ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<ChkMurMur3>:
  endbr64
  mov    %esi,%r9d
  mov    %rdi,%r8
  mov    %edx,%esi
  test   %r9d,%r9d
  lea    0x3(%r9),%edi
  cmovns %r9d,%edi
  sar    $0x2,%edi
  cmp    $0x3,%r9d
  jle    L14
  xor    %ecx,%ecx
  nopw   0x0(%rax,%rax,1)
L39:
  imul   $0xcc9e2d51,(%r8,%rcx,4),%edx
  add    $0x1,%rcx
  mov    %edx,%eax
  sar    $0x11,%edx
  shl    $0xf,%eax
  or     %edx,%eax
  imul   $0x1b873593,%eax,%eax
  xor    %esi,%eax
  mov    %eax,%edx
  sar    $0x13,%eax
  shl    $0xd,%edx
  or     %edx,%eax
  lea    -0x19ab949c(%rax,%rax,4),%esi
  cmp    %ecx,%edi
  jg     L39
L14:
  shl    $0x2,%edi
  mov    %r9d,%eax
  movslq %edi,%rdi
  and    $0x3,%eax
  lea    (%r8,%rdi,4),%rdx
  cmp    $0x2,%eax
  je     L63
  cmp    $0x3,%eax
  je     L51
  cmp    $0x1,%eax
  je     L16
L55:
  xor    %r9d,%esi
  mov    %esi,%eax
  sar    $0x10,%eax
  xor    %esi,%eax
  imul   $0x85ebca6b,%eax,%eax
  mov    %eax,%edx
  sar    $0xd,%edx
  xor    %eax,%edx
  imul   $0xc2b2ae35,%edx,%edx
  mov    %edx,%eax
  sar    $0x10,%eax
  xor    %edx,%eax
  ret
  nop
L16:
  xor    %eax,%eax
L41:
  xor    (%rdx),%eax
  imul   $0xcc9e2d51,%eax,%eax
  mov    %eax,%edx
  sar    $0x11,%eax
  shl    $0xf,%edx
  or     %edx,%eax
  imul   $0x1b873593,%eax,%eax
  xor    %eax,%esi
  jmp    L55
  xchg   %ax,%ax
L51:
  mov    0x8(%rdx),%ecx
  shl    $0x10,%ecx
L37:
  mov    0x4(%rdx),%eax
  shl    $0x8,%eax
  xor    %ecx,%eax
  jmp    L41
L63:
  xor    %ecx,%ecx
  jmp    L37
```

assistant: ```c
uint32_t ChkMurMur3(const void *buf, uint32_t len, uint32_t seed) {
  static const uint32_t c1 = 0xcc9e2d51;
  static const uint32_t c2 = 0x1b873593;
  static const uint32_t r1 = 15;
  static const uint32_t r2 = 13;
  static const uint32_t m = 5;
  static const uint32_t n = 0xe6546b64;
  const uint8_t *tail;
  uint32_t hash = seed, k, k1;
  const int nblocks = len / 4;
  const uint32_t *blocks = (const uint32_t *)buf;
  int i;
  for (i = 0; i < nblocks; i++) {
    k = blocks[i];
    k *= c1;
    k = (k << r1) | (k >> (32 - r1));
    k *= c2;
    hash ^= k;
    hash = ((hash << r2) | (hash >> (32 - r2))) * m + n;
  }
  tail = (const uint8_t *)(((const uint8_t *)buf) + nblocks * 4);
  k1 = 0;
  switch (len & 3) {
  case 3:
    k1 ^= tail[2] << 16;
  case 2:
    k1 ^= tail[1] << 8;
  case 1:
    k1 ^= tail[0];
    k1 *= c1;
    k1 = (k1 << r1) | (k1 >> (32 - r1));
    k1 *= c2;
    hash ^= k1;
  }
  hash ^= len;
  hash ^= (hash >> 16);
  hash *= 0x85ebca6b;
  hash ^= (hash >> 13);
  hash *= 0xc2b2ae35;
  hash ^= (hash >> 16);
  return hash;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ultoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x70,%rsp
  mov    %rdi,-0x58(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %edx,-0x64(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x50(%rbp)
  mov    -0x58(%rbp),%rax
  mov    %rax,-0x48(%rbp)
  cmpl   $0x24,-0x64(%rbp)
  jg     L54
  cmpl   $0x1,-0x64(%rbp)
  jg     L100
L54:
  mov    D14(%rip),%rax
  mov    (%rax),%edx
  mov    D8(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  jmp    L74
L72:
