ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x37,0x8(%rsp)
  movl   $0x4,0xc(%rsp)
  movl   $0x25,0x10(%rsp)
  movl   $0x3,0x14(%rsp)
  movl   $0x2c,0x18(%rsp)
  movl   $0x15,0x1c(%rsp)
  movl   $0x21,0x20(%rsp)
  movl   $0x130,0x24(%rsp)
  movl   $0x2d,0x28(%rsp)
  movl   $0x10,0x2c(%rsp)
  mov    %rsp,%rbx
  mov    $0xb,%edx
  mov    $0x0,%esi
  mov    %rbx,%rdi
  call   <selectSort@plt>
  lea    0x30(%rsp),%r12
  lea    D2(%rip),%rbp
L3:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L3
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    $0x0,%eax
  add    $0x40,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L4:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D2
data_type: string

tool: D2: %d 

assistant: ```c
int main() {
  int arr[] = {1, 2, 55, 4, 37, 3, 44, 21, 33, 304, 45, 16};
  int len = sizeof(arr) / 4;
  selectSort(arr, 0, len - 1);
  int i;
  for (i = 0; i < len; i++) {
    printf("%d ", arr[i]);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xb,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x200000001,%rax
  mov    %rsp,%rbx
  lea    0x30(%rsp),%r12
  mov    %rbx,%rdi
  mov    %rax,(%rsp)
  movabs $0x400000037,%rax
  mov    %rax,0x8(%rsp)
  movabs $0x300000025,%rax
  mov    %rax,0x10(%rsp)
  movabs $0x150000002c,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x13000000021,%rax
  mov    %rax,0x20(%rsp)
  movabs $0x100000002d,%rax
  mov    %rax,0x28(%rsp)
  call   <selectSort@plt>
  nopl   0x0(%rax)
L13:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L13
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D7
datatype: string

tool: D7: %d 

assistant: ```c
int main() {
  int arr[] = {1, 2, 55, 4, 37, 3, 44, 21, 33, 304, 45, 16};
  int len = sizeof(arr) / 4;
  selectSort(arr, 0, len - 1);
  int i;
  for (i = 0; i < len; i++) {
    printf("%d ", arr[i]);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  mov    $0xb,%edx
  xor    %esi,%esi
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  sub    $0x40,%rsp
  movdqa D9(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbx
  lea    0x30(%rsp),%r12
  movaps %xmm0,(%rsp)
  movdqa D15(%rip),%xmm0
  mov    %rbx,%rdi
  movaps %xmm0,0x10(%rsp)
  movdqa D36(%rip),%xmm0
  movaps %xmm0,0x20(%rsp)
  call   <selectSort@plt>
L19:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L19
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x40,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: tag: D2
data_type: string
name: extract_symbolic_data
arguments: tag: D9
data_type: byte[16]
name: extract_symbolic_data
arguments: tag: D15
data_type: byte[16]
name: extract_symbolic_data
arguments: tag: D36
data_type: byte[16]

tool: D2: %d 

tool: {"D9": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x37,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D15": [0x25,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x2c,0x00,0x00,0x00,0x15,0x00,0x00,0x00]}

tool: {"D36": [0x21,0x00,0x00,0x00,0x30,0x01,0x00,0x00,0x2d,0x00,0x00,0x00,0x10,0x00,0x00,0x00]}

