ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<inet6_rth_reverse>:
  endbr64
  cmpl   $0x80,(%rdi)
  jne    L1
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  mov    $0x10,%edx
  mov    %rdi,%rsi
  mov    %rbx,%rdi
  call   <memmove@plt>
  mov    0x0(%rbp),%eax
  lea    0x0(,%rax,8),%r8d
  movslq %r8d,%r8
  shr    $0x2,%r8
  mov    %r8d,%r9d
  sar    %r9d
  cmp    $0x1,%r8d
  jle    L4
  movslq %r8d,%rax
  lea    -0x4(,%rax,4),%rdx
  mov    $0x0,%eax
L24:
  mov    0x8(%rbp),%rcx
  mov    (%rcx,%rax,4),%esi
  mov    0x8(%rbx),%rdi
  mov    (%rcx,%rdx,1),%ecx
  mov    %ecx,(%rdi,%rax,4)
  mov    0x8(%rbx),%rcx
  mov    %esi,(%rcx,%rdx,1)
  add    $0x1,%rax
  sub    $0x4,%rdx
  cmp    %eax,%r9d
  jg     L24
L4:
  mov    %r8d,0x4(%rbx)
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L1:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int inet6_rth_reverse(const void *in, void *out) {
  struct ip6_rthdr *in_rthdr = (struct ip6_rthdr *)in;
  switch (in_rthdr->ip6r_type) {
    struct ip6_rthdr0 *in_rthdr0;
    struct ip6_rthdr0 *out_rthdr0;
  case IPV6_RTHDR_TYPE_0:
    in_rthdr0 = (struct ip6_rthdr0 *)in;
    out_rthdr0 = (struct ip6_rthdr0 *)out;
    memmove(out_rthdr0, in_rthdr0, sizeof(struct ip6_rthdr0));
    int total = in_rthdr0->ip6r0_len * 8 / sizeof(struct in6_addr);
    for (int i = 0; i < total / 2; ++i) {
      struct in6_addr temp = in_rthdr0->ip6r0_addr[i];
      out_rthdr0->ip6r0_addr[i] = in_rthdr0->ip6r0_addr[total - 1 - i];
      out_rthdr0->ip6r0_addr[total - 1 - i] = temp;
    }
    if (total % 2 != 0 && in != out)
      out_rthdr0->ip6r0_addr[total / 2] = in_rthdr0->ip6r0_addr[total / 2];
    out_rthdr0->ip6r0_segleft = total;
    return 0;
  }
  return -1;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<inet6_rth_reverse>:
  endbr64
  cmpl   $0x80,(%rdi)
  jne    L12
  push   %rbp
  mov    $0x10,%edx
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdi,%rsi
  mov    %rbx,%rdi
  sub    $0x8,%rsp
  call   <memmove@plt>
  mov    0x0(%rbp),%eax
  lea    0x0(,%rax,8),%r8d
  movslq %r8d,%r8
  shr    $0x2,%r8
  mov    %r8d,%r11d
  sar    %r11d
  cmp    $0x1,%r8d
  jle    L9
  mov    0x8(%rbp),%r10
  mov    0x8(%rbx),%r9
  movslq %r8d,%rdi
  xor    %edx,%edx
  shl    $0x2,%rdi
  xor    %eax,%eax
  lea    (%r10,%rdi,1),%rbp
  add    %r9,%rdi
L8:
  mov    -0x4(%rbp,%rdx,1),%esi
  mov    (%r10,%rax,4),%ecx
  mov    %esi,(%r9,%rax,4)
  add    $0x1,%rax
  mov    %ecx,-0x4(%rdi,%rdx,1)
  sub    $0x4,%rdx
  cmp    %eax,%r11d
  jg     L8
L9:
  mov    %r8d,0x4(%rbx)
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L12:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int inet6_rth_reverse(const void *in, void *out) {
  struct ip6_rthdr *in_rthdr = (struct ip6_rthdr *)in;
  switch (in_rthdr->ip6r_type) {
    struct ip6_rthdr0 *in_rthdr0;
    struct ip6_rthdr0 *out_rthdr0;
  case IPV6_RTHDR_TYPE_0:
    in_rthdr0 = (struct ip6_rthdr0 *)in;
    out_rthdr0 = (struct ip6_rthdr0 *)out;
    memmove(out_rthdr0, in_rthdr0, sizeof(struct ip6_rthdr0));
