ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Demonstrate the C code that generates the assembly output below:

```asm
<identity>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  lea    D0(%rip),%rdi
  call   <puts@plt>
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  lea    D29(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x0,0x8(%rsp)
  jle    L35
  lea    0x10(%rsp),%r14
  mov    $0x0,%r13d
  lea    D36(%rip),%r12
  jmp    L3
L119:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  add    $0x1,%ebx
  add    $0x4,%rbp
  cmp    %ebx,0xc(%rsp)
  jg     L119
L129:
  add    $0x1,%r13
  add    $0x28,%r14
  cmp    %r13d,0x8(%rsp)
  jle    L35
L3:
  mov    %r14,%rbp
  mov    $0x0,%ebx
  cmpl   $0x0,0xc(%rsp)
  jg     L119
  jmp    L129
L35:
  lea    D27(%rip),%rdi
  call   <puts@plt>
  cmpl   $0x0,0x8(%rsp)
  jle    L65
  lea    0x10(%rsp),%rbp
  mov    $0x0,%r13d
  lea    D32(%rip),%r12
  jmp    L58
L5:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %ebx,0xc(%rsp)
  jg     L5
L48:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r13d
  add    $0x28,%rbp
  cmp    %r13d,0x8(%rsp)
  jle    L65
L58:
  mov    $0x0,%ebx
  cmpl   $0x0,0xc(%rsp)
  jg     L5
  jmp    L48
L65:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L117
  add    $0x1b0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L117:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D0
datatype: string
name: read_data
arguments: data_tag: D1
datatype: string
name: read_data
arguments: data_tag: D29
datatype: string
name: read_data
arguments: data_tag: D36
datatype: string
name: read_data
arguments: data_tag: D27
datatype: string
name: read_data
arguments: data_tag: D32
datatype: string

tool: D0: Enter the order of the matrix A 

tool: D1: %d %d

tool: D29: Enter the elements of matrix A 

tool: D36: %d

tool: D27: MATRIX A is 

tool: D32: %3d

assistant: ```c
void identity() {
  int a[10][10];
  int i, j, row, column, flag = 1;
  printf("Enter the order of the matrix A \n");
  scanf("%d %d", &row, &column);
  printf("Enter the elements of matrix A \n");
  for (i = 0; i < row; i++) {
    for (j = 0; j < column; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  printf("MATRIX A is \n");
  for (i = 0; i < row; i++) {
    for (j = 0; j < column; j++) {
      printf("%3d", a[i][j]);
    }
    printf("\n");
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<identity>:
  endbr64
  push   %r14
  lea    D33(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0x8(%rsp),%rsi
  lea    0xc(%rsp),%rdx
  xor    %eax,%eax
  lea    D56(%rip),%rdi
  call   <scanf@plt>
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%esi
  test   %esi,%esi
  jle    L18
  lea    0x10(%rsp),%r14
  xor    %r13d,%r13d
  lea    D22(%rip),%r12
  nopl   (%rax)
L24:
  mov    0xc(%rsp),%ecx
  mov    %r14,%rbp
  xor    %ebx,%ebx
  test   %ecx,%ecx
  jle    L87
  nopl   (%rax)
L39:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <scanf@plt>
  add    $0x4,%rbp
  cmp    %ebx,0xc(%rsp)
  jg     L39
L87:
  add    $0x1,%r13
  add    $0x28,%r14
  cmp    %r13d,0x8(%rsp)
  jg     L24
L18:
  lea    D29(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%edx
  test   %edx,%edx
  jle    L22
  lea    0x10(%rsp),%rbp
  xor    %r13d,%r13d
  lea    D39(%rip),%r12
  nopl   0x0(%rax)
L48:
  mov    0xc(%rsp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L88
  nopw   0x0(%rax,%rax,1)
L74:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L74
L88:
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    $0x28,%rbp
