ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here?

```asm
<test_assert_mem_eq_pass>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  movl   $0x1,(%rsp)
  movl   $0x2,0x4(%rsp)
  movl   $0x3,0x8(%rsp)
  movl   $0x4,0xc(%rsp)
  movl   $0x5,0x10(%rsp)
  movl   $0x6,0x14(%rsp)
  movl   $0x1,0x20(%rsp)
  movl   $0x2,0x24(%rsp)
  movl   $0x3,0x28(%rsp)
  movl   $0x4,0x2c(%rsp)
  movl   $0x5,0x30(%rsp)
  movl   $0x6,0x34(%rsp)
  lea    0x20(%rsp),%rbp
  mov    %rsp,%rbx
  mov    $0x0,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <assert_mem_eq@plt>
  mov    $0x6,%edx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <assert_mem_eq@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x48,%rsp
  pop    %rbx
  pop    %rbp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_assert_mem_eq_pass(void) {
  int a[6] = {1, 2, 3, 4, 5, 6};
  int b[6] = {1, 2, 3, 4, 5, 6};
  assert_mem_eq(a, b, 0);
  assert_mem_eq(a, b, 6);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<test_assert_mem_eq_pass>:
  endbr64
  movabs $0x400000003,%rdx
  push   %r12
  movabs $0x200000001,%rcx
  push   %rbp
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r12
  mov    %rsp,%rbp
  mov    %rdx,0x8(%rsp)
  movabs $0x600000005,%rax
  mov    %rdx,0x28(%rsp)
  mov    %r12,%rsi
  xor    %edx,%edx
  mov    %rbp,%rdi
  mov    %rcx,(%rsp)
  mov    %rax,0x10(%rsp)
  mov    %rcx,0x20(%rsp)
  mov    %rax,0x30(%rsp)
  call   <assert_mem_eq@plt>
  mov    $0x6,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <assert_mem_eq@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x48,%rsp
  pop    %rbp
  pop    %r12
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void test_assert_mem_eq_pass(void) {
  int a[6] = {1, 2, 3, 4, 5, 6};
  int b[6] = {1, 2, 3, 4, 5, 6};
  assert_mem_eq(a, b, 0);
  assert_mem_eq(a, b, 6);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test_assert_mem_eq_pass>:
  endbr64
  push   %r12
  xor    %edx,%edx
  push   %rbp
  sub    $0x48,%rsp
  movdqa D6(%rip),%xmm0
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%r12
  mov    D13(%rip),%rax
  mov    %rsp,%rbp
  mov    %r12,%rsi
  mov    %rbp,%rdi
  movaps %xmm0,(%rsp)
  mov    %rax,0x10(%rsp)
  mov    %rax,0x30(%rsp)
  movaps %xmm0,0x20(%rsp)
  call   <assert_mem_eq@plt>
  mov    $0x6,%edx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <assert_mem_eq@plt>
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x48,%rsp
  pop    %rbp
  pop    %r12
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D6
data_type: byte[16]
name: extract_symbolic_data
arguments: label: D13
data_type: byte[8]

tool: {"D6": [0x01,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x03,0x00,0x00,0x00,0x04,0x00,0x00,0x00]}

tool: {"D13": [0x05,0x00,0x00,0x00,0x06,0x00,0x00,0x00]}

assistant: ```c
void test_assert_mem_eq_pass(void) {
  int a[6] = {1, 2, 3, 4, 5, 6};
  int b[6] = {1, 2, 3, 4, 5, 6};
  assert_mem_eq(a, b, 0);
  assert_mem_eq(a, b, 6);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<Lut_N>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  subl   $0x1,-0x14(%rbp)
  jmp    L2
L5:
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x10(%rbp),%rax
  add    %rax,%rdx
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x8(%rbp),%rax
  add    %rcx,%rax
  movsd  (%rdx),%xmm0
  movsd  %xmm0,(%rax)
  subl   $0x1,-0x14(%rbp)
L2:
  cmpl   $0x0,-0x14(%rbp)
  jns    L5
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void Lut_N(double *out, double *in, int nc) {
  for (--nc; nc >= 0; nc--)
    out[nc] = in[nc];
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<Lut_N>:
  endbr64
  sub    $0x1,%edx
  js     L12
  movslq %edx,%rdx
L11:
  movsd  (%rsi,%rdx,8),%xmm0
  movsd  %xmm0,(%rdi,%rdx,8)
  sub    $0x1,%rdx
  test   %edx,%edx
  jns    L11
L12:
  ret
```

