ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将以下的汇编代码转换成C语言代码：

```asm
<base64_encode_alloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x20(%rbp),%rax
  add    $0x2,%rax
  movabs $0xaaaaaaaaaaaaaaab,%rdx
  mul    %rdx
  mov    %rdx,%rax
  shr    %rax
  shl    $0x2,%rax
  add    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x20(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jbe    L15
  mov    -0x28(%rbp),%rax
  movq   $0x0,(%rax)
  mov    $0x0,%eax
  jmp    L12
L15:
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    -0x28(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rax
  test   %rax,%rax
  jne    L27
  mov    -0x8(%rbp),%rax
  jmp    L12
L27:
  mov    -0x28(%rbp),%rax
  mov    (%rax),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x20(%rbp),%rsi
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <base64_encode@plt>
  mov    -0x8(%rbp),%rax
  sub    $0x1,%rax
L12:
  leave
  ret
```

assistant: ```c
size_t base64_encode_alloc(const char *in, size_t inlen, char **out) {
  size_t outlen = 1 + ((((inlen) + 2) / 3) * 4);
  if (inlen > outlen) {
    *out = NULL;
    return 0;
  }
  *out = (char *)malloc(outlen);
  if (!*out)
    return outlen;
  base64_encode(in, inlen, *out, outlen);
  return outlen - 1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<base64_encode_alloc>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%r12
  lea    0x2(%rsi),%rdx
  movabs $0xaaaaaaaaaaaaaaab,%rcx
  mov    %rdx,%rax
  mul    %rcx
  shr    %rdx
  lea    0x0(,%rdx,4),%rbx
  lea    0x1(%rbx),%r14
  cmp    %r14,%rsi
  jbe    L39
  movq   $0x0,(%r12)
  mov    $0x0,%ebx
L38:
  mov    %rbx,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L39:
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %r14,%rdi
  call   <malloc@plt>
  mov    %rax,%rdx
  mov    %rax,(%r12)
  test   %rax,%rax
  je     L29
  mov    %r14,%rcx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <base64_encode@plt>
  jmp    L38
L29:
  mov    %r14,%rbx
  jmp    L38
```

assistant: ```c
size_t base64_encode_alloc(const char *in, size_t inlen, char **out) {
  size_t outlen = 1 + ((((inlen) + 2) / 3) * 4);
  if (inlen > outlen) {
    *out = NULL;
    return 0;
  }
  *out = (char *)malloc(outlen);
  if (!*out)
    return outlen;
  base64_encode(in, inlen, *out, outlen);
  return outlen - 1;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<base64_encode_alloc>:
  endbr64
  movabs $0xaaaaaaaaaaaaaaab,%rcx
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  lea    0x2(%rsi),%rdx
  mov    %rdx,%rax
  mul    %rcx
  shr    %rdx
  lea    0x0(,%rdx,4),%r12
  lea    0x1(%r12),%r14
  cmp    %r14,%rsi
  jbe    L15
  xor    %r12d,%r12d
  movq   $0x0,(%rbx)
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L15:
  mov    %rdi,%r13
  mov    %r14,%rdi
  mov    %rsi,%rbp
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rdx
  test   %rax,%rax
  je     L13
  mov    %r14,%rcx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <base64_encode@plt>
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %r14,%r12
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
size_t base64_encode_alloc(const char *in, size_t inlen, char **out) {
  size_t outlen = 1 + ((((inlen) + 2) / 3) * 4);
  if (inlen > outlen) {
    *out = NULL;
    return 0;
  }
  *out = (char *)malloc(outlen);
  if (!*out)
    return outlen;
  base64_encode(in, inlen, *out, outlen);
  return outlen - 1;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<base64_encode_alloc>:
  endbr64
  movabs $0xaaaaaaaaaaaaaaab,%rcx
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rbx
  lea    0x2(%rsi),%rdx
  mov    %rdx,%rax
  mul    %rcx
  shr    %rdx
  lea    0x0(,%rdx,4),%r12
  lea    0x1(%r12),%r14
  cmp    %r14,%rsi
  jbe    L8
  xor    %r12d,%r12d
  movq   $0x0,(%rbx)
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  xchg   %ax,%ax
L8:
  mov    %rdi,%r13
  mov    %r14,%rdi
  mov    %rsi,%rbp
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    %rax,%rdx
  test   %rax,%rax
  je     L0
  mov    %r14,%rcx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <base64_encode@plt>
  pop    %rbx
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L0:
  mov    %r14,%r12
  pop    %rbx
  pop    %rbp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
size_t base64_encode_alloc(const char *in, size_t inlen, char **out) {
  size_t outlen = 1 + ((((inlen) + 2) / 3) * 4);
  if (inlen > outlen) {
    *out = NULL;
    return 0;
  }
  *out = (char *)malloc(outlen);
  if (!*out)
    return outlen;
  base64_encode(in, inlen, *out, outlen);
  return outlen - 1;
}
```
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
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x28(%rbp),%rcx
  lea    -0x2c(%rbp),%rdx
  lea    -0x30(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x2c(%rbp),%eax
  mov    $0x2,%esi
  mov    %eax,%edi
  call   <pow@plt>
  mov    -0x30(%rbp),%ecx
  mov    -0x28(%rbp),%edx
  imul   %ecx,%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x24(%rbp)
  pxor   %xmm0,%xmm0
  comiss -0x24(%rbp),%xmm0
  jbe    L0
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L11
L0:
  mov    -0x24(%rbp),%eax
  movd   %eax,%xmm0
  call   <sqrt@plt>
  mov    -0x2c(%rbp),%edx
  sub    %edx,%eax
  mov    -0x30(%rbp),%edx
  lea    (%rdx,%rdx,1),%ebx
  cltd
  idiv   %ebx
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x20(%rbp)
  mov    -0x2c(%rbp),%eax
  neg    %eax
  mov    %eax,%ebx
  mov    -0x24(%rbp),%eax
  movd   %eax,%xmm0
  call   <sqrt@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  mov    -0x30(%rbp),%edx
  lea    (%rdx,%rdx,1),%esi
  cltd
  idiv   %esi
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x1c(%rbp)
  mov    -0x28(%rbp),%ecx
  mov    -0x2c(%rbp),%edx
  mov    -0x30(%rbp),%eax
  mov    %eax,%esi
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  cvtss2sd -0x1c(%rbp),%xmm0
  pxor   %xmm2,%xmm2
  cvtss2sd -0x20(%rbp),%xmm2
  movq   %xmm2,%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
L11:
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D16
type: string
name: disassemble_data
arguments: tag: D24
type: string
name: disassemble_data
arguments: tag: D6
type: string
name: disassemble_data
arguments: tag: D37
type: string
name: disassemble_data
arguments: tag: D32
type: string