  call   <putchar@plt>
  cmp    %r13d,0x8(%rsp)
  jg     L48
L22:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L64
  add    $0x1b0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L64:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D33
type_infered: string
name: parse_data
arguments: data_tag: D56
type_infered: string
name: parse_data
arguments: data_tag: D19
type_infered: string
name: parse_data
arguments: data_tag: D22
type_infered: string
name: parse_data
arguments: data_tag: D29
type_infered: string
name: parse_data
arguments: data_tag: D39
type_infered: string

tool: D33: Enter the order of the matrix A 

tool: D56: %d %d

tool: D19: Enter the elements of matrix A 

tool: D22: %d

tool: D29: MATRIX A is 

tool: D39: %3d

assistant: ```c
void identity() {
  int a[10][10];
  int i, j, row, column, flag = 1;
  printf("Enter the order of the matrix A \n");
  scanf("%d %d", &row, &column);
  printf("Enter the elements of matrix A \n");
  for (i = 0; i < row; i++) {
    for (j = 0; j < column; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  printf("MATRIX A is \n");
  for (i = 0; i < row; i++) {
    for (j = 0; j < column; j++) {
      printf("%3d", a[i][j]);
    }
    printf("\n");
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<identity>:
  endbr64
  push   %r14
  lea    D31(%rip),%rdi
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1b0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1a8(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    0xc(%rsp),%rdx
  lea    0x8(%rsp),%rsi
  xor    %eax,%eax
  lea    D7(%rip),%rdi
  call   <scanf@plt>
  lea    D30(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%ecx
  test   %ecx,%ecx
  jle    L91
  mov    0xc(%rsp),%eax
  xor    %r14d,%r14d
  xor    %r13d,%r13d
  lea    D4(%rip),%r12
  test   %eax,%eax
  jle    L91
  nopl   0x0(%rax,%rax,1)
L22:
  lea    0x10(%rsp,%r14,4),%rbp
  xor    %ebx,%ebx
  nop
L71:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  add    $0x4,%rbp
  cmp    %ebx,%eax
  jg     L71
  add    $0x1,%r13d
  cmp    0x8(%rsp),%r13d
  jge    L91
  add    $0xa,%r14
  test   %eax,%eax
  jg     L22
L91:
  lea    D33(%rip),%rdi
  call   <puts@plt>
  mov    0x8(%rsp),%edx
  test   %edx,%edx
  jle    L43
  lea    0x10(%rsp),%rbp
  xor    %r13d,%r13d
  lea    D51(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L85:
  mov    0xc(%rsp),%eax
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L56
  nopw   0x0(%rax,%rax,1)
L53:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L53
L56:
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    $0x28,%rbp
  call   <putchar@plt>
  cmp    %r13d,0x8(%rsp)
  jg     L85
L43:
  mov    0x1a8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L77
  add    $0x1b0,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L77:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D31
type_infered: string
name: retrieve_data
arguments: label: D7
type_infered: string
name: retrieve_data
arguments: label: D30
type_infered: string
name: retrieve_data
arguments: label: D4
type_infered: string
name: retrieve_data
arguments: label: D33
type_infered: string
name: retrieve_data
arguments: label: D51
type_infered: string

tool: D31: Enter the order of the matrix A 

tool: D7: %d %d

tool: D30: Enter the elements of matrix A 

tool: D4: %d

tool: D33: MATRIX A is 

tool: D51: %3d

assistant: ```c
void identity() {
  int a[10][10];
  int i, j, row, column, flag = 1;
  printf("Enter the order of the matrix A \n");
  scanf("%d %d", &row, &column);