assistant: ```c
void Lut_N(double *out, double *in, int nc) {
  for (--nc; nc >= 0; nc--)
    out[nc] = in[nc];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<Lut_N>:
  endbr64
  sub    $0x1,%edx
  js     L0
  movslq %edx,%rdx
  nopl   0x0(%rax)
L14:
  movsd  (%rsi,%rdx,8),%xmm0
  movsd  %xmm0,(%rdi,%rdx,8)
  sub    $0x1,%rdx
  test   %edx,%edx
  jns    L14
L0:
  ret
```

assistant: ```c
void Lut_N(double *out, double *in, int nc) {
  for (--nc; nc >= 0; nc--)
    out[nc] = in[nc];
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<Lut_N>:
  endbr64
  mov    %rdi,%rcx
  mov    %edx,%r8d
  sub    $0x1,%edx
  js     L38
  movslq %r8d,%rax
  shl    $0x3,%rax
  lea    -0x10(%rax),%r9
  lea    -0x8(%rdi,%rax,1),%rax
  lea    (%rsi,%r9,1),%r10
  cmp    %rax,%r10
  je     L25
  cmp    $0x2,%edx
  jbe    L25
  add    %rdi,%r9
  mov    %r8d,%edi
  xor    %eax,%eax
  shr    %edi
  neg    %rdi
  shl    $0x4,%rdi
  xchg   %ax,%ax
L31:
  movupd (%r10,%rax,1),%xmm1
  movups %xmm1,(%r9,%rax,1)
  sub    $0x10,%rax
  cmp    %rax,%rdi
  jne    L31
  mov    %r8d,%edi
  mov    %edx,%eax
  and    $0xfffffffe,%edi
  sub    %edi,%eax
  cmp    %edi,%r8d
  je     L38
  cltq
  movsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  ret
L25:
  movslq %edx,%rax
  nopl   0x0(%rax,%rax,1)
L50:
  movsd  (%rsi,%rax,8),%xmm0
  movsd  %xmm0,(%rcx,%rax,8)
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L50
L38:
  ret
```

