ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<SeleccionPadres>:
  endbr64
  push   %r15
  push   %r14
  lea    D13(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  lea    D3(%rip),%rdi
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  lea    0x14(%rsp),%r15
  mov    %r12,%r13
  call   <printf@plt>
L5:
  xor    %eax,%eax
  add    $0x4,%r13
  call   <drand48@plt>
  pxor   %xmm0,%xmm0
  mov    %r14,%rdi
  cvtsi2ss %eax,%xmm0
  mov    $0x1,%eax
  movss  %xmm0,-0x4(%r13)
  cvtss2sd %xmm0,%xmm0
  call   <printf@plt>
  cmp    %r15,%r13
  jne    L5
  mov    $0xa,%edi
  call   <putchar@plt>
  xor    %ecx,%ecx
  mov    $0x5,%esi
L26:
  movss  (%r12,%rcx,1),%xmm0
  mov    $0x4,%eax
  nopl   0x0(%rax)
L34:
  mov    %eax,%edx
  sub    $0x1,%rax
  comiss 0x4(%rbx,%rax,4),%xmm0
  jae    L34
  mov    %esi,%eax
  sub    %edx,%eax
  mov    %eax,0x0(%rbp,%rcx,1)
  add    $0x4,%rcx
  cmp    $0x14,%rcx
  jne    L26
  mov    0x14(%rbp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L23
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L23:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D13
type_guessed: string
name: unpack_data
arguments: data_label: D3
type_guessed: string

tool: D13: %f 

tool: D3: aleatorios

assistant: ```c
int SeleccionPadres(int padres[5], float acumulado[5]) {
  int i, x;
  float Aleatorio[5];
  printf("aleatorios");
  for (i = 0; i < 5; i++) {
    Aleatorio[i] = drand48() * 1;
    printf("%f ", Aleatorio[i]);
  }
  printf("\n");
  for (x = 0; x < 5; x++) {
    i = 5;
    do {
      i--;
    } while (acumulado[i] <= Aleatorio[x]);
    padres[x] = (5 - i);
  }
  return padres[5];
}
```
user: What is the c source code of the assembly code below:

```asm
<primalidade>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L32
L4:
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <geraNumeroMax@plt>
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0xc(%rbp),%ecx
  mov    -0x8(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <expModular@plt>
  mov    %eax,-0x4(%rbp)
  mov    $0x1,%eax
  cltd
  idivl  -0x14(%rbp)
  mov    %edx,%eax
  cmp    %eax,-0x4(%rbp)
  je     L29
  mov    $0x0,%eax
  jmp    L18
L29:
  addl   $0x1,-0x10(%rbp)
L32:
  cmpl   $0x64,-0x10(%rbp)
  jle    L4
  mov    $0x1,%eax
L18:
  leave
  ret
```

assistant: ```c
int primalidade(int n) {
  int b = n - 1;
  int i = 0;
  int a = 0;
  int result;
  while (i <= 100) {
    a = geraNumeroMax(n - 1);
    result = expModular(a, b, n);
    if (result != 1 % n) {
      return 0;
    }
    i++;
  }
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<primalidade>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  lea    -0x1(%rdi),%r13d
  mov    $0x65,%ebp
  mov    $0x1,%r12d
L14:
  mov    %r13d,%edi
  call   <geraNumeroMax@plt>
  mov    %eax,%edi
  mov    %ebx,%edx
  mov    %r13d,%esi
  call   <expModular@plt>
  mov    %eax,%ecx
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  cmp    %ecx,%edx
  jne    L21
  sub    $0x1,%ebp
  jne    L14
  mov    $0x1,%eax
  jmp    L23
L21:
  mov    $0x0,%eax
L23:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int primalidade(int n) {
  int b = n - 1;
  int i = 0;
  int a = 0;
  int result;
  while (i <= 100) {
    a = geraNumeroMax(n - 1);
    result = expModular(a, b, n);
    if (result != 1 % n) {
      return 0;
    }
    i++;
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<primalidade>:
  endbr64
  push   %r13
  lea    -0x1(%rdi),%r13d
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  mov    $0x65,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  jmp    L2