tool: D16: please enter values for a,b,c

tool: D24: %d %d %d

tool: D6: Oops! These numbers give a negative root. Please try again

tool: D37: The quadratic equation is:  %dx^2 +%dx +%d = 0


tool: D32: The solutions are: 
 x1= %.2f
 x2=%.2f


assistant: ```c
main() {
  float x1;
  float x2;
  int a, b, c;
  float shoresh;
  printf("please enter values for a,b,c\n");
  scanf("%d %d %d", &a, &b, &c);
  shoresh = pow(b, 2) - 4 * a * c;
  if (shoresh < 0) {
    printf("Oops! These numbers give a negative root. Please try again\n");
  } else {
    x1 = (-b + sqrt(shoresh)) / (2 * a);
    x2 = (-b - sqrt(shoresh)) / (2 * a);
    printf("The quadratic equation is:  %dx^2 +%dx +%d = 0\n", a, b, c);
    printf("The solutions are: \n x1= %.2f\n x2=%.2f\n", x1, x2);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D49(%rip),%rdi
  call   <puts@plt>
  lea    0x24(%rsp),%rcx
  lea    0x20(%rsp),%rdx
  lea    0x1c(%rsp),%rsi
  lea    D14(%rip),%rdi
  call   <scanf@plt>
  mov    $0x2,%esi
  mov    0x20(%rsp),%edi
  call   <pow@plt>
  mov    0x1c(%rsp),%edx
  imul   0x24(%rsp),%edx
  shl    $0x2,%edx
  sub    %edx,%eax
  pxor   %xmm2,%xmm2
  cvtsi2ss %eax,%xmm2
  movss  %xmm2,0x8(%rsp)
  pxor   %xmm0,%xmm0
  comiss %xmm2,%xmm0
  ja     L16
  movss  0x8(%rsp),%xmm0
  call   <sqrt@plt>
  mov    0x20(%rsp),%ebx
  sub    %ebx,%eax
  mov    0x1c(%rsp),%esi
  lea    (%rsi,%rsi,1),%ecx
  cltd
  idiv   %ecx
  pxor   %xmm3,%xmm3
  cvtsi2ss %eax,%xmm3
  movss  %xmm3,0xc(%rsp)
  movss  0x8(%rsp),%xmm0
  call   <sqrt@plt>
  mov    %eax,%r12d
  mov    0x1c(%rsp),%ebp
  mov    0x24(%rsp),%ecx
  mov    0x20(%rsp),%edx
  mov    %ebp,%esi
  lea    D26(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebx,%eax
  neg    %eax
  sub    %r12d,%eax
  add    %ebp,%ebp
  cltd
  idiv   %ebp
  pxor   %xmm1,%xmm1
  cvtsi2ss %eax,%xmm1
  pxor   %xmm0,%xmm0
  cvtss2sd 0xc(%rsp),%xmm0
  cvtss2sd %xmm1,%xmm1
  lea    D24(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
L12:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L16:
  lea    D47(%rip),%rdi
  call   <puts@plt>
  jmp    L12
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D49
type_guessed: string
name: read_data
arguments: data_tag: D14
type_guessed: string
name: read_data
arguments: data_tag: D26
type_guessed: string
name: read_data
arguments: data_tag: D24
type_guessed: string
name: read_data
arguments: data_tag: D47
type_guessed: string