  mov    -0x64(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x48(%rbp),%rax
  mov    $0x0,%edx
  div    %rcx
  mov    %rdx,%rax
  mov    %rax,-0x38(%rbp)
  mov    -0x64(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x48(%rbp),%rax
  mov    $0x0,%edx
  div    %rsi
  mov    %rax,-0x48(%rbp)
  cmpq   $0x9,-0x38(%rbp)
  jg     L67
  mov    -0x38(%rbp),%rax
  lea    0x30(%rax),%ecx
  mov    -0x50(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x50(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L100
L67:
  mov    -0x38(%rbp),%rax
  lea    0x57(%rax),%ecx
  mov    -0x50(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x50(%rbp)
  mov    %ecx,%edx
  mov    %dl,(%rax)
L100:
  cmpq   $0x0,-0x48(%rbp)
  jne    L72
  lea    -0x30(%rbp),%rax
  cmp    %rax,-0x50(%rbp)
  je     L72
  cmpq   $0x0,-0x60(%rbp)
  jne    L28
  lea    -0x30(%rbp),%rdx
  mov    -0x50(%rbp),%rax
  sub    %rdx,%rax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x60(%rbp)
L28:
  mov    -0x60(%rbp),%rax
  mov    %rax,-0x40(%rbp)
  jmp    L71
L47:
  subq   $0x1,-0x50(%rbp)
  mov    -0x40(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x40(%rbp)
  mov    -0x50(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L71:
  lea    -0x30(%rbp),%rax
  cmp    %rax,-0x50(%rbp)
  ja     L47
  mov    -0x40(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x60(%rbp),%rax
L74:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L59
  call   <__stack_chk_fail@plt>
L59:
  leave
  ret
```

assistant: ```c
char *ultoa(unsigned long value, char *string, int radix) {
  char tmp[33];
  char *tp = tmp;
  long i;
  unsigned long v = value;
  char *sp;
  if (radix > 36 || radix <= 1) {
    errno = EDOM;
    return 0;
  }
  while (v || tp == tmp) {
    i = v % radix;
    v = v / radix;
    if (i < 10)
      *tp++ = i + '0';
    else
      *tp++ = i + 'a' - 10;
  }
  if (string == NULL)
    string = (char *)malloc((tp - tmp) + 1);
  sp = string;
  while (tp > tmp)
    *sp++ = *--tp;
  *sp = 0;
  return string;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ultoa>:
  endbr64
  push   %rbx
  sub    $0x40,%rsp
  mov    %rdi,%rax
  mov    %rsi,%r8
  mov    %fs:0x28,%rcx
  mov    %rcx,0x38(%rsp)
  xor    %ecx,%ecx
  lea    -0x2(%rdx),%edi
  lea    0x11(%rsp),%rcx
  movslq %edx,%rsi
  lea    0x10(%rsp),%r9
  cmp    $0x22,%edi
  ja     L48
L99:
  mov    %rax,%rdi
  mov    $0x0,%edx
  div    %rsi
  cmp    $0x9,%rdx
  jg     L51
  mov    %rcx,%rbx
  add    $0x30,%edx
L96:
  mov    %dl,-0x1(%rcx)
  add    $0x1,%rcx
  cmp    %r9,%rbx
  je     L99
  cmp    %rdi,%rsi
  jbe    L99
  test   %r8,%r8
  je     L15
L14:
  lea    0x10(%rsp),%rax
  cmp    %rax,%rbx
  jbe    L19
  mov    %r8,%rdx
  mov    %rbx,%rax
  lea    0x10(%rsp),%rsi
L86:
  sub    $0x1,%rax
  add    $0x1,%rdx
  movzbl (%rax),%ecx
  mov    %cl,-0x1(%rdx)
  cmp    %rsi,%rax
  jne    L86
  lea    0x40(%rsp),%rax
  sub    %rax,%rbx
  lea    0x30(%rbx,%r8,1),%rax
L81:
  movb   $0x0,(%rax)
  mov    %r8,%rax
L98:
  mov    0x38(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L52
  add    $0x40,%rsp
  pop    %rbx
  ret
L48:
  mov    D19(%rip),%rax
  mov    (%rax),%edx
