ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<chi>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r8
  mov    %rsi,%r14
  mov    %edx,-0x14(%rsp)
  lea    (%rdx,%rdx,4),%eax
  lea    (%rax,%rax,4),%eax
  lea    0x7(%rax),%edx
  test   %eax,%eax
  cmovns %eax,%edx
  sar    $0x3,%edx
  cmp    $0x7,%eax
  jle    L25
  mov    $0x0,%eax
L24:
  movb   $0x0,(%r14,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%edx
  jg     L24
L25:
  mov    -0x14(%rsp),%eax
  lea    (%rax,%rax,4),%eax
  mov    %eax,-0xc(%rsp)
  movl   $0x0,-0x4(%rsp)
  movl   $0x0,-0x8(%rsp)
L56:
  mov    -0x8(%rsp),%edi
  lea    0x1(%rdi),%r13d
  movslq %r13d,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %r13d,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  sub    %eax,%r13d
  mov    -0x14(%rsp),%ebx
  imul   %ebx,%r13d
  lea    0x2(%rdi),%r12d
  movslq %r12d,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x21,%rax
  mov    %r12d,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%eax
  sub    %eax,%r12d
  imul   %ebx,%r12d
  mov    -0x4(%rsp),%eax
  mov    %eax,%ebp
  add    %ebx,%eax
  mov    %eax,-0x4(%rsp)
  mov    %eax,%r10d
  movl   $0x5,-0x10(%rsp)
L41:
  mov    %ebp,%esi
  mov    %r13d,%ebx
  sub    %ebp,%ebx
  mov    %r12d,%r11d
  sub    %ebp,%r11d
  cmpl   $0x0,-0x14(%rsp)
  jle    L16
L31:
  lea    0x7(%rsi),%edx
  test   %esi,%esi
  cmovns %esi,%edx
  sar    $0x3,%edx
  movslq %edx,%rdx
  mov    %esi,%edi
  and    $0x7,%edi
  lea    (%rbx,%rsi,1),%ecx
  lea    (%r11,%rsi,1),%r15d
  lea    0x7(%rcx),%eax
  test   %ecx,%ecx
  cmovns %ecx,%eax
  sar    $0x3,%eax
  cltq
  movzbl (%r8,%rax,1),%eax
  and    $0x7,%ecx
  sar    %cl,%eax
  movzbl (%r8,%rdx,1),%r9d
  mov    %edi,%ecx
  sar    %cl,%r9d
  xor    %r9d,%eax
  xor    $0x1,%eax
  lea    0x7(%r15),%r9d
  test   %r15d,%r15d
  cmovns %r15d,%r9d
  sar    $0x3,%r9d
  movslq %r9d,%r9
  movzbl (%r8,%r9,1),%r9d
  mov    %r15d,%ecx
  and    $0x7,%ecx
  sar    %cl,%r9d
  and    %r9d,%eax
  and    $0x1,%eax
  mov    %edi,%ecx
  shl    %cl,%eax
  or     %al,(%r14,%rdx,1)
  add    $0x1,%esi
  cmp    %r10d,%esi
  jne    L31
L16:
  mov    -0xc(%rsp),%eax
  add    %eax,%ebp
  add    %eax,%r13d
  add    %eax,%r12d
  add    %eax,%r10d
  subl   $0x1,-0x10(%rsp)
  jne    L41
  addl   $0x1,-0x8(%rsp)
  mov    -0x8(%rsp),%eax
  cmp    $0x5,%eax
  jne    L56
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned char *chi(unsigned char *m, unsigned char *mp, int w) {
  for (int i = 0; i < (w * 25 / 8); i++) {
    mp[i] = 0x00;
  }
  for (int x = 0; x < 5; x++) {
    for (int y = 0; y < 5; y++) {
      for (int z = 0; z < w; z++) {
        int ind = w * (5 * y + x) + z;
        mp[ind / 8] |= (((m[ind / 8] >> (ind % 8)) ^ ((m[(w * (5 * y + ((x + 1) % 5)) + z) / 8] >> ((w * (5 * y + ((x + 1) % 5)) + z) % 8)) ^ 1)) & (m[(w * (5 * y + ((x + 2) % 5)) + z) / 8] >> ((w * (5 * y + ((x + 2) % 5)) + z) % 8)) & 1) << (ind % 8);
      }
    }
  }
  return mp;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<chi>:
  endbr64
  push   %r15
  lea    (%rdx,%rdx,4),%eax
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edx,0x10(%rsp)
  lea    (%rax,%rax,4),%edx
  test   %edx,%edx
  lea    0x7(%rdx),%eax
  cmovns %edx,%eax
  sar    $0x3,%eax
  cmp    $0x7,%edx
  jle    L26
  lea    -0x1(%rax),%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  add    $0x1,%rdx
  call   <memset@plt>
L26:
  mov    0x10(%rsp),%eax
  movl   $0x0,0x18(%rsp)
  lea    (%rax,%rax,4),%r12d
  xor    %eax,%eax
L27:
  lea    0x1(%rax),%edx
  mov    $0xcccccccd,%edi
  movl   $0x5,0x14(%rsp)
  mov    %rdx,%rbx
  mov    %edx,0x1c(%rsp)
  imul   %rdi,%rdx
  shr    $0x22,%rdx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%ebx
  lea    0x2(%rax),%edx
  mov    %rdx,%rax
  imul   %rdi,%rdx
  mov    %ebx,%ebp
  mov    0x10(%rsp),%ebx
  mov    %ebx,%edi
  imul   %ebx,%ebp
  shr    $0x22,%rdx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  imul   %ebx,%eax
  mov    %eax,0xc(%rsp)
  mov    0x18(%rsp),%eax
  mov    %eax,%ebx
  add    %edi,%eax
  mov    %eax,0x18(%rsp)
  mov    %eax,%r9d
L10:
  mov    0xc(%rsp),%r10d
  mov    0x10(%rsp),%eax
  mov    %ebp,%r11d
  mov    %ebx,%edx
  sub    %ebx,%r11d
  sub    %ebx,%r10d
  test   %eax,%eax
  jle    L18
  nopl   (%rax)
L37:
  lea    (%r11,%rdx,1),%ecx
  mov    %edx,%esi
