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

```asm
<findUnion>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x338,%rsp
  mov    %rdx,%r8
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  jle    L32
  mov    %esi,%r9d
L19:
  mov    (%rdi,%rax,4),%edx
  mov    %edx,(%rsp,%rax,4)
  add    $0x1,%rax
  cmp    %r9,%rax
  jne    L19
L32:
  lea    (%rsi,%rcx,1),%ebx
  cmp    %ebx,%esi
  jge    L37
  lea    -0x1(%rcx),%eax
  lea    0x4(,%rax,4),%rdi
  movslq %esi,%rsi
  lea    (%rsp,%rsi,4),%rcx
  mov    $0x0,%eax
L31:
  mov    (%r8,%rax,1),%edx
  mov    %edx,(%rcx,%rax,1)
  add    $0x4,%rax
  cmp    %rax,%rdi
  jne    L31
L37:
  mov    %rsp,%rbp
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <sort@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <removeDuplicates@plt>
  mov    %eax,%ebx
  lea    D6(%rip),%rdi
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printSet@plt>
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L30
  mov    $0x0,%eax
  add    $0x338,%rsp
  pop    %rbx
  pop    %rbp
  ret
L30:
  call   <__stack_chk_fail@plt>
```

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

tool: D6: 
Union ::

assistant: ```c
int findUnion(int *set1, int len1, int *set2, int len2) {
  int unionarr[200], i, j, len;
  for (i = 0; i < len1; i++)
    unionarr[i] = set1[i];
  for (i = len1, j = 0; i < (len1 + len2); i++, j++)
    unionarr[i] = set2[j];
  sort(unionarr, (len1 + len2));
  len = removeDuplicates(unionarr, (len1 + len2));
  printf("\nUnion ::");
  printSet(unionarr, len);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<findUnion>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  push   %rbx
  movslq %esi,%rbx
  sub    $0x330,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  test   %ebx,%ebx
  jle    L11
  movslq %ebx,%rdx
  mov    %rdi,%rsi
  mov    %rbp,%rdi
  shl    $0x2,%rdx
  call   <memcpy@plt>
L11:
  lea    (%rbx,%r12,1),%r13d
  cmp    %r13d,%ebx
  jge    L24
  lea    -0x1(%r12),%eax
  lea    0x0(%rbp,%rbx,4),%rdi
  mov    %r14,%rsi
  lea    0x4(,%rax,4),%rdx
  call   <memcpy@plt>
L24:
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <sort@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <removeDuplicates@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%r12d
  call   <printf@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <printSet@plt>
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L14
  add    $0x330,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L14:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D0
type_guessed: string

tool: D0: 
Union ::

assistant: ```c
int findUnion(int *set1, int len1, int *set2, int len2) {
  int unionarr[200], i, j, len;
  for (i = 0; i < len1; i++)
    unionarr[i] = set1[i];
  for (i = len1, j = 0; i < (len1 + len2); i++, j++)
    unionarr[i] = set2[j];
  sort(unionarr, (len1 + len2));
  len = removeDuplicates(unionarr, (len1 + len2));
  printf("\nUnion ::");
  printSet(unionarr, len);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<findUnion>:
  endbr64
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  push   %rbx
  movslq %esi,%rbx
  sub    $0x330,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x328(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  test   %ebx,%ebx
  jle    L20
  movslq %ebx,%rdx
  mov    %rdi,%rsi
  mov    %rbp,%rdi
  shl    $0x2,%rdx
  call   <memcpy@plt>
L20:
  lea    (%rbx,%r12,1),%r13d
  cmp    %r13d,%ebx
  jge    L14
  lea    -0x1(%r12),%eax
  lea    0x0(%rbp,%rbx,4),%rdi
  mov    %r14,%rsi
  lea    0x4(,%rax,4),%rdx
  call   <memcpy@plt>
L14:
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <sort@plt>
  mov    %r13d,%esi
  mov    %rbp,%rdi
  call   <removeDuplicates@plt>
  lea    D0(%rip),%rdi
  mov    %eax,%r12d
  call   <printf@plt>
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <printSet@plt>
  mov    0x328(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  add    $0x330,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D0
type_guessed: string

tool: D0: 
Union ::