    int total = in_rthdr0->ip6r0_len * 8 / sizeof(struct in6_addr);
    for (int i = 0; i < total / 2; ++i) {
      struct in6_addr temp = in_rthdr0->ip6r0_addr[i];
      out_rthdr0->ip6r0_addr[i] = in_rthdr0->ip6r0_addr[total - 1 - i];
      out_rthdr0->ip6r0_addr[total - 1 - i] = temp;
    }
    if (total % 2 != 0 && in != out)
      out_rthdr0->ip6r0_addr[total / 2] = in_rthdr0->ip6r0_addr[total / 2];
    out_rthdr0->ip6r0_segleft = total;
    return 0;
  }
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<inet6_rth_reverse>:
  endbr64
  cmpl   $0x80,(%rdi)
  jne    L28
  push   %rbp
  mov    $0x10,%edx
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  mov    %rdi,%rsi
  mov    %rbx,%rdi
  sub    $0x8,%rsp
  call   <memmove@plt>
  mov    0x0(%rbp),%eax
  lea    0x0(,%rax,8),%r8d
  movslq %r8d,%r8
  shr    $0x2,%r8
  mov    %r8d,%r11d
  sar    %r11d
  cmp    $0x1,%r8d
  jle    L25
  mov    0x8(%rbp),%r10
  mov    0x8(%rbx),%r9
  movslq %r8d,%rdi
  xor    %edx,%edx
  shl    $0x2,%rdi
  xor    %eax,%eax
  lea    (%r10,%rdi,1),%rbp
  add    %r9,%rdi
L1:
  mov    -0x4(%rbp,%rdx,1),%esi
  mov    (%r10,%rax,4),%ecx
  mov    %esi,(%r9,%rax,4)
  add    $0x1,%rax
  mov    %ecx,-0x4(%rdi,%rdx,1)
  sub    $0x4,%rdx
  cmp    %eax,%r11d
  jg     L1
L25:
  mov    %r8d,0x4(%rbx)
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L28:
  mov    $0xffffffff,%eax
  ret
```

assistant: ```c
int inet6_rth_reverse(const void *in, void *out) {
  struct ip6_rthdr *in_rthdr = (struct ip6_rthdr *)in;
  switch (in_rthdr->ip6r_type) {
    struct ip6_rthdr0 *in_rthdr0;
    struct ip6_rthdr0 *out_rthdr0;
  case IPV6_RTHDR_TYPE_0:
    in_rthdr0 = (struct ip6_rthdr0 *)in;
    out_rthdr0 = (struct ip6_rthdr0 *)out;
    memmove(out_rthdr0, in_rthdr0, sizeof(struct ip6_rthdr0));
    int total = in_rthdr0->ip6r0_len * 8 / sizeof(struct in6_addr);
    for (int i = 0; i < total / 2; ++i) {
      struct in6_addr temp = in_rthdr0->ip6r0_addr[i];
      out_rthdr0->ip6r0_addr[i] = in_rthdr0->ip6r0_addr[total - 1 - i];
      out_rthdr0->ip6r0_addr[total - 1 - i] = temp;
    }
    if (total % 2 != 0 && in != out)
      out_rthdr0->ip6r0_addr[total / 2] = in_rthdr0->ip6r0_addr[total / 2];
    out_rthdr0->ip6r0_segleft = total;
    return 0;
  }
  return -1;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pc1_encrypt_buf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L19
L5:
  mov    -0x14(%rbp),%edx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x14(%rbp),%ecx
  mov    -0x30(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    -0x28(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <pc1_encrypt@plt>
  mov    %al,(%rbx)
  addl   $0x1,-0x14(%rbp)
L19:
  mov    -0x14(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jb     L5
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void pc1_encrypt_buf(struct pc1_ctx *pc1, unsigned char *buf, unsigned len) {
  unsigned i;
  for (i = 0; i < len; i++)
    buf[i] = pc1_encrypt(pc1, buf[i]);
}
```
user: What is the c source code of the assembly code below?