  nopl   0x0(%rax)
L4:
  sub    $0x1,%ebp
  je     L16
L2:
  mov    %r13d,%edi
  call   <geraNumeroMax@plt>
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %eax,%edi
  call   <expModular@plt>
  mov    %eax,%r8d
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  cmp    %r8d,%edx
  je     L4
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L16:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int primalidade(int n) {
  int b = n - 1;
  int i = 0;
  int a = 0;
  int result;
  while (i <= 100) {
    a = geraNumeroMax(n - 1);
    result = expModular(a, b, n);
    if (result != 1 % n) {
      return 0;
    }
    i++;
  }
  return 1;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<primalidade>:
  endbr64
  push   %r13
  lea    -0x1(%rdi),%r13d
  push   %r12
  mov    $0x1,%r12d
  push   %rbp
  mov    $0x65,%ebp
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  jmp    L12
  nopl   0x0(%rax)
L18:
  sub    $0x1,%ebp
  je     L3
L12:
  mov    %r13d,%edi
  call   <geraNumeroMax@plt>
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %eax,%edi
  call   <expModular@plt>
  mov    %eax,%r8d
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  cmp    %r8d,%edx
  je     L18
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L3:
  add    $0x8,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int primalidade(int n) {
  int b = n - 1;
  int i = 0;
  int a = 0;
  int result;
  while (i <= 100) {
    a = geraNumeroMax(n - 1);
    result = expModular(a, b, n);
    if (result != 1 % n) {
      return 0;
    }
    i++;
  }
  return 1;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<read_ids>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    $0x0,%eax
  call   <__builtin_ptx_read_laneid@plt>
  mov    %eax,-0x18(%rbp)
  mov    $0x0,%eax
  call   <__builtin_ptx_read_warpid@plt>
  mov    %eax,-0x14(%rbp)
  mov    $0x0,%eax
  call   <__builtin_ptx_read_nwarpid@plt>
  mov    %eax,-0x10(%rbp)
  mov    $0x0,%eax
  call   <__builtin_ptx_read_smid@plt>
  mov    %eax,-0xc(%rbp)
  mov    $0x0,%eax
  call   <__builtin_ptx_read_nsmid@plt>
  mov    %eax,-0x8(%rbp)
  mov    $0x0,%eax
  call   <__builtin_ptx_read_gridid@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%edx
  mov    -0x14(%rbp),%eax
  add    %eax,%edx
  mov    -0x10(%rbp),%eax
  add    %eax,%edx
  mov    -0xc(%rbp),%eax
  add    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  leave
  ret
```

assistant: ```c
int read_ids() {
  int a = __builtin_ptx_read_laneid();
  int b = __builtin_ptx_read_warpid();
  int c = __builtin_ptx_read_nwarpid();
  int d = __builtin_ptx_read_smid();
  int e = __builtin_ptx_read_nsmid();
  int f = __builtin_ptx_read_gridid();
  return a + b + c + d + e + f;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<read_ids>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x0,%eax
  call   <__builtin_ptx_read_laneid@plt>
  mov    %eax,%ebx
  mov    $0x0,%eax
  call   <__builtin_ptx_read_warpid@plt>
  mov    %eax,%r14d
  mov    $0x0,%eax
  call   <__builtin_ptx_read_nwarpid@plt>
  mov    %eax,%r13d
  mov    $0x0,%eax
  call   <__builtin_ptx_read_smid@plt>
  mov    %eax,%r12d
  mov    $0x0,%eax
  call   <__builtin_ptx_read_nsmid@plt>
  mov    %eax,%ebp
  mov    $0x0,%eax
  call   <__builtin_ptx_read_gridid@plt>
  mov    %eax,%edx
  lea    (%rbx,%r14,1),%eax
  add    %r13d,%eax
  add    %r12d,%eax
  add    %ebp,%eax
  add    %edx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int read_ids() {
