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

```asm
<semap_post>:
  endbr64
  push   %r12
  movd   %edx,%xmm1
  movd   %esi,%xmm0
  mov    $0x1,%edx
  punpckldq %xmm1,%xmm0
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rsi
  movq   %xmm0,(%rsp)
  movq   $0x0,0x8(%rsp)
  call   <semop@plt>
  mov    %eax,%r12d
  cmp    $0xffffffff,%eax
  je     L27
L16:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x20,%rsp
  mov    %r12d,%eax
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L27:
  mov    D7(%rip),%rax
  xor    %edi,%edi
  mov    (%rax),%esi
  call   <kill@plt>
  jmp    L16
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int semap_post(int sem_id, int nsem, int cnt) {
  struct sembuf op[1];
  op[0].sem_num = nsem;
  op[0].sem_op = cnt;
  op[0].sem_flg = 0;
  int ret;
  ret = semop(sem_id, op, 1);
  if (ret == -1) {
    kill(0, SIGINT);
  }
  return ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<icmp_checksum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  jmp    L5
L39:
  mov    -0x10(%rbp),%rax
  lea    0x4(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    (%rax),%eax
  add    %eax,-0x14(%rbp)
  subq   $0x2,-0x30(%rbp)
L5:
  cmpq   $0x1,-0x30(%rbp)
  ja     L39
  cmpq   $0x0,-0x30(%rbp)
  je     L22
  movl   $0x0,-0x18(%rbp)
  lea    -0x18(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%eax
  add    %eax,-0x14(%rbp)
L22:
  mov    -0x14(%rbp),%eax
  movzwl %ax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0x10,%eax
  add    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  movzwl %ax,%edx
  mov    -0x14(%rbp),%eax
  sar    $0x10,%eax
  add    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  not    %eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L26
  call   <__stack_chk_fail@plt>
L26:
  leave
  ret
```

assistant: ```c
uint16_t icmp_checksum(struct icmp *i, size_t len) {
  uint32_t sum = 0;
  uint16_t *p = (uint16_t *)i;
  while (len >= 2) {
    sum += *(p++);
    len -= 2;
  }
  if (len) {
    uint16_t x = 0;
    *(char *)&x = *(char *)p;
    sum += x;
  }
  sum = (sum & 0xFFFF) + (sum >> 16);
  sum = (sum & 0xFFFF) + (sum >> 16);
  return ~sum;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<icmp_checksum>:
  endbr64
  cmp    $0x1,%rsi
  jbe    L33
  lea    -0x2(%rsi),%r8
  shr    %r8
  lea    0x1(%r8),%r9
  mov    $0x0,%eax
  mov    $0x0,%edx
L18:
  add    (%rdi,%rax,4),%edx
  mov    %rax,%rcx
  add    $0x1,%rax
  cmp    %r8,%rcx
  jne    L18
  lea    (%rdi,%r9,4),%rdi
  and    $0x1,%esi
L19:
  test   %rsi,%rsi
  je     L4
  movzbl (%rdi),%eax
  lea    (%rax,%rdx,1),%edx
L4:
  movzwl %dx,%eax
  sar    $0x10,%edx
  add    %eax,%edx
  movzwl %dx,%eax
  sar    $0x10,%edx
  add    %edx,%eax
  not    %eax
  ret
L33:
  mov    $0x0,%edx
  jmp    L19
```

assistant: ```c
uint16_t icmp_checksum(struct icmp *i, size_t len) {
  uint32_t sum = 0;
  uint16_t *p = (uint16_t *)i;
  while (len >= 2) {
    sum += *(p++);
    len -= 2;
  }
  if (len) {
    uint16_t x = 0;
    *(char *)&x = *(char *)p;
    sum += x;
  }
  sum = (sum & 0xFFFF) + (sum >> 16);
  sum = (sum & 0xFFFF) + (sum >> 16);
  return ~sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<icmp_checksum>:
  endbr64
  cmp    $0x1,%rsi
  jbe    L25
  lea    -0x2(%rsi),%r8
  xor    %eax,%eax
  xor    %edx,%edx
  shr    %r8
  lea    0x1(%r8),%r9
  nopl   0x0(%rax)
L3:
  mov    %rax,%rcx
  add    (%rdi,%rax,4),%edx
  add    $0x1,%rax
  cmp    %rcx,%r8
  jne    L3
  lea    (%rdi,%r9,4),%rdi
  and    $0x1,%esi
L35:
  test   %rsi,%rsi
  je     L22
  movzbl (%rdi),%eax
  add    %eax,%edx
L22:
  movzwl %dx,%eax
  sar    $0x10,%edx
  add    %eax,%edx
  movzwl %dx,%eax
  sar    $0x10,%edx
  add    %edx,%eax
  not    %eax
  ret
  nopl   0x0(%rax,%rax,1)
L25:
  xor    %edx,%edx
  jmp    L35
```