  lea    (%r10,%rdx,1),%r8d
  mov    %edx,%edi
  mov    %ecx,%eax
  sar    $0x3,%esi
  and    $0x7,%edi
  and    $0x7,%ecx
  sar    $0x3,%eax
  movslq %esi,%rsi
  add    $0x1,%edx
  cltq
  movzbl 0x0(%r13,%rsi,1),%r15d
  movzbl 0x0(%r13,%rax,1),%eax
  sar    %cl,%eax
  mov    %edi,%ecx
  sar    %cl,%r15d
  mov    %r8d,%ecx
  sar    $0x3,%ecx
  xor    %r15d,%eax
  movslq %ecx,%rcx
  xor    $0x1,%eax
  movzbl 0x0(%r13,%rcx,1),%r15d
  mov    %r8d,%ecx
  and    $0x7,%ecx
  sar    %cl,%r15d
  mov    %edi,%ecx
  and    %r15d,%eax
  and    $0x1,%eax
  shl    %cl,%eax
  or     %al,(%r14,%rsi,1)
  cmp    %r9d,%edx
  jne    L37
L18:
  add    %r12d,0xc(%rsp)
  add    %r12d,%ebx
  add    %r12d,%ebp
  add    %r12d,%r9d
  subl   $0x1,0x14(%rsp)
  jne    L10
  mov    0x1c(%rsp),%eax
  cmp    $0x5,%eax
  jne    L27
  add    $0x28,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned char *chi(unsigned char *m, unsigned char *mp, int w) {
  for (int i = 0; i < (w * 25 / 8); i++) {
    mp[i] = 0x00;
  }
  for (int x = 0; x < 5; x++) {
    for (int y = 0; y < 5; y++) {
      for (int z = 0; z < w; z++) {
        int ind = w * (5 * y + x) + z;
        mp[ind / 8] |= (((m[ind / 8] >> (ind % 8)) ^ ((m[(w * (5 * y + ((x + 1) % 5)) + z) / 8] >> ((w * (5 * y + ((x + 1) % 5)) + z) % 8)) ^ 1)) & (m[(w * (5 * y + ((x + 2) % 5)) + z) / 8] >> ((w * (5 * y + ((x + 2) % 5)) + z) % 8)) & 1) << (ind % 8);
      }
    }
  }
  return mp;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<chi>:
  endbr64
  push   %r15
  lea    (%rdx,%rdx,4),%eax
  push   %r14
  mov    %rsi,%r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edx,0x1c(%rsp)
  lea    (%rax,%rax,4),%edx
  test   %edx,%edx
  lea    0x7(%rdx),%eax
  cmovns %edx,%eax
  sar    $0x3,%eax
  cmp    $0x7,%edx
  jle    L15
  lea    -0x1(%rax),%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  add    $0x1,%rdx
  call   <memset@plt>
L15:
  mov    0x1c(%rsp),%eax
  test   %eax,%eax
  jle    L20
  movl   $0x0,0x14(%rsp)
  lea    (%rax,%rax,4),%r12d
  xor    %eax,%eax
L22:
  lea    0x1(%rax),%edx
  mov    $0xcccccccd,%edi
  movl   $0x5,0x10(%rsp)
  mov    %rdx,%rbx
  mov    %edx,0x18(%rsp)
  imul   %rdi,%rdx
  shr    $0x22,%rdx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%ebx
  lea    0x2(%rax),%edx
  mov    %rdx,%rax
  imul   %rdi,%rdx
  mov    %ebx,%ebp
  mov    0x1c(%rsp),%ebx
  mov    %ebx,%edi
  imul   %ebx,%ebp
  shr    $0x22,%rdx
  lea    (%rdx,%rdx,4),%edx
  sub    %edx,%eax
  imul   %ebx,%eax
  mov    %eax,0xc(%rsp)
  mov    0x14(%rsp),%eax
  mov    %eax,%ebx
  add    %edi,%eax
  mov    %eax,0x14(%rsp)
  mov    %eax,%r9d
L39:
  mov    0xc(%rsp),%r10d
  mov    %ebp,%r11d
  mov    %ebx,%edx
  sub    %ebx,%r11d
  sub    %ebx,%r10d
  nopl   (%rax)
L32:
  lea    (%r11,%rdx,1),%ecx
  mov    %edx,%esi
  lea    (%r10,%rdx,1),%r8d
  mov    %edx,%edi
  mov    %ecx,%eax
  sar    $0x3,%esi
  and    $0x7,%edi
  and    $0x7,%ecx
  sar    $0x3,%eax
  movslq %esi,%rsi
  add    $0x1,%edx
  cltq
  movzbl 0x0(%r13,%rsi,1),%r15d
  movzbl 0x0(%r13,%rax,1),%eax
  sar    %cl,%eax
  mov    %edi,%ecx
  sar    %cl,%r15d
  mov    %r8d,%ecx
  sar    $0x3,%ecx
  xor    %r15d,%eax
  movslq %ecx,%rcx
  xor    $0x1,%eax
  movzbl 0x0(%r13,%rcx,1),%r15d
  mov    %r8d,%ecx
  and    $0x7,%ecx
  sar    %cl,%r15d
  mov    %edi,%ecx
  and    %r15d,%eax
  and    $0x1,%eax
  shl    %cl,%eax
  or     %al,(%r14,%rsi,1)
  cmp    %r9d,%edx
  jne    L32
  add    %r12d,0xc(%rsp)
  add    %r12d,%ebx
  lea    (%rdx,%r12,1),%r9d
  add    %r12d,%ebp
  subl   $0x1,0x10(%rsp)
  jne    L39
  mov    0x18(%rsp),%eax
  cmp    $0x5,%eax
  jne    L22
L20:
  add    $0x28,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
unsigned char *chi(unsigned char *m, unsigned char *mp, int w) {
  for (int i = 0; i < (w * 25 / 8); i++) {
    mp[i] = 0x00;
  }
  for (int x = 0; x < 5; x++) {
    for (int y = 0; y < 5; y++) {
      for (int z = 0; z < w; z++) {
        int ind = w * (5 * y + x) + z;
        mp[ind / 8] |= (((m[ind / 8] >> (ind % 8)) ^ ((m[(w * (5 * y + ((x + 1) % 5)) + z) / 8] >> ((w * (5 * y + ((x + 1) % 5)) + z) % 8)) ^ 1)) & (m[(w * (5 * y + ((x + 2) % 5)) + z) / 8] >> ((w * (5 * y + ((x + 2) % 5)) + z) % 8)) & 1) << (ind % 8);
      }
    }
  }
  return mp;
}
```
user: What is the c source code of the assembly code below:

```asm
<chomp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L39
L30:
  addq   $0x1,-0x8(%rbp)
L39:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L27
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  je     L27
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xd,%al
  jne    L30
L27:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *chomp(char *s) {
  char *p = s;
  while (*p && *p != '\n' && *p != '\r')
    p++;
  *p = 0;
  return s;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<chomp>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rdi),%ecx
  cmp    $0xd,%cl
  ja     L36
  mov    %rdi,%rdx
  mov    $0x2401,%esi
  bt     %rcx,%rsi
  jae    L23
  jmp    L49
L36:
  mov    %rdi,%rdx
L23:
  mov    $0x2401,%esi
L29:
  add    $0x1,%rdx
  movzbl (%rdx),%ecx
  cmp    $0xd,%cl
  ja     L29
  bt     %rcx,%rsi
  jae    L29
L49:
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
char *chomp(char *s) {
  char *p = s;
  while (*p && *p != '\n' && *p != '\r')
    p++;
  *p = 0;
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<chomp>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rdi,%rax
  mov    %rdi,%rdx
  cmp    $0xd,%cl
  ja     L18
  mov    $0xffffffffffffdbfe,%rsi
  bt     %rcx,%rsi
  jae    L19
L18:
  mov    $0xffffffffffffdbfe,%rsi
  cs nopw 0x0(%rax,%rax,1)
L10:
  movzbl 0x1(%rdx),%ecx
  add    $0x1,%rdx
  cmp    $0xd,%cl
  ja     L10
  bt     %rcx,%rsi
  jb     L10
L19:
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
char *chomp(char *s) {
  char *p = s;
  while (*p && *p != '\n' && *p != '\r')
    p++;
  *p = 0;
  return s;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<chomp>:
  endbr64
  movzbl (%rdi),%ecx
  mov    %rdi,%rax
  mov    %rdi,%rdx
  cmp    $0xd,%cl
  ja     L36
  mov    $0xffffffffffffdbfe,%rsi
  bt     %rcx,%rsi
  jae    L19
L36:
  mov    $0xffffffffffffdbfe,%rsi
  cs nopw 0x0(%rax,%rax,1)
L17:
  movzbl 0x1(%rdx),%ecx
  add    $0x1,%rdx
  cmp    $0xd,%cl
  ja     L17
  bt     %rcx,%rsi
  jb     L17
L19:
  movb   $0x0,(%rdx)
  ret
```

assistant: ```c
char *chomp(char *s) {
  char *p = s;
  while (*p && *p != '\n' && *p != '\r')
    p++;
  *p = 0;
  return s;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<crc_zap>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  jmp    L14
L16:
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  subl   $0x1,-0xc(%rbp)
  addq   $0x1,-0x8(%rbp)
L14:
  cmpl   $0x0,-0xc(%rbp)
  jg     L16
  nop
  pop    %rbp
  ret
```