assistant: ```c
int main() {
  int arr[] = {1, 2, 55, 4, 37, 3, 44, 21, 33, 304, 45, 16};
  int len = sizeof(arr) / 4;
  selectSort(arr, 0, len - 1);
  int i;
  for (i = 0; i < len; i++) {
    printf("%d ", arr[i]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<ieee_p_1242562249_sub_1081509318381623571_1035706684>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rsi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    0xc(%rcx),%ebp
  test   %ebp,%ebp
  je     L29
  mov    %rsp,%rsi
  call   <ieee_p_1242562249_sub_2177871677178661378_1035706684@plt>
  mov    %rax,%rbp
  mov    0xc(%rsp),%r13d
  mov    %r13d,%edi
  call   <xsi_get_transient_memory@plt>
  mov    %rax,%r12
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    (%rsp),%ecx
  mov    0x4(%rsp),%edx
  mov    0x8(%rsp),%eax
  mov    %edx,%ebp
  sub    %ecx,%ebp
  imul   %eax,%ebp
  add    $0x1,%ebp
L2:
  mov    %ecx,(%rbx)
  mov    %edx,0x4(%rbx)
  mov    %eax,0x8(%rbx)
  mov    %ebp,0xc(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  mov    %r12,%rax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L29:
  mov    0x490(%rdi),%r13
  mov    $0x0,%edi
  call   <xsi_get_transient_memory@plt>
  mov    %rax,%r12
  mov    $0x0,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  call   <memcpy@plt>
  mov    $0x0,%ecx
  mov    $0x1,%edx
  mov    $0xffffffff,%eax
  jmp    L2
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ieee_p_1242562249_sub_1081509318381623571_1035706684(char *t1, char *t2, char *t3, char *t4, int t5) {
  char t7[24];
  char t20[16];
  char *t0;
  char *t8;
  unsigned char t9;
  char *t10;
  char *t11;
  char *t12;
  unsigned int t13;
  unsigned char t14;
  char *t15;
  char *t16;
  char *t17;
  int t18;
  unsigned int t19;
  int t21;
  char *t22;
  int t23;
  char *t24;
  char *t25;
  int t26;
LAB0:
  t8 = (t7 + 4U);
  t9 = (t3 != 0);
  if (t9 == 1)
    goto LAB3;
LAB2:
  t10 = (t7 + 12U);
  *((char **)t10) = t4;
  t11 = (t7 + 20U);
  *((int *)t11) = t5;
  t12 = (t4 + 12U);
  t13 = *((unsigned int *)t12);
  t14 = (t13 < 1);
  if (t14 != 0)
    goto LAB4;
LAB6:
LAB5:
  t12 = ieee_p_1242562249_sub_2177871677178661378_1035706684(t1, t20, t3, t4, t5);
  t15 = (t20 + 12U);
  t13 = *((unsigned int *)t15);
  t13 = (t13 * 1U);
  t0 = xsi_get_transient_memory(t13);
  memcpy(t0, t12, t13);
  t16 = (t20 + 0U);
  t18 = *((int *)t16);
  t17 = (t20 + 4U);
  t21 = *((int *)t17);
  t22 = (t20 + 8U);
  t23 = *((int *)t22);
  t24 = (t2 + 0U);
  t25 = (t24 + 0U);
  *((int *)t25) = t18;
  t25 = (t24 + 4U);
  *((int *)t25) = t21;
  t25 = (t24 + 8U);
  *((int *)t25) = t23;
  t26 = (t21 - t18);
  t19 = (t26 * t23);
  t19 = (t19 + 1);
  t25 = (t24 + 12U);
  *((unsigned int *)t25) = t19;
LAB1:
  return t0;
LAB3:
  *((char **)t8) = t3;
  goto LAB2;
LAB4:
  t15 = (t1 + 1168U);
  t16 = *((char **)t15);
  t0 = xsi_get_transient_memory(0U);
  memcpy(t0, t16, 0U);
  t15 = (t2 + 0U);
  t17 = (t15 + 0U);
  *((int *)t17) = 0;
  t17 = (t15 + 4U);
  *((int *)t17) = 1;
  t17 = (t15 + 8U);
  *((int *)t17) = -1;
  t18 = (1 - 0);
  t19 = (t18 * -1);
  t19 = (t19 + 1);
  t17 = (t15 + 12U);
  *((unsigned int *)t17) = t19;
  goto LAB1;
LAB7:
  goto LAB5;
LAB8:;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<ieee_p_1242562249_sub_1081509318381623571_1035706684>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  sub    $0x28,%rsp
  mov    0xc(%rcx),%ebp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  test   %ebp,%ebp
  je     L17
  mov    %rsp,%rsi
  call   <ieee_p_1242562249_sub_2177871677178661378_1035706684@plt>
  mov    0xc(%rsp),%r13d
  mov    %rax,%rbp
  mov    %r13d,%edi
  call   <xsi_get_transient_memory@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
  mov    0x4(%rsp),%edx
  mov    (%rsp),%ecx
  mov    0x8(%rsp),%eax
  mov    %edx,%ebp
  sub    %ecx,%ebp
  imul   %eax,%ebp
  add    $0x1,%ebp
L10:
  mov    %ecx,(%rbx)
  mov    %edx,0x4(%rbx)
  mov    %eax,0x8(%rbx)
  mov    %ebp,0xc(%rbx)
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x28,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nop
L17:
  mov    0x490(%rdi),%r13
  xor    %edi,%edi
  call   <xsi_get_transient_memory@plt>
  xor    %edx,%edx
  mov    %r13,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memcpy@plt>
  xor    %ecx,%ecx
  mov    $0x1,%edx
  mov    $0xffffffff,%eax
  jmp    L10
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
char *ieee_p_1242562249_sub_1081509318381623571_1035706684(char *t1, char *t2, char *t3, char *t4, int t5) {
  char t7[24];
  char t20[16];
  char *t0;
  char *t8;
  unsigned char t9;
  char *t10;
  char *t11;
  char *t12;
  unsigned int t13;
  unsigned char t14;
  char *t15;
  char *t16;
  char *t17;
  int t18;
  unsigned int t19;
  int t21;
  char *t22;
  int t23;
  char *t24;
  char *t25;
  int t26;
LAB0:
  t8 = (t7 + 4U);
  t9 = (t3 != 0);
  if (t9 == 1)
    goto LAB3;
LAB2:
  t10 = (t7 + 12U);
  *((char **)t10) = t4;
  t11 = (t7 + 20U);
  *((int *)t11) = t5;
  t12 = (t4 + 12U);
  t13 = *((unsigned int *)t12);
  t14 = (t13 < 1);
  if (t14 != 0)
    goto LAB4;
LAB6:
LAB5:
  t12 = ieee_p_1242562249_sub_2177871677178661378_1035706684(t1, t20, t3, t4, t5);
  t15 = (t20 + 12U);
  t13 = *((unsigned int *)t15);
  t13 = (t13 * 1U);
  t0 = xsi_get_transient_memory(t13);
  memcpy(t0, t12, t13);
  t16 = (t20 + 0U);
  t18 = *((int *)t16);
  t17 = (t20 + 4U);
  t21 = *((int *)t17);
  t22 = (t20 + 8U);
  t23 = *((int *)t22);
  t24 = (t2 + 0U);
  t25 = (t24 + 0U);
  *((int *)t25) = t18;
  t25 = (t24 + 4U);
  *((int *)t25) = t21;
  t25 = (t24 + 8U);
  *((int *)t25) = t23;
  t26 = (t21 - t18);
  t19 = (t26 * t23);
  t19 = (t19 + 1);
  t25 = (t24 + 12U);
  *((unsigned int *)t25) = t19;
LAB1:
  return t0;
LAB3:
  *((char **)t8) = t3;
  goto LAB2;
LAB4:
  t15 = (t1 + 1168U);
  t16 = *((char **)t15);
  t0 = xsi_get_transient_memory(0U);
  memcpy(t0, t16, 0U);
  t15 = (t2 + 0U);
  t17 = (t15 + 0U);
  *((int *)t17) = 0;
  t17 = (t15 + 4U);
  *((int *)t17) = 1;
  t17 = (t15 + 8U);
  *((int *)t17) = -1;
  t18 = (1 - 0);
  t19 = (t18 * -1);
  t19 = (t19 + 1);
  t17 = (t15 + 12U);
  *((unsigned int *)t17) = t19;
  goto LAB1;
LAB7:
  goto LAB5;
LAB8:;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<dump>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L13
L6:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L13:
  cmpl   $0x7fff,-0x4(%rbp)
  jle    L6
  nop
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D4
datatype: string