assistant: ```c
int findUnion(int *set1, int len1, int *set2, int len2) {
  int unionarr[200], i, j, len;
  for (i = 0; i < len1; i++)
    unionarr[i] = set1[i];
  for (i = len1, j = 0; i < (len1 + len2); i++, j++)
    unionarr[i] = set2[j];
  sort(unionarr, (len1 + len2));
  len = removeDuplicates(unionarr, (len1 + len2));
  printf("\nUnion ::");
  printSet(unionarr, len);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bucketSort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edx,%ecx
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movslq %esi,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%r8
  and    $0xfffffffffffffff0,%r8
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L84:
  cmp    %rdx,%rsp
  je     L9
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L84
L9:
  mov    %r8,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L73
  orq    $0x0,-0x8(%rsp,%rax,1)
L73:
  mov    %rsp,%r8
  lea    -0x40(%rbp),%rax
  lea    -0x18(%rbp),%r10
L5:
  movl   $0x0,(%rax)
  add    $0x4,%rax
  cmp    %r10,%rax
  jne    L5
  test   %esi,%esi
  jle    L6
  mov    %rdi,%r9
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%r11
L12:
  mov    (%r9),%eax
  cltd
  idiv   %ecx
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%ebx
  sar    $0x1f,%ebx
  sub    %ebx,%edx
  lea    (%rdx,%rdx,4),%ebx
  add    %ebx,%ebx
  sub    %ebx,%eax
  movslq %eax,%rdx
  addl   $0x1,-0x40(%rbp,%rdx,4)
  add    $0x4,%r9
  cmp    %r11,%r9
  jne    L12
L6:
  lea    -0x3c(%rbp),%rax
L25:
  mov    -0x4(%rax),%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %r10,%rax
  jne    L25
  mov    %esi,%r9d
  sub    $0x1,%r9d
  js     L19
  movslq %r9d,%r9
L67:
  mov    (%rdi,%r9,4),%r10d
  mov    %r10d,%eax
  cltd
  idiv   %ecx
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x22,%rdx
  mov    %eax,%r11d
  sar    $0x1f,%r11d
  sub    %r11d,%edx
  lea    (%rdx,%rdx,4),%r11d
  add    %r11d,%r11d
  sub    %r11d,%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp,%rdx,4),%eax
  sub    $0x1,%eax
  movslq %eax,%r11
  mov    %r10d,(%r8,%r11,4)
  mov    %eax,-0x40(%rbp,%rdx,4)
  sub    $0x1,%r9
  test   %r9d,%r9d
  jns    L67
L19:
  test   %esi,%esi
  jle    L39
  mov    %esi,%esi
  mov    $0x0,%eax
L15:
  mov    (%r8,%rax,4),%edx
  mov    %edx,(%rdi,%rax,4)
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L15
L39:
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L34
  mov    -0x8(%rbp),%rbx
  leave
  ret
L34:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bucketSort(int a[], int size, int ex) {
  int count[10];
  int output[size];
  for (int i = 0; i < 10; i++) {
    count[i] = 0;
  }
  for (int i = 0; i < size; i++) {
    count[(a[i] / ex) % 10]++;
  }
  for (int i = 1; i < 10; i++) {
    count[i] = count[i] + count[i - 1];
  }
  for (int i = size - 1; i >= 0; i--) {
    output[count[(a[i] / ex) % 10] - 1] = a[i];
    count[(a[i] / ex) % 10]--;
  }
  for (int i = 0; i < size; i++) {
    a[i] = output[i];
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bucketSort>:
  endbr64
  push   %rbp
  movslq %esi,%r9
  mov    %edx,%ecx
  mov    %r9,%r8
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    0xf(,%r9,4),%rax
  mov    %rsp,%rsi
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rsi
  and    $0xfffffffffffffff0,%rdx
  cmp    %rsi,%rsp
  je     L26
L104:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L104