  printf("Enter the elements of matrix A \n");
  for (i = 0; i < row; i++) {
    for (j = 0; j < column; j++) {
      scanf("%d", &a[i][j]);
    }
  }
  printf("MATRIX A is \n");
  for (i = 0; i < row; i++) {
    for (j = 0; j < column; j++) {
      printf("%3d", a[i][j]);
    }
    printf("\n");
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<strcasecmp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
L26:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %eax,%ebx
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <tolower@plt>
  mov    %eax,%edx
  mov    %ebx,%eax
  sub    %edx,%eax
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L9
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L9
  mov    -0x30(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L36
L9:
  mov    -0x14(%rbp),%eax
  jmp    L45
L36:
  addq   $0x1,-0x28(%rbp)
  addq   $0x1,-0x30(%rbp)
  jmp    L26
L45:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int strcasecmp(const char *str1, const char *str2) {
  while (true) {
    int diff = tolower(*str1) - tolower(*str2);
    if (diff != 0 || !*str1 || !*str2) {
      return diff;
    }
    str1 += 1;
    str2 += 1;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<strcasecmp>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbp
  mov    %rsi,%r12
  jmp    L21
L19:
  add    $0x1,%rbp
  add    $0x1,%r12
L21:
  movsbl 0x0(%rbp),%edi
  call   <tolower@plt>
  mov    %eax,%ebx
  movsbl (%r12),%edi
  call   <tolower@plt>
  sub    %eax,%ebx
  jne    L0
  cmpb   $0x0,0x0(%rbp)
  je     L0
  cmpb   $0x0,(%r12)
  jne    L19
L0:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *str1, const char *str2) {
  while (true) {
    int diff = tolower(*str1) - tolower(*str2);
    if (diff != 0 || !*str1 || !*str2) {
      return diff;
    }
    str1 += 1;
    str2 += 1;
  }
}
```
user: What is the c source code of the assembly code below:

```asm
<strcasecmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  jmp    L20
L30:
  cmpb   $0x0,0x0(%rbp)
  je     L2
  cmpb   $0x0,(%r12)
  je     L2
  add    $0x1,%rbp
  add    $0x1,%r12
L20:
  movsbl 0x0(%rbp),%edi
  call   <tolower@plt>
  movsbl (%r12),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  sub    %eax,%ebx
  je     L30
L2:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *str1, const char *str2) {
  while (true) {
    int diff = tolower(*str1) - tolower(*str2);
    if (diff != 0 || !*str1 || !*str2) {
      return diff;
    }
    str1 += 1;
    str2 += 1;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<strcasecmp>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  jmp    L4
L5:
  cmpb   $0x0,0x0(%rbp)
  je     L29
  cmpb   $0x0,(%r12)
  je     L29
  add    $0x1,%rbp
  add    $0x1,%r12
L4:
  movsbl 0x0(%rbp),%edi
  call   <tolower@plt>
  movsbl (%r12),%edi
  mov    %eax,%ebx
  call   <tolower@plt>
  sub    %eax,%ebx
  je     L5
L29:
  mov    %ebx,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
int strcasecmp(const char *str1, const char *str2) {
  while (true) {
    int diff = tolower(*str1) - tolower(*str2);
    if (diff != 0 || !*str1 || !*str2) {
      return diff;
    }
    str1 += 1;
    str2 += 1;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x1,-0xc(%rbp)
  jmp    L11
L14:
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0xc(%rbp),%xmm0
  movss  D14(%rip),%xmm1
  divss  %xmm1,%xmm0
  movd   %xmm0,%eax
  movd   %eax,%xmm0
  call   <sqr@plt>
  movsd  D8(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,-0x8(%rbp)
  movss  -0x8(%rbp),%xmm0
  movaps %xmm0,%xmm1
