ylfeng/ReF-Decompile-dataset huggingface dataset:
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   %r15
  lea    D21(%rip),%rdi
  push   %r14
  lea    D71(%rip),%r14
  push   %r13
  lea    D23(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  call   <system@plt>
  lea    D5(%rip),%rdi
  call   <puts@plt>
  lea    0x10(%rsp),%rax
  movq   $0x0,(%rsp)
  mov    %rax,0x8(%rsp)
L9:
  mov    0x8(%rsp),%rax
  mov    (%rsp),%rcx
  add    $0x1,%r12d
  xor    %r15d,%r15d
  mov    %rax,%rbp
  lea    (%rax,%rcx,4),%rbx
L41:
  add    $0x1,%r15d
  mov    %r12d,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %r15d,%edx
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    $0x5,%r15d
  jne    L41
  mov    $0xa,%edi
  call   <putchar@plt>
  addq   $0x5,(%rsp)
  cmp    $0x5,%r12d
  jne    L9
  lea    D21(%rip),%rdi
  lea    D67(%rip),%r12
  call   <system@plt>
  lea    D27(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  lea    0x64(%rax),%r13
L12:
  xor    %ebx,%ebx
L48:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x5,%rbx
  jne    L48
  mov    $0xa,%edi
  add    $0x14,%rbp
  call   <putchar@plt>
  cmp    %r13,%rbp
  jne    L12
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rdi
  call   <tamKaremi@plt>
  lea    D12(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L42
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L42:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D21
type_infered: string
name: parse_data
arguments: tag: D71
type_infered: string
name: parse_data
arguments: tag: D23
type_infered: string
name: parse_data
arguments: tag: D5
type_infered: string
name: parse_data
arguments: tag: D67
type_infered: string
name: parse_data
arguments: tag: D27
type_infered: string
name: parse_data
arguments: tag: D12
type_infered: string

tool: D21: clear

tool: D71: %d satir, %d sutun elemani giriniz: 

tool: D23:  %d

tool: D5: 
Lutfen 5x5 matris elemanlarini giriniz!


tool: D67: %d	

tool: D27: ---Matris---


tool: D12: Sonuc=%d


assistant: ```c
int main() {
  system("clear");
  int i, j, sonuc;
  int dizi[5][5];
  printf("\nLutfen 5x5 matris elemanlarini giriniz!\n\n");
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) {
      printf("%d satir, %d sutun elemani giriniz: ", i + 1, j + 1);
      scanf(" %d", &dizi[i][j]);
    }
    printf("\n");
  }
  system("clear");
  printf("---Matris---\n\n");
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) {
      printf("%d\t", dizi[i][j]);
    }
    printf("\n");
  }
  printf("\n");
  sonuc = tamKaremi(dizi);
  printf("Sonuc=%d\n", sonuc);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r15
  lea    D0(%rip),%rdi
  push   %r14
  lea    D12(%rip),%r14
  push   %r13
  lea    D63(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x88,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x78(%rsp)
  xor    %eax,%eax
  call   <system@plt>
  lea    D20(%rip),%rdi
  call   <puts@plt>
  lea    0x10(%rsp),%rax
  movq   $0x0,(%rsp)
  mov    %rax,0x8(%rsp)
L36:
  mov    0x8(%rsp),%rax
  mov    (%rsp),%rcx
  add    $0x1,%r12d
  xor    %r15d,%r15d
  mov    %rax,%rbp
  lea    (%rax,%rcx,4),%rbx
L7:
  add    $0x1,%r15d
  mov    %r12d,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %r15d,%edx
  call   <printf@plt>
  mov    %rbx,%rsi
  mov    %r13,%rdi
  add    $0x4,%rbx
  call   <scanf@plt>
  cmp    $0x5,%r15d
  jne    L7
  mov    $0xa,%edi
  call   <putchar@plt>
  addq   $0x5,(%rsp)
  cmp    $0x5,%r12d
  jne    L36
  lea    D0(%rip),%rdi
  lea    D48(%rip),%r12
  call   <system@plt>
  lea    D43(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%rax
  lea    0x64(%rax),%r13
L10:
  xor    %ebx,%ebx
L19:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x5,%rbx
  jne    L19
  mov    $0xa,%edi
  add    $0x14,%rbp
  call   <putchar@plt>
  cmp    %r13,%rbp
  jne    L10
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x8(%rsp),%rdi
  call   <tamKaremi@plt>
  lea    D30(%rip),%rdi
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x78(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L20
  add    $0x88,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L20:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D0
data_type: string
name: retrieve_data
arguments: data_label: D12
data_type: string
name: retrieve_data
arguments: data_label: D63
data_type: string
name: retrieve_data
arguments: data_label: D20
data_type: string
name: retrieve_data
arguments: data_label: D48
data_type: string
name: retrieve_data
arguments: data_label: D43
data_type: string
name: retrieve_data
arguments: data_label: D30
data_type: string