tool: D4: %f


assistant: ```c
void dump(double *input) {
  int i;
  for (i = 0; i < 32 * 32 * 32; ++i) {
    printf("%f\n", input[i]);
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dump>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  lea    0x40000(%rdi),%r12
  lea    D6(%rip),%rbp
L4:
  movsd  (%rbx),%xmm0
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x8,%rbx
  cmp    %r12,%rbx
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D6
type_infered: string

tool: D6: %f


assistant: ```c
void dump(double *input) {
  int i;
  for (i = 0; i < 32 * 32 * 32; ++i) {
    printf("%f\n", input[i]);
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<dump>:
  endbr64
  push   %r12
  lea    0x40000(%rdi),%r12
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  nopl   0x0(%rax)
L7:
  movsd  (%rbx),%xmm0
  mov    %rbp,%rdi
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L7
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
type: string

tool: D2: %f


assistant: ```c
void dump(double *input) {
  int i;
  for (i = 0; i < 32 * 32 * 32; ++i) {
    printf("%f\n", input[i]);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<dump>:
  endbr64
  push   %r12
  lea    0x40000(%rdi),%r12
  push   %rbp
  lea    D9(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  nopl   0x0(%rax)
L3:
  movsd  (%rbx),%xmm0
  mov    %rbp,%rdi
  add    $0x8,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L3
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

