ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请从给定的汇编指令推导出C源代码：

```asm
<dump_hdr_filter>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpl   $0x0,-0x1c(%rbp)
  je     L20
  cmpl   $0x1,-0x1c(%rbp)
  jne    L1
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L20
L1:
  mov    $0x1,%eax
  jmp    L7
L11:
  movzbl -0x9(%rbp),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <putchar@plt>
L20:
  mov    -0x18(%rbp),%rdx
  lea    -0x9(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fread@plt>
  mov    %eax,-0x10(%rbp)
  cmpl   $0x1,-0x10(%rbp)
  je     L11
  mov    $0x0,%eax
L7:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L27
  call   <__stack_chk_fail@plt>
L27:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type_guessed: string

tool: D2: ===> xtension HDR

assistant: ```c
int dump_hdr_filter(FILE *in, int type) {
  char buf[1];
  int err;
  if (type == 0) {
  } else if (type == 1)
    printf("===> xtension HDR\n");
  else
    return 1;
  while ((err = fread(buf, sizeof(char), 1, in)) == 1) {
    printf("%c", buf[0]);
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<dump_hdr_filter>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  je     L35
  mov    $0x1,%eax
  cmp    $0x1,%esi
  jne    L16
  lea    D6(%rip),%rdi
  call   <puts@plt>
L35:
  lea    0x7(%rsp),%rbp
  jmp    L32
L19:
  movsbl 0x7(%rsp),%edi
  call   <putchar@plt>
L32:
  mov    %rbx,%rcx
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  cmp    $0x1,%eax
  je     L19
  mov    $0x0,%eax
L16:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L48
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
L48:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D6
datatype: string

tool: D6: ===> xtension HDR

assistant: ```c
int dump_hdr_filter(FILE *in, int type) {
  char buf[1];
  int err;
  if (type == 0) {
  } else if (type == 1)
    printf("===> xtension HDR\n");
  else
    return 1;
  while ((err = fread(buf, sizeof(char), 1, in)) == 1) {
    printf("%c", buf[0]);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<dump_hdr_filter>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  je     L16
  mov    $0x1,%eax
  cmp    $0x1,%esi
  je     L15
L59:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L54
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L15:
  lea    D1(%rip),%rdi
  call   <puts@plt>
L16:
  lea    0x7(%rsp),%rbp
  jmp    L38
  nopl   0x0(%rax,%rax,1)
L9:
  movsbl 0x7(%rsp),%edi
  call   <putchar@plt>
L38:
  mov    %rbx,%rcx
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  cmp    $0x1,%eax
  je     L9
  xor    %eax,%eax
  jmp    L59
L54:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D1
type_guessed: string

tool: D1: ===> xtension HDR

assistant: ```c
int dump_hdr_filter(FILE *in, int type) {
  char buf[1];
  int err;
  if (type == 0) {
  } else if (type == 1)
    printf("===> xtension HDR\n");
  else
    return 1;
  while ((err = fread(buf, sizeof(char), 1, in)) == 1) {
    printf("%c", buf[0]);
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<dump_hdr_filter>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  test   %esi,%esi
  je     L6
  mov    $0x1,%eax
  cmp    $0x1,%esi
  je     L34
L29:
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L1
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L34:
  lea    D7(%rip),%rdi
  call   <puts@plt>
L6:
  lea    0x7(%rsp),%rbp
  jmp    L24
  nopl   0x0(%rax,%rax,1)
L2:
  movsbl 0x7(%rsp),%edi
  call   <putchar@plt>
L24:
  mov    %rbx,%rcx
  mov    $0x1,%edx
  mov    $0x1,%esi
  mov    %rbp,%rdi
  call   <fread@plt>
  cmp    $0x1,%eax
  je     L2
  xor    %eax,%eax
  jmp    L29
L1:
  call   <__stack_chk_fail@plt>
```

