ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<good_echo>:
  endbr64
  push   %rbp
  mov    $0x5,%esi
  push   %rbx
  sub    $0x18,%rsp
  mov    D4(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rbp
  mov    (%rbx),%edx
  mov    %rbp,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L11
  mov    D26(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <fputs@plt>
  mov    D14(%rip),%rdx
  cmp    %eax,(%rdx)
  sete   %al
  movzbl %al,%eax
  neg    %eax
L20:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L11:
  mov    (%rbx),%edi
  call   <feof@plt>
  cmp    $0x1,%eax
  sbb    %eax,%eax
  jmp    L20
L15:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int good_echo(void) {
  char s[5];
  char *ret;
  ret = fgets(s, sizeof(s), stdin);
  if (ret == NULL) {
    if (!feof(stdin))
      return -1;
    goto done;
  }
  int out;
  out = fputs(s, stdout);
  if (out == EOF)
    return -1;
done:
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<good_echo>:
  endbr64
  push   %rbp
  mov    $0x5,%esi
  push   %rbx
  sub    $0x18,%rsp
  mov    D26(%rip),%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x3(%rsp),%rbp
  mov    (%rbx),%edx
  mov    %rbp,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  je     L19
  mov    D27(%rip),%rax
  mov    %rbp,%rdi
  mov    (%rax),%esi
  call   <fputs@plt>
  mov    D12(%rip),%rdx
  cmp    %eax,(%rdx)
  sete   %al
  movzbl %al,%eax
  neg    %eax
L0:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L17
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  mov    (%rbx),%edi
  call   <feof@plt>
  cmp    $0x1,%eax
  sbb    %eax,%eax
  jmp    L0
L17:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int good_echo(void) {
  char s[5];
  char *ret;
  ret = fgets(s, sizeof(s), stdin);
  if (ret == NULL) {
    if (!feof(stdin))
      return -1;
    goto done;
  }
  int out;
  out = fputs(s, stdout);
  if (out == EOF)
    return -1;
done:
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0xc(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <balancearCursos@plt>
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  struct Universidad *pucv;
  int sigla = 0;
  balancearCursos(pucv);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <balancearCursos@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  struct Universidad *pucv;
  int sigla = 0;
  balancearCursos(pucv);
  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
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <balancearCursos@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  struct Universidad *pucv;
  int sigla = 0;
  balancearCursos(pucv);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <balancearCursos@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  struct Universidad *pucv;
  int sigla = 0;
  balancearCursos(pucv);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<dumpSS>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L11
L28:
  mov    -0x18(%rbp),%rax
  lea    0x2(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzwl (%rax),%eax
  cwtl
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addq   $0x1,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  and    $0x7,%eax
  test   %rax,%rax
  jne    L11
  mov    $0xa,%edi
  call   <putchar@plt>
L11:
  mov    -0x8(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jl     L28
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D1
data_type: string

tool: D1: %7d 

assistant: ```c
void dumpSS(short *ss, long vm) {
  long index = 0;
  while (index < vm) {
    printf("%7d ", *(ss++));
    if ((++index) % 8 == 0)
      printf("\n");
  }
  printf("\n");
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<dumpSS>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  test   %rsi,%rsi
  jle    L10
  mov    %rdi,%r13
  mov    %rsi,%rbp
  movswl (%rdi),%esi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x1,%ebx
  lea    D3(%rip),%r12
L8:
  cmp    %rbx,%rbp
  je     L10
  movswl 0x0(%r13,%rbx,2),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  test   $0x7,%bl
  jne    L8
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L8
L10:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: label: D3
type: string

tool: D3: %7d 

assistant: ```c
void dumpSS(short *ss, long vm) {
  long index = 0;
  while (index < vm) {
    printf("%7d ", *(ss++));
    if ((++index) % 8 == 0)
      printf("\n");
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dumpSS>:
  endbr64
  test   %rsi,%rsi
  jle    L37
  push   %r13
  lea    D2(%rip),%r13
  xor    %eax,%eax
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  movswl (%rdi),%esi
  mov    %r13,%rdi
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L13:
  cmp    %rbx,%rbp
  je     L18
L23:
  movswl (%r12,%rbx,2),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  test   $0x7,%bl
  jne    L13
  mov    $0xa,%edi
  call   <putchar@plt>
  cmp    %rbx,%rbp
  jne    L23
L18:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L37:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