L26:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L41
L69:
  pxor   %xmm0,%xmm0
  movq   $0x0,-0x20(%rbp)
  mov    %rsp,%rsi
  lea    -0x1(%r8),%r10d
  movaps %xmm0,-0x40(%rbp)
  movaps %xmm0,-0x30(%rbp)
  test   %r8d,%r8d
  jle    L83
  lea    -0x1(%r8),%eax
  mov    %rdi,%r11
  mov    %rax,%r10
  lea    0x4(%rdi,%rax,4),%rbx
  nopl   0x0(%rax)
L45:
  mov    (%r11),%eax
  add    $0x4,%r11
  cltd
  idiv   %ecx
  movslq %eax,%rdx
  mov    %eax,%r12d
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%r12d
  sar    $0x22,%rdx
  sub    %r12d,%edx
  lea    (%rdx,%rdx,4),%r12d
  add    %r12d,%r12d
  sub    %r12d,%eax
  movslq %eax,%rdx
  addl   $0x1,-0x40(%rbp,%rdx,4)
  cmp    %r11,%rbx
  jne    L45
L83:
  lea    -0x3c(%rbp),%rax
  lea    -0x18(%rbp),%r11
  nopl   0x0(%rax)
L55:
  mov    -0x4(%rax),%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %r11,%rax
  jne    L55
  test   %r10d,%r10d
  js     L94
  movslq %r10d,%r10
  xchg   %ax,%ax
L31:
  mov    (%rdi,%r10,4),%r11d
  sub    $0x1,%r10
  mov    %r11d,%eax
  cltd
  idiv   %ecx
  movslq %eax,%rdx
  mov    %eax,%ebx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ebx
  sar    $0x22,%rdx
  sub    %ebx,%edx
  lea    (%rdx,%rdx,4),%ebx
  add    %ebx,%ebx
  sub    %ebx,%eax
  movslq %eax,%rdx
  mov    -0x40(%rbp,%rdx,4),%eax
  sub    $0x1,%eax
  movslq %eax,%rbx
  mov    %eax,-0x40(%rbp,%rdx,4)
  mov    %r11d,(%rsi,%rbx,4)
  test   %r10d,%r10d
  jns    L31
L94:
  test   %r8d,%r8d
  jle    L49
  lea    0x0(,%r9,4),%rdx
  call   <memcpy@plt>
L49:
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L93
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
  xchg   %ax,%ax
L41:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L69
L93:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void bucketSort(int a[], int size, int ex) {
  int count[10];
  int output[size];
  for (int i = 0; i < 10; i++) {
    count[i] = 0;
  }
  for (int i = 0; i < size; i++) {
    count[(a[i] / ex) % 10]++;
  }
  for (int i = 1; i < 10; i++) {
    count[i] = count[i] + count[i - 1];
  }
  for (int i = size - 1; i >= 0; i--) {
    output[count[(a[i] / ex) % 10] - 1] = a[i];
    count[(a[i] / ex) % 10]--;
  }
  for (int i = 0; i < size; i++) {
    a[i] = output[i];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<split_lines>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L24
L2:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L5
  mov    -0x18(%rbp),%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x4(%rbp)
L5:
  addq   $0x1,-0x18(%rbp)
L24:
  mov    -0x18(%rbp),%rax
  cmp    -0x20(%rbp),%rax
  jb     L2
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int split_lines(char *p, const char *plim) {
  int i;
  i = 0;
  while (p < plim) {
    if (*p == '\n') {
      *p = '\0';
      i++;
    }
    p++;
  }
  return (i);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<split_lines>:
  endbr64
  cmp    %rsi,%rdi
  jae    L26
  mov    $0x0,%eax
  jmp    L6
L34:
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  je     L38
L6:
  cmpb   $0xa,(%rdi)
  jne    L34
  movb   $0x0,(%rdi)
  add    $0x1,%eax
  jmp    L34
L38:
  ret
L26:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int split_lines(char *p, const char *plim) {
  int i;
  i = 0;
  while (p < plim) {
    if (*p == '\n') {
      *p = '\0';
      i++;
    }
    p++;
  }
  return (i);
}
```
user: Translate the below assembly code into C programming language:

```asm
<split_lines>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  jae    L5
  nopl   0x0(%rax,%rax,1)
L26:
  cmpb   $0xa,(%rdi)
  jne    L3
  movb   $0x0,(%rdi)
  add    $0x1,%eax
L3:
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  jne    L26
  ret
  nopl   (%rax)
L5:
  ret
```

assistant: ```c
int split_lines(char *p, const char *plim) {
  int i;
  i = 0;
  while (p < plim) {
    if (*p == '\n') {
      *p = '\0';
      i++;
    }
    p++;
  }
  return (i);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<split_lines>:
  endbr64
  xor    %eax,%eax
  cmp    %rsi,%rdi
  jae    L8
  nopl   0x0(%rax,%rax,1)
L5:
  cmpb   $0xa,(%rdi)
  jne    L9
  movb   $0x0,(%rdi)
  add    $0x1,%eax
L9:
  add    $0x1,%rdi
  cmp    %rdi,%rsi
  jne    L5
  ret
  nopl   (%rax)
L8:
  ret
```

assistant: ```c
int split_lines(char *p, const char *plim) {
  int i;
  i = 0;
  while (p < plim) {
    if (*p == '\n') {
      *p = '\0';
      i++;
    }
    p++;
  }
  return (i);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Usage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  nop
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_label: D2
type_guessed: string

tool: D2: usage: propeller-elf-image-size
         [ -v ]            verbose output
         [ -? ]            display a usage message and exit
         <name>            file to load


assistant: ```c
void Usage(void) {
  printf("usage: propeller-elf-image-size\n         [ -v ]            verbose output\n         [ -? ]            display a usage message and exit\n         <name>            file to load\n");
  exit(1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  call   <exit@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: retrieve_data
arguments: label: D3
datatype: string

tool: D3: usage: propeller-elf-image-size
         [ -v ]            verbose output
         [ -? ]            display a usage message and exit
         <name>            file to load


assistant: ```c
void Usage(void) {
  printf("usage: propeller-elf-image-size\n         [ -v ]            verbose output\n         [ -? ]            display a usage message and exit\n         <name>            file to load\n");
  exit(1);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D5(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
type: string

tool: D5: usage: propeller-elf-image-size
         [ -v ]            verbose output
         [ -? ]            display a usage message and exit
         <name>            file to load


assistant: ```c
void Usage(void) {
  printf("usage: propeller-elf-image-size\n         [ -v ]            verbose output\n         [ -? ]            display a usage message and exit\n         <name>            file to load\n");
  exit(1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Usage>:
  endbr64
  sub    $0x8,%rsp
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0x1,%edi
  add    $0x8,%rsp
  jmp    1070 <exit@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D1
type_guessed: string

tool: D1: usage: propeller-elf-image-size
         [ -v ]            verbose output
         [ -? ]            display a usage message and exit
         <name>            file to load


assistant: ```c
void Usage(void) {
  printf("usage: propeller-elf-image-size\n         [ -v ]            verbose output\n         [ -? ]            display a usage message and exit\n         <name>            file to load\n");
  exit(1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<eal_bf_strhash>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  mov    $0x0,%r12d
  mov    -0x28(%rbp),%rbx
  movl   $0x0,-0x14(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L12
L6:
  mov    %r12d,%eax
  shl    $0x5,%eax
  sub    %r12d,%eax
  mov    %eax,%edx
  movzbl (%rbx),%eax
  movzbl %al,%eax
  lea    (%rdx,%rax,1),%r12d
  addl   $0x1,-0x14(%rbp)
  add    $0x1,%rbx
L12:
  mov    -0x14(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jb     L6
  mov    %r12d,%eax
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
```

assistant: ```c
unsigned int eal_bf_strhash(const unsigned char *str, unsigned int len) {
  register unsigned int h = 0;
  register const unsigned char *p = str;
  unsigned int i = 0;
  for (i = 0; i < len; ++i, ++p) {
    h = 31 * h + *p;
  }
  return h;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<eal_bf_strhash>:
  endbr64
  test   %esi,%esi
  je     L0
  mov    %esi,%eax
  lea    (%rdi,%rax,1),%rcx
  mov    $0x0,%eax
L3:
  mov    %eax,%edx
  shl    $0x5,%edx
  sub    %eax,%edx
  movzbl (%rdi),%eax
  add    %edx,%eax
  add    $0x1,%rdi
  cmp    %rcx,%rdi
  jne    L3
  ret
L0:
  mov    %esi,%eax
  ret
```