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

tool: D7: ===> xtension HDR

assistant: ```c
int dump_hdr_filter(FILE *in, int type) {
  char buf[1];
  int err;
  if (type == 0) {
  } else if (type == 1)
    printf("===> xtension HDR\n");
  else
    return 1;
  while ((err = fread(buf, sizeof(char), 1, in)) == 1) {
    printf("%c", buf[0]);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<search_space>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x3,-0x4(%rbp)
  jmp    L16
L0:
  mov    -0x4(%rbp),%edx
  mov    -0x4(%rbp),%ecx
  mov    -0x4(%rbp),%eax
  mov    %ecx,%esi
  mov    %eax,%edi
  call   <search_subspace@plt>
  addl   $0x3,-0x4(%rbp)
L16:
  cmpl   $0x15,-0x4(%rbp)
  jle    L0
  nop
  nop
  leave
  ret
```

assistant: ```c
void search_space(void) {
  for (int c = 3; c <= 21; c += 3) {
    search_subspace(c, c, c);
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<search_space>:
  endbr64
  push   %rbx
  mov    $0x3,%ebx
L1:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %ebx,%edi
  call   <search_subspace@plt>
  add    $0x3,%ebx
  cmp    $0x18,%ebx
  jne    L1
  pop    %rbx
  ret
```

assistant: ```c
void search_space(void) {
  for (int c = 3; c <= 21; c += 3) {
    search_subspace(c, c, c);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<search_space>:
  endbr64
  push   %rbx
  mov    $0x3,%ebx
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %ebx,%edi
  add    $0x3,%ebx
  call   <search_subspace@plt>
  cmp    $0x18,%ebx
  jne    L4
  pop    %rbx
  ret
```

assistant: ```c
void search_space(void) {
  for (int c = 3; c <= 21; c += 3) {
    search_subspace(c, c, c);
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<search_space>:
  endbr64
  push   %rbx
  mov    $0x3,%ebx
  nopw   0x0(%rax,%rax,1)
L4:
  mov    %ebx,%edx
  mov    %ebx,%esi
  mov    %ebx,%edi
  add    $0x3,%ebx
  call   <search_subspace@plt>
  cmp    $0x18,%ebx
  jne    L4
  pop    %rbx
  ret
```

assistant: ```c
void search_space(void) {
  for (int c = 3; c <= 21; c += 3) {
    search_subspace(c, c, c);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<sumsq_mmx>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movq   $0x0,-0x8(%rbp)
  jmp    L10
L40:
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  movswq %ax,%rdx
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  movswq %ax,%rax
  imul   %rdx,%rax
  add    %rax,-0x8(%rbp)
  addq   $0x2,-0x18(%rbp)
  subl   $0x1,-0x1c(%rbp)
L10:
  mov    -0x18(%rbp),%rax
  and    $0x7,%eax
  test   %eax,%eax
  je     L16
  cmpl   $0x0,-0x1c(%rbp)
  jne    L40
L16:
  mov    -0x1c(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <sumsq_mmx_assist@plt>
  add    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  and    $0xfffffff8,%eax
  cltq
  add    %rax,%rax
  add    %rax,-0x18(%rbp)
  andl   $0x7,-0x1c(%rbp)
  jmp    L27
L38:
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  movswq %ax,%rdx
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  movswq %ax,%rax
  imul   %rdx,%rax
  add    %rax,-0x8(%rbp)
  addq   $0x2,-0x18(%rbp)
  subl   $0x1,-0x1c(%rbp)
L27:
  cmpl   $0x0,-0x1c(%rbp)
  jne    L38
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long long sumsq_mmx(signed short *in, int cnt) {
  long long sum = 0;
  while (((int)in & 7) != 0 && cnt != 0) {
    sum += (long)in[0] * in[0];
    in++;
    cnt--;
  }
  sum += sumsq_mmx_assist(in, cnt);
  in += cnt & ~7;
  cnt &= 7;
  while (cnt != 0) {
    sum += (long)in[0] * in[0];
    in++;
    cnt--;
  }
  return sum;
}
```
user: What is the c source code of the assembly code below:

```asm
<sumsq_mmx>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %esi,%ebp
  test   $0x7,%dil
  je     L5
  test   %esi,%esi
  je     L5
  mov    $0x0,%r12d
L44:
  movswq (%rbx),%rax
  imul   %rax,%rax
  add    %rax,%r12
  add    $0x2,%rbx
  sub    $0x1,%ebp
  test   $0x7,%bl
  je     L41
  test   %ebp,%ebp
  jne    L44
L41:
  mov    %ebp,%esi
  mov    %rbx,%rdi
  call   <sumsq_mmx_assist@plt>
  add    %r12,%rax
  mov    %ebp,%edx
  and    $0xfffffff8,%edx
  movslq %edx,%rdx
  lea    (%rbx,%rdx,2),%rcx
  and    $0x7,%ebp
  je     L25
  mov    %ebp,%ebp
  lea    (%rcx,%rbp,2),%rsi
L32:
  movswq (%rcx),%rdx
  imul   %rdx,%rdx
  add    %rdx,%rax
  add    $0x2,%rcx
  cmp    %rsi,%rcx
  jne    L32
L25:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  mov    $0x0,%r12d
  jmp    L41
```

assistant: ```c
long long sumsq_mmx(signed short *in, int cnt) {
  long long sum = 0;
  while (((int)in & 7) != 0 && cnt != 0) {
    sum += (long)in[0] * in[0];
    in++;
    cnt--;
  }
  sum += sumsq_mmx_assist(in, cnt);
  in += cnt & ~7;
  cnt &= 7;
  while (cnt != 0) {
    sum += (long)in[0] * in[0];
    in++;
    cnt--;
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sumsq_mmx>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  and    $0x7,%edi
  push   %rbx
  mov    %esi,%ebx
  jne    L32
  jmp    L31
  nopw   0x0(%rax,%rax,1)
L29:
  movswq 0x0(%rbp),%rax
  add    $0x2,%rbp
  sub    $0x1,%ebx
  imul   %rax,%rax
  add    %rax,%r12
  test   $0x7,%bpl
  je     L31
L32:
  test   %ebx,%ebx
  jne    L29
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <sumsq_mmx_assist@plt>
  add    %rax,%r12
L40:
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L31:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <sumsq_mmx_assist@plt>
  add    %rax,%r12
  mov    %ebx,%eax
  and    $0xfffffff8,%eax
  cltq
  lea    0x0(%rbp,%rax,2),%rdx
  and    $0x7,%ebx
  je     L40
  movslq %ebx,%rbx
  lea    (%rdx,%rbx,2),%rcx
  nopl   (%rax)
L13:
  movswq (%rdx),%rax
  add    $0x2,%rdx
  imul   %rax,%rax
  add    %rax,%r12
  cmp    %rcx,%rdx
  jne    L13
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
long long sumsq_mmx(signed short *in, int cnt) {
  long long sum = 0;
  while (((int)in & 7) != 0 && cnt != 0) {
    sum += (long)in[0] * in[0];
    in++;
    cnt--;
  }
  sum += sumsq_mmx_assist(in, cnt);
  in += cnt & ~7;
  cnt &= 7;
  while (cnt != 0) {
    sum += (long)in[0] * in[0];
    in++;
    cnt--;
  }
  return sum;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<sumsq_mmx>:
  endbr64
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  mov    %rdi,%rbp
  and    $0x7,%edi
  push   %rbx
  mov    %esi,%ebx
  jne    L30
  jmp    L74
  nopw   0x0(%rax,%rax,1)
L2:
  movswq 0x0(%rbp),%rax
  add    $0x2,%rbp
  sub    $0x1,%ebx
  imul   %rax,%rax
  add    %rax,%r12
  test   $0x7,%bpl
  je     L74
L30:
  test   %ebx,%ebx
  jne    L2
  xor    %esi,%esi
  mov    %rbp,%rdi
  call   <sumsq_mmx_assist@plt>
  add    %r12,%rax
L28:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nop
L74:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <sumsq_mmx_assist@plt>
  mov    %ebx,%edx
  and    $0xfffffff8,%edx
  add    %r12,%rax
  movslq %edx,%rdx
  lea    0x0(%rbp,%rdx,2),%rdx
  and    $0x7,%ebx
  je     L28
  movswq (%rdx),%rcx
  imul   %rcx,%rcx
  add    %rcx,%rax
  cmp    $0x1,%ebx
  je     L28
  movswq 0x2(%rdx),%rcx
  imul   %rcx,%rcx
  add    %rcx,%rax
  cmp    $0x2,%ebx
  je     L28
  movswq 0x4(%rdx),%rcx
  imul   %rcx,%rcx
  add    %rcx,%rax
  cmp    $0x3,%ebx
  je     L28
  movswq 0x6(%rdx),%rcx
  imul   %rcx,%rcx
  add    %rcx,%rax
  cmp    $0x4,%ebx
  je     L28
  movswq 0x8(%rdx),%rcx
  imul   %rcx,%rcx
  add    %rcx,%rax
  sub    $0x5,%ebx
  je     L28
  movswq 0xa(%rdx),%rcx
  imul   %rcx,%rcx
  add    %rcx,%rax
  cmp    $0x1,%ebx
  je     L28
  movswq 0xc(%rdx),%rdx
  pop    %rbx
  pop    %rbp
  pop    %r12
  imul   %rdx,%rdx
  add    %rdx,%rax
  ret
```

