ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<ichecksum>:
  endbr64
  push   %r14
  mov    $0x800,%edx
  xor    %esi,%esi
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x810,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0x800(%rsp),%rbx
  mov    %r14,%rdi
  call   <memset@plt>
  nopl   0x0(%rax,%rax,1)
L20:
  mov    $0x800,%ecx
  mov    %ebp,%edx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <readi@plt>
  test   %eax,%eax
  jle    L12
  add    %eax,%ebp
  mov    %r14,%rax
  nopl   0x0(%rax,%rax,1)
L1:
  xor    (%rax),%r12d
  add    $0x4,%rax
  cmp    %rbx,%rax
  jne    L1
  mov    $0x800,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  call   <memset@plt>
  jmp    L20
L12:
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x810,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint ichecksum(struct dinode *din) {
  unsigned int buf[512];
  char *cbuf = (char *)buf;
  uint n = sizeof(buf);
  uint off = 0;
  uint r, i;
  unsigned int checksum = 0;
  memset((void *)cbuf, 0, n);
  unsigned int *bp;
  while ((r = readi(din, cbuf, off, n)) > 0) {
    off += r;
    bp = (unsigned int *)buf;
    for (i = 0; i < sizeof(buf) / sizeof(uint); i++) {
      checksum ^= *bp;
      bp++;
    }
    memset((void *)cbuf, 0, n);
  }
  return checksum;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<ichecksum>:
  endbr64
  push   %r14
  mov    $0x800,%edx
  xor    %esi,%esi
  push   %r13
  mov    %rdi,%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x810,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x808(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  lea    0x800(%rsp),%rbx
  mov    %r14,%rdi
  call   <memset@plt>
  nopl   0x0(%rax,%rax,1)
L9:
  mov    $0x800,%ecx
  mov    %ebp,%edx
  mov    %r14,%rsi
  mov    %r13,%rdi
  call   <readi@plt>
  test   %eax,%eax
  jle    L20
  add    %eax,%ebp
  pxor   %xmm0,%xmm0
  mov    %r14,%rax
  nop
L36:
  pxor   (%rax),%xmm0
  add    $0x10,%rax
  cmp    %rax,%rbx
  jne    L36
  movdqa %xmm0,%xmm1
  mov    $0x800,%edx
  xor    %esi,%esi
  mov    %r14,%rdi
  psrldq $0x8,%xmm1
  pxor   %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  pxor   %xmm1,%xmm0
  movd   %xmm0,%eax
  xor    %eax,%r12d
  call   <memset@plt>
  jmp    L9
L20:
  mov    0x808(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L37
  add    $0x810,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L37:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint ichecksum(struct dinode *din) {
  unsigned int buf[512];
  char *cbuf = (char *)buf;
  uint n = sizeof(buf);
  uint off = 0;
  uint r, i;
  unsigned int checksum = 0;
  memset((void *)cbuf, 0, n);
  unsigned int *bp;
  while ((r = readi(din, cbuf, off, n)) > 0) {
    off += r;
    bp = (unsigned int *)buf;
    for (i = 0; i < sizeof(buf) / sizeof(uint); i++) {
      checksum ^= *bp;
      bp++;
    }
    memset((void *)cbuf, 0, n);
  }
  return checksum;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x5,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D1
type_guessed: string

tool: D1: %d


assistant: ```c
int main() {
  int x = 5;
  int y = x + 1;
  printf("%d\n", y);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%esi
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D6
type_guessed: string

tool: D6: %d


assistant: ```c
int main() {
  int x = 5;
  int y = x + 1;
  printf("%d\n", y);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: tag: D1
datatype: string

tool: D1: %d


assistant: ```c
int main() {
  int x = 5;
  int y = x + 1;
  printf("%d\n", y);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x6,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: disassemble_data
arguments: label: D1
data_type: string