assistant: ```c
unsigned int eal_bf_strhash(const unsigned char *str, unsigned int len) {
  register unsigned int h = 0;
  register const unsigned char *p = str;
  unsigned int i = 0;
  for (i = 0; i < len; ++i, ++p) {
    h = 31 * h + *p;
  }
  return h;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<eal_bf_strhash>:
  endbr64
  test   %esi,%esi
  je     L13
  mov    %esi,%esi
  xor    %eax,%eax
  lea    (%rdi,%rsi,1),%rcx
L18:
  mov    %eax,%edx
  add    $0x1,%rdi
  shl    $0x5,%edx
  sub    %eax,%edx
  movzbl -0x1(%rdi),%eax
  add    %edx,%eax
  cmp    %rdi,%rcx
  jne    L18
  ret
  nopw   0x0(%rax,%rax,1)
L13:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned int eal_bf_strhash(const unsigned char *str, unsigned int len) {
  register unsigned int h = 0;
  register const unsigned char *p = str;
  unsigned int i = 0;
  for (i = 0; i < len; ++i, ++p) {
    h = 31 * h + *p;
  }
  return h;
}
```
user: Translate the below assembly code into C programming language:

```asm
<eal_bf_strhash>:
  endbr64
  test   %esi,%esi
  je     L17
  mov    %esi,%esi
  xor    %eax,%eax
  lea    (%rdi,%rsi,1),%rcx
L15:
  mov    %eax,%edx
  add    $0x1,%rdi
  shl    $0x5,%edx
  sub    %eax,%edx
  movzbl -0x1(%rdi),%eax
  add    %edx,%eax
  cmp    %rdi,%rcx
  jne    L15
  ret
  nopw   0x0(%rax,%rax,1)
L17:
  xor    %eax,%eax
  ret
```

assistant: ```c
unsigned int eal_bf_strhash(const unsigned char *str, unsigned int len) {
  register unsigned int h = 0;
  register const unsigned char *p = str;
  unsigned int i = 0;
  for (i = 0; i < len; ++i, ++p) {
    h = 31 * h + *p;
  }
  return h;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x2,-0x4(%rbp)
  subl   $0x1,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  int a = 2;
  return --a;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int main() {
  int a = 2;
  return --a;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int main() {
  int a = 2;
  return --a;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  mov    $0x1,%eax
  ret
```

assistant: ```c
int main() {
  int a = 2;
  return --a;
}
```
user: Translate the below assembly code into C programming language:

```asm
<av_int2flt>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  add    %eax,%eax
  cmp    $0xff000000,%eax
  jbe    L8
  pxor   %xmm0,%xmm0
  divss  %xmm0,%xmm0
  jmp    L12
L8:
  mov    -0x4(%rbp),%eax
  sar    $0x17,%eax
  movzbl %al,%eax
  lea    -0x96(%rax),%edx
  mov    -0x4(%rbp),%eax
  and    $0x7fffff,%eax
  lea    0x800000(%rax),%ecx
  mov    -0x4(%rbp),%eax
  sar    $0x1f,%eax
  or     $0x1,%eax
  imul   %ecx,%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <ldexp@plt>
L12:
  leave
  ret
```

assistant: ```c
float av_int2flt(int32_t v) {
  if (v + v > 0xFF000000U)
    return (0.0 / 0.0);
  return ldexp(((v & 0x7FFFFF) + (1 << 23)) * (v >> 31 | 1), (v >> 23 & 0xFF) - 150);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<av_int2flt>:
  endbr64
  lea    (%rdi,%rdi,1),%eax
  cmp    $0xff000000,%eax
  jbe    L3
  pxor   %xmm0,%xmm0
  divss  %xmm0,%xmm0
  ret
L3:
  sub    $0x8,%rsp
  mov    %edi,%esi
  sar    $0x17,%esi
  movzbl %sil,%esi
  sub    $0x96,%esi
  mov    %edi,%eax
  and    $0x7fffff,%eax
  add    $0x800000,%eax
  sar    $0x1f,%edi
  or     $0x1,%edi
  imul   %eax,%edi
  call   <ldexp@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
float av_int2flt(int32_t v) {
  if (v + v > 0xFF000000U)
    return (0.0 / 0.0);
  return ldexp(((v & 0x7FFFFF) + (1 << 23)) * (v >> 31 | 1), (v >> 23 & 0xFF) - 150);
}
```
user: Translate the below assembly code into C programming language:

```asm
<av_int2flt>:
  endbr64
  lea    (%rdi,%rdi,1),%eax
  cmp    $0xff000000,%eax
  jbe    L3
  pxor   %xmm0,%xmm0
  divss  %xmm0,%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %edi,%eax
  mov    %edi,%esi
  sar    $0x1f,%edi
  and    $0x7fffff,%eax
  or     $0x1,%edi
  sar    $0x17,%esi
  add    $0x800000,%eax
  movzbl %sil,%esi
  imul   %eax,%edi
  sub    $0x96,%esi
  jmp    1050 <ldexp@plt>
```