  mov    D18(%rip),%rax
  mov    %edx,(%rax)
  mov    $0x0,%eax
  jmp    L98
L51:
  mov    %rcx,%rbx
  add    $0x57,%edx
  jmp    L96
L15:
  lea    0x40(%rsp),%rax
  mov    %rax,0x8(%rsp)
  mov    %ebx,%eax
  sub    0x8(%rsp),%eax
  lea    0x31(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  jmp    L14
L19:
  mov    %r8,%rax
  jmp    L81
L52:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ultoa(unsigned long value, char *string, int radix) {
  char tmp[33];
  char *tp = tmp;
  long i;
  unsigned long v = value;
  char *sp;
  if (radix > 36 || radix <= 1) {
    errno = EDOM;
    return 0;
  }
  while (v || tp == tmp) {
    i = v % radix;
    v = v / radix;
    if (i < 10)
      *tp++ = i + '0';
    else
      *tp++ = i + 'a' - 10;
  }
  if (string == NULL)
    string = (char *)malloc((tp - tmp) + 1);
  sp = string;
  while (tp > tmp)
    *sp++ = *--tp;
  *sp = 0;
  return string;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ultoa>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rcx
  mov    %rcx,0x28(%rsp)
  xor    %ecx,%ecx
  lea    -0x2(%rdx),%ecx
  cmp    $0x22,%ecx
  ja     L75
  mov    %rsp,%r12
  mov    %rdi,%rax
  mov    %rsi,%r8
  movslq %edx,%rcx
  mov    %r12,%rbp
  nopw   0x0(%rax,%rax,1)
L27:
  xor    %edx,%edx
  mov    %rax,%rsi
  add    $0x1,%rbp
  div    %rcx
  lea    0x30(%rdx),%edi
  lea    0x57(%rdx),%ebx
  cmp    $0x9,%rdx
  cmovbe %edi,%ebx
  mov    %bl,-0x1(%rbp)
  cmp    %rsi,%rcx
  jbe    L27
  cmp    %r12,%rbp
  je     L27
  test   %r8,%r8
  je     L17
L106:
  cmp    %r12,%rbp
  jbe    L102
  mov    %r8,%rdx
  mov    %rbp,%rax
  jmp    L90
  nopl   0x0(%rax,%rax,1)
L73:
  movzbl -0x1(%rax),%ebx
L90:
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %bl,-0x1(%rdx)
  cmp    %r12,%rax
  jne    L73
  sub    %rax,%rbp
  add    %r8,%rbp
L65:
  movb   $0x0,0x0(%rbp)
  mov    %r8,%rax
L29:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L68
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L75:
  mov    D16(%rip),%rax
  mov    (%rax),%edx
  mov    D19(%rip),%rax
  mov    %edx,(%rax)
  xor    %eax,%eax
  jmp    L29
L17:
  mov    %rbp,%rdi
  sub    %r12,%rdi
  add    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%r8
  jmp    L106
L102:
  mov    %r8,%rbp
  jmp    L65
L68:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ultoa(unsigned long value, char *string, int radix) {
  char tmp[33];
  char *tp = tmp;
  long i;
  unsigned long v = value;
  char *sp;
  if (radix > 36 || radix <= 1) {
    errno = EDOM;
    return 0;
  }
  while (v || tp == tmp) {
    i = v % radix;
    v = v / radix;
    if (i < 10)
      *tp++ = i + '0';
    else
      *tp++ = i + 'a' - 10;
  }
  if (string == NULL)
    string = (char *)malloc((tp - tmp) + 1);
  sp = string;
  while (tp > tmp)
    *sp++ = *--tp;
  *sp = 0;
  return string;
}
```
user: Translate the below assembly code into C programming language:

```asm
<ultoa>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    -0x2(%rdx),%eax
  cmp    $0x22,%eax
  ja     L112
  mov    %rsp,%rbp
  mov    %rdi,%rcx
  mov    %rsi,%rdi
  movslq %edx,%rsi
  mov    %rbp,%rbx
  jmp    L33
  nopw   0x0(%rax,%rax,1)