```asm
<pc1_encrypt_buf>:
  endbr64
  test   %edx,%edx
  je     L6
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%rbx
  lea    -0x1(%rdx),%eax
  lea    0x1(%rsi,%rax,1),%r12
L16:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  call   <pc1_encrypt@plt>
  mov    %al,(%rbx)
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L6:
  ret
```

assistant: ```c
void pc1_encrypt_buf(struct pc1_ctx *pc1, unsigned char *buf, unsigned len) {
  unsigned i;
  for (i = 0; i < len; i++)
    buf[i] = pc1_encrypt(pc1, buf[i]);
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<pc1_encrypt_buf>:
  endbr64
  test   %edx,%edx
  je     L17
  lea    -0x1(%rdx),%eax
  push   %r12
  lea    0x1(%rsi,%rax,1),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopw   0x0(%rax,%rax,1)
L8:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <pc1_encrypt@plt>
  mov    %al,-0x1(%rbx)
  cmp    %r12,%rbx
  jne    L8
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L17:
  ret
```

assistant: ```c
void pc1_encrypt_buf(struct pc1_ctx *pc1, unsigned char *buf, unsigned len) {
  unsigned i;
  for (i = 0; i < len; i++)
    buf[i] = pc1_encrypt(pc1, buf[i]);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<pc1_encrypt_buf>:
  endbr64
  test   %edx,%edx
  je     L7
  lea    -0x1(%rdx),%eax
  push   %r12
  lea    0x1(%rsi,%rax,1),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopw   0x0(%rax,%rax,1)
L0:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x1,%rbx
  call   <pc1_encrypt@plt>
  mov    %al,-0x1(%rbx)
  cmp    %r12,%rbx
  jne    L0
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L7:
  ret
```

assistant: ```c
void pc1_encrypt_buf(struct pc1_ctx *pc1, unsigned char *buf, unsigned len) {
  unsigned i;
  for (i = 0; i < len; i++)
    buf[i] = pc1_encrypt(pc1, buf[i]);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<procint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtol@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <pushint@plt>
  nop
  leave
  ret
```

assistant: ```c
void procint(char *cmd) {
  long val = strtol(cmd, NULL, 0);
  pushint(val);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<procint>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edx
  mov    $0x0,%esi
  call   <strtol@plt>
  mov    %rax,%rdi
  call   <pushint@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
void procint(char *cmd) {
  long val = strtol(cmd, NULL, 0);
  pushint(val);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<procint>:
  endbr64
  sub    $0x8,%rsp
  xor    %edx,%edx
  xor    %esi,%esi
  call   <strtol@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1060 <pushint@plt>
```

assistant: ```c
void procint(char *cmd) {
  long val = strtol(cmd, NULL, 0);
  pushint(val);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<procint>:
  endbr64
  sub    $0x8,%rsp
  xor    %edx,%edx
  xor    %esi,%esi
  call   <strtol@plt>
  add    $0x8,%rsp
  mov    %rax,%rdi
  jmp    1060 <pushint@plt>
```

assistant: ```c
void procint(char *cmd) {
  long val = strtol(cmd, NULL, 0);
  pushint(val);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<FindAddressByIndex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L21
L10:
  mov    -0x4(%rbp),%eax
  shr    %eax
  cmp    %eax,-0x14(%rbp)
  jne    L38
  mov    -0x8(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  jmp    L32
L38:
  addl   $0x1,-0x4(%rbp)
  mov    -0x8(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
L21:
  mov    -0x8(%rbp),%edx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  jg     L10