  addss  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0xc(%rbp),%xmm0
  mulss  %xmm1,%xmm0
  cvtss2sd %xmm0,%xmm0
  movsd  D37(%rip),%xmm1
  divsd  %xmm1,%xmm0
  cvttsd2si %xmm0,%eax
  mov    %eax,%edi
  call   <sin@plt>
  imul   $0x64,%eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x8(%rbp),%xmm0
  mov    -0x4(%rbp),%eax
  movaps %xmm0,%xmm1
  movd   %eax,%xmm0
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0xc(%rbp)
L11:
  cmpl   $0x3fff,-0xc(%rbp)
  jle    L14
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D14
type: f32
name: extract_symbolic_data
arguments: data_tag: D8
type: f64
name: extract_symbolic_data
arguments: data_tag: D37
type: f64
name: extract_symbolic_data
arguments: data_tag: D32
type: string

tool: D14: 16384

tool: D8: 128

tool: D37: 512

tool: D32: %f %f


assistant: ```c
int main() {
  int i;
  for (i = 1; i < 16384; i++) {
    float j = 128.0 * sqr(((float)i) / 16384.0);
    float v = 100 * sin(2 * j * i / 512.0);
    printf("%f %f\n", v, j);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    $0x1,%ebx
  lea    D9(%rip),%rbp
L3:
  pxor   %xmm3,%xmm3
  cvtsi2ss %ebx,%xmm3
  movss  D15(%rip),%xmm0
  movss  %xmm3,0x8(%rsp)
  mulss  %xmm3,%xmm0
  call   <sqr@plt>
  mulsd  D23(%rip),%xmm0
  cvtsd2ss %xmm0,%xmm0
  movss  %xmm0,0xc(%rsp)
  addss  %xmm0,%xmm0
  mulss  0x8(%rsp),%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  D38(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <sin@plt>
  imul   $0x64,%eax,%eax
  pxor   %xmm0,%xmm0
  cvtsi2ss %eax,%xmm0
  movss  0xc(%rsp),%xmm1
  mov    %rbp,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    $0x4000,%ebx
  jne    L3
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D9
type_infered: string
name: unpack_data
arguments: data_tag: D15
type_infered: byte[4]
name: unpack_data
arguments: data_tag: D23
type_infered: f64
name: unpack_data
arguments: data_tag: D38
type_infered: byte[8]

tool: D9: %f %f


tool: {"D15": [0x00,0x00,0x80,0x38]}

tool: D23: 128

tool: {"D38": [0x00,0x00,0x00,0x00,0x00,0x00,0x60,0x3f]}

assistant: ```c
int main() {
  int i;
  for (i = 1; i < 16384; i++) {
    float j = 128.0 * sqr(((float)i) / 16384.0);
    float v = 100 * sin(2 * j * i / 512.0);
    printf("%f %f\n", v, j);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  lea    D16(%rip),%rbx
  sub    $0x18,%rsp
  cs nopw 0x0(%rax,%rax,1)
L7:
  pxor   %xmm2,%xmm2
  movss  D5(%rip),%xmm0
  cvtsi2ss %ebp,%xmm2
  add    $0x1,%ebp
  mulss  %xmm2,%xmm0
  movss  %xmm2,0xc(%rsp)
  call   <sqr@plt>
  movsd  D18(%rip),%xmm1
  movss  0xc(%rsp),%xmm2
  mulsd  %xmm0,%xmm1
  cvtsd2ss %xmm1,%xmm1
  movaps %xmm1,%xmm0
  movss  %xmm1,0x8(%rsp)
  addss  %xmm1,%xmm0
  mulss  %xmm2,%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  D2(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <sin@plt>
  pxor   %xmm0,%xmm0
  movss  0x8(%rsp),%xmm1
  mov    %rbx,%rdi
  imul   $0x64,%eax,%eax
  cvtsi2ss %eax,%xmm0
  call   <printf@plt>
  cmp    $0x4000,%ebp
  jne    L7
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D16
type_infered: string
name: retrieve_data
arguments: tag: D5
type_infered: byte[4]
name: retrieve_data
arguments: tag: D18
type_infered: f64
name: retrieve_data
arguments: tag: D2
type_infered: byte[8]

tool: D16: %f %f


tool: {"D5": [0x00,0x00,0x80,0x38]}

tool: D18: 128

tool: {"D2": [0x00,0x00,0x00,0x00,0x00,0x00,0x60,0x3f]}

assistant: ```c
int main() {
  int i;
  for (i = 1; i < 16384; i++) {
    float j = 128.0 * sqr(((float)i) / 16384.0);
    float v = 100 * sin(2 * j * i / 512.0);