L81:
  add    $0x30,%edx
  mov    %dl,-0x1(%rbx)
  cmp    %rsi,%rcx
  jae    L129
  cmp    %rbp,%rbx
  jne    L119
L129:
  mov    %rax,%rcx
L33:
  mov    %rcx,%rax
  xor    %edx,%edx
  add    $0x1,%rbx
  div    %rsi
  cmp    $0x9,%rdx
  jbe    L81
  add    $0x57,%edx
  mov    %dl,-0x1(%rbx)
  cmp    %rsi,%rcx
  jae    L129
  cmp    %rbp,%rbx
  je     L129
L119:
  test   %rdi,%rdi
  je     L91
L139:
  cmp    %rbp,%rbx
  jbe    L109
  mov    %rdi,%rdx
  mov    %rbx,%rax
  nopl   (%rax)
L136:
  movzbl -0x1(%rax),%ecx
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %cl,-0x1(%rdx)
  cmp    %rbp,%rax
  jne    L136
  sub    %rax,%rbx
  add    %rdi,%rbx
L67:
  movb   $0x0,(%rbx)
  mov    %rdi,%rax
L25:
  mov    0x28(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L42
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nop
L112:
  mov    D11(%rip),%rax
  mov    (%rax),%edx
  mov    D19(%rip),%rax
  mov    %edx,(%rax)
  xor    %eax,%eax
  jmp    L25
L91:
  mov    %rbx,%rdi
  sub    %rbp,%rdi
  add    $0x1,%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  jmp    L139
L109:
  mov    %rdi,%rbx
  jmp    L67
L42:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ultoa(unsigned long value, char *string, int radix) {
  char tmp[33];
  char *tp = tmp;
  long i;
  unsigned long v = value;
  char *sp;
  if (radix > 36 || radix <= 1) {
    errno = EDOM;
    return 0;
  }
  while (v || tp == tmp) {
    i = v % radix;
    v = v / radix;
    if (i < 10)
      *tp++ = i + '0';
    else
      *tp++ = i + 'a' - 10;
  }
  if (string == NULL)
    string = (char *)malloc((tp - tmp) + 1);
  sp = string;
  while (tp > tmp)
    *sp++ = *--tp;
  *sp = 0;
  return string;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strncpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L9
L27:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
  addq   $0x1,-0x18(%rbp)
  addq   $0x1,-0x20(%rbp)
  subq   $0x1,-0x28(%rbp)
L9:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L1
  cmpq   $0x0,-0x28(%rbp)
  jne    L27
  jmp    L1
L32:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  addq   $0x1,-0x18(%rbp)
L1:
  mov    -0x28(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    %rdx,-0x28(%rbp)
  test   %rax,%rax
  jne    L32
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *strncpy(char *s1, const char *s2, size_t n) {
  char *rs = s1;
  while (*s2 != 0 && 0 < n) {
    *s1 = *s2;
    s1++;
    s2++;
    n--;
  }
  while (0 < n--) {
    *s1 = 0;
    s1++;
  }
  return rs;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strncpy>:
  endbr64
  mov    %rdi,%rax
  movzbl (%rsi),%r8d
  test   %rdx,%rdx
  je     L38
  test   %r8b,%r8b
  je     L38
  mov    %rdi,%rcx
L55:
  mov    %r8b,(%rcx)
  add    $0x1,%rcx
  add    $0x1,%rsi
  movzbl (%rsi),%r8d
  sub    $0x1,%rdx
  je     L8
  test   %r8b,%r8b
  jne    L55
L8:
  test   %rdx,%rdx
  je     L44
  add    %rcx,%rdx
L24:
  movb   $0x0,(%rcx)
  add    $0x1,%rcx
  cmp    %rcx,%rdx
  jne    L24
L44:
  ret
L38:
  mov    %rax,%rcx
  jmp    L8
```

assistant: ```c
char *strncpy(char *s1, const char *s2, size_t n) {
  char *rs = s1;