  int a = __builtin_ptx_read_laneid();
  int b = __builtin_ptx_read_warpid();
  int c = __builtin_ptx_read_nwarpid();
  int d = __builtin_ptx_read_smid();
  int e = __builtin_ptx_read_nsmid();
  int f = __builtin_ptx_read_gridid();
  return a + b + c + d + e + f;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<read_ids>:
  endbr64
  push   %r14
  xor    %eax,%eax
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <__builtin_ptx_read_laneid@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__builtin_ptx_read_warpid@plt>
  mov    %eax,%r14d
  xor    %eax,%eax
  call   <__builtin_ptx_read_nwarpid@plt>
  mov    %eax,%r13d
  xor    %eax,%eax
  call   <__builtin_ptx_read_smid@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <__builtin_ptx_read_nsmid@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <__builtin_ptx_read_gridid@plt>
  mov    %eax,%r8d
  lea    (%rbx,%r14,1),%eax
  pop    %rbx
  add    %r13d,%eax
  add    %r12d,%eax
  add    %ebp,%eax
  pop    %rbp
  pop    %r12
  add    %r8d,%eax
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int read_ids() {
  int a = __builtin_ptx_read_laneid();
  int b = __builtin_ptx_read_warpid();
  int c = __builtin_ptx_read_nwarpid();
  int d = __builtin_ptx_read_smid();
  int e = __builtin_ptx_read_nsmid();
  int f = __builtin_ptx_read_gridid();
  return a + b + c + d + e + f;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_ids>:
  endbr64
  push   %r14
  xor    %eax,%eax
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  call   <__builtin_ptx_read_laneid@plt>
  mov    %eax,%ebx
  xor    %eax,%eax
  call   <__builtin_ptx_read_warpid@plt>
  mov    %eax,%r14d
  xor    %eax,%eax
  call   <__builtin_ptx_read_nwarpid@plt>
  mov    %eax,%r13d
  xor    %eax,%eax
  call   <__builtin_ptx_read_smid@plt>
  mov    %eax,%r12d
  xor    %eax,%eax
  call   <__builtin_ptx_read_nsmid@plt>
  mov    %eax,%ebp
  xor    %eax,%eax
  call   <__builtin_ptx_read_gridid@plt>
  mov    %eax,%r8d
  lea    (%rbx,%r14,1),%eax
  pop    %rbx
  add    %r13d,%eax
  add    %r12d,%eax
  add    %ebp,%eax
  pop    %rbp
  pop    %r12
  add    %r8d,%eax
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int read_ids() {
  int a = __builtin_ptx_read_laneid();
  int b = __builtin_ptx_read_warpid();
  int c = __builtin_ptx_read_nwarpid();
  int d = __builtin_ptx_read_smid();
  int e = __builtin_ptx_read_nsmid();
  int f = __builtin_ptx_read_gridid();
  return a + b + c + d + e + f;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<udp_to_tun>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  mov    %rcx,-0x28(%rbp)
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x14(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <sock_recv_packet@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jns    L13
  mov    $0x0,%eax
  jmp    L18
L13:
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <decrement_ttl@plt>
  test   %eax,%eax
  je     L16
  mov    -0x10(%rbp),%rdx
  mov    -0x20(%rbp),%rcx
  mov    -0x18(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <tun_send_packet@plt>
  jmp    L23
L16:
  nop
L23:
  mov    $0x1,%eax
L18:
  leave
  ret
```

assistant: ```c
int udp_to_tun(int sock, int tun, char *buf, size_t buflen) {
  struct iphdr *iph;
  ssize_t pktlen = sock_recv_packet(sock, buf, buflen);
  if (pktlen < 0)
    return 0;
  iph = (struct iphdr *)buf;
  if (!decrement_ttl(iph)) {
    goto _active;
  }
  tun_send_packet(tun, buf, pktlen);
_active:
  return 1;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<udp_to_tun>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r12d