assistant: ```c
void Lut_N(double *out, double *in, int nc) {
  for (--nc; nc >= 0; nc--)
    out[nc] = in[nc];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<readString>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x24(%rbp)
  jmp    L5
L59:
  mov    -0x24(%rbp),%eax
  cltd
  shr    $0x1b,%edx
  add    %edx,%eax
  and    $0x1f,%eax
  sub    %edx,%eax
  cmp    $0x1f,%eax
  jne    L29
  mov    -0x24(%rbp),%eax
  lea    0x21(%rax),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L0
  mov    -0x20(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L0:
  mov    -0x18(%rbp),%rax
  mov    %rax,-0x20(%rbp)
L29:
  addl   $0x1,-0x24(%rbp)
L5:
  cmpq   $0x0,-0x20(%rbp)
  je     L11
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,(%rbx)
  movzbl (%rbx),%eax
  cmp    $0xa,%al
  jne    L59
L11:
  cmpq   $0x0,-0x20(%rbp)
  je     L21
  mov    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L21:
  mov    -0x20(%rbp),%rax
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char *readString() {
  char *str = malloc(sizeof(char) * 32), *err;
  int pos;
  for (pos = 0; str != NULL && (str[pos] = getchar()) != '\n'; pos++) {
    if (pos % 32 == 32 - 1) {
      if ((err = realloc(str, sizeof(char) * (32 + pos + 1))) == NULL)
        free(str);
      str = err;
    }
  }
  if (str != NULL)
    str[pos] = '\0';
  return str;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<readString>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,%rbp
  mov    $0x0,%ebx
  test   %rax,%rax
  jne    L46
  jmp    L39
L56:
  lea    0x21(%r12),%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  test   %rax,%rax
  je     L55
  mov    %rax,%rbp
L19:
  add    $0x1,%rbx
L46:
  mov    %ebx,%r12d
  lea    0x0(%rbp,%rbx,1),%r13
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %al,0x0(%r13)
  cmp    $0xa,%al
  je     L68
  mov    %ebx,%edx
  sar    $0x1f,%edx
  shr    $0x1b,%edx
  lea    (%rdx,%rbx,1),%eax
  and    $0x1f,%eax
  sub    %edx,%eax
  cmp    $0x1f,%eax
  je     L56
  test   %rbp,%rbp
  jne    L19
  jmp    L39
L55:
  mov    %rbp,%rdi
  call   <free@plt>
L39:
  mov    $0x0,%ebp
  jmp    L41
L68:
  test   %rbp,%rbp
  je     L41
  movslq %ebx,%r14
  movb   $0x0,0x0(%rbp,%r14,1)
L41:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *readString() {
  char *str = malloc(sizeof(char) * 32), *err;
  int pos;
  for (pos = 0; str != NULL && (str[pos] = getchar()) != '\n'; pos++) {
    if (pos % 32 == 32 - 1) {
      if ((err = realloc(str, sizeof(char) * (32 + pos + 1))) == NULL)
        free(str);
      str = err;
    }
  }
  if (str != NULL)
    str[pos] = '\0';
  return str;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<readString>:
  endbr64
  push   %r13
  mov    $0x20,%edi
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L49
  jmp    L12
  nopl   0x0(%rax)
L44:
  add    $0x1,%rbx
L49:
  xor    %eax,%eax
  lea    (%r12,%rbx,1),%rbp
  mov    %ebx,%r13d
  call   <getchar@plt>
  mov    %al,0x0(%rbp)
  cmp    $0xa,%al
  je     L1
  mov    %ebx,%eax
  and    $0x1f,%eax
  cmp    $0x1f,%eax
  jne    L44
  lea    0x21(%r13),%esi
  mov    %r12,%rdi
  call   <realloc@plt>
  test   %rax,%rax
  je     L11
  mov    %rax,%r12
  jmp    L44
  nopl   0x0(%rax)
L1:
  movb   $0x0,0x0(%rbp)
L12:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %r12,%rdi
  xor    %r12d,%r12d
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *readString() {
  char *str = malloc(sizeof(char) * 32), *err;
  int pos;
  for (pos = 0; str != NULL && (str[pos] = getchar()) != '\n'; pos++) {
    if (pos % 32 == 32 - 1) {
      if ((err = realloc(str, sizeof(char) * (32 + pos + 1))) == NULL)
        free(str);
      str = err;
    }
  }
  if (str != NULL)
    str[pos] = '\0';
  return str;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<readString>:
  endbr64
  push   %r13
  mov    $0x20,%edi
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  jne    L5
  jmp    L9
  nopl   0x0(%rax)
L31:
  add    $0x1,%rbx
L5:
  xor    %eax,%eax
  lea    (%r12,%rbx,1),%rbp
  mov    %ebx,%r13d
  call   <getchar@plt>
  mov    %al,0x0(%rbp)
  cmp    $0xa,%al
  je     L1
  mov    %ebx,%eax
  and    $0x1f,%eax
  cmp    $0x1f,%eax
  jne    L31
  lea    0x21(%r13),%esi
  mov    %r12,%rdi
  call   <realloc@plt>
  test   %rax,%rax
  je     L18
  mov    %rax,%r12
  jmp    L31
  nopl   0x0(%rax)
L1:
  movb   $0x0,0x0(%rbp)
L9:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %r12,%rdi
  xor    %r12d,%r12d
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *readString() {
  char *str = malloc(sizeof(char) * 32), *err;
  int pos;
  for (pos = 0; str != NULL && (str[pos] = getchar()) != '\n'; pos++) {
    if (pos % 32 == 32 - 1) {
      if ((err = realloc(str, sizeof(char) * (32 + pos + 1))) == NULL)
        free(str);
      str = err;
    }
  }
  if (str != NULL)
    str[pos] = '\0';
  return str;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rcx
  lea    0x8(%rsp),%rdx
  mov    %rsp,%rsi
  lea    D78(%rip),%rdi
  call   <scanf@plt>
  mov    (%rsp),%rbx
  movq   %rbx,%xmm2
  ucomisd D99(%rip),%xmm2
  jp     L4
  je     L25
L4:
  mov    0x8(%rsp),%rbp
  movq   %rbp,%xmm0
  mulsd  %xmm0,%xmm0
  movq   %rbx,%xmm1
  mulsd  D80(%rip),%xmm1
  mulsd  0x10(%rsp),%xmm1
  subsd  %xmm1,%xmm0
  pxor   %xmm1,%xmm1
  comisd %xmm0,%xmm1
  ja     L49
  call   <sqrt@plt>
  movq   %rbp,%xmm5
  subsd  %xmm5,%xmm0
  mulsd  D14(%rip),%xmm0
  movq   %rbx,%xmm6
  divsd  %xmm6,%xmm0
  lea    D21(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rbx
  mov    (%rsp),%rbp
  movq   %rbx,%xmm0
  mulsd  %xmm0,%xmm0
  movq   %rbp,%xmm1
  mulsd  D80(%rip),%xmm1
  mulsd  0x10(%rsp),%xmm1
  subsd  %xmm1,%xmm0
  call   <sqrt@plt>
  movapd %xmm0,%xmm1
  movq   %rbx,%xmm0
  xorpd  D34(%rip),%xmm0
  subsd  %xmm1,%xmm0
  mulsd  D14(%rip),%xmm0
  movq   %rbp,%xmm5
  divsd  %xmm5,%xmm0
  lea    D6(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
L43:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L22
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L25:
  lea    D105(%rip),%rdi
  call   <puts@plt>
  jmp    L43
L49:
  lea    D105(%rip),%rdi
  call   <puts@plt>
  jmp    L43
L22:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D78
datatype: string
name: disassemble_data
arguments: data_label: D99
datatype: byte[8]
name: disassemble_data
arguments: data_label: D80
datatype: byte[8]
name: disassemble_data
arguments: data_label: D14
datatype: byte[8]
name: disassemble_data
arguments: data_label: D21
datatype: string
name: disassemble_data
arguments: data_label: D34
datatype: byte[16]
name: disassemble_data
arguments: data_label: D6
datatype: string
name: disassemble_data
arguments: data_label: D105
datatype: string