assistant: ```c
long long sumsq_mmx(signed short *in, int cnt) {
  long long sum = 0;
  while (((int)in & 7) != 0 && cnt != 0) {
    sum += (long)in[0] * in[0];
    in++;
    cnt--;
  }
  sum += sumsq_mmx_assist(in, cnt);
  in += cnt & ~7;
  cnt &= 7;
  while (cnt != 0) {
    sum += (long)in[0] * in[0];
    in++;
    cnt--;
  }
  return sum;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<helpmod_percentage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jne    L0
  pxor   %xmm0,%xmm0
  jmp    L18
L0:
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x8(%rbp),%xmm0
  pxor   %xmm2,%xmm2
  cvtsi2ssl -0x4(%rbp),%xmm2
  movaps %xmm0,%xmm1
  divss  %xmm2,%xmm1
  movss  D1(%rip),%xmm0
  mulss  %xmm1,%xmm0
L18:
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D1
type: f32

tool: D1: 100

assistant: ```c
float helpmod_percentage(int larger, int smaller) {
  if (larger == 0)
    return 0.0;
  return 100.0 * ((float)smaller / (float)larger);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<helpmod_percentage>:
  endbr64
  pxor   %xmm0,%xmm0
  test   %edi,%edi
  je     L0
  pxor   %xmm0,%xmm0
  cvtsi2ss %esi,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %edi,%xmm1
  divss  %xmm1,%xmm0
  mulss  D9(%rip),%xmm0
L0:
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
datatype: f32

tool: D9: 100

assistant: ```c
float helpmod_percentage(int larger, int smaller) {
  if (larger == 0)
    return 0.0;
  return 100.0 * ((float)smaller / (float)larger);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<helpmod_percentage>:
  endbr64
  pxor   %xmm0,%xmm0
  test   %edi,%edi
  je     L6
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %esi,%xmm0
  cvtsi2ss %edi,%xmm1
  divss  %xmm1,%xmm0
  mulss  D4(%rip),%xmm0
L6:
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D4
type_infered: f32

tool: D4: 100

assistant: ```c
float helpmod_percentage(int larger, int smaller) {
  if (larger == 0)
    return 0.0;
  return 100.0 * ((float)smaller / (float)larger);
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<helpmod_percentage>:
  endbr64
  pxor   %xmm0,%xmm0
  test   %edi,%edi
  je     L8
  pxor   %xmm0,%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ss %esi,%xmm0
  cvtsi2ss %edi,%xmm1
  divss  %xmm1,%xmm0
  mulss  D2(%rip),%xmm0
L8:
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D2
type_guessed: f32

tool: D2: 100

assistant: ```c
float helpmod_percentage(int larger, int smaller) {
  if (larger == 0)
    return 0.0;
  return 100.0 * ((float)smaller / (float)larger);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<Maxval>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dx,-0x4(%rbp)
  mov    %ax,-0x8(%rbp)
  movzwl -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%ax
  jle    L2
  movzwl -0x4(%rbp),%eax
  jmp    L3
L2:
  movzwl -0x8(%rbp),%eax
L3:
  pop    %rbp
  ret
```

assistant: ```c
short Maxval(short a, short b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<Maxval>:
  endbr64
  cmp    %di,%si
  mov    %edi,%eax
  cmovge %esi,%eax
  ret
```

assistant: ```c
short Maxval(short a, short b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<Maxval>:
  endbr64
  cmp    %di,%si
  mov    %edi,%eax
  cmovge %esi,%eax
  ret
```

assistant: ```c
short Maxval(short a, short b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<Maxval>:
  endbr64
  cmp    %di,%si
  mov    %edi,%eax
  cmovge %esi,%eax
  ret
```