  mov    %rdx,%rbx
  mov    %rcx,%rdx
  mov    %rbx,%rsi
  call   <sock_recv_packet@plt>
  mov    %rax,%rbp
  mov    $0x0,%eax
  test   %rbp,%rbp
  js     L4
  mov    %rbx,%rdi
  call   <decrement_ttl@plt>
  mov    %eax,%edx
  mov    $0x1,%eax
  test   %edx,%edx
  jne    L0
L4:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L0:
  mov    %rbp,%rdx
  mov    %rbx,%rsi
  mov    %r12d,%edi
  call   <tun_send_packet@plt>
  mov    $0x1,%eax
  jmp    L4
```

assistant: ```c
int udp_to_tun(int sock, int tun, char *buf, size_t buflen) {
  struct iphdr *iph;
  ssize_t pktlen = sock_recv_packet(sock, buf, buflen);
  if (pktlen < 0)
    return 0;
  iph = (struct iphdr *)buf;
  if (!decrement_ttl(iph)) {
    goto _active;
  }
  tun_send_packet(tun, buf, pktlen);
_active:
  return 1;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<udp_to_tun>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  mov    %rcx,%rdx
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <sock_recv_packet@plt>
  test   %rax,%rax
  js     L9
  mov    %rbp,%rdi
  mov    %rax,%r12
  mov    $0x1,%r14d
  call   <decrement_ttl@plt>
  test   %eax,%eax
  jne    L13
L9:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L13:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <tun_send_packet@plt>
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int udp_to_tun(int sock, int tun, char *buf, size_t buflen) {
  struct iphdr *iph;
  ssize_t pktlen = sock_recv_packet(sock, buf, buflen);
  if (pktlen < 0)
    return 0;
  iph = (struct iphdr *)buf;
  if (!decrement_ttl(iph)) {
    goto _active;
  }
  tun_send_packet(tun, buf, pktlen);
_active:
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<udp_to_tun>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %esi,%r13d
  push   %r12
  push   %rbp
  mov    %rdx,%rbp
  mov    %rcx,%rdx
  mov    %rbp,%rsi
  sub    $0x8,%rsp
  call   <sock_recv_packet@plt>
  test   %rax,%rax
  js     L5
  mov    %rbp,%rdi
  mov    %rax,%r12
  mov    $0x1,%r14d
  call   <decrement_ttl@plt>
  test   %eax,%eax
  jne    L7
L5:
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L7:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %r13d,%edi
  call   <tun_send_packet@plt>
  add    $0x8,%rsp
  mov    %r14d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
int udp_to_tun(int sock, int tun, char *buf, size_t buflen) {
  struct iphdr *iph;
  ssize_t pktlen = sock_recv_packet(sock, buf, buflen);
  if (pktlen < 0)
    return 0;
  iph = (struct iphdr *)buf;
  if (!decrement_ttl(iph)) {
    goto _active;
  }
  tun_send_packet(tun, buf, pktlen);
_active:
  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
<malloc_image>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *malloc_image(int w, int h) {
  char *img;
  img = (char *)malloc(w * h * sizeof(char));
  return img;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<malloc_image>:
  endbr64
  sub    $0x8,%rsp
  imul   %esi,%edi
  call   <malloc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *malloc_image(int w, int h) {
  char *img;
  img = (char *)malloc(w * h * sizeof(char));
  return img;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<malloc_image>:
  endbr64
  imul   %esi,%edi
  jmp    1050 <malloc@plt>
```

assistant: ```c
char *malloc_image(int w, int h) {
  char *img;
  img = (char *)malloc(w * h * sizeof(char));
  return img;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<malloc_image>:
  endbr64
  imul   %esi,%edi
  jmp    1050 <malloc@plt>