  while (*s2 != 0 && 0 < n) {
    *s1 = *s2;
    s1++;
    s2++;
    n--;
  }
  while (0 < n--) {
    *s1 = 0;
    s1++;
  }
  return rs;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strncpy>:
  endbr64
  push   %r12
  movzbl (%rsi),%eax
  mov    %rdi,%r12
  test   %al,%al
  je     L44
  test   %rdx,%rdx
  jne    L24
  jmp    L33
  nopw   0x0(%rax,%rax,1)
L2:
  test   %rdx,%rdx
  je     L33
L24:
  add    $0x1,%rsi
  mov    %al,(%rdi)
  sub    $0x1,%rdx
  add    $0x1,%rdi
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L2
L44:
  test   %rdx,%rdx
  je     L33
  xor    %esi,%esi
  call   <memset@plt>
L33:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *strncpy(char *s1, const char *s2, size_t n) {
  char *rs = s1;
  while (*s2 != 0 && 0 < n) {
    *s1 = *s2;
    s1++;
    s2++;
    n--;
  }
  while (0 < n--) {
    *s1 = 0;
    s1++;
  }
  return rs;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strncpy>:
  endbr64
  push   %r12
  movzbl (%rsi),%eax
  mov    %rdi,%r12
  test   %al,%al
  je     L26
  test   %rdx,%rdx
  jne    L43
  jmp    L50
  nopw   0x0(%rax,%rax,1)
L55:
  test   %rdx,%rdx
  je     L50
L43:
  add    $0x1,%rsi
  mov    %al,(%rdi)
  sub    $0x1,%rdx
  add    $0x1,%rdi
  movzbl (%rsi),%eax
  test   %al,%al
  jne    L55
L26:
  test   %rdx,%rdx
  je     L50
  xor    %esi,%esi
  call   <memset@plt>
L50:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *strncpy(char *s1, const char *s2, size_t n) {
  char *rs = s1;
  while (*s2 != 0 && 0 < n) {
    *s1 = *s2;
    s1++;
    s2++;
    n--;
  }
  while (0 < n--) {
    *s1 = 0;
    s1++;
  }
  return rs;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<is_atg>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <is_a@plt>
  test   %rax,%rax
  je     L10
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <is_t@plt>
  test   %rax,%rax
  je     L10
  mov    -0xc(%rbp),%eax
  lea    0x2(%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <is_g@plt>
  test   %rax,%rax
  jne    L35
L10:
  mov    $0x0,%eax
  jmp    L12
L35:
  mov    $0x1,%eax
L12:
  leave
  ret
```

assistant: ```c
int is_atg(unsigned char *seq, int n) {
  if (is_a(seq, n) == 0 || is_t(seq, n + 1) == 0 || is_g(seq, n + 2) == 0)
    return 0;
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<is_atg>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  call   <is_a@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  jne    L27
L17:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L27:
  lea    0x1(%rbx),%esi
  mov    %rbp,%rdi
  call   <is_t@plt>
  mov    %rax,%rdx
  mov    $0x0,%eax
  test   %rdx,%rdx
  je     L17
  lea    0x2(%rbx),%esi
  mov    %rbp,%rdi
  call   <is_g@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  jmp    L17
```

assistant: ```c
int is_atg(unsigned char *seq, int n) {
  if (is_a(seq, n) == 0 || is_t(seq, n + 1) == 0 || is_g(seq, n + 2) == 0)
    return 0;
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<is_atg>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <is_a@plt>
  test   %rax,%rax
  jne    L16
L13:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L16:
  lea    0x1(%rbx),%esi
  mov    %rbp,%rdi