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

tool: D2: %7d 

assistant: ```c
void dumpSS(short *ss, long vm) {
  long index = 0;
  while (index < vm) {
    printf("%7d ", *(ss++));
    if ((++index) % 8 == 0)
      printf("\n");
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<dumpSS>:
  endbr64
  test   %rsi,%rsi
  jle    L32
  push   %r13
  lea    D9(%rip),%r13
  xor    %eax,%eax
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x8,%rsp
  movswl (%rdi),%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    $0x1,%rbp
  je     L23
  nopl   0x0(%rax,%rax,1)
L29:
  movswl (%r12,%rbx,2),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  test   $0x7,%bl
  jne    L17
  mov    $0xa,%edi
  call   <putchar@plt>
L17:
  cmp    %rbx,%rbp
  jne    L29
L23:
  add    $0x8,%rsp
  mov    $0xa,%edi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L32:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: parse_data
arguments: tag: D9
type: string

tool: D9: %7d 

assistant: ```c
void dumpSS(short *ss, long vm) {
  long index = 0;
  while (index < vm) {
    printf("%7d ", *(ss++));
    if ((++index) % 8 == 0)
      printf("\n");
  }
  printf("\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<transfer_VYU888_to_YUVA8888>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  mov    -0x10(%rbp),%rdx
  movzbl 0x1(%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl 0x2(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x2(%rax),%rdx
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  add    $0x3,%rax
  movb   $0xff,(%rax)
  mov    -0x8(%rbp),%rax
  mov    (%rax),%rax
  lea    0x4(%rax),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void transfer_VYU888_to_YUVA8888(unsigned char *(*output), unsigned char *input) {
  (*output)[0] = input[1];
  (*output)[1] = input[2];
  (*output)[2] = input[0];
  (*output)[3] = 0xff;
  (*output) += 4;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<transfer_VYU888_to_YUVA8888>:
  endbr64
  mov    (%rdi),%rax
  movzbl 0x1(%rsi),%edx
  mov    %dl,(%rax)
  mov    (%rdi),%rax
  movzbl 0x2(%rsi),%edx
  mov    %dl,0x1(%rax)
  mov    (%rdi),%rax
  movzbl (%rsi),%edx
  mov    %dl,0x2(%rax)
  mov    (%rdi),%rax
  movb   $0xff,0x3(%rax)
  addq   $0x4,(%rdi)
  ret
```

assistant: ```c
void transfer_VYU888_to_YUVA8888(unsigned char *(*output), unsigned char *input) {
  (*output)[0] = input[1];
  (*output)[1] = input[2];
  (*output)[2] = input[0];
  (*output)[3] = 0xff;
  (*output) += 4;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<transfer_VYU888_to_YUVA8888>:
  endbr64
  mov    (%rdi),%rax
  movzbl 0x1(%rsi),%edx
  mov    %dl,(%rax)
  mov    (%rdi),%rax
  movzbl 0x2(%rsi),%edx
  mov    %dl,0x1(%rax)
  mov    (%rdi),%rax
  movzbl (%rsi),%edx
  mov    %dl,0x2(%rax)
  mov    (%rdi),%rax
  movb   $0xff,0x3(%rax)
  addq   $0x4,(%rdi)
  ret
```