assistant: ```c
short Maxval(short a, short b) {
  if (a > b)
    return a;
  else
    return b;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <getpid@plt>
  mov    %eax,%edi
  lea    -0xc(%rbp),%rdx
  lea    -0x10(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x0,%edx
  mov    %rax,%rsi
  call   <wait4@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
int main(void) {
  int status;
  struct rusage rusage;
  return wait4(getpid(), &status, 0, &rusage);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <getpid@plt>
  mov    %eax,%edi
  lea    0x4(%rsp),%rcx
  mov    %rsp,%rsi
  mov    $0x0,%edx
  call   <wait4@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int status;
  struct rusage rusage;
  return wait4(getpid(), &status, 0, &rusage);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <getpid@plt>
  xor    %edx,%edx
  lea    0x4(%rsp),%rcx
  mov    %rsp,%rsi
  mov    %eax,%edi
  call   <wait4@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int status;
  struct rusage rusage;
  return wait4(getpid(), &status, 0, &rusage);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  call   <getpid@plt>
  xor    %edx,%edx
  lea    0x4(%rsp),%rcx
  mov    %rsp,%rsi
  mov    %eax,%edi
  call   <wait4@plt>
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main(void) {
  int status;
  struct rusage rusage;
  return wait4(getpid(), &status, 0, &rusage);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x80000000,-0x24(%rbp)
  movl   $0x1,-0x20(%rbp)
  mov    -0x24(%rbp),%eax
  mov    -0x20(%rbp),%edx
  sub    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x24(%rbp)
  movl   $0x80000064,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  sub    $0x400,%eax
  mov    %eax,-0x24(%rbp)
  movl   $0x80000000,-0x24(%rbp)
  movl   $0xffffffff,-0x20(%rbp)
  mov    -0x24(%rbp),%edx
  mov    -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x1c(%rbp)
  movl   $0x800003e8,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  sub    $0x100000,%eax
  mov    %eax,-0x20(%rbp)
  movabs $0x8000000000000000,%rcx
  mov    %rcx,-0x18(%rbp)
  movq   $0x1,-0x10(%rbp)
  mov    -0x18(%rbp),%rax
  mov    -0x10(%rbp),%rdx
  sub    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x18(%rbp)
  movabs $0x8000000000000064,%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  sub    $0x400,%rax
  mov    %rax,-0x18(%rbp)
  mov    %rcx,-0x18(%rbp)
  movq   $0xffffffffffffffff,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movabs $0x80000000000003e8,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  sub    $0x100000,%rax
  mov    %rax,-0x10(%rbp)
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
  volatile int j = (-0x7fffffff - 1);
  volatile int i = 1;
  volatile int k = j - i;
  j--;
  j = (-0x7fffffff - 1) + 100;
  j -= (1 << 10);
  j = (-0x7fffffff - 1);
  i = -1;
  k = j - -i;
  i = (-0x7fffffff - 1) + 1000;
  i -= (1 << 20);
  volatile long int l = (-0x7fffffffffffffffL - 1L);
  volatile long int m = 1;
  volatile long int n = l - m;
  l--;
  l = (-0x7fffffffffffffffL - 1L) + 100;
  l -= (1 << 10);
  l = (-0x7fffffffffffffffL - 1L);
  m = -1;
  n = l - -m;
  m = (-0x7fffffffffffffffL - 1L) + 1000;
  m -= (1 << 20);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  movl   $0x80000000,-0x24(%rsp)
  movl   $0x1,-0x20(%rsp)
  mov    -0x24(%rsp),%eax
  mov    -0x20(%rsp),%edx
  sub    %edx,%eax
  mov    %eax,-0x1c(%rsp)
  mov    -0x24(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x24(%rsp)
  movl   $0x80000064,-0x24(%rsp)
  mov    -0x24(%rsp),%eax
  sub    $0x400,%eax