tool: D49: please enter values for a,b,c

tool: D14: %d %d %d

tool: D26: The quadratic equation is:  %dx^2 +%dx +%d = 0


tool: D24: The solutions are: 
 x1= %.2f
 x2=%.2f


tool: D47: Oops! These numbers give a negative root. Please try again

assistant: ```c
main() {
  float x1;
  float x2;
  int a, b, c;
  float shoresh;
  printf("please enter values for a,b,c\n");
  scanf("%d %d %d", &a, &b, &c);
  shoresh = pow(b, 2) - 4 * a * c;
  if (shoresh < 0) {
    printf("Oops! These numbers give a negative root. Please try again\n");
  } else {
    x1 = (-b + sqrt(shoresh)) / (2 * a);
    x2 = (-b - sqrt(shoresh)) / (2 * a);
    printf("The quadratic equation is:  %dx^2 +%dx +%d = 0\n", a, b, c);
    printf("The solutions are: \n x1= %.2f\n x2=%.2f\n", x1, x2);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r12
  lea    D18(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x20(%rsp),%rdx
  lea    0x24(%rsp),%rcx
  lea    0x1c(%rsp),%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    0x20(%rsp),%edi
  mov    $0x2,%esi
  call   <pow@plt>
  mov    0x1c(%rsp),%edx
  imul   0x24(%rsp),%edx
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  shl    $0x2,%edx
  sub    %edx,%eax
  cvtsi2ss %eax,%xmm0
  comiss %xmm0,%xmm1
  ja     L18
  movss  %xmm0,0xc(%rsp)
  call   <sqrt@plt>
  mov    0x20(%rsp),%ebx
  mov    0x1c(%rsp),%esi
  pxor   %xmm2,%xmm2
  movss  0xc(%rsp),%xmm0
  sub    %ebx,%eax
  lea    (%rsi,%rsi,1),%ecx
  cltd
  idiv   %ecx
  cvtsi2ss %eax,%xmm2
  movss  %xmm2,0x8(%rsp)
  call   <sqrt@plt>
  mov    0x1c(%rsp),%ebp
  mov    0x20(%rsp),%edx
  lea    D49(%rip),%rdi
  mov    0x24(%rsp),%ecx
  mov    %eax,%r12d
  xor    %eax,%eax
  mov    %ebp,%esi
  add    %ebp,%ebp
  call   <printf@plt>
  mov    %ebx,%eax
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  neg    %eax
  movss  0x8(%rsp),%xmm2
  lea    D43(%rip),%rdi
  sub    %r12d,%eax
  cltd
  cvtss2sd %xmm2,%xmm0
  idiv   %ebp
  cvtsi2ss %eax,%xmm1
  mov    $0x2,%eax
  cvtss2sd %xmm1,%xmm1
  call   <printf@plt>
L11:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L16
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  jmp    L11
L16:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D18
type_guessed: string
name: parse_data
arguments: tag: D15
type_guessed: string
name: parse_data
arguments: tag: D49
type_guessed: string
name: parse_data
arguments: tag: D43
type_guessed: string
name: parse_data
arguments: tag: D17
type_guessed: string