assistant: ```c
void transfer_VYU888_to_YUVA8888(unsigned char *(*output), unsigned char *input) {
  (*output)[0] = input[1];
  (*output)[1] = input[2];
  (*output)[2] = input[0];
  (*output)[3] = 0xff;
  (*output) += 4;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<transfer_VYU888_to_YUVA8888>:
  endbr64
  mov    (%rdi),%rax
  movzbl 0x1(%rsi),%edx
  mov    %dl,(%rax)
  mov    (%rdi),%rax
  movzbl 0x2(%rsi),%edx
  mov    %dl,0x1(%rax)
  mov    (%rdi),%rax
  movzbl (%rsi),%edx
  mov    %dl,0x2(%rax)
  mov    (%rdi),%rax
  movb   $0xff,0x3(%rax)
  addq   $0x4,(%rdi)
  ret
```

assistant: ```c
void transfer_VYU888_to_YUVA8888(unsigned char *(*output), unsigned char *input) {
  (*output)[0] = input[1];
  (*output)[1] = input[2];
  (*output)[2] = input[0];
  (*output)[3] = 0xff;
  (*output) += 4;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strcasecmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x20,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%r12
  mov    -0x30(%rbp),%rbx
  cmp    %rbx,%r12
  jne    L78
  mov    $0x0,%eax
  jmp    L5
L78:
  movzbl (%r12),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isupper@plt>
  test   %rax,%rax
  je     L34
  movzbl (%r12),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  jmp    L24
L34:
  movzbl (%r12),%eax
L24:
  mov    %al,-0x12(%rbp)
  movzbl (%rbx),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <isupper@plt>
  test   %rax,%rax
  je     L42
  movzbl (%rbx),%eax
  movzbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  jmp    L18
L42:
  movzbl (%rbx),%eax
L18:
  mov    %al,-0x11(%rbp)
  cmpb   $0x0,-0x12(%rbp)
  je     L17
  add    $0x1,%r12
  add    $0x1,%rbx
  movzbl -0x12(%rbp),%eax
  cmp    -0x11(%rbp),%al
  je     L78
  jmp    L75
L17:
  nop
L75:
  movzbl -0x12(%rbp),%eax
  movzbl -0x11(%rbp),%edx
  sub    %edx,%eax
L5:
  add    $0x20,%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
int strcasecmp(const char *s1, const char *s2) {
  register const unsigned char *p1 = (const unsigned char *)s1;
  register const unsigned char *p2 = (const unsigned char *)s2;
  unsigned char c1, c2;
  if (p1 == p2 || 0)
    return 0;
  do {
    c1 = (isupper(*p1) ? tolower(*p1) : (*p1));
    c2 = (isupper(*p2) ? tolower(*p2) : (*p2));
    if (0 || c1 == '\0')
      break;
    ++p1;
    ++p2;
  } while (c1 == c2);
  return c1 - c2;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<strcasecmp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  cmp    %rsi,%rdi
  jne    L4
  mov    $0x0,%eax
  jmp    L70
L38:
  movzbl 0x0(%rbp),%edi
  call   <tolower@plt>
  mov    %eax,%r12d
  jmp    L78
L24:
  movzbl (%rbx),%edx
L41:
  test   %r12b,%r12b
  je     L29
  add    $0x1,%rbp
  add    $0x1,%rbx
  cmp    %dl,%r12b
  jne    L29
L4:
  movzbl 0x0(%rbp),%edi
  call   <isupper@plt>
  test   %rax,%rax
  jne    L38
  movzbl 0x0(%rbp),%r12d
L78:
  movzbl (%rbx),%edi
  call   <isupper@plt>
  test   %rax,%rax
  je     L24
  movzbl (%rbx),%edi
  call   <tolower@plt>
  mov    %eax,%edx
  jmp    L41
L29:
  movzbl %r12b,%eax
  movzbl %dl,%edx
  sub    %edx,%eax
L70:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *s1, const char *s2) {
  register const unsigned char *p1 = (const unsigned char *)s1;
  register const unsigned char *p2 = (const unsigned char *)s2;
  unsigned char c1, c2;
  if (p1 == p2 || 0)
    return 0;
  do {
    c1 = (isupper(*p1) ? tolower(*p1) : (*p1));
    c2 = (isupper(*p2) ? tolower(*p2) : (*p2));
    if (0 || c1 == '\0')
      break;
    ++p1;
    ++p2;
  } while (c1 == c2);
  return c1 - c2;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<strcasecmp>:
  endbr64
  cmp    %rsi,%rdi
  je     L61
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L35
  nopl   0x0(%rax)
L15:
  movzbl (%rbx),%edi
  movzbl 0x0(%rbp),%r12d
  call   <isupper@plt>
  test   %rax,%rax
  jne    L20
L60:
  movzbl (%rbx),%edx
  test   %r12b,%r12b
  je     L21
L10:
  add    $0x1,%rbp
  add    $0x1,%rbx
  cmp    %dl,%r12b
  jne    L3
L35:
  movzbl 0x0(%rbp),%edi
  call   <isupper@plt>
  test   %rax,%rax
  je     L15
  movzbl 0x0(%rbp),%edi
  call   <tolower@plt>
  movzbl (%rbx),%edi
  mov    %eax,%r12d
  call   <isupper@plt>
  test   %rax,%rax
  je     L60
L20:
  movzbl (%rbx),%edi
  call   <tolower@plt>
  movzbl %al,%edx
  test   %r12b,%r12b
  jne    L10
L21:
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  sub    %edx,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  movzbl %r12b,%eax
  pop    %rbx
  pop    %rbp
  sub    %edx,%eax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L61:
  xor    %eax,%eax
  ret
```

