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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x120,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x11c(%rbp),%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x0,-0x118(%rbp)
  jmp    L71
L31:
  movl   $0x0,-0x114(%rbp)
  jmp    L120
L111:
  lea    -0x110(%rbp),%rdx
  mov    -0x114(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x118(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rcx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x114(%rbp)
L120:
  mov    -0x11c(%rbp),%eax
  cmp    %eax,-0x114(%rbp)
  jl     L111
  addl   $0x1,-0x118(%rbp)
L71:
  mov    -0x11c(%rbp),%eax
  cmp    %eax,-0x118(%rbp)
  jl     L31
  movl   $0x0,-0x118(%rbp)
  jmp    L119
L53:
  mov    -0x118(%rbp),%eax
  mov    %eax,-0x114(%rbp)
  jmp    L8
L81:
  lea    -0x110(%rbp),%rdx
  mov    -0x118(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x114(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rcx,%rax
  shl    $0x2,%rax
  add    %rax,%rdx
  lea    -0x110(%rbp),%rcx
  mov    -0x114(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x118(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rsi,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
  addl   $0x1,-0x114(%rbp)
L8:
  mov    -0x11c(%rbp),%eax
  cmp    %eax,-0x114(%rbp)
  jl     L81
  addl   $0x1,-0x118(%rbp)
L119:
  mov    -0x11c(%rbp),%eax
  cmp    %eax,-0x118(%rbp)
  jl     L53
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x0,-0x118(%rbp)
  jmp    L116
L1:
  movl   $0x0,-0x114(%rbp)
  jmp    L76
L129:
  mov    -0x114(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x118(%rbp),%eax
  cltq
  shl    $0x3,%rax
  add    %rdx,%rax
  mov    -0x110(%rbp,%rax,4),%eax
  mov    %eax,%esi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x114(%rbp)
L76:
  mov    -0x11c(%rbp),%eax
  cmp    %eax,-0x114(%rbp)
  jl     L129
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x118(%rbp)
L116:
  mov    -0x11c(%rbp),%eax
  cmp    %eax,-0x118(%rbp)
  jl     L1
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D2
datatype: string
name: parse_data
arguments: data_tag: D25
datatype: string

tool: D2: %d

tool: D25: %d 

assistant: ```c
int main() {
  int T[8][8];
  int scale;
  scanf("%d", &scale);
  int i, j;
  for (i = 0; i < scale; i++)
    for (j = 0; j < scale; j++)
      scanf("%d", &T[i][j]);
  for (i = 0; i < scale; i++)
    for (j = i; j < scale; j++)
      swap(&T[i][j], &T[j][i]);
  printf("\n");
  for (i = 0; i < scale; i++) {
    for (j = 0; j < scale; j++)
      printf("%d ", T[i][j]);
    printf("\n");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x128,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  lea    D11(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L67
  lea    0x10(%rsp),%r13
  mov    %r13,%r15
  mov    $0x0,%r14d
  lea    D11(%rip),%r12
  jmp    L184
L162:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0xc(%rsp)
  jg     L162
L33:
  mov    0xc(%rsp),%eax
  add    $0x1,%r14
  add    $0x20,%r15
  cmp    %r14d,%eax
  jle    L86
L184:
  mov    %r15,%rbp
  mov    $0x0,%ebx
  cmpl   $0x0,0xc(%rsp)
  jg     L162
  jmp    L33
L86:
  test   %eax,%eax
  jle    L67
  mov    $0x0,%r14d
  jmp    L144
L109:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <swap@plt>
  add    $0x1,%r12d
  add    $0x4,%rbx
  add    $0x20,%rbp
  cmp    %r12d,0xc(%rsp)
  jg     L109
L50:
  add    $0x1,%r14
  add    $0x24,%r13
  cmp    %r14d,0xc(%rsp)
  jle    L67
L144:
  mov    %r14d,%r12d
  mov    %r13,%rbp
  mov    %r13,%rbx
  cmp    %r14d,0xc(%rsp)
  jg     L109
  jmp    L50
L67:
  mov    $0xa,%edi
  call   <putchar@plt>
  cmpl   $0x0,0xc(%rsp)
  jle    L18
  lea    0x10(%rsp),%rbp
  mov    $0x0,%r13d
  lea    D6(%rip),%r12
  jmp    L41
L134:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,0xc(%rsp)
  jg     L134
L48:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r13d
  add    $0x20,%rbp
  cmp    %r13d,0xc(%rsp)
  jle    L18
L41:
  mov    $0x0,%ebx
  cmpl   $0x0,0xc(%rsp)
  jg     L134
  jmp    L48
L18:
  mov    0x118(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L172
  mov    $0x0,%eax
  add    $0x128,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L172:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D11
type_infered: string
name: parse_data
arguments: label: D6
type_infered: string