  call   <is_t@plt>
  test   %rax,%rax
  je     L13
  lea    0x2(%rbx),%esi
  mov    %rbp,%rdi
  call   <is_g@plt>
  test   %rax,%rax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int is_atg(unsigned char *seq, int n) {
  if (is_a(seq, n) == 0 || is_t(seq, n + 1) == 0 || is_g(seq, n + 2) == 0)
    return 0;
  return 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<is_atg>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <is_a@plt>
  test   %rax,%rax
  jne    L10
L11:
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L10:
  lea    0x1(%rbx),%esi
  mov    %rbp,%rdi
  call   <is_t@plt>
  test   %rax,%rax
  je     L11
  lea    0x2(%rbx),%esi
  mov    %rbp,%rdi
  call   <is_g@plt>
  test   %rax,%rax
  setne  %al
  add    $0x8,%rsp
  movzbl %al,%eax
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int is_atg(unsigned char *seq, int n) {
  if (is_a(seq, n) == 0 || is_t(seq, n + 1) == 0 || is_g(seq, n + 2) == 0)
    return 0;
  return 1;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<tim_substr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L23
L14:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    D12(%rip),%rcx
  mov    %dl,(%rax,%rcx,1)
  addl   $0x1,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L23:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L7
  mov    -0x8(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L14
L7:
  mov    -0x8(%rbp),%eax
  cltq
  lea    D12(%rip),%rdx
  movb   $0x0,(%rax,%rdx,1)
  lea    D12(%rip),%rax
  leave
  ret
```

assistant: ```c
char *tim_substr(string, start, how_many)
char *string;
int start, how_many;
{
  int x, cnt;
  int str_len = strlen(string);
  static char ret_str[100];
  for (x = start, cnt = 0; x < str_len && cnt < how_many; cnt++, x++)
    ret_str[cnt] = string[x];
  ret_str[cnt] = '\0';
  return ret_str;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<tim_substr>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%r12d
  mov    %edx,%ebp
  call   <strlen@plt>
  cmp    %eax,%r12d
  jge    L12
  mov    %eax,%r8d
  test   %ebp,%ebp
  jle    L12
  lea    0x1(%r12),%eax
  cltq
  mov    $0x0,%edx
  movslq %r12d,%r12
  lea    D2(%rip),%rdi
  sub    %r12,%rdi
L23:
  movzbl -0x1(%rbx,%rax,1),%ecx
  mov    %cl,-0x1(%rdi,%rax,1)
  add    $0x1,%edx
  cmp    %eax,%r8d
  setg   %sil
  cmp    %edx,%ebp
  setg   %cl
  add    $0x1,%rax
  test   %cl,%sil
  jne    L23
L15:
  lea    D2(%rip),%rax
  movslq %edx,%rdx
  movb   $0x0,(%rax,%rdx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L12:
  mov    $0x0,%edx
  jmp    L15
```

assistant: ```c
char *tim_substr(string, start, how_many)
char *string;
int start, how_many;
{
  int x, cnt;
  int str_len = strlen(string);
  static char ret_str[100];
  for (x = start, cnt = 0; x < str_len && cnt < how_many; cnt++, x++)
    ret_str[cnt] = string[x];
  ret_str[cnt] = '\0';
  return ret_str;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<tim_substr>:
  endbr64
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cmp    %eax,%r12d
  jge    L18
  test   %ebp,%ebp
  jle    L18
  lea    D11(%rip),%r9
  mov    %eax,%edi
  lea    0x1(%r12),%eax
