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

```asm
<oblRozwKwad>:
  endbr64
  push   %rbp
  movapd %xmm0,%xmm3
  movapd %xmm1,%xmm0
  mov    %rsi,%rbp
  push   %rbx
  mulsd  %xmm1,%xmm0
  movapd %xmm3,%xmm4
  mov    %rdi,%rbx
  sub    $0x28,%rsp
  mulsd  D6(%rip),%xmm4
  movsd  %xmm1,0x18(%rsp)
  movsd  %xmm3,0x10(%rsp)
  mulsd  %xmm2,%xmm4
  subsd  %xmm4,%xmm0
  movsd  %xmm0,0x8(%rsp)
  call   <fabs@plt>
  movsd  0x8(%rsp),%xmm0
  movsd  0x10(%rsp),%xmm3
  test   %eax,%eax
  movsd  0x18(%rsp),%xmm1
  jle    L16
  pxor   %xmm2,%xmm2
  xor    %eax,%eax
  comisd %xmm0,%xmm2
  jbe    L19
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L16:
  addsd  %xmm3,%xmm3
  xorpd  D4(%rip),%xmm1
  mov    $0x1,%eax
  divsd  %xmm3,%xmm1
  movsd  %xmm1,(%rbx)
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L19:
  movsd  %xmm1,0x10(%rsp)
  movsd  %xmm3,0x8(%rsp)
  movsd  %xmm0,0x18(%rsp)
  call   <sqrt@plt>
  pxor   %xmm2,%xmm2
  movsd  0x8(%rsp),%xmm3
  movsd  0x10(%rsp),%xmm1
  cvtsi2sd %rax,%xmm2
  movsd  0x18(%rsp),%xmm0
  addsd  %xmm3,%xmm3
  subsd  %xmm1,%xmm2
  movsd  %xmm3,0x8(%rsp)
  divsd  %xmm3,%xmm2
  movsd  %xmm2,(%rbx)
  call   <sqrt@plt>
  pxor   %xmm0,%xmm0
  movsd  0x10(%rsp),%xmm1
  xorpd  D4(%rip),%xmm1
  cvtsi2sd %rax,%xmm0
  movsd  0x8(%rsp),%xmm3
  mov    $0x2,%eax
  subsd  %xmm0,%xmm1
  divsd  %xmm3,%xmm1
  movsd  %xmm1,0x0(%rbp)
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D6
type_guessed: f64
name: disassemble_data
arguments: tag: D4
type_guessed: byte[16]

tool: D6: 4

tool: {"D4": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

assistant: ```c
int oblRozwKwad(double a, double b, double c, double *t1, double *t2) {
  double delta = b * b - 4 * a * c;
  if (fabs(delta) < 1e-5) {
    *t1 = (-1) * (b) / (2 * a);
    return 1;
  } else if (delta < 0) {
    return 0;
  } else {
    *t1 = ((-1) * (b) + sqrt(delta)) / (2 * a);
    *t2 = ((-1) * (b)-sqrt(delta)) / (2 * a);
    return 2;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<freetrix>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L5
L6:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <free@plt>
  addl   $0x1,-0x4(%rbp)
L5:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L6
  movq   $0x0,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  nop
  leave
  ret
```

assistant: ```c
void freetrix(float **m_a, int size) {
  for (int i = 0; i < size; i++) {
    free(m_a[i]);
  }
  m_a = 0;
  free(m_a);
}
```
user: What is the c source code of the assembly code below?

```asm
<freetrix>:
  endbr64
  test   %esi,%esi
  jle    L12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x8(%rdi,%rax,8),%rbp
L8:
  mov    (%rbx),%rdi
  call   <free@plt>
  add    $0x8,%rbx
  cmp    %rbp,%rbx
  jne    L8
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L12:
  ret
```

assistant: ```c
void freetrix(float **m_a, int size) {
  for (int i = 0; i < size; i++) {
    free(m_a[i]);
  }
  m_a = 0;
  free(m_a);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<freetrix>:
  endbr64
  test   %esi,%esi
  jle    L7
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x8(%rdi,%rax,8),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L18:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <free@plt>
  cmp    %rbp,%rbx
  jne    L18
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L7:
  ret
```

assistant: ```c
void freetrix(float **m_a, int size) {
  for (int i = 0; i < size; i++) {
    free(m_a[i]);
  }
  m_a = 0;
  free(m_a);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<freetrix>:
  endbr64
  test   %esi,%esi
  jle    L12
  push   %rbp
  lea    -0x1(%rsi),%eax
  push   %rbx
  lea    0x8(%rdi,%rax,8),%rbp
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L2:
  mov    (%rbx),%rdi
  add    $0x8,%rbx
  call   <free@plt>
  cmp    %rbp,%rbx
  jne    L2
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  ret
```

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