tool: D11: %d

tool: D6: %d 

assistant: ```c
int main() {
  int T[8][8];
  int scale;
  scanf("%d", &scale);
  int i, j;
  for (i = 0; i < scale; i++)
    for (j = 0; j < scale; j++)
      scanf("%d", &T[i][j]);
  for (i = 0; i < scale; i++)
    for (j = i; j < scale; j++)
      swap(&T[i][j], &T[j][i]);
  printf("\n");
  for (i = 0; i < scale; i++) {
    for (j = 0; j < scale; j++)
      printf("%d ", T[i][j]);
    printf("\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D11(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x128,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L43
  lea    0x10(%rsp),%r13
  xor    %r12d,%r12d
  mov    %r13,%r14
  nopl   0x0(%rax,%rax,1)
L55:
  mov    %r14,%r15
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L101
  nopl   0x0(%rax)
L77:
  mov    %r15,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  add    $0x4,%r15
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    %ebx,%eax
  jg     L77
L101:
  add    $0x1,%r12
  add    $0x20,%r14
  cmp    %r12d,%eax
  jg     L55
  test   %eax,%eax
  jle    L43
  xor    %r14d,%r14d
  cs nopw 0x0(%rax,%rax,1)
L19:
  mov    %r14d,%r12d
  mov    %r13,%rbp
  mov    %r13,%rbx
  cmp    %r14d,%eax
  jle    L71
  xchg   %ax,%ax
L58:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  add    $0x1,%r12d
  add    $0x4,%rbx
  call   <swap@plt>
  mov    0xc(%rsp),%eax
  add    $0x20,%rbp
  cmp    %r12d,%eax
  jg     L58
L71:
  add    $0x1,%r14
  add    $0x24,%r13
  cmp    %r14d,%eax
  jg     L19
L43:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L61
  lea    0x10(%rsp),%rbp
  xor    %r13d,%r13d
  lea    D16(%rip),%r12
  xchg   %ax,%ax
L8:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L70:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L70
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    $0x20,%rbp
  call   <putchar@plt>
  cmp    %r13d,0xc(%rsp)
  jg     L8
L61:
  mov    0x118(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L52
  add    $0x128,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L52:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D11
data_type: string
name: unpack_data
arguments: data_label: D16
data_type: string

tool: D11: %d

tool: D16: %d 

assistant: ```c
int main() {
  int T[8][8];
  int scale;
  scanf("%d", &scale);
  int i, j;
  for (i = 0; i < scale; i++)
    for (j = 0; j < scale; j++)
      scanf("%d", &T[i][j]);
  for (i = 0; i < scale; i++)
    for (j = i; j < scale; j++)
      swap(&T[i][j], &T[j][i]);
  printf("\n");
  for (i = 0; i < scale; i++) {
    for (j = 0; j < scale; j++)
      printf("%d ", T[i][j]);
    printf("\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D11(%rip),%rbp
  push   %rbx
  mov    %rbp,%rdi
  sub    $0x128,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x118(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%rsi
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L63
  xor    %r14d,%r14d
  xor    %r13d,%r13d
  lea    0x10(%rsp),%r12
  nop
L2:
  lea    (%r12,%r14,4),%r15
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L37:
  mov    %r15,%rsi
  mov    %rbp,%rdi
  add    $0x1,%ebx
  add    $0x4,%r15
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    %ebx,%eax
  jg     L37
  add    $0x1,%r13d
  cmp    %r13d,%eax
  jle    L60
  add    $0x8,%r14
  test   %eax,%eax
  jg     L2
L63:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L68
  lea    0x10(%rsp),%rbp
  xor    %r13d,%r13d
  lea    D6(%rip),%r12
L20:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L14:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L14
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    $0x20,%rbp
  call   <putchar@plt>
  cmp    %r13d,0xc(%rsp)
  jg     L20
L68:
  mov    0x118(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L107
  add    $0x128,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L60:
  test   %eax,%eax
  jle    L63
  mov    %r12,%r14
  xor    %r13d,%r13d
  xchg   %ax,%ax
L71:
  mov    %r13d,%r12d
  mov    %r14,%rbp
  mov    %r14,%rbx
  cmp    %r13d,%eax
  jle    L70
  xchg   %ax,%ax
L11:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  add    $0x1,%r12d
  add    $0x4,%rbx
  call   <swap@plt>
  mov    0xc(%rsp),%eax
  add    $0x20,%rbp
  cmp    %r12d,%eax
  jg     L11
L70:
  add    $0x1,%r13
  add    $0x24,%r14
  cmp    %r13d,%eax
  jg     L71
  jmp    L63
L107:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D11
datatype: string
name: parse_data
arguments: label: D6
datatype: string