assistant: ```c
void crc_zap(buf, bufsz) char *buf;
int bufsz;
{
  for (; bufsz > 0; bufsz--, buf++)
    *buf = 0;
  return;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<crc_zap>:
  endbr64
  test   %esi,%esi
  jle    L9
  mov    %esi,%esi
  lea    (%rdi,%rsi,1),%rax
L18:
  movb   $0x0,(%rdi)
  add    $0x1,%rdi
  cmp    %rax,%rdi
  jne    L18
L9:
  ret
```

assistant: ```c
void crc_zap(buf, bufsz) char *buf;
int bufsz;
{
  for (; bufsz > 0; bufsz--, buf++)
    *buf = 0;
  return;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<crc_zap>:
  endbr64
  test   %esi,%esi
  jle    L2
  movslq %esi,%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L2:
  ret
```

assistant: ```c
void crc_zap(buf, bufsz) char *buf;
int bufsz;
{
  for (; bufsz > 0; bufsz--, buf++)
    *buf = 0;
  return;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<crc_zap>:
  endbr64
  test   %esi,%esi
  jle    L6
  movslq %esi,%rdx
  xor    %esi,%esi
  jmp    1050 <memset@plt>
  nopw   0x0(%rax,%rax,1)
L6:
  ret
```

assistant: ```c
void crc_zap(buf, bufsz) char *buf;
int bufsz;
{
  for (; bufsz > 0; bufsz--, buf++)
    *buf = 0;
  return;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <bar@plt>
  mov    $0x0,%eax
  call   <foo@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  bar();
  foo();
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

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

assistant: ```c
int main() {
  bar();
  foo();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

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

assistant: ```c
int main() {
  bar();
  foo();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

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

assistant: ```c
int main() {
  bar();
  foo();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  lea    D1(%rip),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D1
type_infered: string

tool: D1: hello, linux!

assistant: ```c
int main() {
  char *s = "hello, linux!";
  write(1, s, strlen(s));
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbx
  lea    D8(%rip),%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    %rbx,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  mov    $0x0,%eax
  pop    %rbx
  ret
```

