ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %rsi,-0x30(%rbp)
  cmpl   $0x4,-0x24(%rbp)
  je     L26
  mov    -0x30(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L8
L26:
  mov    -0x30(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %rax,-0x18(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x10,%rax
  mov    (%rax),%rax
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x30(%rbp),%rax
  add    $0x18,%rax
  mov    (%rax),%rax
  mov    $0x10,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <strtoul@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x20(%rbp)
  jmp    L0
L31:
  mov    -0x20(%rbp),%rax
  mov    %rax,%rsi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  add    %rax,-0x20(%rbp)
L0:
  mov    -0x20(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jbe    L31
  mov    $0x0,%eax
L8:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D13
type: string
name: extract_symbolic_data
arguments: data_label: D9
type: string

tool: D13: Counts in hexadecimal.  Usage:

	%s start skip end


tool: D9: %08lx


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long start;
  unsigned long skip;
  unsigned long end;
  unsigned long i;
  if (argc != 4) {
    printf("Counts in hexadecimal.  Usage:\n\n\t%s start skip end\n", argv[0]);
    return -1;
  }
  start = strtoul(argv[1], NULL, 16);
  skip = strtoul(argv[2], NULL, 16);
  end = strtoul(argv[3], NULL, 16);
  for (i = start; i <= end; i += skip) {
    printf("%08lx\n", i);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rsi,%rbp
  cmp    $0x4,%edi
  jne    L16
  mov    0x8(%rsi),%rdi
  mov    $0x10,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  mov    %rax,%rbx
  mov    0x10(%rbp),%rdi
  mov    $0x10,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  mov    %rax,%r12
  mov    0x18(%rbp),%rdi
  mov    $0x10,%edx
  mov    $0x0,%esi
  call   <strtoul@plt>
  mov    %rax,%rbp
  cmp    %rax,%rbx
  ja     L32
  lea    D3(%rip),%r13
L31:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    %r12,%rbx
  cmp    %rbx,%rbp
  jae    L31
  mov    $0x0,%eax
L49:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L16:
  mov    (%rsi),%rsi
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0xffffffff,%eax
  jmp    L49
L32:
  mov    $0x0,%eax
  jmp    L49
```

tool_calls: name: unpack_data
arguments: label: D3
data_type: string
name: unpack_data
arguments: label: D11
data_type: string

tool: D3: %08lx


tool: D11: Counts in hexadecimal.  Usage:

	%s start skip end


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long start;
  unsigned long skip;
  unsigned long end;
  unsigned long i;
  if (argc != 4) {
    printf("Counts in hexadecimal.  Usage:\n\n\t%s start skip end\n", argv[0]);
    return -1;
  }
  start = strtoul(argv[1], NULL, 16);
  skip = strtoul(argv[2], NULL, 16);
  end = strtoul(argv[3], NULL, 16);
  for (i = start; i <= end; i += skip) {
    printf("%08lx\n", i);
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x4,%edi
  jne    L7
  mov    0x8(%rsi),%rdi
  mov    $0x10,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  mov    0x10(%rbp),%rdi
  mov    $0x10,%edx
  xor    %esi,%esi
  mov    %rax,%rbx
  call   <strtoul@plt>
  mov    0x18(%rbp),%rdi
  mov    $0x10,%edx
  xor    %esi,%esi
  mov    %rax,%r12
  call   <strtoul@plt>
  mov    %rax,%rbp
  cmp    %rax,%rbx
  ja     L6
  lea    D1(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L32:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    %r12,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jae    L32
L6:
  xor    %eax,%eax
L3:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  mov    (%rsi),%rsi
  lea    D17(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L3
```

tool_calls: name: read_data
arguments: data_label: D1
type: string
name: read_data
arguments: data_label: D17
type: string

tool: D1: %08lx


tool: D17: Counts in hexadecimal.  Usage:

	%s start skip end


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long start;
  unsigned long skip;
  unsigned long end;
  unsigned long i;
  if (argc != 4) {
    printf("Counts in hexadecimal.  Usage:\n\n\t%s start skip end\n", argv[0]);
    return -1;
  }
  start = strtoul(argv[1], NULL, 16);
  skip = strtoul(argv[2], NULL, 16);
  end = strtoul(argv[3], NULL, 16);
  for (i = start; i <= end; i += skip) {
    printf("%08lx\n", i);
  }
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  cmp    $0x4,%edi
  jne    L22
  mov    0x8(%rsi),%rdi
  mov    $0x10,%edx
  xor    %esi,%esi
  call   <strtoul@plt>
  mov    0x10(%rbp),%rdi
  mov    $0x10,%edx
  xor    %esi,%esi
  mov    %rax,%rbx
  call   <strtoul@plt>
  mov    0x18(%rbp),%rdi
  mov    $0x10,%edx
  xor    %esi,%esi
  mov    %rax,%r12
  call   <strtoul@plt>
  mov    %rax,%rbp
  cmp    %rax,%rbx
  ja     L30
  lea    D8(%rip),%r13
  nopl   0x0(%rax,%rax,1)
L5:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    %r12,%rbx
  call   <printf@plt>
  cmp    %rbx,%rbp
  jae    L5
L30:
  xor    %eax,%eax
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L22:
  mov    (%rsi),%rsi
  lea    D4(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  or     $0xffffffff,%eax
  jmp    L18
```

tool_calls: name: retrieve_data
arguments: data_label: D8
type_infered: string
name: retrieve_data
arguments: data_label: D4
type_infered: string

tool: D8: %08lx


tool: D4: Counts in hexadecimal.  Usage:

	%s start skip end


assistant: ```c
int main(int argc, char *argv[]) {
  unsigned long start;
  unsigned long skip;
  unsigned long end;
  unsigned long i;
  if (argc != 4) {
    printf("Counts in hexadecimal.  Usage:\n\n\t%s start skip end\n", argv[0]);
    return -1;
  }
  start = strtoul(argv[1], NULL, 16);
  skip = strtoul(argv[2], NULL, 16);
  end = strtoul(argv[3], NULL, 16);
  for (i = start; i <= end; i += skip) {
    printf("%08lx\n", i);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<to_signed_int>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  shr    $0xf,%eax
  test   %eax,%eax
  je     L9
  orl    $0xffff8000,-0x4(%rbp)
L9:
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int to_signed_int(unsigned int value) {
  int signed_value = value;
  if (value >> 15)
    signed_value |= -1 << 15;
  return signed_value;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<to_signed_int>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%edx
  or     $0xffff8000,%edx
  mov    %edi,%ecx
  shr    $0xf,%ecx
  cmovne %edx,%eax
  ret
```

assistant: ```c
int to_signed_int(unsigned int value) {
  int signed_value = value;
  if (value >> 15)
    signed_value |= -1 << 15;
  return signed_value;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<to_signed_int>:
  endbr64
  mov    %edi,%edx
  mov    %edi,%ecx
  mov    %edi,%eax
  or     $0xffff8000,%edx
  shr    $0xf,%ecx
  cmovne %edx,%eax
  ret
```

assistant: ```c
int to_signed_int(unsigned int value) {
  int signed_value = value;
  if (value >> 15)
    signed_value |= -1 << 15;
  return signed_value;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<to_signed_int>:
  endbr64
  mov    %edi,%edx
  mov    %edi,%ecx
  mov    %edi,%eax
  or     $0xffff8000,%edx
  shr    $0xf,%ecx
  cmovne %edx,%eax
  ret
```

assistant: ```c
int to_signed_int(unsigned int value) {
  int signed_value = value;
  if (value >> 15)
    signed_value |= -1 << 15;
  return signed_value;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%eax
  call   <test_spr@plt>
  test   %rax,%rax
  je     L6
  mov    $0x0,%eax
  call   <abort@plt>
L6:
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
  if (test_spr())
    abort();
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <test_spr@plt>
  test   %rax,%rax
  jne    L2
L0:
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
L2:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L0
```

assistant: ```c
int main() {
  if (test_spr())
    abort();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_spr@plt>
  test   %rax,%rax
  jne    L1
L19:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L1:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L19
```

assistant: ```c
int main() {
  if (test_spr())
    abort();
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <test_spr@plt>
  test   %rax,%rax
  jne    L17
L19:
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
L17:
  xor    %eax,%eax
  call   <abort@plt>
  jmp    L19
```

assistant: ```c
int main() {
  if (test_spr())
    abort();
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<stringclone>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L16
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <getmem@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
  jmp    L5
L16:
  mov    $0x0,%eax
L5:
  leave
  ret
```