assistant: ```c
uint16_t icmp_checksum(struct icmp *i, size_t len) {
  uint32_t sum = 0;
  uint16_t *p = (uint16_t *)i;
  while (len >= 2) {
    sum += *(p++);
    len -= 2;
  }
  if (len) {
    uint16_t x = 0;
    *(char *)&x = *(char *)p;
    sum += x;
  }
  sum = (sum & 0xFFFF) + (sum >> 16);
  sum = (sum & 0xFFFF) + (sum >> 16);
  return ~sum;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<icmp_checksum>:
  endbr64
  mov    %rdi,%rdx
  cmp    $0x1,%rsi
  jbe    L22
  lea    -0x2(%rsi),%rax
  mov    %rax,%rdi
  shr    %rdi
  add    $0x1,%rdi
  cmp    $0x5,%rax
  jbe    L46
  mov    %rdi,%r8
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  shr    $0x2,%r8
  cs nopw 0x0(%rax,%rax,1)
L83:
  mov    %rax,%rcx
  add    $0x1,%rax
  shl    $0x4,%rcx
  movdqu (%rdx,%rcx,1),%xmm1
  paddd  %xmm1,%xmm0
  cmp    %rax,%r8
  jne    L83
  movdqa %xmm0,%xmm1
  mov    %rdi,%r8
  psrldq $0x8,%xmm1
  and    $0xfffffffffffffffc,%r8
  paddd  %xmm1,%xmm0
  mov    %r8,%rcx
  lea    (%rdx,%r8,4),%r9
  movdqa %xmm0,%xmm1
  neg    %rcx
  psrldq $0x4,%xmm1
  lea    (%rsi,%rcx,2),%rcx
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %r8,%rdi
  je     L72
L55:
  lea    -0x2(%rcx),%r8
  add    (%r9),%eax
  cmp    $0x1,%r8
  jbe    L72
  sub    $0x4,%rcx
  add    0x4(%r9),%eax
  cmp    $0x1,%rcx
  jbe    L72
  add    0x8(%r9),%eax
L72:
  lea    (%rdx,%rdi,4),%rdx
  and    $0x1,%esi
L28:
  test   %rsi,%rsi
  je     L67
  movzbl (%rdx),%edx
  add    %edx,%eax
L67:
  movzwl %ax,%edx
  sar    $0x10,%eax
  add    %eax,%edx
  movzwl %dx,%eax
  sar    $0x10,%edx
  add    %edx,%eax
  not    %eax
  ret
  nopl   0x0(%rax)
L22:
  xor    %eax,%eax
  jmp    L28
L46:
  mov    %rdx,%r9
  mov    %rsi,%rcx
  xor    %eax,%eax
  jmp    L55
```

assistant: ```c
uint16_t icmp_checksum(struct icmp *i, size_t len) {
  uint32_t sum = 0;
  uint16_t *p = (uint16_t *)i;
  while (len >= 2) {
    sum += *(p++);
    len -= 2;
  }
  if (len) {
    uint16_t x = 0;
    *(char *)&x = *(char *)p;
    sum += x;
  }
  sum = (sum & 0xFFFF) + (sum >> 16);
  sum = (sum & 0xFFFF) + (sum >> 16);
  return ~sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L19
  mov    -0x10(%rbp),%rax
  jmp    L5
L19:
  mov    -0x10(%rbp),%rax
  imul   -0x8(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    $0x1,%rdx
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <factorial>
L5:
  leave
  ret
```

assistant: ```c
long factorial(long num, long acc) {
  if (num == 0)
    return acc;
  else
    return factorial(num - 1, acc * num);
}
```
user: Translate the below assembly code into C programming language:

```asm
<factorial>:
  endbr64
  mov    %rsi,%rax
  test   %rdi,%rdi
  jne    L3
  ret
L3:
  sub    $0x8,%rsp
  imul   %rdi,%rax
  mov    %rax,%rsi
  sub    $0x1,%rdi
  call   <factorial>
  add    $0x8,%rsp
  ret
```

assistant: ```c
long factorial(long num, long acc) {
  if (num == 0)
    return acc;
  else
    return factorial(num - 1, acc * num);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<factorial>:
  endbr64
  mov    %rsi,%rax
  test   %rdi,%rdi
  je     L14
  nopl   0x0(%rax)
L3:
  imul   %rdi,%rax
  sub    $0x1,%rdi
  jne    L3
L14:
  ret
```

assistant: ```c
long factorial(long num, long acc) {
  if (num == 0)
    return acc;
  else
    return factorial(num - 1, acc * num);
}
```
user: What is the c source code of the assembly code below?

```asm
<factorial>:
  endbr64
  mov    %rsi,%rax
  test   %rdi,%rdi
  je     L3
  nopl   0x0(%rax)
L17:
  imul   %rdi,%rax
  sub    $0x1,%rdi
  jne    L17
L3:
  ret
```

assistant: ```c
long factorial(long num, long acc) {
  if (num == 0)
    return acc;
  else
    return factorial(num - 1, acc * num);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<inportb>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %ax,-0x14(%rbp)
  movzwl -0x14(%rbp),%eax
  mov    %eax,%edx
  in     (%dx),%al
  mov    %al,-0x1(%rbp)
  movzbl -0x1(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned char inportb(unsigned short port) {
  unsigned char rv;
  asm volatile("inb %1, %0" : "=a"(rv) : "dN"(port));
  return rv;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<inportb>:
  endbr64
  mov    %edi,%edx
  in     (%dx),%al
  ret
```

assistant: ```c
unsigned char inportb(unsigned short port) {
  unsigned char rv;
  asm volatile("inb %1, %0" : "=a"(rv) : "dN"(port));
  return rv;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<inportb>:
  endbr64
  mov    %edi,%edx
  in     (%dx),%al
  ret
```

assistant: ```c
unsigned char inportb(unsigned short port) {
  unsigned char rv;
  asm volatile("inb %1, %0" : "=a"(rv) : "dN"(port));
  return rv;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<inportb>:
  endbr64
  mov    %edi,%edx
  in     (%dx),%al
  ret
```

assistant: ```c
unsigned char inportb(unsigned short port) {
  unsigned char rv;
  asm volatile("inb %1, %0" : "=a"(rv) : "dN"(port));
  return rv;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_linea>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x8(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <fgets@plt>
  test   %rax,%rax
  jne    L18
  mov    $0x1,%eax
  jmp    L1
L18:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    $0x0,%eax
L1:
  leave
  ret
```

assistant: ```c
int read_linea(FILE *stream, char linea[1000]) {
  if (fgets(linea, 1000, stream) == NULL)
    return 1;
  linea[strlen(linea) - 1] = '\0';
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_linea>:
  endbr64
  push   %rbx
  mov    %rdi,%rdx
  mov    %rsi,%rbx
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  je     L5
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%rbx,%rax,1)
  mov    $0x0,%eax
L5:
  pop    %rbx
  ret
```

assistant: ```c
int read_linea(FILE *stream, char linea[1000]) {
  if (fgets(linea, 1000, stream) == NULL)
    return 1;
  linea[strlen(linea) - 1] = '\0';
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_linea>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdi,%rdx
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L3
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%rbx,%rax,1)
  xor    %eax,%eax
L3:
  pop    %rbx
  ret
```

assistant: ```c
int read_linea(FILE *stream, char linea[1000]) {
  if (fgets(linea, 1000, stream) == NULL)
    return 1;
  linea[strlen(linea) - 1] = '\0';
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<read_linea>:
  endbr64
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdi,%rdx
  mov    $0x3e8,%esi
  mov    %rbx,%rdi
  call   <fgets@plt>
  mov    %rax,%r8
  mov    $0x1,%eax
  test   %r8,%r8
  je     L2
  mov    %rbx,%rdi
  call   <strlen@plt>
  cltq
  movb   $0x0,-0x1(%rbx,%rax,1)
  xor    %eax,%eax
L2:
  pop    %rbx
  ret
```