  mov    $0x0,%eax
L32:
  leave
  ret
```

assistant: ```c
char *FindAddressByIndex(unsigned int aIndex, char *aAddresses) {
  unsigned int lCurrentIndex = 0;
  unsigned int lConnectionFoundIndex = 0;
  while (strlen(aAddresses + lCurrentIndex) > 0) {
    if ((lConnectionFoundIndex / 2) == aIndex) {
      return aAddresses + lCurrentIndex;
    }
    lConnectionFoundIndex++;
    lCurrentIndex += strlen(aAddresses + lCurrentIndex) + 1;
  }
  return NULL;
}
```
user: Translate the below assembly code into C programming language:

```asm
<FindAddressByIndex>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%r14d
  mov    %rsi,%r13
  mov    $0x0,%r12d
  mov    $0x0,%ebp
  jmp    L38
L8:
  add    $0x1,%r12d
  lea    0x1(%rbp,%rax,1),%ebp
L38:
  mov    %ebp,%ebx
  add    %r13,%rbx
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  jle    L32
  mov    %r12d,%edx
  shr    %edx
  cmp    %r14d,%edx
  jne    L8
  jmp    L12
L32:
  mov    $0x0,%ebx
L12:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *FindAddressByIndex(unsigned int aIndex, char *aAddresses) {
  unsigned int lCurrentIndex = 0;
  unsigned int lConnectionFoundIndex = 0;
  while (strlen(aAddresses + lCurrentIndex) > 0) {
    if ((lConnectionFoundIndex / 2) == aIndex) {
      return aAddresses + lCurrentIndex;
    }
    lConnectionFoundIndex++;
    lCurrentIndex += strlen(aAddresses + lCurrentIndex) + 1;
  }
  return NULL;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<FindAddressByIndex>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L26
  nopl   0x0(%rax)
L20:
  mov    %r14d,%edx
  shr    %edx
  cmp    %ebp,%edx
  je     L0
  add    $0x1,%r14d
  lea    0x1(%r13,%rax,1),%r13d
L26:
  mov    %r13d,%r12d
  add    %rbx,%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  jne    L20
  xor    %r12d,%r12d
L0:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *FindAddressByIndex(unsigned int aIndex, char *aAddresses) {
  unsigned int lCurrentIndex = 0;
  unsigned int lConnectionFoundIndex = 0;
  while (strlen(aAddresses + lCurrentIndex) > 0) {
    if ((lConnectionFoundIndex / 2) == aIndex) {
      return aAddresses + lCurrentIndex;
    }
    lConnectionFoundIndex++;
    lCurrentIndex += strlen(aAddresses + lCurrentIndex) + 1;
  }
  return NULL;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<FindAddressByIndex>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %edi,%ebp
  push   %rbx
  mov    %rsi,%rbx
  jmp    L0
  nopl   0x0(%rax)
L26:
  mov    %r14d,%edx
  shr    %edx
  cmp    %ebp,%edx
  je     L1
  add    $0x1,%r14d
  lea    0x1(%r13,%rax,1),%r13d
L0:
  mov    %r13d,%r12d
  add    %rbx,%r12
  mov    %r12,%rdi
  call   <strlen@plt>
  test   %rax,%rax
  jne    L26
  xor    %r12d,%r12d
L1:
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *FindAddressByIndex(unsigned int aIndex, char *aAddresses) {
  unsigned int lCurrentIndex = 0;
  unsigned int lConnectionFoundIndex = 0;
  while (strlen(aAddresses + lCurrentIndex) > 0) {
    if ((lConnectionFoundIndex / 2) == aIndex) {
      return aAddresses + lCurrentIndex;
    }
    lConnectionFoundIndex++;
    lCurrentIndex += strlen(aAddresses + lCurrentIndex) + 1;
  }
  return NULL;
}
```
user: What is the c source code of the assembly code below:

```asm
<vnum_test8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    0xc(%rax),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    0x10(%rax),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x1,-0x14(%rbp)
  jmp    L11
L1:
  mov    -0x28(%rbp),%rax