assistant: ```c
char *stringclone(char *s) {
  if (s) {
    char *p = (char *)getmem(strlen(s) + 1);
    strcpy(p, s);
    return p;
  }
  return NULL;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<stringclone>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  test   %rdi,%rdi
  je     L4
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <getmem@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L4:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
char *stringclone(char *s) {
  if (s) {
    char *p = (char *)getmem(strlen(s) + 1);
    strcpy(p, s);
    return p;
  }
  return NULL;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<stringclone>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L8
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <getmem@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
L8:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *stringclone(char *s) {
  if (s) {
    char *p = (char *)getmem(strlen(s) + 1);
    strcpy(p, s);
    return p;
  }
  return NULL;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<stringclone>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L1
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <getmem@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
L1:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *stringclone(char *s) {
  if (s) {
    char *p = (char *)getmem(strlen(s) + 1);
    strcpy(p, s);
    return p;
  }
  return NULL;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<piececount>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L2
L15:
  mov    -0x1c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x1c(%rbp)
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x4(%rbp)
L2:
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jle    L15
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
piececount(player, startrow, endrow) int *player, startrow, endrow;
{
  int sum;
  sum = 0;
  while (startrow <= endrow)
    sum += player[startrow++];
  return (sum);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<piececount>:
  endbr64
  cmp    %edx,%esi
  jg     L15
  movslq %esi,%rcx
  lea    (%rdi,%rcx,4),%rax
  sub    %esi,%edx
  add    %rcx,%rdx
  lea    0x4(%rdi,%rdx,4),%rcx
  mov    $0x0,%edx
L9:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L9
L12:
  mov    %edx,%eax
  ret
L15:
  mov    $0x0,%edx
  jmp    L12
```

assistant: ```c
piececount(player, startrow, endrow) int *player, startrow, endrow;
{
  int sum;
  sum = 0;
  while (startrow <= endrow)
    sum += player[startrow++];
  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
<piececount>:
  endbr64
  cmp    %edx,%esi
  jg     L2
  movslq %esi,%rcx
  sub    %esi,%edx
  xor    %r8d,%r8d
  add    %rcx,%rdx
  lea    (%rdi,%rcx,4),%rax
  lea    0x4(%rdi,%rdx,4),%rdx
  nopl   0x0(%rax)
L13:
  add    (%rax),%r8d
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L13
  mov    %r8d,%eax
  ret
L2:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
piececount(player, startrow, endrow) int *player, startrow, endrow;
{
  int sum;
  sum = 0;
  while (startrow <= endrow)
    sum += player[startrow++];
  return (sum);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<piececount>:
  endbr64
  mov    %rdi,%r8
  mov    %esi,%ecx
  mov    %edx,%edi
  cmp    %edx,%esi
  jg     L10
  mov    %edx,%eax
  sub    %esi,%eax
  lea    0x1(%rax),%r9d
  cmp    $0x2,%eax
  jbe    L54
  movslq %esi,%rax
  pxor   %xmm0,%xmm0
  lea    (%r8,%rax,4),%rdx
  mov    %r9d,%eax
  shr    $0x2,%eax
  lea    -0x1(%rax),%esi
  lea    0x10(%rdx),%rax
  shl    $0x4,%rsi
  add    %rax,%rsi
  jmp    L30
  nopl   (%rax)
L42:
  add    $0x10,%rax
L30:
  movdqu (%rdx),%xmm1
  mov    %rax,%rdx
  paddd  %xmm1,%xmm0
  cmp    %rsi,%rax
  jne    L42
  movdqa %xmm0,%xmm1
  mov    %r9d,%edx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%edx
  paddd  %xmm1,%xmm0
  add    %edx,%ecx
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  cmp    %edx,%r9d
  je     L66
L53:
  movslq %ecx,%rdx
  lea    0x0(,%rdx,4),%rsi
  add    (%r8,%rdx,4),%eax
  lea    0x1(%rcx),%edx
  cmp    %edx,%edi
  jl     L33
  add    $0x2,%ecx
  add    0x4(%r8,%rsi,1),%eax
  cmp    %ecx,%edi
  jl     L33
  add    0x8(%r8,%rsi,1),%eax
  ret
  nop
L10:
  xor    %eax,%eax
L33:
  ret
  nopl   0x0(%rax,%rax,1)
L66:
  ret
L54:
  xor    %eax,%eax
  jmp    L53
```