    printf("%f %f\n", v, j);
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %rbp
  mov    $0x1,%ebp
  push   %rbx
  lea    D2(%rip),%rbx
  sub    $0x18,%rsp
  cs nopw 0x0(%rax,%rax,1)
L6:
  pxor   %xmm2,%xmm2
  movss  D13(%rip),%xmm0
  cvtsi2ss %ebp,%xmm2
  add    $0x1,%ebp
  mulss  %xmm2,%xmm0
  movss  %xmm2,0xc(%rsp)
  call   <sqr@plt>
  movsd  D0(%rip),%xmm1
  movss  0xc(%rsp),%xmm2
  mulsd  %xmm0,%xmm1
  cvtsd2ss %xmm1,%xmm1
  movaps %xmm1,%xmm0
  movss  %xmm1,0x8(%rsp)
  addss  %xmm1,%xmm0
  mulss  %xmm2,%xmm0
  cvtss2sd %xmm0,%xmm0
  mulsd  D25(%rip),%xmm0
  cvttsd2si %xmm0,%edi
  call   <sin@plt>
  pxor   %xmm0,%xmm0
  movss  0x8(%rsp),%xmm1
  mov    %rbx,%rdi
  imul   $0x64,%eax,%eax
  cvtsi2ss %eax,%xmm0
  call   <printf@plt>
  cmp    $0x4000,%ebp
  jne    L6
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D2
data_type: string
name: disassemble_data
arguments: tag: D13
data_type: byte[4]
name: disassemble_data
arguments: tag: D0
data_type: f64
name: disassemble_data
arguments: tag: D25
data_type: byte[8]

tool: D2: %f %f


tool: {"D13": [0x00,0x00,0x80,0x38]}

tool: D0: 128

tool: {"D25": [0x00,0x00,0x00,0x00,0x00,0x00,0x60,0x3f]}

assistant: ```c
int main() {
  int i;
  for (i = 1; i < 16384; i++) {
    float j = 128.0 * sqr(((float)i) / 16384.0);
    float v = 100 * sin(2 * j * i / 512.0);
    printf("%f %f\n", v, j);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<TO_LE16>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x18(%rbp)
  lea    -0x18(%rbp),%rax
  mov    %rax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  jmp    L24
L16:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    -0x24(%rbp),%edx
  mov    %dl,(%rax)
  addl   $0x1,-0x14(%rbp)
  sarl   $0x8,-0x24(%rbp)
L24:
  mov    -0x14(%rbp),%eax
  cmp    $0x3,%eax
  jbe    L16
  mov    -0x18(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

assistant: ```c
uint16_t TO_LE16(uint16_t in) {
  uint16_t out = 0;
  unsigned char *pout = (unsigned char *)&out;
  int i;
  for (i = 0; i < sizeof(in); ++i, in >>= 8)
    pout[i] = in & 0xff;
  return out;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<TO_LE16>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rax
  lea    0x8(%rsp),%rdx
L10:
  mov    %dil,(%rax)
  sar    $0x8,%edi
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L10
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L19
  add    $0x18,%rsp
  ret
L19:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint16_t TO_LE16(uint16_t in) {
  uint16_t out = 0;
  unsigned char *pout = (unsigned char *)&out;
  int i;
  for (i = 0; i < sizeof(in); ++i, in >>= 8)
    pout[i] = in & 0xff;
  return out;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<TO_LE16>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rax
  lea    0x8(%rsp),%rdx
L6:
  mov    %dil,(%rax)
  add    $0x1,%rax
  sar    $0x8,%edi
  cmp    %rdx,%rax
  jne    L6
  mov    0x4(%rsp),%eax
  mov    0x8(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
uint16_t TO_LE16(uint16_t in) {
  uint16_t out = 0;
  unsigned char *pout = (unsigned char *)&out;
  int i;
  for (i = 0; i < sizeof(in); ++i, in >>= 8)
    pout[i] = in & 0xff;
  return out;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<TO_LE16>:
  endbr64
  mov    %edi,%eax
  ret
```

assistant: ```c
uint16_t TO_LE16(uint16_t in) {
  uint16_t out = 0;
  unsigned char *pout = (unsigned char *)&out;
  int i;
  for (i = 0; i < sizeof(in); ++i, in >>= 8)
    pout[i] = in & 0xff;
  return out;
}
```
user: What is the c source code of the assembly code below?