  mov    0x8(%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movl   $0x2,(%rax)
  mov    -0x10(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0x10(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,-0xc(%rbp)
  addl   $0x1,-0x14(%rbp)
L11:
  mov    -0x14(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L1
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void vnum_test8(int *data) {
  int i;
  int stop = data[3];
  int m = data[4];
  int n = m;
  for (i = 1; i < stop; i++) {
    int k = data[2];
    data[k] = 2;
    data[0] = m - n;
    k = data[1];
    m = m + k;
    n = n + k;
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<vnum_test8>:
  endbr64
  mov    0xc(%rdi),%r8d
  mov    0x10(%rdi),%eax
  cmp    $0x1,%r8d
  jle    L10
  mov    %eax,%ecx
  mov    $0x1,%edx
L5:
  movslq 0x8(%rdi),%rsi
  movl   $0x2,(%rdi,%rsi,4)
  mov    %eax,%esi
  add    0x4(%rdi),%eax
  mov    %ecx,%r9d
  add    0x4(%rdi),%ecx
  add    $0x1,%edx
  cmp    %edx,%r8d
  jne    L5
  sub    %r9d,%esi
  mov    %esi,(%rdi)
L10:
  ret
```

assistant: ```c
void vnum_test8(int *data) {
  int i;
  int stop = data[3];
  int m = data[4];
  int n = m;
  for (i = 1; i < stop; i++) {
    int k = data[2];
    data[k] = 2;
    data[0] = m - n;
    k = data[1];
    m = m + k;
    n = n + k;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<vnum_test8>:
  endbr64
  mov    0xc(%rdi),%ecx
  cmp    $0x1,%ecx
  jle    L14
  mov    $0x1,%eax
  nopl   0x0(%rax)
L9:
  movslq 0x8(%rdi),%rdx
  add    $0x1,%eax
  movl   $0x2,(%rdi,%rdx,4)
  cmp    %eax,%ecx
  jne    L9
  movl   $0x0,(%rdi)
L14:
  ret
```

assistant: ```c
void vnum_test8(int *data) {
  int i;
  int stop = data[3];
  int m = data[4];
  int n = m;
  for (i = 1; i < stop; i++) {
    int k = data[2];
    data[k] = 2;
    data[0] = m - n;
    k = data[1];
    m = m + k;
    n = n + k;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<vnum_test8>:
  endbr64
  mov    0xc(%rdi),%ecx
  cmp    $0x1,%ecx
  jle    L8
  mov    $0x1,%eax
  nopl   0x0(%rax)
L18:
  movslq 0x8(%rdi),%rdx
  add    $0x1,%eax
  movl   $0x2,(%rdi,%rdx,4)
  cmp    %eax,%ecx
  jne    L18
  movl   $0x0,(%rdi)
L8:
  ret
```

assistant: ```c
void vnum_test8(int *data) {
  int i;
  int stop = data[3];
  int m = data[4];
  int n = m;
  for (i = 1; i < stop; i++) {
    int k = data[2];
    data[k] = 2;
    data[0] = m - n;
    k = data[1];
    m = m + k;
    n = n + k;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<convert_ivector_vector>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %rcx,-0x30(%rbp)
  mov    -0x20(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  jmp    L11
L7:
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%rax
  lea    0x0(,%rax,4),%rcx
  mov    -0x30(%rbp),%rax
  add    %rcx,%rax
  pxor   %xmm0,%xmm0
  cvtsi2ss %edx,%xmm0
  movss  %xmm0,(%rax)
  addq   $0x1,-0x8(%rbp)
L11:
  mov    -0x8(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jle    L7
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void convert_ivector_vector(int *X, long nl, long nh, float *Y) {
  long i;
  for (i = nl; i <= nh; i++)
    Y[i] = (float)X[i];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<convert_ivector_vector>:
  endbr64
  cmp    %rdx,%rsi
  jg     L13
  add    $0x1,%rdx
L4:
  pxor   %xmm0,%xmm0
  cvtsi2ssl (%rdi,%rsi,4),%xmm0
  movss  %xmm0,(%rcx,%rsi,4)
  add    $0x1,%rsi
  cmp    %rdx,%rsi
  jne    L4
L13:
  ret
```

assistant: ```c
void convert_ivector_vector(int *X, long nl, long nh, float *Y) {
  long i;
  for (i = nl; i <= nh; i++)