tool: D78: %lf %lf %lf

tool: {"D99": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: {"D80": [0x00,0x00,0x00,0x00,0x00,0x00,0x10,0x40]}

tool: {"D14": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: D21: R1 = %.5lf


tool: {"D34": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D6: R2 = %.5lf


tool: D105: Impossivel calcular

assistant: ```c
int main() {
  double a, b, c;
  scanf("%lf %lf %lf", &a, &b, &c);
  if (a == 0)
    printf("Impossivel calcular\n");
  else if (b * b - 4 * a * c < 0)
    printf("Impossivel calcular\n");
  else {
    printf("R1 = %.5lf\n", (-b + sqrt(b * b - 4 * a * c)) / 2 / a);
    printf("R2 = %.5lf\n", (-b - sqrt(b * b - 4 * a * c)) / 2 / a);
  }
  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
<main>:
  endbr64
  sub    $0x38,%rsp
  lea    D77(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rcx
  lea    0x18(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  call   <scanf@plt>
  movsd  0x10(%rsp),%xmm1
  pxor   %xmm3,%xmm3
  ucomisd %xmm3,%xmm1
  jp     L36
  je     L26
L36:
  movsd  0x18(%rsp),%xmm4
  movsd  D86(%rip),%xmm2
  movsd  %xmm1,(%rsp)
  mulsd  %xmm1,%xmm2
  movapd %xmm4,%xmm0
  movsd  %xmm4,0x8(%rsp)
  mulsd  %xmm4,%xmm0
  mulsd  0x20(%rsp),%xmm2
  subsd  %xmm2,%xmm0
  comisd %xmm0,%xmm3
  ja     L26
  call   <sqrt@plt>
  movsd  0x8(%rsp),%xmm4
  movsd  (%rsp),%xmm1
  lea    D60(%rip),%rdi
  mov    $0x1,%eax
  subsd  %xmm4,%xmm0
  mulsd  D70(%rip),%xmm0
  divsd  %xmm1,%xmm0
  call   <printf@plt>
  movsd  0x18(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm3
  movsd  D86(%rip),%xmm2
  movapd %xmm1,%xmm0
  movsd  %xmm1,0x8(%rsp)
  mulsd  %xmm3,%xmm2
  movsd  %xmm3,(%rsp)
  mulsd  %xmm1,%xmm0
  mulsd  0x20(%rsp),%xmm2
  subsd  %xmm2,%xmm0
  call   <sqrt@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  (%rsp),%xmm3
  lea    D10(%rip),%rdi
  xorpd  D28(%rip),%xmm1
  mov    $0x1,%eax
  subsd  %xmm0,%xmm1
  mulsd  D70(%rip),%xmm1
  divsd  %xmm3,%xmm1
  movapd %xmm1,%xmm0
  call   <printf@plt>
  jmp    L48
L26:
  lea    D44(%rip),%rdi
  call   <puts@plt>
L48:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L43
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L43:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D77
datatype: string
name: read_data
arguments: label: D86
datatype: byte[8]
name: read_data
arguments: label: D60
datatype: string
name: read_data
arguments: label: D70
datatype: byte[8]
name: read_data
arguments: label: D10
datatype: string
name: read_data
arguments: label: D28
datatype: byte[16]
name: read_data
arguments: label: D44
datatype: string