tool: D1: %d


assistant: ```c
int main() {
  int x = 5;
  int y = x + 1;
  printf("%d\n", y);
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<v_putcad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %edx,-0xc(%rbp)
  mov    %ecx,-0x10(%rbp)
  mov    %r8,-0x18(%rbp)
  jmp    L14
L5:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%edi
  mov    -0x10(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%esi
  mov    -0x4(%rbp),%eax
  mov    %edi,%r8d
  mov    %eax,%edi
  call   <v_putchar@plt>
  addq   $0x1,-0x18(%rbp)
  addl   $0x1,-0x4(%rbp)
L14:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L5
  nop
  nop
  leave
  ret
```

assistant: ```c
void v_putcad(int x, int y, unsigned color, unsigned textback, char *cad) {
  while (cad[0] != 0) {
    v_putchar(x, y, color, textback, cad[0]);
    cad++;
    x++;
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<v_putcad>:
  endbr64
  mov    %r8,%rax
  movzbl (%r8),%r8d
  test   %r8b,%r8b
  je     L5
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%r12d
  mov    %edx,%r13d
  mov    %ecx,%r14d
  mov    %rax,%rbx
  sub    %eax,%edi
  mov    %edi,%ebp
L18:
  lea    0x0(%rbp,%rbx,1),%edi
  movsbl %r8b,%r8d
  mov    %r14d,%ecx
  mov    %r13d,%edx
  mov    %r12d,%esi
  call   <v_putchar@plt>
  add    $0x1,%rbx
  movzbl (%rbx),%r8d
  test   %r8b,%r8b
  jne    L18
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L5:
  ret
```

