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

```asm
<strStr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <buildNext@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x1c(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x18(%rbp)
  jmp    L65
L8:
  cmpl   $0x0,-0x18(%rbp)
  js     L20
  mov    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x28(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  jne    L49
L20:
  addl   $0x1,-0x1c(%rbp)
  addl   $0x1,-0x18(%rbp)
  jmp    L65
L49:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x18(%rbp)
L65:
  mov    -0x18(%rbp),%eax
  cmp    -0x10(%rbp),%eax
  jge    L2
  mov    -0x1c(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jl     L8
L2:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x1c(%rbp),%eax
  sub    -0x18(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x14(%rbp)
  jge    L64
  mov    $0xffffffff,%eax
  jmp    L0
L64:
  mov    -0xc(%rbp),%eax
L0:
  leave
  ret
```

assistant: ```c
int strStr(char *haystack, char *needle) {
  int *next = buildNext(needle);
  int n = strlen(haystack);
  int i = 0;
  int m = strlen(needle);
  int j = 0;
  while (j < m && i < n) {
    if (j < 0 || needle[j] == haystack[i]) {
      i++;
      j++;
    } else {
      j = next[j];
    }
  }
  free(next);
  int ret = i - j;
  if (n < ret + m)
    return -1;
  else
    return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strStr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%r15
  mov    %rsi,%r14
  mov    %rsi,%rdi
  call   <buildNext@plt>
  mov    %rax,0x8(%rsp)
  mov    %r15,%rdi
  call   <strlen@plt>
  mov    %eax,%r13d
  mov    %r14,%rdi
  call   <strlen@plt>
  mov    %eax,%r12d
  test   %eax,%eax
  jle    L50
  test   %r13d,%r13d
  jle    L50
  mov    $0x0,%ebx
  mov    $0x0,%ebp
  jmp    L24
L17:
  add    $0x1,%ebp
  add    $0x1,%ebx
L55:
  cmp    %ebx,%r12d
  jle    L65
  cmp    %r13d,%ebp
  jge    L65
L24:
  test   %ebx,%ebx
  js     L17
  movslq %ebx,%rax
  movslq %ebp,%rdx
  movzbl (%r15,%rdx,1),%ecx
  cmp    %cl,(%r14,%rax,1)
  je     L17
  mov    0x8(%rsp),%rsi
  mov    (%rsi,%rax,4),%ebx
  jmp    L55
L50:
  mov    $0x0,%ebx
  mov    $0x0,%ebp
L65:
  mov    0x8(%rsp),%rdi
  call   <free@plt>
  mov    %ebp,%eax
  sub    %ebx,%eax
  add    %eax,%r12d
  cmp    %r13d,%r12d
  mov    $0xffffffff,%edx
  cmovg  %edx,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int strStr(char *haystack, char *needle) {
  int *next = buildNext(needle);
  int n = strlen(haystack);
  int i = 0;
  int m = strlen(needle);
  int j = 0;
  while (j < m && i < n) {
    if (j < 0 || needle[j] == haystack[i]) {
      i++;
      j++;
    } else {
      j = next[j];
    }
  }
  free(next);
  int ret = i - j;
  if (n < ret + m)
    return -1;
  else
    return ret;
}
```
user: What is the c source code of the assembly code below:

```asm
<strStr>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  call   <buildNext@plt>
  mov    %r13,%rdi
  mov    %rax,%r14
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L48
  test   %ebp,%ebp
  jle    L48
  xor    %eax,%eax
  xor    %edx,%edx
  jmp    L79
  nop
L54:
  movslq %edx,%rsi
  movslq %eax,%rcx
  movzbl 0x0(%r13,%rsi,1),%edi
  cmp    %dil,(%r12,%rcx,1)
  je     L13
  mov    (%r14,%rcx,4),%eax
  cmp    %ebx,%eax
  jge    L40
L12:
  cmp    %edx,%ebp
  jle    L40
L79:
  test   %eax,%eax
  jns    L54
L13:
  add    $0x1,%eax
  add    $0x1,%edx
  cmp    %ebx,%eax
  jl     L12
L40:
  sub    %eax,%edx
  mov    %edx,%r12d
  add    %edx,%ebx
L75:
  mov    %r14,%rdi
  call   <free@plt>
  cmp    %ebx,%ebp
  mov    $0xffffffff,%eax
  pop    %rbx
  cmovl  %eax,%r12d
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L48:
  xor    %r12d,%r12d
  jmp    L75
```

assistant: ```c
int strStr(char *haystack, char *needle) {
  int *next = buildNext(needle);
  int n = strlen(haystack);
  int i = 0;
  int m = strlen(needle);
  int j = 0;
  while (j < m && i < n) {
    if (j < 0 || needle[j] == haystack[i]) {
      i++;
      j++;
    } else {
      j = next[j];
    }
  }
  free(next);
  int ret = i - j;
  if (n < ret + m)
    return -1;
  else
    return ret;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<strStr>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  call   <buildNext@plt>
  mov    %r13,%rdi
  mov    %rax,%r14
  call   <strlen@plt>
  mov    %r12,%rdi
  mov    %eax,%ebp
  call   <strlen@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L22
  test   %ebp,%ebp
  jle    L22
  xor    %eax,%eax
  xor    %edx,%edx
  jmp    L59
  nop
L19:
  movslq %edx,%rsi
  movslq %eax,%rcx
  movzbl 0x0(%r13,%rsi,1),%edi
  cmp    %dil,(%r12,%rcx,1)
  je     L88
  mov    (%r14,%rcx,4),%eax
  cmp    %ebx,%eax
  jge    L32
L6:
  cmp    %edx,%ebp
  jle    L32
L59:
  test   %eax,%eax
  jns    L19
L88:
  add    $0x1,%eax
  add    $0x1,%edx
  cmp    %ebx,%eax
  jl     L6
L32:
  sub    %eax,%edx
  mov    %edx,%r12d
  add    %edx,%ebx
L36:
  mov    %r14,%rdi
  call   <free@plt>
  cmp    %ebx,%ebp
  mov    $0xffffffff,%eax
  pop    %rbx
  cmovl  %eax,%r12d
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L22:
  xor    %r12d,%r12d
  jmp    L36
```

assistant: ```c
int strStr(char *haystack, char *needle) {
  int *next = buildNext(needle);
  int n = strlen(haystack);
  int i = 0;
  int m = strlen(needle);
  int j = 0;
  while (j < m && i < n) {
    if (j < 0 || needle[j] == haystack[i]) {
      i++;
      j++;
    } else {
      j = next[j];
    }
  }
  free(next);
  int ret = i - j;
  if (n < ret + m)
    return -1;
  else
    return ret;
}
```
user: What is the c source code of the assembly code below?