tool: D77: %lf %lf %lf

tool: {"D86": [0x00,0x00,0x00,0x00,0x00,0x00,0x10,0x40]}

tool: D60: R1 = %.5lf


tool: {"D70": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: D10: R2 = %.5lf


tool: {"D28": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D44: Impossivel calcular

assistant: ```c
int main() {
  double a, b, c;
  scanf("%lf %lf %lf", &a, &b, &c);
  if (a == 0)
    printf("Impossivel calcular\n");
  else if (b * b - 4 * a * c < 0)
    printf("Impossivel calcular\n");
  else {
    printf("R1 = %.5lf\n", (-b + sqrt(b * b - 4 * a * c)) / 2 / a);
    printf("R2 = %.5lf\n", (-b - sqrt(b * b - 4 * a * c)) / 2 / a);
  }
  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
<main>:
  endbr64
  sub    $0x38,%rsp
  lea    D32(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    0x20(%rsp),%rcx
  lea    0x18(%rsp),%rdx
  lea    0x10(%rsp),%rsi
  call   <scanf@plt>
  movsd  0x10(%rsp),%xmm1
  pxor   %xmm3,%xmm3
  ucomisd %xmm3,%xmm1
  jp     L18
  je     L45
L18:
  movsd  0x18(%rsp),%xmm4
  movsd  D82(%rip),%xmm2
  movsd  %xmm1,(%rsp)
  mulsd  %xmm1,%xmm2
  movapd %xmm4,%xmm0
  movsd  %xmm4,0x8(%rsp)
  mulsd  %xmm4,%xmm0
  mulsd  0x20(%rsp),%xmm2
  subsd  %xmm2,%xmm0
  comisd %xmm0,%xmm3
  ja     L45
  call   <sqrt@plt>
  movsd  0x8(%rsp),%xmm4
  movsd  (%rsp),%xmm1
  lea    D37(%rip),%rdi
  mov    $0x1,%eax
  subsd  %xmm4,%xmm0
  mulsd  D20(%rip),%xmm0
  divsd  %xmm1,%xmm0
  call   <printf@plt>
  movsd  0x18(%rsp),%xmm1
  movsd  0x10(%rsp),%xmm3
  movsd  D82(%rip),%xmm2
  movapd %xmm1,%xmm0
  movsd  %xmm1,0x8(%rsp)
  mulsd  %xmm3,%xmm2
  movsd  %xmm3,(%rsp)
  mulsd  %xmm1,%xmm0
  mulsd  0x20(%rsp),%xmm2
  subsd  %xmm2,%xmm0
  call   <sqrt@plt>
  movsd  0x8(%rsp),%xmm1
  movsd  (%rsp),%xmm3
  lea    D71(%rip),%rdi
  xorpd  D17(%rip),%xmm1
  mov    $0x1,%eax
  subsd  %xmm0,%xmm1
  mulsd  D20(%rip),%xmm1
  divsd  %xmm3,%xmm1
  movapd %xmm1,%xmm0
  call   <printf@plt>
  jmp    L31
L45:
  lea    D4(%rip),%rdi
  call   <puts@plt>
L31:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  xor    %eax,%eax
  add    $0x38,%rsp
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D32
type_infered: string
name: read_data
arguments: data_label: D82
type_infered: byte[8]
name: read_data
arguments: data_label: D37
type_infered: string
name: read_data
arguments: data_label: D20
type_infered: byte[8]
name: read_data
arguments: data_label: D71
type_infered: string
name: read_data
arguments: data_label: D17
type_infered: byte[16]
name: read_data
arguments: data_label: D4
type_infered: string

tool: D32: %lf %lf %lf

tool: {"D82": [0x00,0x00,0x00,0x00,0x00,0x00,0x10,0x40]}

tool: D37: R1 = %.5lf


tool: {"D20": [0x00,0x00,0x00,0x00,0x00,0x00,0xe0,0x3f]}

tool: D71: R2 = %.5lf


tool: {"D17": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D4: Impossivel calcular