  xor    %edx,%edx
  mov    %r9,%r8
  cltq
  sub    %r12,%r8
  cs nopw 0x0(%rax,%rax,1)
L25:
  movzbl -0x1(%rbx,%rax,1),%ecx
  add    $0x1,%edx
  cmp    %eax,%edi
  setg   %sil
  cmp    %edx,%ebp
  mov    %cl,-0x1(%r8,%rax,1)
  setg   %cl
  add    $0x1,%rax
  test   %cl,%sil
  jne    L25
  movslq %edx,%rdx
  mov    %r9,%rax
  movb   $0x0,(%r9,%rdx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L18:
  xor    %edx,%edx
  lea    D11(%rip),%r9
  movslq %edx,%rdx
  mov    %r9,%rax
  movb   $0x0,(%r9,%rdx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *tim_substr(string, start, how_many)
char *string;
int start, how_many;
{
  int x, cnt;
  int str_len = strlen(string);
  static char ret_str[100];
  for (x = start, cnt = 0; x < str_len && cnt < how_many; cnt++, x++)
    ret_str[cnt] = string[x];
  ret_str[cnt] = '\0';
  return ret_str;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<tim_substr>:
  endbr64
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %edx,%ebp
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  cmp    %eax,%r12d
  jge    L27
  test   %ebp,%ebp
  jle    L27
  lea    D9(%rip),%r9
  mov    %eax,%edi
  lea    0x1(%r12),%eax
  xor    %edx,%edx
  mov    %r9,%r8
  cltq
  sub    %r12,%r8
  cs nopw 0x0(%rax,%rax,1)
L11:
  movzbl -0x1(%rbx,%rax,1),%ecx
  add    $0x1,%edx
  cmp    %eax,%edi
  setg   %sil
  cmp    %edx,%ebp
  mov    %cl,-0x1(%r8,%rax,1)
  setg   %cl
  add    $0x1,%rax
  test   %cl,%sil
  jne    L11
  movslq %edx,%rdx
  mov    %r9,%rax
  movb   $0x0,(%r9,%rdx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L27:
  xor    %edx,%edx
  lea    D9(%rip),%r9
  movslq %edx,%rdx
  mov    %r9,%rax
  movb   $0x0,(%r9,%rdx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *tim_substr(string, start, how_many)
char *string;
int start, how_many;
{
  int x, cnt;
  int str_len = strlen(string);
  static char ret_str[100];
  for (x = start, cnt = 0; x < str_len && cnt < how_many; cnt++, x++)
    ret_str[cnt] = string[x];
  ret_str[cnt] = '\0';
  return ret_str;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<wots_gen_pk>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L1
L17:
  mov    -0x4(%rbp),%edx
  mov    -0x30(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <set_chain_addr@plt>
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x30(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <wots_gen_sk@plt>
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rsi
  mov    -0x4(%rbp),%eax
  shl    $0x4,%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x30(%rbp),%rcx
  mov    -0x28(%rbp),%rdx
  mov    %rcx,%r9
  mov    %rdx,%r8
  mov    $0xf,%ecx
  mov    $0x0,%edx
  mov    %rax,%rdi
  call   <gen_chain@plt>
  addl   $0x1,-0x4(%rbp)
L1:
  cmpl   $0x22,-0x4(%rbp)
  jle    L17
  nop
  nop
  leave
  ret
```

assistant: ```c
void wots_gen_pk(unsigned char *pk, const unsigned char *sk_seed, const unsigned char *pub_seed, uint32_t addr[8]) {
  uint32_t i;
  for (i = 0; i < ((8 * 16 / 4) + 3); i++) {
    set_chain_addr(addr, i);
    wots_gen_sk(pk + i * 16, sk_seed, addr);
    gen_chain(pk + i * 16, pk + i * 16, 0, 16 - 1, pub_seed, addr);
  }
}
```