assistant: ```c
piececount(player, startrow, endrow) int *player, startrow, endrow;
{
  int sum;
  sum = 0;
  while (startrow <= endrow)
    sum += player[startrow++];
  return (sum);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <func@plt>
  mov    $0x14,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <__ESBMC_assume@plt>
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void main() {
  func();
  char *test = malloc(20);
  __ESBMC_assume(test);
  free(test);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%eax
  call   <func@plt>
  mov    $0x14,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rax,%rdi
  call   <__ESBMC_assume@plt>
  mov    %rbx,%rdi
  call   <free@plt>
  pop    %rbx
  ret
```

assistant: ```c
void main() {
  func();
  char *test = malloc(20);
  __ESBMC_assume(test);
  free(test);
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <func@plt>
  mov    $0x14,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <__ESBMC_assume@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1080 <free@plt>
```

assistant: ```c
void main() {
  func();
  char *test = malloc(20);
  __ESBMC_assume(test);
  free(test);
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbp
  xor    %eax,%eax
  call   <func@plt>
  mov    $0x14,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    %rax,%rdi
  call   <__ESBMC_assume@plt>
  mov    %rbp,%rdi
  pop    %rbp
  jmp    1080 <free@plt>
```

assistant: ```c
void main() {
  func();
  char *test = malloc(20);
  __ESBMC_assume(test);
  free(test);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<hashString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  movq   $0x1505,-0x8(%rbp)
  jmp    L9
L14:
  mov    -0x8(%rbp),%rax
  shl    $0x5,%rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  mov    -0xc(%rbp),%eax
  cltq
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
L9:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jne    L14
  mov    -0x8(%rbp),%rax
  pop    %rbp
  ret
```

assistant: ```c
unsigned long hashString(char *str) {
  unsigned long hash = 5381;
  int c;
  while ((c = *str++))
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: Translate the below assembly code into C programming language:

```asm
<hashString>:
  endbr64
  lea    0x1(%rdi),%rcx
  movzbl (%rdi),%edx
  test   %dl,%dl
  je     L2
  mov    $0x1505,%eax
L7:
  mov    %rax,%rsi
  shl    $0x5,%rsi
  add    %rsi,%rax
  movsbq %dl,%rdx
  add    %rdx,%rax
  add    $0x1,%rcx
  movzbl -0x1(%rcx),%edx
  test   %dl,%dl
  jne    L7
  ret
L2:
  mov    $0x1505,%eax
  ret
```

assistant: ```c
unsigned long hashString(char *str) {
  unsigned long hash = 5381;
  int c;
  while ((c = *str++))
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hashString>:
  endbr64
  movsbq (%rdi),%rdx
  lea    0x1(%rdi),%rcx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L13
  nopl   (%rax)
L18:
  mov    %rax,%rsi
  add    $0x1,%rcx
  shl    $0x5,%rsi
  add    %rsi,%rax
  add    %rdx,%rax
  movsbq -0x1(%rcx),%rdx
  test   %dl,%dl
  jne    L18
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  ret
```

assistant: ```c
unsigned long hashString(char *str) {
  unsigned long hash = 5381;
  int c;
  while ((c = *str++))
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<hashString>:
  endbr64
  movsbq (%rdi),%rdx
  lea    0x1(%rdi),%rcx
  mov    $0x1505,%eax
  test   %dl,%dl
  je     L13
  nopl   (%rax)
L6:
  mov    %rax,%rsi
  add    $0x1,%rcx
  shl    $0x5,%rsi
  add    %rsi,%rax
  add    %rdx,%rax
  movsbq -0x1(%rcx),%rdx
  test   %dl,%dl
  jne    L6
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  ret
```

