, (i + 1) * sizeof(int));
      (*list)[i] = atof(str);
    };
    i++;
  };
  *string = orig_string;
  return i;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<str2intlist>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r12
  mov    %rsi,%r15
  mov    %rdx,%r14
  mov    (%rdi),%rax
  mov    %rax,0x8(%rsp)
  mov    $0x0,%r13d
  jmp    L19
L15:
  lea    0x0(,%rbp,4),%esi
  mov    (%r14),%rdi
  call   <realloc@plt>
  mov    %rax,(%r14)
  lea    -0x4(%rax,%rbp,4),%r13
  mov    %rbx,%rdi
  call   <atof@plt>
  mov    %eax,0x0(%r13)
  mov    %rbp,%r13
L19:
  mov    (%r12),%rdi
  mov    %r12,%rdx
  mov    %r15,%rsi
  call   <strtok_r@plt>
  mov    %rax,%rbx
  lea    0x1(%r13),%rbp
  test   %rax,%rax
  jne    L15
  mov    0x8(%rsp),%rax
  mov    %rax,(%r12)
  mov    %r13d,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned int str2intlist(char **string, const char *sep, int **list) {
  int i = 0;
  char *str = (char *)NULL;
  char *orig_string = *string;
  ;
  while (1) {
    str = *string;
    str = strtok_r(str, sep, string);
    ;
    if (str == NULL)
      break;
    else {
      *list = (int *)realloc(*list, (i + 1) * sizeof(int));
      (*list)[i] = atof(str);
    };
    i++;
  };
  *string = orig_string;
  return i;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<str2intlist>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    (%rdi),%rax
  mov    %rax,0x8(%rsp)
  mov    %rax,%rdi
  jmp    L9
  nopw   0x0(%rax,%rax,1)
L3:
  mov    (%r12),%rdi
  lea    0x0(,%rbx,4),%esi
  call   <realloc@plt>
  mov    %r15,%rdi
  mov    %rax,(%r12)
  lea    -0x4(%rax,%rbx,4),%r14
  call   <atof@plt>
  mov    0x0(%rbp),%rdi
  mov    %eax,(%r14)
L9:
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %ebx,%r14d
  add    $0x1,%rbx
  call   <strtok_r@plt>
  mov    %rax,%r15
  test   %rax,%rax
  jne    L3
  mov    0x8(%rsp),%rax
  mov    %rax,0x0(%rbp)
  add    $0x18,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned int str2intlist(char **string, const char *sep, int **list) {
  int i = 0;
  char *str = (char *)NULL;
  char *orig_string = *string;
  ;
  while (1) {
    str = *string;
    str = strtok_r(str, sep, string);
    ;
    if (str == NULL)
      break;
    else {
      *list = (int *)realloc(*list, (i + 1) * sizeof(int));
      (*list)[i] = atof(str);
    };
    i++;
  };
  *string = orig_string;
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<str2intlist>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rsi,%r13
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x18,%rsp
  mov    (%rdi),%rax
  mov    %rax,0x8(%rsp)
  mov    %rax,%rdi
  jmp    L19
  nopw   0x0(%rax,%rax,1)
L9:
  mov    (%r12),%rdi
  lea    0x0(,%rbx,4),%esi
  call   <realloc@plt>
  mov    %r15,%rdi
  mov    %rax,(%r12)
  lea    -0x4(%rax,%rbx,4),%r14
  call   <atof@plt>
  mov    0x0(%rbp),%rdi
  mov    %eax,(%r14)
L19:
  mov    %rbp,%rdx
  mov    %r13,%rsi
  mov    %ebx,%r14d
  add    $0x1,%rbx
  call   <strtok_r@plt>
  mov    %rax,%r15
  test   %rax,%rax
  jne    L9
  mov    0x8(%rsp),%rax
  mov    %rax,0x0(%rbp)
  add    $0x18,%rsp
  mov    %r14d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned int str2intlist(char **string, const char *sep, int **list) {
  int i = 0;
  char *str = (char *)NULL;
  char *orig_string = *string;
  ;
  while (1) {
    str = *string;
    str = strtok_r(str, sep, string);
    ;
    if (str == NULL)
      break;
    else {
      *list = (int *)realloc(*list, (i + 1) * sizeof(int));
      (*list)[i] = atof(str);
    };
    i++;
  };
  *string = orig_string;
  return i;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L23
L14:
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  -0x14(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x14(%rbp)
  addl   $0x1,-0x10(%rbp)
L23:
  mov    -0x1c(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L14
  mov    -0x1c(%rbp),%eax
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  movss  -0x14(%rbp),%xmm0
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  mov    -0x14(%rbp),%eax
  movd   %eax,%xmm0
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0xc(%rbp),%eax
  movd   %eax,%xmm0
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D27
type_infered: string
name: extract_symbolic_data
arguments: label: D29
type_infered: string
name: extract_symbolic_data
arguments: label: D22
type_infered: string