assistant: ```c
int strcasecmp(const char *s1, const char *s2) {
  register const unsigned char *p1 = (const unsigned char *)s1;
  register const unsigned char *p2 = (const unsigned char *)s2;
  unsigned char c1, c2;
  if (p1 == p2 || 0)
    return 0;
  do {
    c1 = (isupper(*p1) ? tolower(*p1) : (*p1));
    c2 = (isupper(*p2) ? tolower(*p2) : (*p2));
    if (0 || c1 == '\0')
      break;
    ++p1;
    ++p2;
  } while (c1 == c2);
  return c1 - c2;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<strcasecmp>:
  endbr64
  cmp    %rsi,%rdi
  je     L29
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L74
  nopl   0x0(%rax)
L67:
  movzbl (%rbx),%edi
  movzbl 0x0(%rbp),%r12d
  call   <isupper@plt>
  test   %rax,%rax
  jne    L15
L3:
  movzbl (%rbx),%edx
  test   %r12b,%r12b
  je     L52
L69:
  add    $0x1,%rbp
  add    $0x1,%rbx
  cmp    %dl,%r12b
  jne    L54
L74:
  movzbl 0x0(%rbp),%edi
  call   <isupper@plt>
  test   %rax,%rax
  je     L67
  movzbl 0x0(%rbp),%edi
  call   <tolower@plt>
  movzbl (%rbx),%edi
  mov    %eax,%r12d
  call   <isupper@plt>
  test   %rax,%rax
  je     L3
L15:
  movzbl (%rbx),%edi
  call   <tolower@plt>
  movzbl %al,%edx
  test   %r12b,%r12b
  jne    L69
L52:
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  sub    %edx,%eax
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L54:
  movzbl %r12b,%eax
  pop    %rbx
  pop    %rbp
  sub    %edx,%eax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L29:
  xor    %eax,%eax
  ret
```

assistant: ```c
int strcasecmp(const char *s1, const char *s2) {
  register const unsigned char *p1 = (const unsigned char *)s1;
  register const unsigned char *p2 = (const unsigned char *)s2;
  unsigned char c1, c2;
  if (p1 == p2 || 0)
    return 0;
  do {
    c1 = (isupper(*p1) ? tolower(*p1) : (*p1));
    c2 = (isupper(*p2) ? tolower(*p2) : (*p2));
    if (0 || c1 == '\0')
      break;
    ++p1;
    ++p2;
  } while (c1 == c2);
  return c1 - c2;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<my_sleep>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  lea    -0x20(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <nanosleep@plt>
  test   %rax,%rax
  jns    L26
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L7
L26:
  mov    $0x0,%eax
L7:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L14
  call   <__stack_chk_fail@plt>
L14:
  leave
  ret
```

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

tool: D5: Nano sleep system call failed 


assistant: ```c
int my_sleep(int seconds) {
  struct timespec t_req, t_rem;
  t_req.tv_sec = seconds;
  t_req.tv_nsec = 0;
  if (nanosleep(&t_req, &t_rem) < 0) {
    printf("Nano sleep system call failed \n");
    return -1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<my_sleep>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %edi,(%rsp)
  movq   $0x0,0x8(%rsp)
  lea    0x10(%rsp),%rsi
  mov    %rsp,%rdi
  call   <nanosleep@plt>
  test   %rax,%rax
  js     L6
  mov    $0x0,%eax
L29:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L18
  add    $0x38,%rsp
  ret
L6:
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L29
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D4
type_infered: string

tool: D4: Nano sleep system call failed 


assistant: ```c
int my_sleep(int seconds) {
  struct timespec t_req, t_rem;
  t_req.tv_sec = seconds;
  t_req.tv_nsec = 0;
  if (nanosleep(&t_req, &t_rem) < 0) {
    printf("Nano sleep system call failed \n");
    return -1;
  }
  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
<my_sleep>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  mov    %edi,(%rsp)
  mov    %rsp,%rdi
  movq   $0x0,0x8(%rsp)
  call   <nanosleep@plt>
  test   %rax,%rax
  js     L28
  xor    %eax,%eax
L21:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x38,%rsp
  ret
L28:
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L21
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D6
type_guessed: string