```asm
<RSA_read_privatekey>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  lea    D5(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <fopen@plt>
  mov    %rax,-0x10(%rbp)
  cmpq   $0x0,-0x10(%rbp)
  jne    L7
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
L7:
  mov    $0x0,%eax
  call   <RSA_new@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <PEM_read_RSAPrivateKey@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D5
type_guessed: string
name: retrieve_data
arguments: label: D0
type_guessed: string

tool: D5: rb

tool: D0: 
open file [ %s ] error!



assistant: ```c
RSA *RSA_read_privatekey(char *pszSource_file) {
  FILE *pfkey;
  RSA *pRsa;
  if ((pfkey = fopen(pszSource_file, "rb")) == NULL) {
    printf("\nopen file [ %s ] error!\n\n", pszSource_file);
    exit(0);
  }
  pRsa = RSA_new();
  pRsa = PEM_read_RSAPrivateKey(pfkey, NULL, NULL, NULL);
  return pRsa;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<RSA_read_privatekey>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  lea    D6(%rip),%rsi
  call   <fopen@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  je     L8
L12:
  mov    $0x0,%eax
  call   <RSA_new@plt>
  mov    $0x0,%ecx
  mov    $0x0,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <PEM_read_RSAPrivateKey@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  mov    %rbp,%rsi
  lea    D16(%rip),%rdi
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  jmp    L12
```

tool_calls: name: parse_data
arguments: label: D6
data_type: string
name: parse_data
arguments: label: D16
data_type: string

tool: D6: rb

tool: D16: 
open file [ %s ] error!



assistant: ```c
RSA *RSA_read_privatekey(char *pszSource_file) {
  FILE *pfkey;
  RSA *pRsa;
  if ((pfkey = fopen(pszSource_file, "rb")) == NULL) {
    printf("\nopen file [ %s ] error!\n\n", pszSource_file);
    exit(0);
  }
  pRsa = RSA_new();
  pRsa = PEM_read_RSAPrivateKey(pfkey, NULL, NULL, NULL);
  return pRsa;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<RSA_read_privatekey>:
  endbr64
  push   %r12
  lea    D19(%rip),%rsi
  mov    %rdi,%r12
  push   %rbp
  sub    $0x8,%rsp
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L2
L9:
  xor    %eax,%eax
  call   <RSA_new@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  xor    %ecx,%ecx
  pop    %rbp
  xor    %edx,%edx
  xor    %esi,%esi
  pop    %r12
  jmp    10a0 <PEM_read_RSAPrivateKey@plt>
  xchg   %ax,%ax
L2:
  mov    %r12,%rsi
  lea    D12(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L9
```

tool_calls: name: parse_data
arguments: tag: D19
data_type: string
name: parse_data
arguments: tag: D12
data_type: string

tool: D19: rb

tool: D12: 
open file [ %s ] error!



assistant: ```c
RSA *RSA_read_privatekey(char *pszSource_file) {
  FILE *pfkey;
  RSA *pRsa;
  if ((pfkey = fopen(pszSource_file, "rb")) == NULL) {
    printf("\nopen file [ %s ] error!\n\n", pszSource_file);
    exit(0);
  }
  pRsa = RSA_new();
  pRsa = PEM_read_RSAPrivateKey(pfkey, NULL, NULL, NULL);
  return pRsa;
}
```
user: Translate the below assembly code into C programming language:

```asm
<RSA_read_privatekey>:
  endbr64
  push   %r12
  lea    D8(%rip),%rsi
  mov    %rdi,%r12
  push   %rbp
  sub    $0x8,%rsp
  call   <fopen@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L2
L9:
  xor    %eax,%eax
  call   <RSA_new@plt>
  add    $0x8,%rsp
  mov    %rbp,%rdi
  xor    %ecx,%ecx
  pop    %rbp
  xor    %edx,%edx
  xor    %esi,%esi
  pop    %r12
  jmp    10a0 <PEM_read_RSAPrivateKey@plt>
  xchg   %ax,%ax
L2:
  mov    %r12,%rsi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  xor    %edi,%edi
  call   <exit@plt>
  jmp    L9
```

tool_calls: name: read_data
arguments: tag: D8
datatype: string
name: read_data
arguments: tag: D1
datatype: string

tool: D8: rb

tool: D1: 
open file [ %s ] error!



assistant: ```c
RSA *RSA_read_privatekey(char *pszSource_file) {
  FILE *pfkey;
  RSA *pRsa;
  if ((pfkey = fopen(pszSource_file, "rb")) == NULL) {
    printf("\nopen file [ %s ] error!\n\n", pszSource_file);
    exit(0);
  }
  pRsa = RSA_new();
  pRsa = PEM_read_RSAPrivateKey(pfkey, NULL, NULL, NULL);
  return pRsa;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<shuffle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %rdi,-0x48(%rbp)
  mov    %esi,-0x4c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x4c(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L29:
  cmp    %rdx,%rsp
  je     L57
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L29
L57:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L54
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L54:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x34(%rbp)
  mov    -0x4c(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  jmp    L21
L25:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idivl  -0x4c(%rbp)
  mov    %edx,-0x30(%rbp)
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x48(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x34(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x34(%rbp)
  mov    (%rcx),%ecx
  mov    -0x20(%rbp),%rdx
  cltq
  mov    %ecx,(%rdx,%rax,4)
  jmp    L34
L53:
  mov    -0x30(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x48(%rbp),%rax
  add    %rdx,%rax
  mov    -0x30(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x48(%rbp),%rdx
  add    %rcx,%rdx
  mov    (%rax),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x30(%rbp)
L34:
  mov    -0x4c(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0x30(%rbp)
  jl     L53
  subl   $0x1,-0x4c(%rbp)
L21:
  cmpl   $0x0,-0x4c(%rbp)
  jne    L25
  movl   $0x0,-0x34(%rbp)
  jmp    L82
L76:
  mov    -0x34(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x48(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x20(%rbp),%rax
  mov    -0x34(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%eax
  mov    %eax,(%rcx)
  addl   $0x1,-0x34(%rbp)
L82:
  mov    -0x34(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jl     L76
  mov    %rbx,%rsp
  nop
  mov    -0x18(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L41
  call   <__stack_chk_fail@plt>
L41:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void shuffle(int *deck, int n) {
  int scratch[n];
  int i = 0;
  int size = n;
  int k;
  srand(time(NULL));
  while (n) {
    k = rand() % n;
    scratch[i++] = deck[k];
    for (; k < n - 1; k++)
      deck[k] = deck[k + 1];
    n--;
  }
  for (i = 0; i < size; i++)
    deck[i] = scratch[i];
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<shuffle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,%r12
  mov    %esi,-0x44(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  mov    %esi,%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L39:
  cmp    %rdx,%rsp
  je     L59
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L39
L59:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L30
  orq    $0x0,-0x8(%rsp,%rax,1)
L30:
  mov    %rsp,%r14
  mov    $0x0,%edi
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  mov    -0x44(%rbp),%ebx
  test   %ebx,%ebx
  je     L36
  mov    %r14,%r13
  mov    $0xfffffffe,%r15d
L84:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idiv   %ebx
  movslq %edx,%rcx
  lea    (%r12,%rcx,4),%rax
  mov    (%rax),%esi
  mov    %esi,0x0(%r13)
  mov    %ebx,%esi
  sub    $0x1,%ebx
  cmp    %ebx,%edx
  jge    L7
  mov    %r15d,%edi
  sub    %edx,%edi
  lea    (%rdi,%rsi,1),%edx
  add    %rcx,%rdx
  lea    0x4(%r12,%rdx,4),%rcx
L31:
  mov    0x4(%rax),%edx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L31
L7:
  add    $0x4,%r13
  test   %ebx,%ebx
  jne    L84
  mov    -0x44(%rbp),%eax
  test   %eax,%eax