assistant: ```c
void v_putcad(int x, int y, unsigned color, unsigned textback, char *cad) {
  while (cad[0] != 0) {
    v_putchar(x, y, color, textback, cad[0]);
    cad++;
    x++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<v_putcad>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %r8,%rbx
  movsbl (%r8),%r8d
  test   %r8b,%r8b
  je     L15
  sub    %ebx,%edi
  mov    %esi,%ebp
  mov    %edx,%r12d
  mov    %ecx,%r13d
  mov    %edi,%r14d
  nopl   (%rax)
L9:
  lea    (%r14,%rbx,1),%edi
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  call   <v_putchar@plt>
  movsbl 0x1(%rbx),%r8d
  add    $0x1,%rbx
  test   %r8b,%r8b
  jne    L9
L15:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void v_putcad(int x, int y, unsigned color, unsigned textback, char *cad) {
  while (cad[0] != 0) {
    v_putchar(x, y, color, textback, cad[0]);
    cad++;
    x++;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<v_putcad>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %r8,%rbx
  movsbl (%r8),%r8d
  test   %r8b,%r8b
  je     L9
  sub    %ebx,%edi
  mov    %esi,%ebp
  mov    %edx,%r12d
  mov    %ecx,%r13d
  mov    %edi,%r14d
  nopl   (%rax)
L14:
  lea    (%r14,%rbx,1),%edi
  mov    %r13d,%ecx
  mov    %r12d,%edx
  mov    %ebp,%esi
  call   <v_putchar@plt>
  movsbl 0x1(%rbx),%r8d
  add    $0x1,%rbx
  test   %r8b,%r8b
  jne    L14
L9:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
void v_putcad(int x, int y, unsigned color, unsigned textback, char *cad) {
  while (cad[0] != 0) {
    v_putchar(x, y, color, textback, cad[0]);
    cad++;
    x++;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<reverseVowels>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  call   <strlen@plt>
  test   %rax,%rax
  je     L83
  lea    -0x1(%rax),%rbx
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rax,%r12
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  test   %rbx,%rbx
  je     L19
  mov    $0x0,%edx
  movabs $0x10411100104111,%r8
  jmp    L28
L60:
  sub    $0x1,%rbx
L109:
  cmp    %rdx,%rbx
  jbe    L40
  lea    (%rbx,%rbp,1),%rcx
  movzbl (%rcx),%esi
  lea    -0x41(%rsi),%eax
  cmp    $0x34,%al
  ja     L60
  bt     %rax,%r8
  jae    L60
  movzbl (%rdi),%eax
  mov    %sil,(%rdi)
  mov    %al,(%rcx)
  sub    $0x1,%rbx
  test   %al,%al
  je     L109
L40:
  add    $0x1,%rdx
  cmp    %rdx,%rbx
  jbe    L19
L28:
  lea    0x0(%rbp,%rdx,1),%rdi
  movzbl (%rdi),%eax
  sub    $0x41,%eax
  cmp    $0x34,%al
  ja     L40
  bt     %rax,%r8
  jae    L40
  jmp    L109
L83:
  lea    0x1(%rax),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L19:
  mov    %r12,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *reverseVowels(char *s) {
  char *retval = NULL;
  size_t slen = strlen(s);
  size_t i = 0;
  size_t j = slen == 0 ? 0 : slen - 1;
  char swap = '\0';
  if (!s) {
    return NULL;
  }
  retval = (char *)malloc(sizeof(char) * (slen + 1));
  strcpy(retval, s);
  if (!slen) {
    return retval;
  }
  while (i < j) {
    switch (retval[i]) {
    case 'a':
    case 'e':
    case 'i':
    case 'o':
    case 'u':
    case 'A':
    case 'E':
    case 'I':
    case 'O':
    case 'U':
      while (j > i) {
        switch (retval[j]) {
        case 'a':
        case 'e':
        case 'i':
        case 'o':
        case 'u':
        case 'A':
        case 'E':
        case 'I':
        case 'O':
        case 'U':
          swap = retval[i];
          retval[i] = retval[j];
          retval[j] = swap;
          break;
        }
        --j;
        if (swap) {
          swap = '\0';
          break;
        }
      }
      break;
    }
    ++i;
  }
  return retval;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<reverseVowels>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %rax,%rax
  jne    L31
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <strcpy@plt>
L13:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L31:
  lea    0x1(%rax),%edi
  lea    -0x1(%rax),%rbx
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %rax,%rdi
  mov    %rax,%rbp
  mov    %rax,%r13
  call   <strcpy@plt>
  test   %rbx,%rbx
  je     L13
  movabs $0x10411100104111,%r8
  xor    %edx,%edx
  jmp    L6
  nopl   0x0(%rax,%rax,1)
L116:
  add    $0x1,%rdx
  cmp    %rdx,%rbx
  jbe    L13
L6:
  lea    (%rdx,%rbp,1),%rdi
  movzbl (%rdi),%eax
  sub    $0x41,%eax
  cmp    $0x34,%al
  ja     L116
  bt     %rax,%r8
  jae    L116
  jmp    L57
  nop
L68:
  movzbl (%rdi),%eax
  mov    %sil,(%rdi)
  mov    %al,(%rcx)
  test   %al,%al
  jne    L116
L46:
  cmp    %rbx,%rdx
  jae    L116
L57:
  lea    0x0(%rbp,%rbx,1),%rcx
  sub    $0x1,%rbx
  movzbl (%rcx),%esi
  lea    -0x41(%rsi),%eax
  cmp    $0x34,%al
  ja     L46
  bt     %rax,%r8
  jae    L46
  jmp    L68
```

assistant: ```c
char *reverseVowels(char *s) {
  char *retval = NULL;
  size_t slen = strlen(s);
  size_t i = 0;
  size_t j = slen == 0 ? 0 : slen - 1;
  char swap = '\0';
  if (!s) {
    return NULL;
  }
  retval = (char *)malloc(sizeof(char) * (slen + 1));
  strcpy(retval, s);
  if (!slen) {
    return retval;
  }
  while (i < j) {
    switch (retval[i]) {
    case 'a':
    case 'e':
    case 'i':
    case 'o':
    case 'u':
    case 'A':
    case 'E':
    case 'I':
    case 'O':
    case 'U':
      while (j > i) {
        switch (retval[j]) {
        case 'a':
        case 'e':
        case 'i':
        case 'o':
        case 'u':
        case 'A':
        case 'E':
        case 'I':
        case 'O':
        case 'U':
          swap = retval[i];
          retval[i] = retval[j];
          retval[j] = swap;
          break;
        }
        --j;
        if (swap) {
          swap = '\0';
          break;
        }
      }
      break;
    }
    ++i;
  }
  return retval;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<reverseVowels>:
  endbr64
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %rax,%rax
  jne    L53
  mov    $0x1,%edi
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %rax,%rdi
  mov    %rax,%r13
  call   <strcpy@plt>
L3:
  add    $0x8,%rsp
  mov    %r13,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L53:
  lea    0x1(%rax),%edi
  lea    -0x1(%rax),%rbx
  call   <malloc@plt>
  mov    %r12,%rsi
  mov    %rax,%rdi
  mov    %rax,%rbp
  mov    %rax,%r13
  call   <strcpy@plt>
  test   %rbx,%rbx
  je     L3
  movabs $0x10411100104111,%r8
  xor    %edx,%edx
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L83:
  add    $0x1,%rdx
  cmp    %rdx,%rbx
  jbe    L3
L9:
  lea    (%rdx,%rbp,1),%rdi
  movzbl (%rdi),%eax
  sub    $0x41,%eax
  cmp    $0x34,%al
  ja     L83
  bt     %rax,%r8
  jae    L83
  jmp    L75
  nop
L84:
  movzbl (%rdi),%eax
  mov    %sil,(%rdi)
  mov    %al,(%rcx)
  test   %al,%al
  jne    L83
L33:
  cmp    %rdx,%rbx
  jbe    L83
L75:
  lea    0x0(%rbp,%rbx,1),%rcx
  sub    $0x1,%rbx
  movzbl (%rcx),%esi
  lea    -0x41(%rsi),%eax
  cmp    $0x34,%al
  ja     L33
  bt     %rax,%r8
  jae    L33
  jmp    L84
```