```asm
<queue_work>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int queue_work(struct workqueue_struct *wq, struct work_struct *work) {
  int a;
  return a;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<queue_work>:
  endbr64
  mov    $0x0,%eax
  ret
```

assistant: ```c
int queue_work(struct workqueue_struct *wq, struct work_struct *work) {
  int a;
  return a;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<queue_work>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int queue_work(struct workqueue_struct *wq, struct work_struct *work) {
  int a;
  return a;
}
```
user: What is the c source code of the assembly code below:

```asm
<queue_work>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int queue_work(struct workqueue_struct *wq, struct work_struct *work) {
  int a;
  return a;
}
```
user: What is the c source code of the assembly code below?

```asm
<plusOne>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpl   $0x0,-0x1c(%rbp)
  jne    L63
  mov    -0x28(%rbp),%rax
  movl   $0x1,(%rax)
  mov    -0x8(%rbp),%rax
  movl   $0x1,(%rax)
  mov    -0x8(%rbp),%rax
  jmp    L37
L63:
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
  jmp    L16
L33:
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  sub    $0xa,%edx
  mov    %edx,(%rax)
  subl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  add    $0x1,%edx
  mov    %edx,(%rax)
L16:
  cmpl   $0x0,-0xc(%rbp)
  jle    L25
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0x9,%eax
  jg     L33
L25:
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x9,%eax
  jle    L66
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  lea    -0xa(%rax),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%rax
  lea    0x4(%rax),%rcx
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  mov    %rcx,%rdi
  call   <memcpy@plt>
  mov    -0x8(%rbp),%rax
  movl   $0x1,(%rax)
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    -0x28(%rbp),%rax
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
  jmp    L37
L66:
  mov    -0x1c(%rbp),%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    -0x28(%rbp),%rax
  mov    -0x1c(%rbp),%edx
  mov    %edx,(%rax)
  mov    -0x8(%rbp),%rax
L37:
  leave
  ret
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int *res, i;
  res = malloc((digitsSize + 1) * sizeof(int));
  if (!digitsSize) {
    *returnSize = 1;
    res[0] = 1;
    return (res);
  }
  i = digitsSize - 1;
  digits[i]++;
  while (i > 0 && digits[i] >= 10) {
    digits[i] -= 10;
    i--;
    digits[i]++;
  }
  if (digits[0] >= 10) {
    digits[0] -= 10;
    memcpy(res + 1, digits, digitsSize * sizeof(int));
    res[0] = 1;
    *returnSize = digitsSize + 1;
    return (res);
  }
  memcpy(res, digits, digitsSize * sizeof(int));
  *returnSize = digitsSize;
  return (res);
}
```
user: Translate the below assembly code into C programming language:

```asm
<plusOne>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %rdx,%r13
  lea    0x1(%rsi),%r14d
  lea    0x0(,%r14,4),%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %ebx,%ebx
  je     L15
  lea    -0x1(%rbx),%ecx
  movslq %ecx,%rdx
  shl    $0x2,%rdx
  lea    0x0(%rbp,%rdx,1),%rax
  addl   $0x1,(%rax)
  test   %ecx,%ecx
  jle    L34
  lea    -0x4(%rbp,%rdx,1),%rcx
  lea    -0x2(%rbx),%edx
  shl    $0x2,%rdx
  sub    %rdx,%rcx
L60:
  mov    (%rax),%edx
  cmp    $0x9,%edx
  jle    L34
  sub    $0xa,%edx
  mov    %edx,(%rax)
  addl   $0x1,-0x4(%rax)
  sub    $0x4,%rax
  cmp    %rcx,%rax
  jne    L60
L34:
  mov    0x0(%rbp),%eax
  cmp    $0x9,%eax
  jg     L52
  lea    0x0(,%rbx,4),%edx
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <memcpy@plt>
  mov    %ebx,0x0(%r13)
L51:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L15:
  movl   $0x1,0x0(%r13)
  movl   $0x1,(%rax)
  jmp    L51
L52:
  sub    $0xa,%eax
  mov    %eax,0x0(%rbp)
  lea    0x0(,%rbx,4),%edx