  jle    L36
  mov    %eax,%ecx
  mov    $0x0,%eax
L3:
  mov    (%r14,%rax,4),%edx
  mov    %edx,(%r12,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%rcx
  jne    L3
L36:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void shuffle(int *deck, int n) {
  int scratch[n];
  int i = 0;
  int size = n;
  int k;
  srand(time(NULL));
  while (n) {
    k = rand() % n;
    scratch[i++] = deck[k];
    for (; k < n - 1; k++)
      deck[k] = deck[k + 1];
    n--;
  }
  for (i = 0; i < size; i++)
    deck[i] = scratch[i];
}
```
user: Translate the below assembly code into C programming language:

```asm
<shuffle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %esi,%rax
  mov    %rsp,%rcx
  mov    %rax,-0x48(%rbp)
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L58
L43:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L43
L58:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L7
L70:
  xor    %edi,%edi
  mov    %rsp,%rbx
  mov    %rbx,-0x50(%rbp)
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %r14d,%r14d
  je     L60
  mov    %r14d,%r15d
  mov    $0xfffffffe,%r13d
  nopl   0x0(%rax,%rax,1)
L61:
  xor    %eax,%eax
  call   <rand@plt>
  cltd
  idiv   %r15d
  movslq %edx,%rax
  shl    $0x2,%rax
  lea    (%r12,%rax,1),%rdi
  mov    (%rdi),%esi
  mov    %esi,(%rbx)
  mov    %r15d,%esi
  sub    $0x1,%r15d
  cmp    %r15d,%edx
  jge    L62
  mov    %r13d,%r8d
  sub    %edx,%r8d
  lea    (%r8,%rsi,1),%edx
  lea    0x4(%r12,%rax,1),%rsi
  lea    0x4(,%rdx,4),%rdx
  call   <memmove@plt>
L62:
  add    $0x4,%rbx
  test   %r15d,%r15d
  jne    L61
  test   %r14d,%r14d
  jle    L60
  mov    -0x48(%rbp),%rdx
  mov    -0x50(%rbp),%rsi
  mov    %r12,%rdi
  shl    $0x2,%rdx
  call   <memcpy@plt>
L60:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L7:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L70
L26:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void shuffle(int *deck, int n) {
  int scratch[n];
  int i = 0;
  int size = n;
  int k;
  srand(time(NULL));
  while (n) {
    k = rand() % n;
    scratch[i++] = deck[k];
    for (; k < n - 1; k++)
      deck[k] = deck[k + 1];
    n--;
  }
  for (i = 0; i < size; i++)
    deck[i] = scratch[i];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<shuffle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  mov    %esi,%r14d
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %esi,%rax
  mov    %rsp,%rcx
  mov    %rax,-0x48(%rbp)
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L85
L82:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L82
L85:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L7
L2:
  xor    %edi,%edi
  mov    %rsp,%rbx
  mov    %rbx,-0x50(%rbp)
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  test   %r14d,%r14d
  je     L53
  mov    %r14d,%r15d
  mov    $0xfffffffe,%r13d
  nopl   0x0(%rax,%rax,1)
L68:
  xor    %eax,%eax
  call   <rand@plt>
  cltd
  idiv   %r15d
  movslq %edx,%rax
  shl    $0x2,%rax
  lea    (%r12,%rax,1),%rdi
  mov    (%rdi),%esi
  mov    %esi,(%rbx)
  mov    %r15d,%esi
  sub    $0x1,%r15d
  cmp    %r15d,%edx
  jge    L18
  mov    %r13d,%r8d
  sub    %edx,%r8d
  lea    (%r8,%rsi,1),%edx
  lea    0x4(%r12,%rax,1),%rsi
  lea    0x4(,%rdx,4),%rdx
  call   <memmove@plt>
L18:
  add    $0x4,%rbx
  test   %r15d,%r15d
  jne    L68
  test   %r14d,%r14d
  jle    L53
  mov    -0x48(%rbp),%rdx
  mov    -0x50(%rbp),%rsi
  mov    %r12,%rdi
  shl    $0x2,%rdx
  call   <memcpy@plt>
L53:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L31
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L7:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L2
L31:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void shuffle(int *deck, int n) {
  int scratch[n];
  int i = 0;
  int size = n;
  int k;
  srand(time(NULL));
  while (n) {
    k = rand() % n;
    scratch[i++] = deck[k];
    for (; k < n - 1; k++)
      deck[k] = deck[k + 1];
    n--;
  }
  for (i = 0; i < size; i++)
    deck[i] = scratch[i];
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ret32word>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movq   $0x0,-0x18(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x3(%rax),%rdx
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  lea    0x2(%rax),%rdx
  mov    -0x28(%rbp),%rax
  movzbl 0x1(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  movzbl 0x2(%rax),%eax
  mov    %al,(%rdx)
  mov    -0x28(%rbp),%rax
  movzbl 0x3(%rax),%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L9
  call   <__stack_chk_fail@plt>
L9:
  leave
  ret
```