assistant: ```c
int read_linea(FILE *stream, char linea[1000]) {
  if (fgets(linea, 1000, stream) == NULL)
    return 1;
  linea[strlen(linea) - 1] = '\0';
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<memcpy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L4
L14:
  mov    -0x8(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x10(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L4:
  mov    -0x24(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x24(%rbp)
  test   %eax,%eax
  jne    L14
  mov    -0x18(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
void *memcpy(void *_dst, const void *_src, unsigned int len) {
  unsigned char *dst = _dst;
  const unsigned char *src = _src;
  while (len-- > 0) {
    *dst++ = *src++;
  }
  return _dst;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<memcpy>:
  endbr64
  mov    %rdi,%rax
  lea    -0x1(%rdx),%edi
  test   %edx,%edx
  je     L13
  mov    %edi,%edi
  mov    $0x0,%edx
L15:
  movzbl (%rsi,%rdx,1),%ecx
  mov    %cl,(%rax,%rdx,1)
  mov    %rdx,%rcx
  add    $0x1,%rdx
  cmp    %rcx,%rdi
  jne    L15
L13:
  ret
```

assistant: ```c
void *memcpy(void *_dst, const void *_src, unsigned int len) {
  unsigned char *dst = _dst;
  const unsigned char *src = _src;
  while (len-- > 0) {
    *dst++ = *src++;
  }
  return _dst;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<memcpy>:
  endbr64
  mov    %rdi,%rax
  test   %edx,%edx
  je     L6
  mov    %edx,%edx
  xor    %ecx,%ecx
  nop
L1:
  movzbl (%rsi,%rcx,1),%r8d
  mov    %r8b,(%rax,%rcx,1)
  add    $0x1,%rcx
  cmp    %rdx,%rcx
  jne    L1
L6:
  ret
```

assistant: ```c
void *memcpy(void *_dst, const void *_src, unsigned int len) {
  unsigned char *dst = _dst;
  const unsigned char *src = _src;
  while (len-- > 0) {
    *dst++ = *src++;
  }
  return _dst;
}
```
user: What is the c source code of the assembly code below?

```asm
<memcpy>:
  endbr64
  mov    %rdi,%r8
  mov    %rsi,%rcx
  lea    -0x1(%rdx),%r10d
  test   %edx,%edx
  je     L68
  lea    0x1(%rsi),%rsi
  mov    %rdi,%rax
  sub    %rsi,%rax
  cmp    $0xe,%rax
  jbe    L120
  cmp    $0x6,%r10d
  jbe    L120
  cmp    $0xe,%r10d
  jbe    L2
  mov    %edx,%esi
  xor    %eax,%eax
  shr    $0x4,%esi
  shl    $0x4,%rsi
  nopl   0x0(%rax)
L77:
  movdqu (%rcx,%rax,1),%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rsi
  jne    L77
  mov    %edx,%r9d
  and    $0xfffffff0,%r9d
  mov    %r9d,%edi
  sub    %r9d,%r10d
  lea    (%r8,%rdi,1),%rax
  lea    (%rcx,%rdi,1),%rsi
  cmp    %r9d,%edx
  je     L68
  sub    %r9d,%edx
  lea    -0x1(%rdx),%r9d
  cmp    $0x6,%r9d
  jbe    L28
L132:
  mov    (%rcx,%rdi,1),%rcx
  mov    %rcx,(%r8,%rdi,1)
  mov    %edx,%ecx
  and    $0xfffffff8,%ecx
  mov    %ecx,%edi
  sub    %ecx,%r10d
  add    %rdi,%rax
  add    %rdi,%rsi
  cmp    %ecx,%edx
  je     L68
L28:
  movzbl (%rsi),%edx
  mov    %dl,(%rax)
  test   %r10d,%r10d
  je     L68
  movzbl 0x1(%rsi),%edx
  mov    %dl,0x1(%rax)
  cmp    $0x1,%r10d
  je     L68
  movzbl 0x2(%rsi),%edx
  mov    %dl,0x2(%rax)
  cmp    $0x2,%r10d
  je     L68
  movzbl 0x3(%rsi),%edx
  mov    %dl,0x3(%rax)
  cmp    $0x3,%r10d
  je     L68
  movzbl 0x4(%rsi),%edx
  mov    %dl,0x4(%rax)
  cmp    $0x4,%r10d
  je     L68
  movzbl 0x5(%rsi),%edx
  mov    %dl,0x5(%rax)
  cmp    $0x5,%r10d
  je     L68
  movzbl 0x6(%rsi),%edx
  mov    %dl,0x6(%rax)
L68:
  mov    %r8,%rax
  ret
  nopl   0x0(%rax)
L120:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L71:
  movzbl (%rcx,%rax,1),%esi
  mov    %sil,(%r8,%rax,1)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L71
  mov    %r8,%rax
  ret
L2:
  mov    %rdi,%rax
  mov    %rcx,%rsi
  xor    %edi,%edi
  jmp    L132
```

assistant: ```c
void *memcpy(void *_dst, const void *_src, unsigned int len) {
  unsigned char *dst = _dst;
  const unsigned char *src = _src;
  while (len-- > 0) {
    *dst++ = *src++;
  }
  return _dst;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<strchr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  jmp    L41
L39:
  addq   $0x1,-0x8(%rbp)
L41:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L34
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    -0xc(%rbp),%edx
  cmp    %dl,%al
  jne    L39
L34:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  mov    -0xc(%rbp),%edx
  cmp    %dl,%al
  jne    L1
  mov    -0x8(%rbp),%rax
  jmp    L20
L1:
  mov    $0x0,%eax
L20:
  pop    %rbp
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  for (; *s && *s != (char)c; s++)
    ;
  return (*s == (char)c ? (char *)s : 0);
}
```
user: Translate the below assembly code into C programming language:

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  test   %al,%al
  je     L18
L14:
  cmp    %sil,%al
  je     L7
  add    $0x1,%rdi
  movzbl (%rdi),%eax
  test   %al,%al
  jne    L14
L18:
  test   %sil,%sil
  mov    $0x0,%eax
  cmove  %rdi,%rax
  ret
L7:
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  for (; *s && *s != (char)c; s++)
    ;
  return (*s == (char)c ? (char *)s : 0);
}
```
user: What is the c source code of the assembly code below:

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  test   %al,%al
  jne    L14
  jmp    L25
  nop
L5:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L25
L14:
  cmp    %al,%dl
  jne    L5
  mov    %rdi,%rax
  ret
  nopl   0x0(%rax)
L25:
  xor    %eax,%eax
  test   %sil,%sil
  cmove  %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  for (; *s && *s != (char)c; s++)
    ;
  return (*s == (char)c ? (char *)s : 0);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<strchr>:
  endbr64
  movzbl (%rdi),%eax
  mov    %esi,%edx
  test   %al,%al
  jne    L21
  jmp    L6
  nop
L12:
  movzbl 0x1(%rdi),%eax
  add    $0x1,%rdi
  test   %al,%al
  je     L6
L21:
  cmp    %al,%dl
  jne    L12
  mov    %rdi,%rax
  ret
  nopl   0x0(%rax)
L6:
  xor    %eax,%eax
  test   %sil,%sil
  cmove  %rdi,%rax
  ret
```

assistant: ```c
char *strchr(const char *s, int c) {
  for (; *s && *s != (char)c; s++)
    ;
  return (*s == (char)c ? (char *)s : 0);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<t0s>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dl,-0x14(%rbp)
  mov    %al,-0x18(%rbp)
  movsbl -0x14(%rbp),%eax
  mov    %eax,%edx
  lea    D44(%rip),%rax
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl -0x18(%rbp),%eax
  mov    %eax,%edx
  lea    D44(%rip),%rax
  mov    %rax,%rsi
  lea    D38(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl -0x14(%rbp),%edx
  movsbl -0x18(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  movzbl -0x14(%rbp),%edx
  movzbl -0x18(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D44
type_guessed: string
name: parse_data
arguments: label: D30
type_guessed: string
name: parse_data
arguments: label: D38
type_guessed: string
name: parse_data
arguments: label: D27
type_guessed: string

tool: D44: t0s

tool: D30: %s: a: %d


tool: D38: %s: b: %d


tool: D27: %d


assistant: ```c
char t0s(char a, char b) {
  printf("%s: a: %d\n", __func__, a);
  printf("%s: b: %d\n", __func__, b);
  printf("%d\n", a + b);
  int x = (int)a;
  return a + b;
}
```
user: What is the c source code of the assembly code below?