tool: D18: please enter values for a,b,c

tool: D15: %d %d %d

tool: D49: The quadratic equation is:  %dx^2 +%dx +%d = 0


tool: D43: The solutions are: 
 x1= %.2f
 x2=%.2f


tool: D17: Oops! These numbers give a negative root. Please try again

assistant: ```c
main() {
  float x1;
  float x2;
  int a, b, c;
  float shoresh;
  printf("please enter values for a,b,c\n");
  scanf("%d %d %d", &a, &b, &c);
  shoresh = pow(b, 2) - 4 * a * c;
  if (shoresh < 0) {
    printf("Oops! These numbers give a negative root. Please try again\n");
  } else {
    x1 = (-b + sqrt(shoresh)) / (2 * a);
    x2 = (-b - sqrt(shoresh)) / (2 * a);
    printf("The quadratic equation is:  %dx^2 +%dx +%d = 0\n", a, b, c);
    printf("The solutions are: \n x1= %.2f\n x2=%.2f\n", x1, x2);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r12
  lea    D15(%rip),%rdi
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x20(%rsp),%rdx
  lea    0x24(%rsp),%rcx
  lea    0x1c(%rsp),%rsi
  lea    D39(%rip),%rdi
  call   <scanf@plt>
  mov    0x20(%rsp),%edi
  mov    $0x2,%esi
  call   <pow@plt>
  mov    0x1c(%rsp),%edx
  imul   0x24(%rsp),%edx
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  shl    $0x2,%edx
  sub    %edx,%eax
  cvtsi2ss %eax,%xmm0
  comiss %xmm0,%xmm1
  ja     L9
  movss  %xmm0,0xc(%rsp)
  call   <sqrt@plt>
  mov    0x20(%rsp),%ebx
  mov    0x1c(%rsp),%esi
  pxor   %xmm2,%xmm2
  movss  0xc(%rsp),%xmm0
  sub    %ebx,%eax
  lea    (%rsi,%rsi,1),%ecx
  cltd
  idiv   %ecx
  cvtsi2ss %eax,%xmm2
  movss  %xmm2,0x8(%rsp)
  call   <sqrt@plt>
  mov    0x1c(%rsp),%ebp
  mov    0x20(%rsp),%edx
  lea    D19(%rip),%rdi
  mov    0x24(%rsp),%ecx
  mov    %eax,%r12d
  xor    %eax,%eax
  mov    %ebp,%esi
  add    %ebp,%ebp
  call   <printf@plt>
  mov    %ebx,%eax
  pxor   %xmm1,%xmm1
  pxor   %xmm0,%xmm0
  neg    %eax
  movss  0x8(%rsp),%xmm2
  lea    D1(%rip),%rdi
  sub    %r12d,%eax
  cltd
  cvtss2sd %xmm2,%xmm0
  idiv   %ebp
  cvtsi2ss %eax,%xmm1
  mov    $0x2,%eax
  cvtss2sd %xmm1,%xmm1
  call   <printf@plt>
L29:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x30,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L9:
  lea    D21(%rip),%rdi
  call   <puts@plt>
  jmp    L29
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D15
data_type: string
name: parse_data
arguments: data_tag: D39
data_type: string
name: parse_data
arguments: data_tag: D19
data_type: string
name: parse_data
arguments: data_tag: D1
data_type: string
name: parse_data
arguments: data_tag: D21
data_type: string