assistant: ```c
float av_int2flt(int32_t v) {
  if (v + v > 0xFF000000U)
    return (0.0 / 0.0);
  return ldexp(((v & 0x7FFFFF) + (1 << 23)) * (v >> 31 | 1), (v >> 23 & 0xFF) - 150);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<av_int2flt>:
  endbr64
  lea    (%rdi,%rdi,1),%eax
  cmp    $0xff000000,%eax
  jbe    L3
  pxor   %xmm0,%xmm0
  divss  %xmm0,%xmm0
  ret
  nopw   0x0(%rax,%rax,1)
L3:
  mov    %edi,%eax
  mov    %edi,%esi
  sar    $0x1f,%edi
  and    $0x7fffff,%eax
  or     $0x1,%edi
  sar    $0x17,%esi
  add    $0x800000,%eax
  movzbl %sil,%esi
  imul   %eax,%edi
  sub    $0x96,%esi
  jmp    1050 <ldexp@plt>
```

assistant: ```c
float av_int2flt(int32_t v) {
  if (v + v > 0xFF000000U)
    return (0.0 / 0.0);
  return ldexp(((v & 0x7FFFFF) + (1 << 23)) * (v >> 31 | 1), (v >> 23 & 0xFF) - 150);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<PICC_get_err_message>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  cmpl   $0x2,-0x4(%rbp)
  je     L6
  cmpl   $0x2,-0x4(%rbp)
  jg     L27
  cmpl   $0x0,-0x4(%rbp)
  je     L69
  cmpl   $0x1,-0x4(%rbp)
  je     L79
  jmp    L27
L69:
  lea    D17(%rip),%rax
  jmp    L19
L79:
  lea    D36(%rip),%rax
  jmp    L19
L6:
  lea    D3(%rip),%rax
  jmp    L19
L27:
  lea    D19(%rip),%rax
L19:
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D17
data_type: string
name: retrieve_data
arguments: tag: D36
data_type: string
name: retrieve_data
arguments: tag: D3
data_type: string
name: retrieve_data
arguments: tag: D19
data_type: string

tool: D17: memory allocation

tool: D36: pthread creation

tool: D3: a deadlock occured

tool: D19: unknwon error

assistant: ```c
char *PICC_get_err_message(int errcode) {
  switch (errcode) {
  case 0:
    return "memory allocation";
  case 1:
    return "pthread creation";
  case 2:
    return "a deadlock occured";
  default:
    return "unknwon error";
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<PICC_get_err_message>:
  endbr64
  lea    D34(%rip),%rax
  cmp    $0x1,%edi
  je     L2
  lea    D39(%rip),%rax
  cmp    $0x2,%edi
  je     L2
  test   %edi,%edi
  lea    D30(%rip),%rax
  lea    D11(%rip),%rdx
  cmove  %rdx,%rax
L2:
  ret
```

tool_calls: name: retrieve_data
arguments: label: D34
data_type: string
name: retrieve_data
arguments: label: D39
data_type: string
name: retrieve_data
arguments: label: D30
data_type: string
name: retrieve_data
arguments: label: D11
data_type: string

tool: D34: pthread creation

tool: D39: a deadlock occured

tool: D30: unknwon error

tool: D11: memory allocation

assistant: ```c
char *PICC_get_err_message(int errcode) {
  switch (errcode) {
  case 0:
    return "memory allocation";
  case 1:
    return "pthread creation";
  case 2:
    return "a deadlock occured";
  default:
    return "unknwon error";
  }
}
```