tool: D0: clear

tool: D12: %d satir, %d sutun elemani giriniz: 

tool: D63:  %d

tool: D20: 
Lutfen 5x5 matris elemanlarini giriniz!


tool: D48: %d	

tool: D43: ---Matris---


tool: D30: Sonuc=%d


assistant: ```c
int main() {
  system("clear");
  int i, j, sonuc;
  int dizi[5][5];
  printf("\nLutfen 5x5 matris elemanlarini giriniz!\n\n");
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) {
      printf("%d satir, %d sutun elemani giriniz: ", i + 1, j + 1);
      scanf(" %d", &dizi[i][j]);
    }
    printf("\n");
  }
  system("clear");
  printf("---Matris---\n\n");
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) {
      printf("%d\t", dizi[i][j]);
    }
    printf("\n");
  }
  printf("\n");
  sonuc = tamKaremi(dizi);
  printf("Sonuc=%d\n", sonuc);
  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
  push   %rbp
  mov    %rsp,%rbp
  mov    $0x0,%edi
  call   <thr1@plt>
  mov    $0x0,%edi
  call   <thr2@plt>
  mov    $0x0,%edi
  call   <thr3@plt>
  mov    $0x0,%edi
  call   <thr4@plt>
  mov    $0x0,%edi
  call   <thr5@plt>
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  thr1(0);
__CPROVER_ASYNC_1:
  thr2(0);
__CPROVER_ASYNC_2:
  thr3(0);
__CPROVER_ASYNC_3:
  thr4(0);
__CPROVER_ASYNC_4:
  thr5(0);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <thr1@plt>
  mov    $0x0,%edi
  call   <thr2@plt>
  mov    $0x0,%edi
  call   <thr3@plt>
  mov    $0x0,%edi
  call   <thr4@plt>
  mov    $0x0,%edi
  call   <thr5@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  thr1(0);
__CPROVER_ASYNC_1:
  thr2(0);
__CPROVER_ASYNC_2:
  thr3(0);
__CPROVER_ASYNC_3:
  thr4(0);
__CPROVER_ASYNC_4:
  thr5(0);
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <thr1@plt>
  xor    %edi,%edi
  call   <thr2@plt>
  xor    %edi,%edi
  call   <thr3@plt>
  xor    %edi,%edi
  call   <thr4@plt>
  xor    %edi,%edi
  call   <thr5@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  thr1(0);
__CPROVER_ASYNC_1:
  thr2(0);
__CPROVER_ASYNC_2:
  thr3(0);
__CPROVER_ASYNC_3:
  thr4(0);
__CPROVER_ASYNC_4:
  thr5(0);
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <thr1@plt>
  xor    %edi,%edi
  call   <thr2@plt>
  xor    %edi,%edi
  call   <thr3@plt>
  xor    %edi,%edi
  call   <thr4@plt>
  xor    %edi,%edi
  call   <thr5@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main() {
__CPROVER_ASYNC_0:
  thr1(0);
__CPROVER_ASYNC_1:
  thr2(0);
__CPROVER_ASYNC_2:
  thr3(0);
__CPROVER_ASYNC_3:
  thr4(0);
__CPROVER_ASYNC_4:
  thr5(0);
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<select>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %edx,-0x20(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L41
L22:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L10
L4:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    (%rax),%eax
  cmp    %eax,%edx
  jge    L20
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L20:
  addl   $0x1,-0x4(%rbp)
L10:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L4
  mov    -0xc(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <swap@plt>
  addl   $0x1,-0xc(%rbp)
L41:
  mov    -0xc(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jl     L22
  mov    -0x20(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  leave
  ret
```

assistant: ```c
int select(int *x, int len, int n) {
  int i, k, j;
  for (i = 0; i < n; i++) {
    k = i;
    for (j = i + 1; j < len; j++)
      if (x[j] < x[k])
        k = j;
    swap(x + k, x + i);
  }
  return (x[n - 1]);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<select>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %edx,%r15d
  test   %edx,%edx
  jle    L39
  mov    %esi,%ebp
  mov    %rdi,%r13
  lea    0x1(%rdx),%r14d
  mov    $0x1,%r12d
L13:
  lea    -0x1(%r12),%edx
  cmp    %r12d,%ebp
  jle    L23
  mov    %r12,%rax
L37:
  movslq %edx,%rcx
  mov    (%rbx,%rcx,4),%esi
  cmp    %esi,(%rbx,%rax,4)
  cmovl  %eax,%edx
  add    $0x1,%rax
  cmp    %eax,%ebp
  jg     L37
L23:
  movslq %edx,%rdx
  lea    (%rbx,%rdx,4),%rdi
  mov    %r13,%rsi
  call   <swap@plt>
  add    $0x4,%r13
  add    $0x1,%r12
  cmp    %r14,%r12
  jne    L13
L39:
  movslq %r15d,%r15
  mov    -0x4(%rbx,%r15,4),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
int select(int *x, int len, int n) {
  int i, k, j;
  for (i = 0; i < n; i++) {
    k = i;
    for (j = i + 1; j < len; j++)
      if (x[j] < x[k])
        k = j;
    swap(x + k, x + i);
  }
  return (x[n - 1]);
}
```
user: What is the c source code of the assembly code below:

```asm
<select>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  push   %rbx
  movslq %edx,%rbx
  test   %ebx,%ebx
  jle    L15
  mov    %esi,%ebp
  mov    %rdi,%r8
  xor    %r13d,%r13d
  xchg   %ax,%ax
L23:
  mov    %r13d,%esi
  add    $0x1,%r13d
  cmp    %ebp,%r13d
  jge    L13
  lea    0x4(%r8),%r14
  mov    %r13d,%eax
  mov    %r14,%rdx
  cs nopw 0x0(%rax,%rax,1)
L32:
  movslq %esi,%rcx
  lea    (%r12,%rcx,4),%rdi
  mov    (%rdi),%ecx
  cmp    %ecx,(%rdx)
  jge    L58
  mov    %rdx,%rdi
  mov    %eax,%esi
L58:
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    %eax,%ebp
  jne    L32
  mov    %r8,%rsi
  call   <swap@plt>
  mov    %r14,%r8
  cmp    %r13d,%ebx
  jne    L23
L15:
  mov    -0x4(%r12,%rbx,4),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  mov    %r8,%rdi
  lea    0x4(%r8),%r14
  mov    %r8,%rsi
  call   <swap@plt>
  mov    %r14,%r8
  cmp    %r13d,%ebx
  jne    L23
  jmp    L15
```

assistant: ```c
int select(int *x, int len, int n) {
  int i, k, j;
  for (i = 0; i < n; i++) {
    k = i;
    for (j = i + 1; j < len; j++)
      if (x[j] < x[k])
        k = j;
    swap(x + k, x + i);
  }
  return (x[n - 1]);
}
```
user: Translate the below assembly code into C programming language:

```asm
<select>:
  endbr64
  push   %r14
  push   %r13
  mov    %rdi,%r13
  push   %r12
  push   %rbp
  push   %rbx
  movslq %edx,%rbx
  test   %ebx,%ebx
  jle    L14
  mov    %esi,%ebp
  mov    %rdi,%r9
  xor    %r12d,%r12d
  xchg   %ax,%ax
L64:
  mov    %r12d,%r8d
  add    $0x1,%r12d
  cmp    %ebp,%r12d
  jge    L18
  lea    0x4(%r9),%r14
  mov    (%r9),%esi
  mov    %r12d,%edx
  mov    %r14,%rax
  jmp    L34
  nopl   0x0(%rax,%rax,1)
L46:
  movslq %r8d,%rcx
  add    $0x1,%edx
  add    $0x4,%rax
  lea    0x0(%r13,%rcx,4),%rdi
  cmp    %ebp,%edx
  je     L68
L34:
  mov    (%rax),%ecx
  mov    %rax,%rdi
  cmp    %esi,%ecx
  jge    L46
  mov    %edx,%r8d
  add    $0x1,%edx
  mov    %ecx,%esi
  add    $0x4,%rax
  cmp    %ebp,%edx
  jne    L34
L68:
  mov    %r9,%rsi
  call   <swap@plt>
  mov    %r14,%r9
  cmp    %r12d,%ebx
  jne    L64
L14:
  mov    -0x4(%r13,%rbx,4),%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %r9,%rdi
  lea    0x4(%r9),%r14
  mov    %r9,%rsi
  call   <swap@plt>
  mov    %r14,%r9
  cmp    %r12d,%ebx
  jne    L64
  jmp    L14
```

assistant: ```c
int select(int *x, int len, int n) {
  int i, k, j;
  for (i = 0; i < n; i++) {
    k = i;
    for (j = i + 1; j < len; j++)
      if (x[j] < x[k])
        k = j;
    swap(x + k, x + i);
  }
  return (x[n - 1]);
}
```
user: What is the c source code of the assembly code below?

```asm
<flipped_double>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movsd  D0(%rip),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  test   %eax,%eax
  jne    L0
  mov    -0x4(%rbp),%eax
  test   %eax,%eax
  je     L0
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L34
L0:
  mov    -0x8(%rbp),%eax
  test   %eax,%eax
  je     L34
  mov    -0x4(%rbp),%eax
  test   %eax,%eax
  jne    L34
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L34:
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D0
type: byte[8]
name: disassemble_data
arguments: label: D37
type: string
name: disassemble_data
arguments: label: D26
type: string
name: disassemble_data
arguments: label: D12
type: string

tool: {"D0": [0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x40]}

tool: D37: #define double_wordorder_bigendian_p 0


tool: D26: #define double_wordorder_bigendian_p 1


tool: D12: 


assistant: ```c
void flipped_double(void) {
  typedef struct {
    unsigned lo, hi;
  } dfloat;
  union {
    dfloat eksplicit;
    double machine_double;
  } x;
  x.machine_double = 2;
  dfloat test = x.eksplicit;
  if (test.lo == 0 && test.hi != 0) {
    printf("#define double_wordorder_bigendian_p 0\n");
  } else if (test.lo != 0 && test.hi == 0) {
    printf("#define double_wordorder_bigendian_p 1\n");
  } else {
  }
  printf("\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<flipped_double>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    D18(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type_infered: string
name: extract_symbolic_data
arguments: label: D18
type_infered: string

tool: D9: #define double_wordorder_bigendian_p 0


tool: D18: 


assistant: ```c
void flipped_double(void) {
  typedef struct {
    unsigned lo, hi;
  } dfloat;
  union {
    dfloat eksplicit;
    double machine_double;
  } x;
  x.machine_double = 2;
  dfloat test = x.eksplicit;
  if (test.lo == 0 && test.hi != 0) {
    printf("#define double_wordorder_bigendian_p 0\n");
  } else if (test.lo != 0 && test.hi == 0) {
    printf("#define double_wordorder_bigendian_p 1\n");
  } else {
  }
  printf("\n");
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<flipped_double>:
  endbr64
  sub    $0x8,%rsp
  lea    D9(%rip),%rdi
  call   <printf@plt>
  lea    D17(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D9
type: string
name: parse_data
arguments: data_tag: D17
type: string

tool: D9: #define double_wordorder_bigendian_p 0


tool: D17: 


assistant: ```c
void flipped_double(void) {
  typedef struct {
    unsigned lo, hi;
  } dfloat;
  union {
    dfloat eksplicit;
    double machine_double;
  } x;
  x.machine_double = 2;
  dfloat test = x.eksplicit;
  if (test.lo == 0 && test.hi != 0) {
    printf("#define double_wordorder_bigendian_p 0\n");
  } else if (test.lo != 0 && test.hi == 0) {
    printf("#define double_wordorder_bigendian_p 1\n");
  } else {
  }
  printf("\n");
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<flipped_double>:
  endbr64
  sub    $0x8,%rsp
  lea    D3(%rip),%rdi
  call   <printf@plt>
  lea    D10(%rip),%rdi
  add    $0x8,%rsp
  jmp    1050 <printf@plt>
```