    Y[i] = (float)X[i];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<convert_ivector_vector>:
  endbr64
  cmp    %rdx,%rsi
  jg     L17
  add    $0x1,%rdx
  nopl   (%rax)
L6:
  pxor   %xmm0,%xmm0
  cvtsi2ssl (%rdi,%rsi,4),%xmm0
  movss  %xmm0,(%rcx,%rsi,4)
  add    $0x1,%rsi
  cmp    %rdx,%rsi
  jne    L6
L17:
  ret
```

assistant: ```c
void convert_ivector_vector(int *X, long nl, long nh, float *Y) {
  long i;
  for (i = nl; i <= nh; i++)
    Y[i] = (float)X[i];
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<convert_ivector_vector>:
  endbr64
  cmp    %rdx,%rsi
  jg     L7
  sub    %rsi,%rdx
  push   %rbx
  mov    %rsi,%r11
  lea    0x1(%rdx),%r10
  cmp    $0x2,%rdx
  jbe    L24
  mov    %r10,%r9
  lea    0x0(,%rsi,4),%r8
  xor    %eax,%eax
  shr    $0x2,%r9
  lea    (%rdi,%r8,1),%rbx
  add    %rcx,%r8
  shl    $0x4,%r9
  xchg   %ax,%ax
L32:
  movdqu (%rbx,%rax,1),%xmm1
  cvtdq2ps %xmm1,%xmm0
  movups %xmm0,(%r8,%rax,1)
  add    $0x10,%rax
  cmp    %r9,%rax
  jne    L32
  mov    %r10,%rax
  and    $0xfffffffffffffffc,%rax
  add    %rax,%rsi
  cmp    %rax,%r10
  je     L35
L10:
  sub    %rax,%r10
  cmp    %rax,%rdx
  je     L68
  add    %r11,%rax
  movq   (%rdi,%rax,4),%xmm0
  cvtdq2ps %xmm0,%xmm0
  movlps %xmm0,(%rcx,%rax,4)
  mov    %r10,%rax
  and    $0xfffffffffffffffe,%rax
  add    %rax,%rsi
  cmp    %rax,%r10
  je     L35
L68:
  pxor   %xmm0,%xmm0
  cvtsi2ssl (%rdi,%rsi,4),%xmm0
  movss  %xmm0,(%rcx,%rsi,4)
L35:
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  ret
L24:
  xor    %eax,%eax
  jmp    L10
```

assistant: ```c
void convert_ivector_vector(int *X, long nl, long nh, float *Y) {
  long i;
  for (i = nl; i <= nh; i++)
    Y[i] = (float)X[i];
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<vec_swap_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    %ecx,-0x2c(%rbp)
  mov    %r8d,-0x30(%rbp)
  mov    %r9d,-0x34(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L31
  cmpq   $0x0,-0x20(%rbp)
  je     L31
  cmpq   $0x0,-0x28(%rbp)
  je     L31
  cmpl   $0x0,-0x2c(%rbp)
  je     L31
  mov    $0x1,%eax
  jmp    L43
L31:
  mov    $0x0,%eax
L43:
  mov    %eax,%edi
  call   <assert@plt>
  mov    -0x2c(%rbp),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <vec_expand_@plt>
  mov    D2(%rip),%rax
  mov    (%rax),%rdx
  mov    D15(%rip),%rax
  mov    (%rax),%rax
  cmp    %rax,%rdx
  je     L63
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  imul   -0x2c(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  cltq
  lea    (%rdx,%rax,1),%rcx
  mov    -0x2c(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x34(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  cltq
  lea    (%rdx,%rax,1),%rsi
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  cltq
  lea    (%rdx,%rax,1),%rcx
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edx
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x34(%rbp),%eax
  imul   -0x2c(%rbp),%eax
  cltq
  lea    (%rdx,%rax,1),%rcx
  mov    -0x2c(%rbp),%edx
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  jmp    L48
L63:
  nop
L48:
  leave
  ret
```

assistant: ```c
void vec_swap_(char **data, int *length, int *capacity, int memsz, int idx1, int idx2) {
  assert(data && length && capacity && memsz);
  char *tmp;
  vec_expand_(data, length, capacity, memsz);
  if (errno == ENOMEM)
    return;
  tmp = *data + *length * memsz;
  memcpy(tmp, *data + (idx1 * memsz), memsz);
  memcpy(*data + (idx1 * memsz), *data + (idx2 * memsz), memsz);
  memcpy(*data + (idx2 * memsz), tmp, memsz);
}
```