assistant: ```c
unsigned long ret32word(workspace)
unsigned char *workspace;
{
  unsigned long foo;
  unsigned char *bleh;
  foo = 0L;
  bleh = (unsigned char *)&foo;
  bleh[3] = workspace[0];
  bleh[2] = workspace[1];
  bleh[1] = workspace[2];
  bleh[0] = workspace[3];
  return (foo);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ret32word>:
  endbr64
  movzbl (%rdi),%eax
  movzbl 0x1(%rdi),%edx
  shl    $0x10,%rdx
  shl    $0x18,%rax
  or     %rdx,%rax
  movzbl 0x2(%rdi),%ecx
  mov    %cl,%ah
  mov    0x3(%rdi),%al
  ret
```

assistant: ```c
unsigned long ret32word(workspace)
unsigned char *workspace;
{
  unsigned long foo;
  unsigned char *bleh;
  foo = 0L;
  bleh = (unsigned char *)&foo;
  bleh[3] = workspace[0];
  bleh[2] = workspace[1];
  bleh[1] = workspace[2];
  bleh[0] = workspace[3];
  return (foo);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<ret32word>:
  endbr64
  mov    (%rdi),%eax
  bswap  %eax
  mov    %eax,%eax
  ret
```

assistant: ```c
unsigned long ret32word(workspace)
unsigned char *workspace;
{
  unsigned long foo;
  unsigned char *bleh;
  foo = 0L;
  bleh = (unsigned char *)&foo;
  bleh[3] = workspace[0];
  bleh[2] = workspace[1];
  bleh[1] = workspace[2];
  bleh[0] = workspace[3];
  return (foo);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ret32word>:
  endbr64
  mov    (%rdi),%eax
  bswap  %eax
  mov    %eax,%eax
  ret
```

assistant: ```c
unsigned long ret32word(workspace)
unsigned char *workspace;
{
  unsigned long foo;
  unsigned char *bleh;
  foo = 0L;
  bleh = (unsigned char *)&foo;
  bleh[3] = workspace[0];
  bleh[2] = workspace[1];
  bleh[1] = workspace[2];
  bleh[0] = workspace[3];
  return (foo);
}
```
user: What is the c source code of the assembly code below:

```asm
<uart_putchar>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  nop
L8:
  mov    $0x20000020,%eax
  mov    (%rax),%eax
  and    $0x2,%eax
  test   %eax,%eax
  je     L8
  mov    $0x20000000,%edx
  mov    -0x4(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void uart_putchar(int value) {
  while (((*(volatile unsigned int *)(0x20000020)) & 0x02) == 0)
    ;
  (*(volatile unsigned int *)(0x20000000) = (value));
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<uart_putchar>:
  endbr64
L9:
  mov    0x20000020,%eax
  test   $0x2,%al
  je     L9
  mov    %edi,0x20000000
  ret
```

assistant: ```c
void uart_putchar(int value) {
  while (((*(volatile unsigned int *)(0x20000020)) & 0x02) == 0)
    ;
  (*(volatile unsigned int *)(0x20000000) = (value));
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<uart_putchar>:
  endbr64
  nopl   0x0(%rax)
L9:
  mov    0x20000020,%eax
  test   $0x2,%al
  je     L9
  mov    %edi,0x20000000
  ret
```

assistant: ```c
void uart_putchar(int value) {
  while (((*(volatile unsigned int *)(0x20000020)) & 0x02) == 0)
    ;
  (*(volatile unsigned int *)(0x20000000) = (value));
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<uart_putchar>:
  endbr64
  nopl   0x0(%rax)
L1:
  mov    0x20000020,%eax
  test   $0x2,%al
  je     L1
  mov    %edi,0x20000000
  ret
```

assistant: ```c
void uart_putchar(int value) {
  while (((*(volatile unsigned int *)(0x20000020)) & 0x02) == 0)
    ;
  (*(volatile unsigned int *)(0x20000000) = (value));
}
```