```asm
<test_00>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L6:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L6
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int test_00(int *x, int c) {
  int s = 0;
  int i;
  for (i = 0; i < c; ++i)
    s += x[i];
  return s;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<test_00>:
  endbr64
  test   %esi,%esi
  jle    L28
  mov    %rdi,%rax
  lea    -0x1(%rsi),%edx
  lea    0x4(%rdi,%rdx,4),%rcx
  mov    $0x0,%edx
L23:
  add    (%rax),%edx
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L23
L7:
  mov    %edx,%eax
  ret
L28:
  mov    $0x0,%edx
  jmp    L7
```

assistant: ```c
int test_00(int *x, int c) {
  int s = 0;
  int i;
  for (i = 0; i < c; ++i)
    s += x[i];
  return s;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<test_00>:
  endbr64
  test   %esi,%esi
  jle    L15
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rdx
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L8:
  add    (%rdi),%eax
  add    $0x4,%rdi
  cmp    %rdx,%rdi
  jne    L8
  ret
  nopl   0x0(%rax)
L15:
  xor    %eax,%eax
  ret
```

assistant: ```c
int test_00(int *x, int c) {
  int s = 0;
  int i;
  for (i = 0; i < c; ++i)
    s += x[i];
  return s;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<test_00>:
  endbr64
  mov    %rdi,%rdx
  test   %esi,%esi
  jle    L69
  lea    -0x1(%rsi),%eax
  cmp    $0x2,%eax
  jbe    L56
  mov    %esi,%ecx
  mov    %rdi,%rax
  pxor   %xmm0,%xmm0
  shr    $0x2,%ecx
  sub    $0x1,%ecx
  shl    $0x4,%rcx
  lea    0x10(%rdi,%rcx,1),%rcx
  nopl   0x0(%rax,%rax,1)
L1:
  movdqu (%rax),%xmm2
  add    $0x10,%rax
  paddd  %xmm2,%xmm0
  cmp    %rcx,%rax
  jne    L1
  movdqa %xmm0,%xmm1
  mov    %esi,%ecx
  psrldq $0x8,%xmm1
  and    $0xfffffffc,%ecx
  paddd  %xmm1,%xmm0
  movdqa %xmm0,%xmm1
  psrldq $0x4,%xmm1
  paddd  %xmm1,%xmm0
  movd   %xmm0,%eax
  test   $0x3,%sil
  je     L57
L41:
  movslq %ecx,%rdi
  lea    0x0(,%rdi,4),%r8
  add    (%rdx,%rdi,4),%eax
  lea    0x1(%rcx),%edi
  cmp    %edi,%esi
  jle    L18
  add    $0x2,%ecx
  add    0x4(%rdx,%r8,1),%eax
  cmp    %ecx,%esi
  jle    L18
  add    0x8(%rdx,%r8,1),%eax
  ret
  nopl   0x0(%rax)
L69:
  xor    %eax,%eax
L18:
  ret
  nopl   0x0(%rax,%rax,1)
L57:
  ret
L56:
  xor    %ecx,%ecx
  xor    %eax,%eax
  jmp    L41
```

assistant: ```c
int test_00(int *x, int c) {
  int s = 0;
  int i;
  for (i = 0; i < c; ++i)
    s += x[i];
  return s;
}
```