tool: D27: %d

tool: D29: Sum: %.0f


tool: D22: Avg: %.2f

assistant: ```c
int main() {
  float e;
  int n, i, f;
  float k;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &f);
    e = e + f;
  }
  k = e / n;
  printf("Sum: %.0f\n", e);
  printf("Avg: %.2f", k);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  lea    D24(%rip),%rdi
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  test   %eax,%eax
  jle    L20
  mov    $0x1,%ebx
  lea    D24(%rip),%rbp
L9:
  lea    0x14(%rsp),%rsi
  mov    %rbp,%rdi
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2ssl 0x14(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  add    $0x1,%ebx
  mov    0x10(%rsp),%eax
  cmp    %ebx,%eax
  jge    L9
L20:
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  0xc(%rsp),%xmm2
  movaps %xmm2,%xmm3
  divss  %xmm0,%xmm3
  movd   %xmm3,%ebx
  movaps %xmm2,%xmm0
  lea    D3(%rip),%rdi
  call   <printf@plt>
  movd   %ebx,%xmm0
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D24
data_type: string
name: retrieve_data
arguments: label: D3
data_type: string
name: retrieve_data
arguments: label: D4
data_type: string

tool: D24: %d

tool: D3: Sum: %.0f


tool: D4: Avg: %.2f

assistant: ```c
int main() {
  float e;
  int n, i, f;
  float k;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &f);
    e = e + f;
  }
  k = e / n;
  printf("Sum: %.0f\n", e);
  printf("Avg: %.2f", k);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  lea    D9(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  call   <scanf@plt>
  mov    0x10(%rsp),%eax
  test   %eax,%eax
  jle    L13
  mov    $0x1,%ebx
  lea    0x14(%rsp),%r12
  nopw   0x0(%rax,%rax,1)
L21:
  mov    %r12,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  call   <scanf@plt>
  pxor   %xmm0,%xmm0
  mov    0x10(%rsp),%eax
  cvtsi2ssl 0x14(%rsp),%xmm0
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  cmp    %ebx,%eax
  jge    L21
L13:
  pxor   %xmm0,%xmm0
  movss  0xc(%rsp),%xmm3
  lea    D23(%rip),%rdi
  cvtsi2ss %eax,%xmm0
  movaps %xmm3,%xmm1
  divss  %xmm0,%xmm1
  movaps %xmm3,%xmm0
  movss  %xmm1,0xc(%rsp)
  call   <printf@plt>
  movss  0xc(%rsp),%xmm1
  lea    D22(%rip),%rdi
  movaps %xmm1,%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D9
data_type: string
name: parse_data
arguments: data_tag: D23
data_type: string
name: parse_data
arguments: data_tag: D22
data_type: string

tool: D9: %d

tool: D23: Sum: %.0f


tool: D22: Avg: %.2f

assistant: ```c
int main() {
  float e;
  int n, i, f;
  float k;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &f);
    e = e + f;
  }
  k = e / n;
  printf("Sum: %.0f\n", e);
  printf("Avg: %.2f", k);
  return 0;
}
```