assistant: ```c
int main() {
  double a, b, c;
  scanf("%lf %lf %lf", &a, &b, &c);
  if (a == 0)
    printf("Impossivel calcular\n");
  else if (b * b - 4 * a * c < 0)
    printf("Impossivel calcular\n");
  else {
    printf("R1 = %.5lf\n", (-b + sqrt(b * b - 4 * a * c)) / 2 / a);
    printf("R2 = %.5lf\n", (-b - sqrt(b * b - 4 * a * c)) / 2 / a);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<write_buffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x8(%rbp)
  cmpl   $0x14,-0x8(%rbp)
  jle    L16
  mov    -0x8(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x4(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write_large_buffer@plt>
  jmp    L12
L16:
  mov    -0x8(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x4(%rbp),%eax
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <write@plt>
L12:
  leave
  ret
```

assistant: ```c
int write_buffer(int fd, unsigned char *buf, int len) {
  if (len > 20) {
    return write_large_buffer(fd, buf, len);
  }
  return write(fd, buf, len);
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<write_buffer>:
  endbr64
  sub    $0x8,%rsp
  cmp    $0x14,%edx
  jg     L3
  call   <write@plt>
L9:
  add    $0x8,%rsp
  ret
L3:
  call   <write_large_buffer@plt>
  jmp    L9
```

assistant: ```c
int write_buffer(int fd, unsigned char *buf, int len) {
  if (len > 20) {
    return write_large_buffer(fd, buf, len);
  }
  return write(fd, buf, len);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<write_buffer>:
  endbr64
  cmp    $0x14,%edx
  jg     L4
  jmp    1060 <write@plt>
  xchg   %ax,%ax
L4:
  jmp    1070 <write_large_buffer@plt>
```

assistant: ```c
int write_buffer(int fd, unsigned char *buf, int len) {
  if (len > 20) {
    return write_large_buffer(fd, buf, len);
  }
  return write(fd, buf, len);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<write_buffer>:
  endbr64
  cmp    $0x14,%edx
  jg     L5
  jmp    1060 <write@plt>
  xchg   %ax,%ax
L5:
  jmp    1070 <write_large_buffer@plt>
```

assistant: ```c
int write_buffer(int fd, unsigned char *buf, int len) {
  if (len > 20) {
    return write_large_buffer(fd, buf, len);
  }
  return write(fd, buf, len);
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xa,-0x14(%rbp)
  movabs $0x216f6c6c6548,%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L1
  call   <__stack_chk_fail@plt>
L1:
  leave
  ret
```

assistant: ```c
int main() {
  int x = 10;
  char c[8] = "Hello!";
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

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

assistant: ```c
int main() {
  int x = 10;
  char c[8] = "Hello!";
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
int main() {
  int x = 10;
  char c[8] = "Hello!";
  return 0;
}
```
user: What is the c source code of the assembly code below:

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

assistant: ```c
int main() {
  int x = 10;
  char c[8] = "Hello!";
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<checker>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L26
L1:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jne    L17
  mov    $0x0,%eax
  jmp    L39
L17:
  addl   $0x1,-0x4(%rbp)
L26:
  mov    -0x4(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L1
  mov    $0x1,%eax
L39:
  pop    %rbp
  ret
```

assistant: ```c
int checker(int a[], int n, int len) {
  int i = 0;
  for (; i < len; i++) {
    if (a[i] == n) {
      return 0;
    }
  }
  return 1;
}
```
user: What is the c source code of the assembly code below?

```asm
<checker>:
  endbr64
  test   %edx,%edx
  jle    L9
  mov    %rdi,%rax
  lea    -0x1(%rdx),%edx
  lea    0x4(%rdi,%rdx,4),%rdx
L4:
  cmp    %esi,(%rax)
  je     L16
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L4
  mov    $0x1,%eax
  ret
L9:
  mov    $0x1,%eax
  ret
L16:
  mov    $0x0,%eax
  ret
```

assistant: ```c
int checker(int a[], int n, int len) {
  int i = 0;
  for (; i < len; i++) {
    if (a[i] == n) {
      return 0;
    }
  }
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<checker>:
  endbr64
  test   %edx,%edx
  jle    L6
  lea    -0x1(%rdx),%eax
  lea    0x4(%rdi,%rax,4),%rax
  jmp    L36
  nopw   0x0(%rax,%rax,1)
L34:
  add    $0x4,%rdi
  cmp    %rax,%rdi
  je     L6
L36:
  cmp    %esi,(%rdi)
  jne    L34
  xor    %eax,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L6:
  mov    $0x1,%eax
  ret
```

assistant: ```c
int checker(int a[], int n, int len) {
  int i = 0;
  for (; i < len; i++) {
    if (a[i] == n) {
      return 0;
    }
  }
  return 1;
}
```