assistant: ```c
char *reverseVowels(char *s) {
  char *retval = NULL;
  size_t slen = strlen(s);
  size_t i = 0;
  size_t j = slen == 0 ? 0 : slen - 1;
  char swap = '\0';
  if (!s) {
    return NULL;
  }
  retval = (char *)malloc(sizeof(char) * (slen + 1));
  strcpy(retval, s);
  if (!slen) {
    return retval;
  }
  while (i < j) {
    switch (retval[i]) {
    case 'a':
    case 'e':
    case 'i':
    case 'o':
    case 'u':
    case 'A':
    case 'E':
    case 'I':
    case 'O':
    case 'U':
      while (j > i) {
        switch (retval[j]) {
        case 'a':
        case 'e':
        case 'i':
        case 'o':
        case 'u':
        case 'A':
        case 'E':
        case 'I':
        case 'O':
        case 'U':
          swap = retval[i];
          retval[i] = retval[j];
          retval[j] = swap;
          break;
        }
        --j;
        if (swap) {
          swap = '\0';
          break;
        }
      }
      break;
    }
    ++i;
  }
  return retval;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L11
L18:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movsd  -0x8(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L11:
  mov    -0xc(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L18
  movsd  -0x8(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  pop    %rbp
  ret
```

assistant: ```c
double sum(double *t, int size) {
  double sum = 0;
  for (int i = 0; i < size; i++)
    sum += t[i];
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<sum>:
  endbr64
  test   %esi,%esi
  jle    L16
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x8(%rdi,%rdx,8),%rdx
  pxor   %xmm0,%xmm0
L7:
  addsd  (%rax),%xmm0
  add    $0x8,%rax
  cmp    %rdx,%rax
  jne    L7
  ret
L16:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double sum(double *t, int size) {
  double sum = 0;
  for (int i = 0; i < size; i++)
    sum += t[i];
  return sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<sum>:
  endbr64
  test   %esi,%esi
  jle    L8
  lea    -0x1(%rsi),%eax
  pxor   %xmm0,%xmm0
  lea    0x8(%rdi,%rax,8),%rax
  nopl   0x0(%rax)
L15:
  addsd  (%rdi),%xmm0
  add    $0x8,%rdi
  cmp    %rax,%rdi
  jne    L15
  ret
  cs nopw 0x0(%rax,%rax,1)
L8:
  pxor   %xmm0,%xmm0
  ret
```

assistant: ```c
double sum(double *t, int size) {
  double sum = 0;
  for (int i = 0; i < size; i++)
    sum += t[i];
  return sum;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<sum>:
  endbr64
  test   %esi,%esi
  jle    L11
  cmp    $0x1,%esi
  je     L7
  mov    %esi,%edx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    %edx
  sub    $0x1,%edx
  shl    $0x4,%rdx
  lea    0x10(%rdi,%rdx,1),%rdx
  nopl   0x0(%rax)
L45:
  movsd  (%rax),%xmm1
  add    $0x10,%rax
  addsd  %xmm1,%xmm0
  movsd  -0x8(%rax),%xmm1
  addsd  %xmm1,%xmm0
  cmp    %rdx,%rax
  jne    L45
  mov    %esi,%eax
  and    $0xfffffffe,%eax
  and    $0x1,%esi
  je     L38
L33:
  cltq
  addsd  (%rdi,%rax,8),%xmm0
  ret
  nopl   0x0(%rax)
L38:
  ret
  nopl   0x0(%rax)
L11:
  pxor   %xmm0,%xmm0
  ret
L7:
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  jmp    L33
```

assistant: ```c
double sum(double *t, int size) {
  double sum = 0;
  for (int i = 0; i < size; i++)
    sum += t[i];
  return sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<reverseEndian>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x10(%rbp)
  jmp    L19
L16:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x9(%rbp)
  mov    -0x1c(%rbp),%eax
  sub    -0x10(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x10(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x1c(%rbp),%eax
  sub    -0x10(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x9(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x10(%rbp)
L19:
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x10(%rbp)
  jl     L16
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L17
  call   <__stack_chk_fail@plt>
L17:
  leave
  ret
```