tool_calls: name: disassemble_data
arguments: data_label: D8
type: string

tool: D8: hello, linux!

assistant: ```c
int main() {
  char *s = "hello, linux!";
  write(1, s, strlen(s));
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D0(%rip),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    $0x1,%edi
  mov    %eax,%edx
  call   <write@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
type: string

tool: D0: hello, linux!

assistant: ```c
int main() {
  char *s = "hello, linux!";
  write(1, s, strlen(s));
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rbp
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    %rbp,%rsi
  mov    $0x1,%edi
  mov    %eax,%edx
  call   <write@plt>
  xor    %eax,%eax
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D9
type_guessed: string

tool: D9: hello, linux!

assistant: ```c
int main() {
  char *s = "hello, linux!";
  write(1, s, strlen(s));
}
```
user: What is the c source code of the assembly code below:

```asm
<getDistance>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %rcx,-0x18(%rbp)
  mov    -0xc(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jne    L11
  mov    -0x4(%rbp),%eax
  jmp    L5
L11:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
L5:
  pop    %rbp
  ret
```

assistant: ```c
int getDistance(int value, int startVertex, int start, int *shortestPath) {
  if (start == startVertex) {
    return value;
  } else {
    return shortestPath[start] + value;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<getDistance>:
  endbr64
  mov    %edi,%eax
  cmp    %esi,%edx
  je     L0
  movslq %edx,%rdx
  add    (%rcx,%rdx,4),%eax
L0:
  ret
```

assistant: ```c
int getDistance(int value, int startVertex, int start, int *shortestPath) {
  if (start == startVertex) {
    return value;
  } else {
    return shortestPath[start] + value;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<getDistance>:
  endbr64
  mov    %edi,%eax
  cmp    %esi,%edx
  je     L7
  movslq %edx,%rdx
  add    (%rcx,%rdx,4),%eax
L7:
  ret
```

assistant: ```c
int getDistance(int value, int startVertex, int start, int *shortestPath) {
  if (start == startVertex) {
    return value;
  } else {
    return shortestPath[start] + value;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<getDistance>:
  endbr64
  mov    %edi,%eax
  cmp    %esi,%edx
  je     L5
  movslq %edx,%rdx
  add    (%rcx,%rdx,4),%eax
L5:
  ret
```

assistant: ```c
int getDistance(int value, int startVertex, int start, int *shortestPath) {
  if (start == startVertex) {
    return value;
  } else {
    return shortestPath[start] + value;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0xc(%rbp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  shr    $0x20,%rax
  sar    $0x7,%eax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  test   %eax,%eax
  jne    L26
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L15
L26:
  mov    -0xc(%rbp),%eax
  and    $0x3,%eax
  test   %eax,%eax
  jne    L13
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L15
L13:
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L15:
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_label: D34
datatype: string
name: parse_data
arguments: data_label: D23
datatype: string
name: parse_data
arguments: data_label: D29
datatype: string