tool: D11: %d

tool: D6: %d 

assistant: ```c
int main() {
  int T[8][8];
  int scale;
  scanf("%d", &scale);
  int i, j;
  for (i = 0; i < scale; i++)
    for (j = 0; j < scale; j++)
      scanf("%d", &T[i][j]);
  for (i = 0; i < scale; i++)
    for (j = i; j < scale; j++)
      swap(&T[i][j], &T[j][i]);
  printf("\n");
  for (i = 0; i < scale; i++) {
    for (j = 0; j < scale; j++)
      printf("%d ", T[i][j]);
    printf("\n");
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x38(%rbp),%eax
  cltq
  mov    -0x30(%rbp,%rax,4),%edx
  mov    -0x24(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x34(%rbp)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L2
  call   <__stack_chk_fail@plt>
L2:
  leave
  ret
```

assistant: ```c
void main(void) {
  int x[10];
  int y;
  int w;
  w = x[y] + x[3];
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main(void) {
  int x[10];
  int y;
  int w;
  w = x[y] + x[3];
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main(void) {
  int x[10];
  int y;
  int w;
  w = x[y] + x[3];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  ret
```

assistant: ```c
void main(void) {
  int x[10];
  int y;
  int w;
  w = x[y] + x[3];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<areaOfRectangle>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x14(%rbp),%eax
  imul   -0x18(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int areaOfRectangle(int length, int bredth) {
  int area;
  area = length * bredth;
  return area;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<areaOfRectangle>:
  endbr64
  mov    %edi,%eax
  imul   %esi,%eax
  ret
```

assistant: ```c
int areaOfRectangle(int length, int bredth) {
  int area;
  area = length * bredth;
  return area;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<areaOfRectangle>:
  endbr64
  mov    %edi,%eax
  imul   %esi,%eax
  ret
```

assistant: ```c
int areaOfRectangle(int length, int bredth) {
  int area;
  area = length * bredth;
  return area;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<areaOfRectangle>:
  endbr64
  mov    %edi,%eax
  imul   %esi,%eax
  ret
```

assistant: ```c
int areaOfRectangle(int length, int bredth) {
  int area;
  area = length * bredth;
  return area;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<quicksort>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L3
  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   <quicksort>
  mov    -0x4(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    -0x20(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <quicksort>
L3:
  nop
  leave
  ret
```

assistant: ```c
void quicksort(int data[], int lower, int upper) {
  int pivot;
  if (upper > lower) {
    pivot = partition(data, lower, upper);
    quicksort(data, lower, pivot - 1);
    quicksort(data, pivot + 1, upper);
  }
}
```
user: What is the c source code of the assembly code below?