assistant: ```c
void reverseEndian(char *arr, int length) {
  char temp[1];
  for (int i = 0; i < length / 2; i++) {
    memcpy(&temp, &arr[i], 1);
    memcpy(&arr[i], &arr[length - i - 1], 1);
    memcpy(&arr[length - i - 1], &temp, 1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverseEndian>:
  endbr64
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  cmp    $0x1,%esi
  jle    L2
  movslq %esi,%rsi
  lea    -0x1(%rdi,%rsi,1),%rdx
  mov    $0x0,%eax
L1:
  movzbl (%rdi,%rax,1),%ecx
  movzbl (%rdx),%esi
  mov    %sil,(%rdi,%rax,1)
  mov    %cl,(%rdx)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%r8d
  jg     L1
L2:
  ret
```

assistant: ```c
void reverseEndian(char *arr, int length) {
  char temp[1];
  for (int i = 0; i < length / 2; i++) {
    memcpy(&temp, &arr[i], 1);
    memcpy(&arr[i], &arr[length - i - 1], 1);
    memcpy(&arr[length - i - 1], &temp, 1);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<reverseEndian>:
  endbr64
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  cmp    $0x1,%esi
  jle    L11
  movslq %esi,%rsi
  xor    %eax,%eax
  lea    -0x1(%rdi,%rsi,1),%rdx
L8:
  movzbl (%rdx),%esi
  movzbl (%rdi,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%rdi,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%r8d
  jg     L8
L11:
  ret
```

assistant: ```c
void reverseEndian(char *arr, int length) {
  char temp[1];
  for (int i = 0; i < length / 2; i++) {
    memcpy(&temp, &arr[i], 1);
    memcpy(&arr[i], &arr[length - i - 1], 1);
    memcpy(&arr[length - i - 1], &temp, 1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<reverseEndian>:
  endbr64
  mov    %esi,%r8d
  shr    $0x1f,%r8d
  add    %esi,%r8d
  sar    %r8d
  cmp    $0x1,%esi
  jle    L2
  movslq %esi,%rsi
  xor    %eax,%eax
  lea    -0x1(%rdi,%rsi,1),%rdx
L5:
  movzbl (%rdx),%esi
  movzbl (%rdi,%rax,1),%ecx
  sub    $0x1,%rdx
  mov    %sil,(%rdi,%rax,1)
  add    $0x1,%rax
  mov    %cl,0x1(%rdx)
  cmp    %eax,%r8d
  jg     L5
L2:
  ret
```

assistant: ```c
void reverseEndian(char *arr, int length) {
  char temp[1];
  for (int i = 0; i < length / 2; i++) {
    memcpy(&temp, &arr[i], 1);
    memcpy(&arr[i], &arr[length - i - 1], 1);
    memcpy(&arr[length - i - 1], &temp, 1);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<assign_picture>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    %ecx,-0x24(%rbp)
  mov    %r8d,-0x28(%rbp)
  mov    %r9,-0x30(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L124
  cmpq   $0x0,-0x30(%rbp)
  jne    L78
L124:
  mov    $0xffffffff,%eax
  jmp    L20
L78:
  cmpl   $0x320,-0x1c(%rbp)
  jg     L100
  cmpl   $0x1e0,-0x20(%rbp)
  jle    L4
L100:
  mov    $0xfffffffe,%eax
  jmp    L20
L4:
  mov    -0x1c(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  cmp    $0x320,%eax
  jg     L11
  mov    -0x20(%rbp),%edx
  mov    -0x28(%rbp),%eax
  add    %edx,%eax
  cmp    $0x1e0,%eax
  jle    L91
L11:
  mov    $0xfffffffd,%eax
  jmp    L20
L91:
  mov    -0x20(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L42
L114:
  mov    -0xc(%rbp),%eax
  imul   $0x640,%eax,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L29
L101:
  mov    -0x10(%rbp),%eax
  lea    (%rax,%rax,1),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x30(%rbp),%rax
  lea    0x2(%rax),%rdx
  mov    %rdx,-0x30(%rbp)
  mov    -0x4(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzwl (%rax),%eax
  mov    %ax,(%rdx)
  addl   $0x1,-0x10(%rbp)
L29:
  mov    -0x1c(%rbp),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x10(%rbp)
  jl     L101
  addl   $0x1,-0xc(%rbp)
L42:
  mov    -0x20(%rbp),%edx
  mov    -0x28(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L114
  mov    $0x0,%eax
L20:
  pop    %rbp
  ret
```

assistant: ```c
int assign_picture(char *fb, int x, int y, int xoft, int yoft, const unsigned short *Image) {
  int x1, y1;
  int xres, yres;
  if (!fb || !Image)
    return -1;
  if (x > 800 || y > 480)
    return -2;
  if (x + xoft > 800 || y + yoft > 480)
    return -3;
  for (y1 = y; y1 < y + yoft; y1++) {
    yres = y1 * 1600;
    for (x1 = x; x1 < x + xoft; x1++) {
      xres = x1 * 2 + yres;
      *((unsigned short *)(fb + xres)) = *Image++;
    }
  }
  return 0;
}
```