```

assistant: ```c
char *malloc_image(int w, int h) {
  char *img;
  img = (char *)malloc(w * h * sizeof(char));
  return img;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<convert_to_hexa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x4(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L3
L19:
  mov    -0x14(%rbp),%ecx
  movslq %ecx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %ecx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%edx
  mov    %edx,%eax
  shl    $0x2,%eax
  add    %edx,%eax
  add    %eax,%eax
  sub    %eax,%ecx
  mov    %ecx,%edx
  mov    %edx,%eax
  imul   -0x4(%rbp),%eax
  add    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x14(%rbp)
  shll   $0x4,-0x4(%rbp)
L3:
  cmpl   $0x0,-0x14(%rbp)
  jg     L19
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int convert_to_hexa(int dec) {
  int result;
  int base;
  base = 1;
  result = 0;
  while (dec > 0) {
    result += (dec % 10) * base;
    dec /= 10;
    base *= 16;
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<convert_to_hexa>:
  endbr64
  test   %edi,%edi
  jle    L14
  mov    $0x1,%esi
  mov    $0x0,%r8d
L3:
  movslq %edi,%rax
  imul   $0x66666667,%rax,%rax
  sar    $0x22,%rax
  mov    %edi,%edx
  sar    $0x1f,%edx
  sub    %edx,%eax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  mov    %edi,%edx
  sub    %ecx,%edx
  imul   %esi,%edx
  add    %edx,%r8d
  mov    %edi,%edx
  mov    %eax,%edi
  shl    $0x4,%esi
  cmp    $0x9,%edx
  jg     L3
L2:
  mov    %r8d,%eax
  ret
L14:
  mov    $0x0,%r8d
  jmp    L2
```

assistant: ```c
int convert_to_hexa(int dec) {
  int result;
  int base;
  base = 1;
  result = 0;
  while (dec > 0) {
    result += (dec % 10) * base;
    dec /= 10;
    base *= 16;
  }
  return result;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<convert_to_hexa>:
  endbr64
  test   %edi,%edi
  jle    L0
  mov    $0x1,%esi
  xor    %r8d,%r8d
  mov    $0xcccccccd,%r9d
  cs nopw 0x0(%rax,%rax,1)
L17:
  mov    %edi,%eax
  mov    %edi,%edx
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  imul   %esi,%edx
  shl    $0x4,%esi
  add    %edx,%r8d
  mov    %edi,%edx
  mov    %eax,%edi
  cmp    $0x9,%edx
  jg     L17
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L0:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int convert_to_hexa(int dec) {
  int result;
  int base;
  base = 1;
  result = 0;
  while (dec > 0) {
    result += (dec % 10) * base;
    dec /= 10;
    base *= 16;
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<convert_to_hexa>:
  endbr64
  test   %edi,%edi
  jle    L13
  mov    $0x1,%esi
  xor    %r8d,%r8d
  mov    $0xcccccccd,%r9d
  cs nopw 0x0(%rax,%rax,1)
L5:
  mov    %edi,%eax
  mov    %edi,%edx
  imul   %r9,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ecx
  add    %ecx,%ecx
  sub    %ecx,%edx
  imul   %esi,%edx
  shl    $0x4,%esi
  add    %edx,%r8d
  mov    %edi,%edx
  mov    %eax,%edi
  cmp    $0x9,%edx
  jg     L5
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L13:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int convert_to_hexa(int dec) {
  int result;
  int base;
  base = 1;
  result = 0;
  while (dec > 0) {
    result += (dec % 10) * base;
    dec /= 10;
    base *= 16;
  }
  return result;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isvalue>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xfe0,%rsp
  mov    %rdi,-0xfd8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movw   $0x0,-0xfc2(%rbp)
  movq   $0x61,-0xfc0(%rbp)
  movq   $0x0,-0xfb8(%rbp)