  lea    0x4(%r12),%rdi
  mov    %rbp,%rsi
  call   <memcpy@plt>
  movl   $0x1,(%r12)
  mov    %r14d,0x0(%r13)
  jmp    L51
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int *res, i;
  res = malloc((digitsSize + 1) * sizeof(int));
  if (!digitsSize) {
    *returnSize = 1;
    res[0] = 1;
    return (res);
  }
  i = digitsSize - 1;
  digits[i]++;
  while (i > 0 && digits[i] >= 10) {
    digits[i] -= 10;
    i--;
    digits[i]++;
  }
  if (digits[0] >= 10) {
    digits[0] -= 10;
    memcpy(res + 1, digits, digitsSize * sizeof(int));
    res[0] = 1;
    *returnSize = digitsSize + 1;
    return (res);
  }
  memcpy(res, digits, digitsSize * sizeof(int));
  *returnSize = digitsSize;
  return (res);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<plusOne>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  lea    0x1(%rsi),%ebp
  push   %rbx
  lea    0x0(,%rbp,4),%r15d
  mov    %rdx,%rbx
  mov    %r15d,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r14
  test   %r12d,%r12d
  je     L68
  lea    -0x1(%r12),%edx
  movslq %edx,%rsi
  lea    0x0(%r13,%rsi,4),%rcx
  mov    (%rcx),%eax
  add    $0x1,%eax
  mov    %eax,(%rcx)
  test   %edx,%edx
  jle    L6
  movslq %r12d,%rdx
  lea    -0x4(%r13,%rsi,4),%rcx
  lea    -0xc(%r13,%rdx,4),%rsi
  lea    -0x2(%r12),%edx
  shl    $0x2,%rdx
  sub    %rdx,%rsi
  jmp    L42
  nopl   (%rax)
L56:
  sub    $0xa,%eax
  sub    $0x4,%rcx
  mov    %eax,0x8(%rcx)
  mov    0x4(%rcx),%eax
  add    $0x1,%eax
  mov    %eax,0x4(%rcx)
  cmp    %rsi,%rcx
  je     L6
L42:
  cmp    $0x9,%eax
  jg     L56
L6:
  mov    0x0(%r13),%eax
  lea    -0x4(%r15),%edx
  cmp    $0x9,%eax
  jg     L30
  mov    %r13,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %r12d,(%rbx)
L8:
  add    $0x8,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L30:
  sub    $0xa,%eax
  lea    0x4(%r14),%rdi
  mov    %r13,%rsi
  mov    %eax,0x0(%r13)
  call   <memcpy@plt>
  movl   $0x1,(%r14)
  mov    %ebp,(%rbx)
  jmp    L8
  xchg   %ax,%ax
L68:
  movl   $0x1,(%rbx)
  movl   $0x1,(%rax)
  jmp    L8
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int *res, i;
  res = malloc((digitsSize + 1) * sizeof(int));
  if (!digitsSize) {
    *returnSize = 1;
    res[0] = 1;
    return (res);
  }
  i = digitsSize - 1;
  digits[i]++;
  while (i > 0 && digits[i] >= 10) {
    digits[i] -= 10;
    i--;
    digits[i]++;
  }
  if (digits[0] >= 10) {
    digits[0] -= 10;
    memcpy(res + 1, digits, digitsSize * sizeof(int));
    res[0] = 1;
    *returnSize = digitsSize + 1;
    return (res);
  }
  memcpy(res, digits, digitsSize * sizeof(int));
  *returnSize = digitsSize;
  return (res);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<plusOne>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  lea    0x1(%rsi),%ebp
  push   %rbx
  lea    0x0(,%rbp,4),%r15d
  mov    %rdx,%rbx
  mov    %r15d,%edi
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r14
  test   %r12d,%r12d
  je     L47
  lea    -0x1(%r12),%ecx
  movslq %ecx,%rdx
  shl    $0x2,%rdx
  lea    0x0(%r13,%rdx,1),%rax
  addl   $0x1,(%rax)
  test   %ecx,%ecx
  jle    L55
  lea    -0x4(%r13,%rdx,1),%rsi
  lea    -0x2(%r12),%edx
  shl    $0x2,%rdx
  sub    %rdx,%rsi
  jmp    L56
  nopl   (%rax)
L10:
  sub    $0xa,%ecx
  addl   $0x1,-0x4(%rax)
  sub    $0x4,%rax
  mov    %ecx,0x4(%rax)
  cmp    %rsi,%rax
  je     L55
L56:
  mov    (%rax),%ecx
  cmp    $0x9,%ecx
  jg     L10
L55:
  mov    0x0(%r13),%eax
  lea    -0x4(%r15),%edx
  cmp    $0x9,%eax
  jg     L5
  mov    %r13,%rsi
  mov    %r14,%rdi
  call   <memcpy@plt>
  mov    %r12d,(%rbx)
L72:
  add    $0x8,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L5:
  sub    $0xa,%eax
  lea    0x4(%r14),%rdi
  mov    %r13,%rsi
  mov    %eax,0x0(%r13)
  call   <memcpy@plt>
  movl   $0x1,(%r14)
  mov    %ebp,(%rbx)
  jmp    L72
  xchg   %ax,%ax
L47:
  movl   $0x1,(%rbx)
  movl   $0x1,(%rax)
  jmp    L72
```