assistant: ```c
unsigned long hashString(char *str) {
  unsigned long hash = 5381;
  int c;
  while ((c = *str++))
    hash = ((hash << 5) + hash) + c;
  return hash;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<decToBinUl>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movq   $0x1,-0x8(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L56
L3:
  shlq   -0x8(%rbp)
  addl   $0x1,-0xc(%rbp)
L56:
  mov    -0x8(%rbp),%rax
  cmp    -0x18(%rbp),%rax
  jbe    L3
  mov    -0x8(%rbp),%rax
  shr    %rax
  mov    %rax,-0x8(%rbp)
  subl   $0x1,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  jmp    L41
L33:
  mov    -0x18(%rbp),%rax
  sub    -0x8(%rbp),%rax
  test   %rax,%rax
  js     L13
  mov    -0x8(%rbp),%rax
  sub    %rax,-0x18(%rbp)
  mov    -0xc(%rbp),%eax
  sub    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movl   $0x1,(%rax)
  jmp    L10
L13:
  mov    -0xc(%rbp),%eax
  sub    -0x10(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movl   $0x0,(%rax)
L10:
  mov    -0x8(%rbp),%rax
  shr    %rax
  mov    %rax,-0x8(%rbp)
  subl   $0x1,-0x10(%rbp)
L41:
  cmpl   $0x0,-0x10(%rbp)
  jns    L33
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int decToBinUl(unsigned long long n, int binary[]) {
  unsigned long long i = 1;
  int y = 0, x = 0;
  while (i <= n) {
    i *= 2;
    x++;
  }
  i /= 2;
  x--;
  for (y = x; y >= 0; y--) {
    if ((long)(n - i) >= 0) {
      n -= i;
      binary[x - y] = 1;
    } else {
      binary[x - y] = 0;
    }
    i /= 2;
  }
  return (int)x;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<decToBinUl>:
  endbr64
  test   %rdi,%rdi
  je     L61
  mov    $0x0,%ecx
  mov    $0x1,%edx
L17:
  mov    %rdx,%rax
  add    %rdx,%rdx
  mov    %ecx,%r8d
  add    $0x1,%ecx
  cmp    %rdx,%rdi
  jae    L17
  btr    $0x3f,%rax
  test   %r8d,%r8d
  js     L3
  mov    %r8d,%r9d
  mov    $0x0,%edx
  jmp    L1
L21:
  movslq %edx,%rdi
  movl   $0x1,(%rsi,%rdi,4)
  mov    %rcx,%rdi
L66:
  shr    %rax
  lea    0x1(%rdx),%ecx
  cmp    %r9d,%edx
  je     L3
  mov    %ecx,%edx
L1:
  mov    %rdi,%rcx
  sub    %rax,%rcx
  jns    L21
  movslq %edx,%rcx
  movl   $0x0,(%rsi,%rcx,4)
  jmp    L66
L61:
  mov    $0xffffffff,%r8d
L3:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int decToBinUl(unsigned long long n, int binary[]) {
  unsigned long long i = 1;
  int y = 0, x = 0;
  while (i <= n) {
    i *= 2;
    x++;
  }
  i /= 2;
  x--;
  for (y = x; y >= 0; y--) {
    if ((long)(n - i) >= 0) {
      n -= i;
      binary[x - y] = 1;
    } else {
      binary[x - y] = 0;
    }
    i /= 2;
  }
  return (int)x;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<decToBinUl>:
  endbr64
  test   %rdi,%rdi
  je     L27
  xor    %edx,%edx
  mov    $0x1,%eax
L3:
  add    %rax,%rax
  mov    %edx,%r8d
  lea    0x1(%rdx),%edx
  cmp    %rax,%rdi
  jae    L3
  movslq %r8d,%rdx
  shr    %rax
  lea    0x4(%rsi,%rdx,4),%r9
  nopl   0x0(%rax)
L7:
  mov    %rdi,%rcx
  sub    %rax,%rcx
  mov    %rcx,%rdx
  sar    $0x3f,%rdx
  add    $0x1,%edx
  test   %rcx,%rcx
  cmovns %rcx,%rdi
  mov    %edx,(%rsi)
  add    $0x4,%rsi
  shr    %rax
  cmp    %rsi,%r9
  jne    L7
  mov    %r8d,%eax
  ret
  nopl   0x0(%rax)
L27:
  mov    $0xffffffff,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int decToBinUl(unsigned long long n, int binary[]) {
  unsigned long long i = 1;
  int y = 0, x = 0;
  while (i <= n) {
    i *= 2;
    x++;
  }
  i /= 2;
  x--;
  for (y = x; y >= 0; y--) {
    if ((long)(n - i) >= 0) {
      n -= i;
      binary[x - y] = 1;
    } else {
      binary[x - y] = 0;
    }
    i /= 2;
  }
  return (int)x;
}
```