  mov    %eax,-0x24(%rsp)
  movl   $0x80000000,-0x24(%rsp)
  movl   $0xffffffff,-0x20(%rsp)
  mov    -0x24(%rsp),%eax
  mov    -0x20(%rsp),%edx
  add    %edx,%eax
  mov    %eax,-0x1c(%rsp)
  movl   $0x800003e8,-0x20(%rsp)
  mov    -0x20(%rsp),%eax
  sub    $0x100000,%eax
  mov    %eax,-0x20(%rsp)
  movabs $0x8000000000000000,%rdx
  mov    %rdx,-0x18(%rsp)
  movq   $0x1,-0x10(%rsp)
  mov    -0x18(%rsp),%rax
  mov    -0x10(%rsp),%rcx
  sub    %rcx,%rax
  mov    %rax,-0x8(%rsp)
  mov    -0x18(%rsp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x18(%rsp)
  movabs $0x8000000000000064,%rax
  mov    %rax,-0x18(%rsp)
  mov    -0x18(%rsp),%rax
  sub    $0x400,%rax
  mov    %rax,-0x18(%rsp)
  mov    %rdx,-0x18(%rsp)
  movq   $0xffffffffffffffff,-0x10(%rsp)
  mov    -0x18(%rsp),%rax
  mov    -0x10(%rsp),%rdx
  add    %rdx,%rax
  mov    %rax,-0x8(%rsp)
  movabs $0x80000000000003e8,%rax
  mov    %rax,-0x10(%rsp)
  mov    -0x10(%rsp),%rax
  sub    $0x100000,%rax
  mov    %rax,-0x10(%rsp)
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
  volatile int j = (-0x7fffffff - 1);
  volatile int i = 1;
  volatile int k = j - i;
  j--;
  j = (-0x7fffffff - 1) + 100;
  j -= (1 << 10);
  j = (-0x7fffffff - 1);
  i = -1;
  k = j - -i;
  i = (-0x7fffffff - 1) + 1000;
  i -= (1 << 20);
  volatile long int l = (-0x7fffffffffffffffL - 1L);
  volatile long int m = 1;
  volatile long int n = l - m;
  l--;
  l = (-0x7fffffffffffffffL - 1L) + 100;
  l -= (1 << 10);
  l = (-0x7fffffffffffffffL - 1L);
  m = -1;
  n = l - -m;
  m = (-0x7fffffffffffffffL - 1L) + 1000;
  m -= (1 << 20);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  movl   $0x80000000,-0x24(%rsp)
  movl   $0x1,-0x20(%rsp)
  mov    -0x24(%rsp),%eax
  mov    -0x20(%rsp),%edx
  sub    %edx,%eax
  mov    %eax,-0x1c(%rsp)
  mov    -0x24(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x24(%rsp)
  movl   $0x80000064,-0x24(%rsp)
  mov    -0x24(%rsp),%eax
  sub    $0x400,%eax
  mov    %eax,-0x24(%rsp)
  movl   $0x80000000,-0x24(%rsp)
  movl   $0xffffffff,-0x20(%rsp)
  mov    -0x24(%rsp),%eax
  mov    -0x20(%rsp),%edx
  add    %edx,%eax
  movabs $0x8000000000000000,%rdx
  mov    %eax,-0x1c(%rsp)
  movl   $0x800003e8,-0x20(%rsp)
  mov    -0x20(%rsp),%eax
  sub    $0x100000,%eax
  mov    %eax,-0x20(%rsp)
  mov    %rdx,-0x18(%rsp)
  movq   $0x1,-0x10(%rsp)
  mov    -0x18(%rsp),%rax
  mov    -0x10(%rsp),%rcx
  sub    %rcx,%rax
  mov    %rax,-0x8(%rsp)
  mov    -0x18(%rsp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x18(%rsp)
  movabs $0x8000000000000064,%rax
  mov    %rax,-0x18(%rsp)
  mov    -0x18(%rsp),%rax
  sub    $0x400,%rax
  mov    %rax,-0x18(%rsp)
  mov    %rdx,-0x18(%rsp)
  movq   $0xffffffffffffffff,-0x10(%rsp)
  mov    -0x18(%rsp),%rax
  mov    -0x10(%rsp),%rdx
  add    %rdx,%rax
  mov    %rax,-0x8(%rsp)
  movabs $0x80000000000003e8,%rax
  mov    %rax,-0x10(%rsp)
  mov    -0x10(%rsp),%rax
  sub    $0x100000,%rax
  mov    %rax,-0x10(%rsp)
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
  volatile int j = (-0x7fffffff - 1);
  volatile int i = 1;
  volatile int k = j - i;
  j--;
  j = (-0x7fffffff - 1) + 100;
  j -= (1 << 10);
  j = (-0x7fffffff - 1);
  i = -1;
  k = j - -i;
  i = (-0x7fffffff - 1) + 1000;
  i -= (1 << 20);
  volatile long int l = (-0x7fffffffffffffffL - 1L);
  volatile long int m = 1;
  volatile long int n = l - m;
  l--;
  l = (-0x7fffffffffffffffL - 1L) + 100;
  l -= (1 << 10);
  l = (-0x7fffffffffffffffL - 1L);
  m = -1;
  n = l - -m;
  m = (-0x7fffffffffffffffL - 1L) + 1000;
  m -= (1 << 20);
  return 0;
}
```