tool: D6: Nano sleep system call failed 


assistant: ```c
int my_sleep(int seconds) {
  struct timespec t_req, t_rem;
  t_req.tv_sec = seconds;
  t_req.tv_nsec = 0;
  if (nanosleep(&t_req, &t_rem) < 0) {
    printf("Nano sleep system call failed \n");
    return -1;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<my_sleep>:
  endbr64
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rsi
  mov    %edi,(%rsp)
  mov    %rsp,%rdi
  movq   $0x0,0x8(%rsp)
  call   <nanosleep@plt>
  test   %rax,%rax
  js     L17
  xor    %eax,%eax
L27:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L15
  add    $0x38,%rsp
  ret
L17:
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L27
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type: string

tool: D9: Nano sleep system call failed 


assistant: ```c
int my_sleep(int seconds) {
  struct timespec t_req, t_rem;
  t_req.tv_sec = seconds;
  t_req.tv_nsec = 0;
  if (nanosleep(&t_req, &t_rem) < 0) {
    printf("Nano sleep system call failed \n");
    return -1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<multiplicar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  imul   -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D1
type_guessed: string

tool: D1: 
La multiplicacion es: %d


assistant: ```c
int multiplicar(int a, int b) {
  a *= b;
  printf("\nLa multiplicacion es: %d\n", a);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<multiplicar>:
  endbr64
  sub    $0x8,%rsp
  imul   %edi,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type_guessed: string

tool: D1: 
La multiplicacion es: %d


assistant: ```c
int multiplicar(int a, int b) {
  a *= b;
  printf("\nLa multiplicacion es: %d\n", a);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<multiplicar>:
  endbr64
  imul   %edi,%esi
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: label: D9
data_type: string

tool: D9: 
La multiplicacion es: %d


assistant: ```c
int multiplicar(int a, int b) {
  a *= b;
  printf("\nLa multiplicacion es: %d\n", a);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<multiplicar>:
  endbr64
  imul   %edi,%esi
  sub    $0x8,%rsp
  lea    D8(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D8
type: string

tool: D8: 
La multiplicacion es: %d


assistant: ```c
int multiplicar(int a, int b) {
  a *= b;
  printf("\nLa multiplicacion es: %d\n", a);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L1
L6:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L1:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  ucomisd %xmm0,%xmm0
  jnp    L6
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double sum(double in[]) {
  double out = 0;
  for (int i = 0; !isnan(in[i]); i++)
    out += in[i];
  return out;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<sum>:
  endbr64
  movsd  (%rdi),%xmm0
  ucomisd %xmm0,%xmm0
  jp     L27
  add    $0x8,%rdi
  pxor   %xmm1,%xmm1
L3:
  addsd  %xmm0,%xmm1
  add    $0x8,%rdi
  movsd  -0x8(%rdi),%xmm0
  ucomisd %xmm0,%xmm0
  jnp    L3
L25:
  movapd %xmm1,%xmm0
  ret
L27:
  pxor   %xmm1,%xmm1
  jmp    L25
```

assistant: ```c
double sum(double in[]) {
  double out = 0;
  for (int i = 0; !isnan(in[i]); i++)
    out += in[i];
  return out;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sum>:
  endbr64
  movsd  (%rdi),%xmm0
  ucomisd %xmm0,%xmm0
  jp     L6
  add    $0x8,%rdi
  pxor   %xmm1,%xmm1
  cs nopw 0x0(%rax,%rax,1)
L11:
  add    $0x8,%rdi
  addsd  %xmm0,%xmm1
  movsd  -0x8(%rdi),%xmm0
  ucomisd %xmm0,%xmm0
  jnp    L11
  movapd %xmm1,%xmm0
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  pxor   %xmm1,%xmm1
  movapd %xmm1,%xmm0
  ret
```

assistant: ```c
double sum(double in[]) {
  double out = 0;
  for (int i = 0; !isnan(in[i]); i++)
    out += in[i];
  return out;
}
```