```asm
<t0s>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  mov    %esi,%ebx
  movsbl %dil,%r12d
  mov    %r12d,%edx
  lea    D12(%rip),%r14
  mov    %r14,%rsi
  lea    D18(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movsbl %bl,%r13d
  mov    %r13d,%edx
  mov    %r14,%rsi
  lea    D16(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    (%r12,%r13,1),%esi
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: label: D12
type_guessed: string
name: read_data
arguments: label: D18
type_guessed: string
name: read_data
arguments: label: D16
type_guessed: string
name: read_data
arguments: label: D19
type_guessed: string

tool: D12: t0s

tool: D18: %s: a: %d


tool: D16: %s: b: %d


tool: D19: %d


assistant: ```c
char t0s(char a, char b) {
  printf("%s: a: %d\n", __func__, a);
  printf("%s: b: %d\n", __func__, b);
  printf("%d\n", a + b);
  int x = (int)a;
  return a + b;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<t0s>:
  endbr64
  push   %r14
  lea    D26(%rip),%r14
  xor    %eax,%eax
  push   %r13
  push   %r12
  movsbl %dil,%r12d
  push   %rbp
  mov    %r12d,%edx
  mov    %edi,%ebp
  lea    D1(%rip),%rdi
  push   %rbx
  mov    %esi,%ebx
  mov    %r14,%rsi
  movsbl %bl,%r13d
  call   <printf@plt>
  mov    %r13d,%edx
  mov    %r14,%rsi
  xor    %eax,%eax
  lea    D17(%rip),%rdi
  call   <printf@plt>
  lea    (%r12,%r13,1),%esi
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D26
type_guessed: string
name: extract_symbolic_data
arguments: label: D1
type_guessed: string
name: extract_symbolic_data
arguments: label: D17
type_guessed: string
name: extract_symbolic_data
arguments: label: D28
type_guessed: string

tool: D26: t0s

tool: D1: %s: a: %d


tool: D17: %s: b: %d


tool: D28: %d


assistant: ```c
char t0s(char a, char b) {
  printf("%s: a: %d\n", __func__, a);
  printf("%s: b: %d\n", __func__, b);
  printf("%d\n", a + b);
  int x = (int)a;
  return a + b;
}
```
user: Translate the below assembly code into C programming language:

```asm
<t0s>:
  endbr64
  push   %r14
  lea    D26(%rip),%r14
  xor    %eax,%eax
  push   %r13
  push   %r12
  movsbl %dil,%r12d
  push   %rbp
  mov    %r12d,%edx
  mov    %edi,%ebp
  lea    D14(%rip),%rdi
  push   %rbx
  mov    %esi,%ebx
  mov    %r14,%rsi
  movsbl %bl,%r13d
  call   <printf@plt>
  mov    %r13d,%edx
  mov    %r14,%rsi
  xor    %eax,%eax
  lea    D38(%rip),%rdi
  call   <printf@plt>
  lea    (%r12,%r13,1),%esi
  lea    D15(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    (%rbx,%rbp,1),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: disassemble_data
arguments: label: D26
datatype: string
name: disassemble_data
arguments: label: D14
datatype: string
name: disassemble_data
arguments: label: D38
datatype: string
name: disassemble_data
arguments: label: D15
datatype: string

tool: D26: t0s

tool: D14: %s: a: %d


tool: D38: %s: b: %d


tool: D15: %d


assistant: ```c
char t0s(char a, char b) {
  printf("%s: a: %d\n", __func__, a);
  printf("%s: b: %d\n", __func__, b);
  printf("%d\n", a + b);
  int x = (int)a;
  return a + b;
}
```
user: What is the c source code of the assembly code below:

```asm
<safeStrCpy_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpq   $0x0,-0x20(%rbp)
  je     L20
  jmp    L28
L19:
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x4(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rcx),%edx
  mov    %dl,(%rax)
L28:
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x4(%rbp)
  jl     L19
L20:
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void safeStrCpy_(char *dst, const char *src, int dstSize) {
  int i = 0;
  if (src)
    for (; i < dstSize - 1; *(dst++) = src[i++])
      ;
  *dst = '\0';
}
```