  lea    -0xfb0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x7b,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movq   $0x62,-0xbd0(%rbp)
  movq   $0x0,-0xbc8(%rbp)
  lea    -0xbc0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x7b,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movq   $0x63,-0x7e0(%rbp)
  movq   $0x0,-0x7d8(%rbp)
  lea    -0x7d0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x7b,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  movq   $0x64,-0x3f0(%rbp)
  movq   $0x0,-0x3e8(%rbp)
  lea    -0x3e0(%rbp),%rdx
  mov    $0x0,%eax
  mov    $0x7b,%ecx
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  lea    -0xfc0(%rbp),%rdx
  mov    -0xfd8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L83
  movw   $0x1,-0xfc2(%rbp)
  jmp    L38
L83:
  lea    -0xbd0(%rbp),%rdx
  mov    -0xfd8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L5
  movw   $0x2,-0xfc2(%rbp)
  jmp    L38
L5:
  lea    -0x7e0(%rbp),%rdx
  mov    -0xfd8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L46
  movw   $0x3,-0xfc2(%rbp)
  jmp    L38
L46:
  lea    -0x3f0(%rbp),%rdx
  mov    -0xfd8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  jne    L27
  movw   $0x4,-0xfc2(%rbp)
  jmp    L38
L27:
  movw   $0x0,-0xfc2(%rbp)
L38:
  movzwl -0xfc2(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
unsigned short int isvalue(char string[1000]) {
  unsigned short int ret = 0;
  char val_a[1000] = "a";
  char val_b[1000] = "b";
  char val_c[1000] = "c";
  char val_d[1000] = "d";
  if (strcmp(string, val_a) == 0)
    ret = 1;
  else if (strcmp(string, val_b) == 0)
    ret = 2;
  else if (strcmp(string, val_c) == 0)
    ret = 3;
  else if (strcmp(string, val_d) == 0)
    ret = 4;
  else
    ret = 0;
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<isvalue>:
  endbr64
  push   %rbx
  sub    $0xfc0,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0xfb8(%rsp)
  xor    %eax,%eax
  movq   $0x61,(%rsp)
  movq   $0x0,0x8(%rsp)
  lea    0x10(%rsp),%rdi
  mov    $0x7b,%ecx
  rep stos %rax,%es:(%rdi)
  movq   $0x62,0x3f0(%rsp)
  movq   $0x0,0x3f8(%rsp)
  lea    0x400(%rsp),%rdi
  mov    $0x7b,%ecx
  rep stos %rax,%es:(%rdi)
  movq   $0x63,0x7e0(%rsp)
  movq   $0x0,0x7e8(%rsp)
  lea    0x7f0(%rsp),%rdi
  mov    $0x7b,%ecx
  rep stos %rax,%es:(%rdi)
  movq   $0x64,0xbd0(%rsp)
  movq   $0x0,0xbd8(%rsp)
  lea    0xbe0(%rsp),%rdi
  mov    $0x7b,%ecx
  rep stos %rax,%es:(%rdi)
  mov    %rsp,%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x1,%eax
  test   %rdx,%rdx
  jne    L49
L30:
  mov    0xfb8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  add    $0xfc0,%rsp
  pop    %rbx
  ret
L49:
  lea    0x3f0(%rsp),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x2,%eax
  test   %rdx,%rdx
  je     L30
  lea    0x7e0(%rsp),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  mov    %rax,%rdx
  mov    $0x3,%eax
  test   %rdx,%rdx
  je     L30
  lea    0xbd0(%rsp),%rsi
  mov    %rbx,%rdi
  call   <strcmp@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  shl    $0x2,%eax
  jmp    L30
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
unsigned short int isvalue(char string[1000]) {
  unsigned short int ret = 0;
  char val_a[1000] = "a";
  char val_b[1000] = "b";
  char val_c[1000] = "c";
  char val_d[1000] = "d";
  if (strcmp(string, val_a) == 0)
    ret = 1;
  else if (strcmp(string, val_b) == 0)
    ret = 2;
  else if (strcmp(string, val_c) == 0)
    ret = 3;
  else if (strcmp(string, val_d) == 0)
    ret = 4;
  else
    ret = 0;
  return ret;
}
```