assistant: ```c
int *plusOne(int *digits, int digitsSize, int *returnSize) {
  int *res, i;
  res = malloc((digitsSize + 1) * sizeof(int));
  if (!digitsSize) {
    *returnSize = 1;
    res[0] = 1;
    return (res);
  }
  i = digitsSize - 1;
  digits[i]++;
  while (i > 0 && digits[i] >= 10) {
    digits[i] -= 10;
    i--;
    digits[i]++;
  }
  if (digits[0] >= 10) {
    digits[0] -= 10;
    memcpy(res + 1, digits, digitsSize * sizeof(int));
    res[0] = 1;
    *returnSize = digitsSize + 1;
    return (res);
  }
  memcpy(res, digits, digitsSize * sizeof(int));
  *returnSize = digitsSize;
  return (res);
}
```
user: What is the c source code of the assembly code below:

```asm
<think>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D7
type_infered: string

tool: D7: P   %ld pensando...


assistant: ```c
void think(unsigned long i) {
  printf("P   %ld pensando...\n", i);
  sleep(1);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<think>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <sleep@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_guessed: string

tool: D9: P   %ld pensando...


assistant: ```c
void think(unsigned long i) {
  printf("P   %ld pensando...\n", i);
  sleep(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
<think>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <sleep@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D5
type: string

tool: D5: P   %ld pensando...


assistant: ```c
void think(unsigned long i) {
  printf("P   %ld pensando...\n", i);
  sleep(1);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<think>:
  endbr64
  sub    $0x8,%rsp
  mov    %rdi,%rsi
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <sleep@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D9
type: string

tool: D9: P   %ld pensando...


assistant: ```c
void think(unsigned long i) {
  printf("P   %ld pensando...\n", i);
  sleep(1);
}
```
user: What is the c source code of the assembly code below?

```asm
<selection>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  jmp    L19
L12:
  mov    -0x8(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <min@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
  addl   $0x1,-0x8(%rbp)
L19:
  mov    -0x8(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L12
  nop
  nop
  leave
  ret
```

assistant: ```c
void selection(int a[], int n) {
  int i, index;
  for (i = 0; i < n; i++) {
    index = min(a, n, i);
    swap(&a[index], &a[i]);
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<selection>:
  endbr64
  test   %esi,%esi
  jle    L17
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%r12d
  mov    %esi,%r13d
  mov    $0x0,%ebx
L8:
  mov    %ebx,%edx
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <min@plt>
  lea    0x0(%rbp,%rbx,4),%rsi
  cltq
  lea    0x0(%rbp,%rax,4),%rdi
  call   <swap@plt>
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L8
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  ret
```

assistant: ```c
void selection(int a[], int n) {
  int i, index;
  for (i = 0; i < n; i++) {
    index = min(a, n, i);
    swap(&a[index], &a[i]);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<selection>:
  endbr64
  test   %esi,%esi
  jle    L11
  push   %r13
  mov    %esi,%r13d
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   (%rax)
L2:
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <min@plt>
  lea    0x0(%rbp,%rbx,4),%rsi
  add    $0x1,%rbx
  cltq
  lea    0x0(%rbp,%rax,4),%rdi
  call   <swap@plt>
  cmp    %r12,%rbx
  jne    L2
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  ret
```

assistant: ```c
void selection(int a[], int n) {
  int i, index;
  for (i = 0; i < n; i++) {
    index = min(a, n, i);
    swap(&a[index], &a[i]);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<selection>:
  endbr64
  test   %esi,%esi
  jle    L18
  push   %r13
  mov    %esi,%r13d
  push   %r12
  movslq %esi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  nopl   (%rax)
L8:
  mov    %ebx,%edx
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <min@plt>
  lea    0x0(%rbp,%rbx,4),%rsi
  add    $0x1,%rbx
  cltq
  lea    0x0(%rbp,%rax,4),%rdi
  call   <swap@plt>
  cmp    %r12,%rbx
  jne    L8
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  ret
```

assistant: ```c
void selection(int a[], int n) {
  int i, index;
  for (i = 0; i < n; i++) {
    index = min(a, n, i);
    swap(&a[index], &a[i]);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<stringCopy>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
char *stringCopy(char *s) {
  char *r = s;
  return r;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<stringCopy>:
  endbr64
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *stringCopy(char *s) {
  char *r = s;
  return r;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stringCopy>:
  endbr64
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *stringCopy(char *s) {
  char *r = s;
  return r;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<stringCopy>:
  endbr64
  mov    %rdi,%rax
  ret
```

assistant: ```c
char *stringCopy(char *s) {
  char *r = s;
  return r;
}
```