```asm
<quicksort>:
  endbr64
  cmp    %esi,%edx
  jg     L6
  ret
L6:
  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   <quicksort>
  lea    0x1(%rbp),%esi
  mov    %ebx,%edx
  mov    %r12,%rdi
  call   <quicksort>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void quicksort(int data[], int lower, int upper) {
  int pivot;
  if (upper > lower) {
    pivot = partition(data, lower, upper);
    quicksort(data, lower, pivot - 1);
    quicksort(data, pivot + 1, upper);
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<seek_string>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x40,%rsp
  mov    %rdi,-0x2038(%rbp)
  mov    %rsi,-0x2040(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x2040(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x201c(%rbp)
  movl   $0x0,-0x2024(%rbp)
  mov    -0x2038(%rbp),%rax
  mov    %rax,%rdi
  call   <ftell@plt>
  mov    %eax,-0x2020(%rbp)
  jmp    L40
L93:
  mov    -0x2038(%rbp),%rdx
  lea    -0x2010(%rbp),%rax
  mov    %rdx,%rcx
  mov    $0x2000,%edx
  mov    $0x1,%esi
  mov    %rax,%rdi
  call   <fread@plt>
  mov    %eax,-0x2018(%rbp)
  cmpl   $0x0,-0x2018(%rbp)
  jle    L33
  mov    -0x201c(%rbp),%eax
  sub    -0x2024(%rbp),%eax
  mov    %eax,%edx
  mov    -0x2024(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x2040(%rbp),%rax
  lea    (%rcx,%rax,1),%rdi
  mov    -0x2018(%rbp),%esi
  lea    -0x2010(%rbp),%rax
  mov    %edx,%ecx
  mov    %rdi,%rdx
  mov    %rax,%rdi
  call   <find_partial@plt>
  mov    %eax,-0x2014(%rbp)
  cmpl   $0xffffffff,-0x2014(%rbp)
  je     L94
  mov    -0x2018(%rbp),%eax
  sub    -0x2014(%rbp),%eax
  mov    %eax,%edx
  mov    -0x2024(%rbp),%eax
  add    %edx,%eax
  cmp    %eax,-0x201c(%rbp)
  jg     L6
  mov    D7(%rip),%rax
  mov    (%rax),%edx
  mov    -0x2020(%rbp),%ecx
  mov    -0x2014(%rbp),%eax
  add    %ecx,%eax
  sub    -0x2024(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x201c(%rbp),%eax
  add    %eax,%ecx
  mov    -0x2038(%rbp),%rax
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <fseek@plt>
  test   %rax,%rax
  je     L86
  mov    $0x0,%eax
  jmp    L18
L86:
  mov    $0x1,%eax
  jmp    L18
L6:
  mov    -0x2018(%rbp),%eax
  sub    -0x2014(%rbp),%eax
  add    %eax,-0x2024(%rbp)
L94:
  mov    -0x2018(%rbp),%eax
  add    %eax,-0x2020(%rbp)
L40:
  mov    -0x2038(%rbp),%rax
  mov    %rax,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L93
  jmp    L13
L33:
  nop
L13:
  mov    $0x0,%eax
L18:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L53
  call   <__stack_chk_fail@plt>
L53:
  leave
  ret
```

assistant: ```c
int seek_string(FILE *fp, const char *str) {
  char block[8192];
  int str_at;
  int str_len;
  int size;
  int ret;
  int pos;
  str_len = strlen(str);
  str_at = 0;
  pos = ftell(fp);
  while (!feof(fp)) {
    size = fread(block, 1, 8192, fp);
    if (size < 1)
      break;
    ret = find_partial(block, size, str + str_at, str_len - str_at);
    if (ret != -1) {
      if (str_at + (size - ret) >= str_len) {
        if (fseek(fp, pos + ret - str_at + str_len, SEEK_SET) != 0)
          return 0;
        return 1;
      } else
        str_at += (size - ret);
    }
    pos += size;
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<seek_string>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x28,%rsp
  mov    %rdi,%r13
  mov    %rsi,(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x2018(%rsp)
  xor    %eax,%eax
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%r15d
  mov    %r13,%rdi
  call   <ftell@plt>
  mov    %eax,0xc(%rsp)
  mov    $0x0,%r12d
  lea    0x10(%rsp),%r14
  jmp    L3
L45:
  mov    0xc(%rsp),%esi
  add    %eax,%esi
  sub    %r12d,%esi