tool: D15: please enter values for a,b,c

tool: D39: %d %d %d

tool: D19: The quadratic equation is:  %dx^2 +%dx +%d = 0


tool: D1: The solutions are: 
 x1= %.2f
 x2=%.2f


tool: D21: Oops! These numbers give a negative root. Please try again

assistant: ```c
main() {
  float x1;
  float x2;
  int a, b, c;
  float shoresh;
  printf("please enter values for a,b,c\n");
  scanf("%d %d %d", &a, &b, &c);
  shoresh = pow(b, 2) - 4 * a * c;
  if (shoresh < 0) {
    printf("Oops! These numbers give a negative root. Please try again\n");
  } else {
    x1 = (-b + sqrt(shoresh)) / (2 * a);
    x2 = (-b - sqrt(shoresh)) / (2 * a);
    printf("The quadratic equation is:  %dx^2 +%dx +%d = 0\n", a, b, c);
    printf("The solutions are: \n x1= %.2f\n x2=%.2f\n", x1, x2);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<quick_sort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x1c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jge    L1
  mov    -0x20(%rbp),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <partition@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick_sort>
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quick_sort>
L1:
  nop
  leave
  ret
```

assistant: ```c
void quick_sort(int a[], int l, int r) {
  int j;
  if (l < r) {
    j = partition(a, l, r);
    quick_sort(a, l, j - 1);
    quick_sort(a, j + 1, r);
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<quick_sort>:
  endbr64
  cmp    %edx,%esi
  jl     L2
  ret
L2:
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r12
  mov    %esi,%r13d
  mov    %edx,%ebx
  call   <partition@plt>
  mov    %eax,%ebp
  lea    -0x1(%rax),%edx
  mov    %r13d,%esi
  mov    %r12,%rdi
  call   <quick_sort>
  lea    0x1(%rbp),%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <quick_sort>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void quick_sort(int a[], int l, int r) {
  int j;
  if (l < r) {
    j = partition(a, l, r);
    quick_sort(a, l, j - 1);
    quick_sort(a, j + 1, r);
  }
}
```
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
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movabs $0x77202c6f6c6c6548,%rax
  mov    %rax,-0x26(%rbp)
  movl   $0x646c726f,-0x1e(%rbp)
  movw   $0x21,-0x1a(%rbp)
  movq   $0xe,-0x40(%rbp)
  mov    D9(%rip),%rax
  mov    (%rax),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  movslq %eax,%rdx
  mov    %rdx,%rsi
  mov    $0x0,%edi
  cltq
  mov    $0x10,%edx
  sub    $0x1,%rdx
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L10:
  cmp    %rdx,%rsp
  je     L37
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L10
L37:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L19
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L19:
  mov    %rsp,%rax
  add    $0x0,%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x30(%rbp),%rdx
  mov    -0x40(%rbp),%rcx
  lea    -0x26(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <SHA1@plt>
  mov    %rbx,%rsp
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L33
  call   <__stack_chk_fail@plt>
L33:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int main() {
  char data[] = "Hello, world!";
  size_t length = sizeof(data);
  unsigned char hash[SHA_DIGEST_LENGTH];
  SHA1(data, length, hash);
}
```
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    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x77202c6f6c6c6548,%rax
  mov    %rax,-0x16(%rbp)
  movl   $0x646c726f,-0xe(%rbp)
  movw   $0x21,-0xa(%rbp)
  mov    D1(%rip),%rax
  movslq (%rax),%rax
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L15:
  cmp    %rdx,%rsp
  je     L24
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L15
L24:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L34
  orq    $0x0,-0x8(%rsp,%rax,1)
L34:
  lea    -0x16(%rbp),%rdi
  mov    %rsp,%rdx
  mov    $0xe,%esi
  call   <SHA1@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L21
  mov    $0x0,%eax
  leave
  ret
L21:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  char data[] = "Hello, world!";
  size_t length = sizeof(data);
  unsigned char hash[SHA_DIGEST_LENGTH];
  SHA1(data, length, hash);
}
```
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
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  movabs $0x77202c6f6c6c6548,%rax
  mov    %rsp,%rcx
  movl   $0x646c726f,-0xe(%rbp)
  mov    %rax,-0x16(%rbp)
  mov    $0x21,%eax
  mov    %ax,-0xa(%rbp)
  mov    D9(%rip),%rax
  movslq (%rax),%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L34
L43:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L43
L34:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L44
L14:
  lea    -0x16(%rbp),%rdi
  mov    %rsp,%rdx
  mov    $0xe,%esi
  call   <SHA1@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L47
  leave
  xor    %eax,%eax
  ret
L44:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L14
L47:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  char data[] = "Hello, world!";
  size_t length = sizeof(data);
  unsigned char hash[SHA_DIGEST_LENGTH];
  SHA1(data, length, hash);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  movabs $0x77202c6f6c6c6548,%rax
  mov    %rsp,%rcx
  movl   $0x646c726f,-0xe(%rbp)
  mov    %rax,-0x16(%rbp)
  mov    $0x21,%eax
  mov    %ax,-0xa(%rbp)
  mov    D1(%rip),%rax
  movslq (%rax),%rax
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L9
L30:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L30
L9:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L17
L1:
  lea    -0x16(%rbp),%rdi
  mov    %rsp,%rdx
  mov    $0xe,%esi
  call   <SHA1@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  leave
  xor    %eax,%eax
  ret
L17:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L1
L18:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  char data[] = "Hello, world!";
  size_t length = sizeof(data);
  unsigned char hash[SHA_DIGEST_LENGTH];
  SHA1(data, length, hash);
}
```