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

tool: D9: %f


assistant: ```c
void dump(double *input) {
  int i;
  for (i = 0; i < 32 * 32 * 32; ++i) {
    printf("%f\n", input[i]);
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<rev>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %rdi,-0x88(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x88(%rbp),%rdx
  lea    -0x70(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x88(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  sub    $0x1,%eax
  mov    %eax,-0x78(%rbp)
  movl   $0x0,-0x74(%rbp)
  jmp    L4
L12:
  mov    -0x74(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x88(%rbp),%rax
  add    %rax,%rdx
  mov    -0x78(%rbp),%eax
  cltq
  movzbl -0x70(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
  subl   $0x1,-0x78(%rbp)
  addl   $0x1,-0x74(%rbp)
L4:
  cmpl   $0x0,-0x78(%rbp)
  jns    L12
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

assistant: ```c
void rev(char a[]) {
  int i, j;
  char a2[100];
  strcpy(a2, a);
  for (i = strlen(a) - 1, j = 0; i >= 0; i--, j++) {
    a[j] = a2[i];
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<rev>:
  endbr64
  push   %rbx
  sub    $0x70,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rdi
  mov    %rbx,%rsi
  call   <strcpy@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L4
  movslq %eax,%rdx
  lea    (%rsp,%rdx,1),%rcx
  mov    %rbx,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rsi
L20:
  movzbl -0x1(%rcx),%eax
  mov    %al,(%rdx)
  sub    $0x1,%rcx
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  jne    L20
L4:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x70,%rsp
  pop    %rbx
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void rev(char a[]) {
  int i, j;
  char a2[100];
  strcpy(a2, a);
  for (i = strlen(a) - 1, j = 0; i >= 0; i--, j++) {
    a[j] = a2[i];
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<rev>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L10
  movslq %eax,%rdx
  sub    $0x1,%eax
  mov    %rbx,%rdi
  add    %rbp,%rdx
  lea    0x1(%rbx,%rax,1),%rcx
  nopl   0x0(%rax,%rax,1)
L22:
  movzbl -0x1(%rdx),%eax
  add    $0x1,%rdi
  sub    $0x1,%rdx
  mov    %al,-0x1(%rdi)
  cmp    %rcx,%rdi
  jne    L22
L10:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void rev(char a[]) {
  int i, j;
  char a2[100];
  strcpy(a2, a);
  for (i = strlen(a) - 1, j = 0; i >= 0; i--, j++) {
    a[j] = a2[i];
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<rev>:
  endbr64
  push   %rbp
  mov    %rdi,%rsi
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rdi
  call   <strcpy@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  test   %eax,%eax
  jle    L14
  movslq %eax,%rdx
  sub    $0x1,%eax
  mov    %rbx,%rdi
  add    %rbp,%rdx
  lea    0x1(%rbx,%rax,1),%rcx
  nopl   0x0(%rax,%rax,1)
L27:
  movzbl -0x1(%rdx),%eax
  add    $0x1,%rdi
  sub    $0x1,%rdx
  mov    %al,-0x1(%rdi)
  cmp    %rcx,%rdi
  jne    L27
L14:
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x78,%rsp
  pop    %rbx
  pop    %rbp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void rev(char a[]) {
  int i, j;
  char a2[100];
  strcpy(a2, a);
  for (i = strlen(a) - 1, j = 0; i >= 0; i--, j++) {
    a[j] = a2[i];
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edx
  sub    $0x1,%rdx
  mov    %rdx,-0x28(%rbp)
  mov    %eax,%edx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  mov    %eax,%edx
  mov    %rdx,%rsi
  mov    $0x0,%edi
  mov    %eax,%eax
  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
L30:
  cmp    %rdx,%rsp
  je     L11
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L30
L11:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L43
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L43:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x20(%rbp)
  cmpl   $0x1869f,-0x2c(%rbp)
  ja     L22
  cmpl   $0xa,-0x2c(%rbp)
  jbe    L22
  mov    $0x1,%eax
  jmp    L61
L22:
  mov    $0x0,%eax
L61:
  mov    %eax,%edi
  call   <__ESBMC_assume@plt>
  mov    -0x20(%rbp),%rax
  movl   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  movl   $0x1,0x4(%rax)
  mov    -0x20(%rbp),%rax
  movl   $0x2,0x8(%rax)
  mov    -0x20(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x1,%eax
  sete   %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <assert@plt>
  mov    %rbx,%rsp
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main() {
  unsigned n;
  int a[n];
  __ESBMC_assume(n < 100000 && n > 10);
  a[0] = 0;
  a[1] = 1;
  a[2] = 2;
  assert(a[0] == 1);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <__ESBMC_assume@plt>
  mov    $0x0,%edi
  call   <assert@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  unsigned n;
  int a[n];
  __ESBMC_assume(n < 100000 && n > 10);
  a[0] = 0;
  a[1] = 1;
  a[2] = 2;
  assert(a[0] == 1);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <__ESBMC_assume@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  unsigned n;
  int a[n];
  __ESBMC_assume(n < 100000 && n > 10);
  a[0] = 0;
  a[1] = 1;
  a[2] = 2;
  assert(a[0] == 1);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <__ESBMC_assume@plt>
  xor    %edi,%edi
  call   <assert@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
  unsigned n;
  int a[n];
  __ESBMC_assume(n < 100000 && n > 10);
  a[0] = 0;
  a[1] = 1;
  a[2] = 2;
  assert(a[0] == 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<factorial>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movl   $0x1,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L6
L8:
  mov    -0x8(%rbp),%eax
  imul   -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  subl   $0x1,-0x4(%rbp)
L6:
  cmpl   $0x1,-0x4(%rbp)
  ja     L8
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
unsigned int factorial(unsigned int n) {
  unsigned int sum = 1;
  for (unsigned int i = n; i > 1; i--) {
    sum *= i;
  }
  return sum;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<factorial>:
  endbr64
  cmp    $0x1,%edi
  jbe    L13
  mov    $0x1,%eax
L12:
  imul   %edi,%eax
  sub    $0x1,%edi
  cmp    $0x1,%edi
  jne    L12
  ret
L13:
  mov    $0x1,%eax
  ret
```

assistant: ```c
unsigned int factorial(unsigned int n) {
  unsigned int sum = 1;
  for (unsigned int i = n; i > 1; i--) {
    sum *= i;
  }
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<factorial>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jbe    L16
  xchg   %ax,%ax
L0:
  imul   %edi,%eax
  sub    $0x1,%edi
  cmp    $0x1,%edi
  jne    L0
  ret
  nopl   0x0(%rax)
L16:
  ret
```