  add    %r15d,%esi
  mov    D6(%rip),%rax
  mov    (%rax),%edx
  mov    %r13,%rdi
  call   <fseek@plt>
  test   %rax,%rax
  sete   %bpl
  movzbl %bpl,%ebp
  jmp    L24
L33:
  add    %ebx,0xc(%rsp)
L3:
  mov    %r13,%rdi
  call   <feof@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  jne    L19
  mov    %r13,%rcx
  mov    $0x2000,%edx
  mov    $0x1,%esi
  mov    %r14,%rdi
  call   <fread@plt>
  mov    %eax,%ebx
  test   %eax,%eax
  jle    L24
  mov    %r15d,%ecx
  sub    %r12d,%ecx
  movslq %r12d,%rdx
  add    (%rsp),%rdx
  mov    %eax,%esi
  mov    %r14,%rdi
  call   <find_partial@plt>
  cmp    $0xffffffff,%eax
  je     L33
  mov    %ebx,%edx
  sub    %eax,%edx
  add    %r12d,%edx
  cmp    %r15d,%edx
  jge    L45
  mov    %edx,%r12d
  jmp    L33
L19:
  mov    $0x0,%ebp
L24:
  mov    0x2018(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L51
  mov    %ebp,%eax
  add    $0x2028,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L51:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int seek_string(FILE *fp, const char *str) {
  char block[8192];
  int str_at;
  int str_len;
  int size;
  int ret;
  int pos;
  str_len = strlen(str);
  str_at = 0;
  pos = ftell(fp);
  while (!feof(fp)) {
    size = fread(block, 1, 8192, fp);
    if (size < 1)
      break;
    ret = find_partial(block, size, str + str_at, str_len - str_at);
    if (ret != -1) {
      if (str_at + (size - ret) >= str_len) {
        if (fseek(fp, pos + ret - str_at + str_len, SEEK_SET) != 0)
          return 0;
        return 1;
      } else
        str_at += (size - ret);
    }
    pos += size;
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<seek_string>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x28,%rsp
  xor    %ebx,%ebx
  mov    %rsi,0x8(%rsp)
  mov    %rdi,%rbp
  mov    %rsi,%rdi
  lea    0x10(%rsp),%r13
  mov    %fs:0x28,%rax
  mov    %rax,0x2018(%rsp)
  xor    %eax,%eax
  call   <strlen@plt>
  mov    %rbp,%rdi
  mov    %eax,%r14d
  call   <ftell@plt>
  mov    %eax,%r12d
  jmp    L53
  nopl   0x0(%rax)
L37:
  mov    %rbp,%rcx
  mov    $0x2000,%edx
  mov    $0x1,%esi
  mov    %r13,%rdi
  call   <fread@plt>
  mov    %eax,%r15d
  test   %eax,%eax
  jle    L58
  mov    %r14d,%ecx
  movslq %ebx,%rdx
  mov    %eax,%esi
  add    0x8(%rsp),%rdx
  sub    %ebx,%ecx
  mov    %r13,%rdi
  call   <find_partial@plt>
  cmp    $0xffffffff,%eax
  je     L13
  mov    %r15d,%edx
  sub    %eax,%edx
  add    %ebx,%edx
  cmp    %r14d,%edx
  jge    L25
  mov    %edx,%ebx
L13:
  add    %r15d,%r12d
L53:
  mov    %rbp,%rdi
  call   <feof@plt>
  test   %eax,%eax
  je     L37
L58:
  xor    %eax,%eax
L2:
  mov    0x2018(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L56
  add    $0x2028,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L25:
  lea    (%r12,%rax,1),%esi
  mov    D4(%rip),%rax
  mov    %rbp,%rdi
  sub    %ebx,%esi
  mov    (%rax),%edx
  lea    (%rsi,%r14,1),%esi
  call   <fseek@plt>
  test   %rax,%rax
  sete   %al
  movzbl %al,%eax
  jmp    L2
L56:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int seek_string(FILE *fp, const char *str) {
  char block[8192];
  int str_at;
  int str_len;
  int size;
  int ret;
  int pos;
  str_len = strlen(str);
  str_at = 0;
  pos = ftell(fp);
  while (!feof(fp)) {
    size = fread(block, 1, 8192, fp);
    if (size < 1)
      break;
    ret = find_partial(block, size, str + str_at, str_len - str_at);
    if (ret != -1) {
      if (str_at + (size - ret) >= str_len) {
        if (fseek(fp, pos + ret - str_at + str_len, SEEK_SET) != 0)
          return 0;
        return 1;
      } else
        str_at += (size - ret);
    }
    pos += size;
  }
  return 0;
}
```