tool_calls: name: read_data
arguments: tag: D3
type: string
name: read_data
arguments: tag: D10
type: string

tool: D3: #define double_wordorder_bigendian_p 0


tool: D10: 


assistant: ```c
void flipped_double(void) {
  typedef struct {
    unsigned lo, hi;
  } dfloat;
  union {
    dfloat eksplicit;
    double machine_double;
  } x;
  x.machine_double = 2;
  dfloat test = x.eksplicit;
  if (test.lo == 0 && test.hi != 0) {
    printf("#define double_wordorder_bigendian_p 0\n");
  } else if (test.lo != 0 && test.hi == 0) {
    printf("#define double_wordorder_bigendian_p 1\n");
  } else {
  }
  printf("\n");
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<calc_reverse>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %rax,-0x8(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L18
L4:
  mov    -0x14(%rbp),%eax
  cltq
  neg    %rax
  mov    %rax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x30(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x14(%rbp)
L18:
  mov    -0x14(%rbp),%eax
  cltq
  cmp    %rax,-0x10(%rbp)
  ja     L4
  mov    -0x30(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  nop
  leave
  ret
```

assistant: ```c
void calc_reverse(char *input, char *output) {
  size_t len_input = strlen(input);
  char *last_char = input + len_input - 1;
  for (int i = 0; i < len_input; i++) {
    output[i] = *(last_char - i);
  }
  output[len_input] = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<calc_reverse>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %rsi,%rbp
  call   <strlen@plt>
  test   %rax,%rax
  je     L8
  lea    -0x1(%rax),%rdx
  add    %rbx,%rdx
  mov    %rbp,%rcx
L14:
  movzbl (%rdx),%esi
  mov    %sil,(%rcx)
  mov    %rdx,%rsi
  sub    $0x1,%rdx
  add    $0x1,%rcx
  cmp    %rbx,%rsi
  jne    L14
L8:
  movb   $0x0,0x0(%rbp,%rax,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void calc_reverse(char *input, char *output) {
  size_t len_input = strlen(input);
  char *last_char = input + len_input - 1;
  for (int i = 0; i < len_input; i++) {
    output[i] = *(last_char - i);
  }
  output[len_input] = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<calc_reverse>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %rax,%rax
  je     L11
  lea    -0x1(%rax),%rdx
  mov    %rbp,%rcx
  add    %rbx,%rdx
  nopl   0x0(%rax)
L9:
  movzbl (%rdx),%r8d
  add    $0x1,%rcx
  mov    %r8b,-0x1(%rcx)
  mov    %rdx,%r8
  sub    $0x1,%rdx
  cmp    %rbx,%r8
  jne    L9
L11:
  movb   $0x0,0x0(%rbp,%rax,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void calc_reverse(char *input, char *output) {
  size_t len_input = strlen(input);
  char *last_char = input + len_input - 1;
  for (int i = 0; i < len_input; i++) {
    output[i] = *(last_char - i);
  }
  output[len_input] = '\0';
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<calc_reverse>:
  endbr64
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  test   %rax,%rax
  je     L17
  lea    -0x1(%rax),%rdx
  mov    %rbp,%rcx
  add    %rbx,%rdx
  nopl   0x0(%rax)
L16:
  movzbl (%rdx),%r8d
  add    $0x1,%rcx
  mov    %r8b,-0x1(%rcx)
  mov    %rdx,%r8
  sub    $0x1,%rdx
  cmp    %rbx,%r8
  jne    L16
L17:
  movb   $0x0,0x0(%rbp,%rax,1)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void calc_reverse(char *input, char *output) {
  size_t len_input = strlen(input);
  char *last_char = input + len_input - 1;
  for (int i = 0; i < len_input; i++) {
    output[i] = *(last_char - i);
  }
  output[len_input] = '\0';
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  lea    D28(%rip),%rsi
  mov    $0x1,%edi
  call   <ece391_fdputs@plt>
  lea    D13(%rip),%rbp
  lea    D5(%rip),%r12
  lea    D53(%rip),%r13
L7:
  mov    %rsp,%rbx
L51:
  mov    %rbp,%rsi
  mov    $0x1,%edi
  call   <ece391_fdputs@plt>
  mov    $0x3ff,%edx
  mov    %rbx,%rsi
  mov    $0x0,%edi
  call   <ece391_read@plt>
  cmp    $0xffffffff,%eax
  je     L63
  test   %eax,%eax
  jle    L60
  lea    -0x1(%rax),%edx
  movslq %edx,%rcx
  cmpb   $0xa,(%rsp,%rcx,1)
  cmove  %edx,%eax
L60:
  cltq
  movb   $0x0,(%rsp,%rax,1)
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <ece391_strcmp@plt>
  test   %rax,%rax
  je     L24
  cmpb   $0x0,(%rsp)
  je     L51
  mov    %rsp,%rdi
  call   <ece391_execute@plt>
  cmp    $0xffffffff,%eax
  je     L71
  cmp    $0x100,%eax
  je     L90
  test   %eax,%eax
  je     L7
  mov    %r13,%rsi
  mov    $0x1,%edi
  call   <ece391_fdputs@plt>
  jmp    L7
L63:
  lea    D51(%rip),%rsi
  mov    $0x1,%edi
  call   <ece391_fdputs@plt>
  mov    $0x3,%eax
L113:
  mov    0x408(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x418,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L71:
  lea    D22(%rip),%rsi
  mov    $0x1,%edi
  call   <ece391_fdputs@plt>
  jmp    L7
L90:
  lea    D31(%rip),%rsi
  mov    $0x1,%edi
  call   <ece391_fdputs@plt>
  jmp    L7
L24:
  mov    $0x0,%eax
  jmp    L113
L8:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D28
data_type: string
name: read_data
arguments: data_label: D13
data_type: string
name: read_data
arguments: data_label: D5
data_type: string
name: read_data
arguments: data_label: D53
data_type: string
name: read_data
arguments: data_label: D51
data_type: string
name: read_data
arguments: data_label: D22
data_type: string
name: read_data
arguments: data_label: D31
data_type: string

tool: D28: Starting 391 Shell


tool: D13: 391OS> 

tool: D5: exit

tool: D53: program terminated abnormally


tool: D51: read from keyboard failed


tool: D22: no such command


tool: D31: program terminated by exception


assistant: ```c
int main() {
  int32_t cnt, rval;
  uint8_t buf[1024];
  ece391_fdputs(1, (uint8_t *)"Starting 391 Shell\n");
  while (1) {
    ece391_fdputs(1, (uint8_t *)"391OS> ");
    if (-1 == (cnt = ece391_read(0, buf, 1024 - 1))) {
      ece391_fdputs(1, (uint8_t *)"read from keyboard failed\n");
      return 3;
    }
    if (cnt > 0 && '\n' == buf[cnt - 1])
      cnt--;
    buf[cnt] = '\0';
    if (0 == ece391_strcmp(buf, (uint8_t *)"exit"))
      return 0;
    if ('\0' == buf[0])
      continue;
    rval = ece391_execute(buf);
    if (-1 == rval)
      ece391_fdputs(1, (uint8_t *)"no such command\n");
    else if (256 == rval)
      ece391_fdputs(1, (uint8_t *)"program terminated by exception\n");
    else if (0 != rval)
      ece391_fdputs(1, (uint8_t *)"program terminated abnormally\n");
  }
}
```