tool: D34: %d

tool: D23: Yes

tool: D29: No

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  if (a % 400 == 0)
    printf("Yes");
  else if (a % 4 == 0)
    printf("Yes");
  else
    printf("No");
  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    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  lea    D18(%rip),%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  movslq %edx,%rax
  imul   $0x51eb851f,%rax,%rax
  sar    $0x27,%rax
  mov    %edx,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%eax
  imul   $0x190,%eax,%eax
  cmp    %eax,%edx
  je     L7
  test   $0x3,%dl
  jne    L42
  lea    D20(%rip),%rdi
  call   <printf@plt>
L10:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L29
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L7:
  lea    D20(%rip),%rdi
  call   <printf@plt>
  jmp    L10
L42:
  lea    D27(%rip),%rdi
  call   <printf@plt>
  jmp    L10
L29:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D18
type: string
name: parse_data
arguments: data_label: D20
type: string
name: parse_data
arguments: data_label: D27
type: string

tool: D18: %d

tool: D20: Yes

tool: D27: No

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  if (a % 400 == 0)
    printf("Yes");
  else if (a % 4 == 0)
    printf("Yes");
  else
    printf("No");
  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    $0x18,%rsp
  lea    D20(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  imul   $0xc28f5c29,%edx,%eax
  add    $0x51eb850,%eax
  ror    $0x4,%eax
  cmp    $0xa3d70a,%eax
  jbe    L25
  and    $0x3,%edx
  jne    L16
L25:
  lea    D26(%rip),%rdi
  call   <printf@plt>
L7:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L16:
  lea    D27(%rip),%rdi
  call   <printf@plt>
  jmp    L7
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D20
datatype: string
name: parse_data
arguments: data_label: D26
datatype: string
name: parse_data
arguments: data_label: D27
datatype: string

tool: D20: %d

tool: D26: Yes

tool: D27: No

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  if (a % 400 == 0)
    printf("Yes");
  else if (a % 4 == 0)
    printf("Yes");
  else
    printf("No");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  lea    D6(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rsi
  call   <scanf@plt>
  mov    0x4(%rsp),%edx
  imul   $0xc28f5c29,%edx,%eax
  add    $0x51eb850,%eax
  ror    $0x4,%eax
  cmp    $0xa3d70a,%eax
  jbe    L12
  and    $0x3,%edx
  jne    L0
L12:
  lea    D0(%rip),%rdi
  call   <printf@plt>
L17:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L0:
  lea    D27(%rip),%rdi
  call   <printf@plt>
  jmp    L17
L27:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D6
type_infered: string
name: retrieve_data
arguments: label: D0
type_infered: string
name: retrieve_data
arguments: label: D27
type_infered: string

tool: D6: %d

tool: D0: Yes

tool: D27: No

assistant: ```c
int main() {
  int a;
  scanf("%d", &a);
  if (a % 400 == 0)
    printf("Yes");
  else if (a % 4 == 0)
    printf("Yes");
  else
    printf("No");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L10
  jmp    L22
L11:
  mov    -0x4(%rbp),%eax
  sub    %eax,-0x8(%rbp)
L22:
  cmpl   $0x0,-0x8(%rbp)
  jns    L11
L10:
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main() {
  int x;
  int y;
  x = __VERIFIER_nondet_int();
  y = __VERIFIER_nondet_int();
  if (y < 1) {
    while (x >= 0) {
      x = x - y;
    }
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <__VERIFIER_nondet_int@plt>
  test   %eax,%eax
  jle    L26
L20:
  mov    $0x0,%eax
  pop    %rbx
  ret
L26:
  test   %ebx,%ebx
  js     L20
L35:
  sub    %eax,%ebx
  jns    L35
  jmp    L20
```

assistant: ```c
int main() {
  int x;
  int y;
  x = __VERIFIER_nondet_int();
  y = __VERIFIER_nondet_int();
  if (y < 1) {
    while (x >= 0) {
      x = x - y;
    }
  }
  return 0;
}
```