assistant: ```c
unsigned int factorial(unsigned int n) {
  unsigned int sum = 1;
  for (unsigned int i = n; i > 1; i--) {
    sum *= i;
  }
  return sum;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<factorial>:
  endbr64
  mov    $0x1,%eax
  cmp    $0x1,%edi
  jbe    L6
  xchg   %ax,%ax
L3:
  imul   %edi,%eax
  sub    $0x1,%edi
  cmp    $0x1,%edi
  jne    L3
  ret
  nopl   0x0(%rax)
L6:
  ret
```

assistant: ```c
unsigned int factorial(unsigned int n) {
  unsigned int sum = 1;
  for (unsigned int i = n; i > 1; i--) {
    sum *= i;
  }
  return sum;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<proc_stat_after_cmd>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    $0x29,%esi
  mov    %rax,%rdi
  call   <strrchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L7
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x0,(%rax)
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *proc_stat_after_cmd(char *p) {
  p = strrchr(p, ')');
  if (p)
    *p++ = '\0';
  return p;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<proc_stat_after_cmd>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x29,%esi
  call   <strrchr@plt>
  mov    %rax,%rdx
  test   %rax,%rax
  je     L1
  lea    0x1(%rax),%rdx
  movb   $0x0,(%rax)
L1:
  mov    %rdx,%rax
  add    $0x8,%rsp
  ret
```

assistant: ```c
char *proc_stat_after_cmd(char *p) {
  p = strrchr(p, ')');
  if (p)
    *p++ = '\0';
  return p;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<proc_stat_after_cmd>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x29,%esi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L6
  lea    0x1(%rax),%r8
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  mov    %r8,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L6:
  xor    %r8d,%r8d
  add    $0x8,%rsp
  mov    %r8,%rax
  ret
```

assistant: ```c
char *proc_stat_after_cmd(char *p) {
  p = strrchr(p, ')');
  if (p)
    *p++ = '\0';
  return p;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<proc_stat_after_cmd>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x29,%esi
  call   <strrchr@plt>
  test   %rax,%rax
  je     L0
  lea    0x1(%rax),%r8
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  mov    %r8,%rax
  ret
  cs nopw 0x0(%rax,%rax,1)
L0:
  xor    %r8d,%r8d
  add    $0x8,%rsp
  mov    %r8,%rax
  ret
```

assistant: ```c
char *proc_stat_after_cmd(char *p) {
  p = strrchr(p, ')');
  if (p)
    *p++ = '\0';
  return p;
}
```