ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r15
  xor    %edi,%edi
  lea    D15(%rip),%r15
  push   %r14
  lea    D17(%rip),%r14
  push   %r13
  lea    D14(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  call   <time@plt>
  mov    %eax,%edi
  call   <srand@plt>
  xor    %eax,%eax
  call   <print_version@plt>
  movl   $0x3,0x8(%rsp)
  movl   $0x1,0xc(%rsp)
  nopl   0x0(%rax)
L15:
  mov    0xc(%rsp),%ebp
  mov    0x8(%rsp),%eax
  mov    $0xa,%edi
  mov    %ebp,%esi
  lea    -0x3(%rax),%r12d
  call   <pow@plt>
  lea    -0x1(%rbp),%esi
  mov    $0xa,%edi
  mov    %eax,%ebx
  call   <pow@plt>
  mov    %eax,%r8d
  mov    $0xf4240,%eax
  cltd
  idiv   %r8d
  mov    %eax,%ebp
L11:
  mov    %ebx,%edx
  mov    %r15,%rsi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %ebx,%esi
  mov    %ebp,%edi
  add    $0x1,%r12d
  call   <benchmark@plt>
  mov    %ebp,%edi
  call   <print_results@plt>
  xor    %eax,%eax
  mov    %r12d,%esi
  mov    %r13,%rdi
  call   <printf@plt>
  cmp    0x8(%rsp),%r12d
  jne    L11
  lea    0x3(%r12),%eax
  addl   $0x1,0xc(%rsp)
  mov    %eax,0x8(%rsp)
  cmp    $0x12,%r12d
  jne    L15
  mov    $0x12,%esi
  mov    $0x12,%edi
  call   <print_summary@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D15
type_infered: string
name: retrieve_data
arguments: tag: D17
type_infered: string
name: retrieve_data
arguments: tag: D14
type_infered: string

tool: D15: dvariancepn

tool: D17: # c::%s:len=%d


tool: D14: ok %d benchmark finished


assistant: ```c
int main(void) {
  double elapsed;
  int count;
  int iter;
  int len;
  int i;
  int j;
  srand(time(NULL));
  print_version();
  count = 0;
  for (i = 1; i <= 6; i++) {
    len = pow(10, i);
    iter = 1000000 / pow(10, i - 1);
    for (j = 0; j < 3; j++) {
      count += 1;
      printf("# c::%s:len=%d\n", "dvariancepn", len);
      elapsed = benchmark(iter, len);
      print_results(iter, elapsed);
      printf("ok %d benchmark finished\n", count);
    }
  }
  print_summary(count, count);
}
```
user: Translate the below assembly code into C programming language:

```asm
<checksum>:
  endbr64
  mov    %rdi,%rdx
  mov    %rsi,%r10
  cmp    $0xf,%rsi
  jbe    L59
  lea    -0x10(%rsi),%rax
  and    $0xfffffffffffffff0,%rax
  lea    0x10(%rdi,%rax,1),%rsi
  mov    $0x0,%r9d
  mov    $0x0,%edi
  mov    $0x0,%r8d
  mov    $0x0,%eax
L85:
  movzbl (%rdx),%ecx
  movzbl 0x4(%rdx),%r11d
  add    %r11d,%ecx
  movzbl 0x8(%rdx),%r11d
  add    %r11d,%ecx
  movzbl 0xc(%rdx),%r11d
  add    %r11d,%ecx
  add    %ecx,%eax
  movzbl 0x1(%rdx),%ecx
  movzbl 0x5(%rdx),%r11d
  add    %r11d,%ecx
  movzbl 0x9(%rdx),%r11d
  add    %r11d,%ecx
  movzbl 0xd(%rdx),%r11d
  add    %r11d,%ecx
  add    %ecx,%r8d
  movzbl 0x2(%rdx),%ecx
  movzbl 0x6(%rdx),%r11d
  add    %r11d,%ecx
  movzbl 0xa(%rdx),%r11d
  add    %r11d,%ecx
  movzbl 0xe(%rdx),%r11d
  add    %r11d,%ecx
  add    %ecx,%edi
  movzbl 0x3(%rdx),%ecx
  movzbl 0x7(%rdx),%r11d
  add    %r11d,%ecx
  movzbl 0xb(%rdx),%r11d
  add    %r11d,%ecx
  movzbl 0xf(%rdx),%r11d
  add    %r11d,%ecx
  add    %ecx,%r9d
  add    $0x10,%rdx
  cmp    %rsi,%rdx
  jne    L85
  and    $0xf,%r10d
L11:
  mov    %rsi,%rdx
  cmp    $0x3,%r10
  jbe    L52
  lea    -0x4(%r10),%rdx
  shr    $0x2,%rdx
  lea    0x4(%rsi,%rdx,4),%rdx
L37:
  movzbl (%rsi),%ecx
  add    %ecx,%eax
  movzbl 0x1(%rsi),%ecx
  add    %ecx,%r8d
  movzbl 0x2(%rsi),%ecx
  add    %ecx,%edi
  movzbl 0x3(%rsi),%ecx
  add    %ecx,%r9d
  add    $0x4,%rsi
  cmp    %rdx,%rsi
  jne    L37
  and    $0x3,%r10d
L52:
  shl    $0x8,%edi
  shl    $0x10,%r8d
  add    %r8d,%edi
  add    %r9d,%edi
  shl    $0x18,%eax
  add    %edi,%eax
  cmp    $0x2,%r10
  je     L4
  cmp    $0x3,%r10
  je     L51
  cmp    $0x1,%r10
  jne    L21
  jmp    L66
L59:
  mov    %rdi,%rsi
  mov    $0x0,%r9d
  mov    $0x0,%edi
  mov    $0x0,%r8d
  mov    $0x0,%eax
  jmp    L11
L51:
  movzbl 0x2(%rdx),%ecx
  shl    $0x8,%ecx
  add    %ecx,%eax
L4:
  movzbl 0x1(%rdx),%ecx
  shl    $0x10,%ecx
  add    %ecx,%eax
L66:
  movzbl (%rdx),%edx
  shl    $0x18,%edx
  add    %edx,%eax
L21:
  ret
```

assistant: ```c
unsigned int checksum(const char *zIn, size_t N) {
  const unsigned char *z = (const unsigned char *)zIn;
  unsigned sum0 = 0;
  unsigned sum1 = 0;
  unsigned sum2 = 0;
  unsigned sum3 = 0;
  while (N >= 16) {
    sum0 += ((unsigned)z[0] + z[4] + z[8] + z[12]);
    sum1 += ((unsigned)z[1] + z[5] + z[9] + z[13]);
    sum2 += ((unsigned)z[2] + z[6] + z[10] + z[14]);
    sum3 += ((unsigned)z[3] + z[7] + z[11] + z[15]);
    z += 16;
    N -= 16;
  }
  while (N >= 4) {
    sum0 += z[0];
    sum1 += z[1];
    sum2 += z[2];
    sum3 += z[3];
    z += 4;
    N -= 4;
  }
  sum3 += (sum2 << 8) + (sum1 << 16) + (sum0 << 24);
  switch (N) {
  case 3:
    sum3 += (z[2] << 8);
  case 2:
    sum3 += (z[1] << 16);
  case 1:
    sum3 += (z[0] << 24);
  default:;
  }
  return sum3;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<checksum>:
  endbr64
  mov    %rdi,%rdx
  mov    %rsi,%r10
  cmp    $0xf,%rsi
  jbe    L10
  lea    -0x10(%rsi),%rax
  xor    %r9d,%r9d
  xor    %r8d,%r8d
  and    $0xfffffffffffffff0,%rax
  lea    0x10(%rdi,%rax,1),%rsi
  xor    %edi,%edi
  xor    %eax,%eax
  nopl   0x0(%rax,%rax,1)
L75:
  movzbl 0x4(%rdx),%r11d
  movzbl (%rdx),%ecx
  add    $0x10,%rdx
  add    %r11d,%ecx
  movzbl -0x8(%rdx),%r11d
  add    %r11d,%ecx
  movzbl -0x4(%rdx),%r11d
  add    %r11d,%ecx
  movzbl -0xb(%rdx),%r11d
  add    %ecx,%eax
  movzbl -0xf(%rdx),%ecx
  add    %r11d,%ecx
  movzbl -0x7(%rdx),%r11d
  add    %r11d,%ecx
  movzbl -0x3(%rdx),%r11d
  add    %r11d,%ecx
  movzbl -0xa(%rdx),%r11d
  add    %ecx,%r8d
  movzbl -0xe(%rdx),%ecx
  add    %r11d,%ecx
  movzbl -0x6(%rdx),%r11d
  add    %r11d,%ecx
  movzbl -0x2(%rdx),%r11d
  add    %r11d,%ecx
  movzbl -0x9(%rdx),%r11d
  add    %ecx,%edi
  movzbl -0xd(%rdx),%ecx
  add    %r11d,%ecx
  movzbl -0x5(%rdx),%r11d
  add    %r11d,%ecx
  movzbl -0x1(%rdx),%r11d
  add    %r11d,%ecx
  add    %ecx,%r9d
  cmp    %rsi,%rdx
  jne    L75
  and    $0xf,%r10d
L51:
  mov    %rsi,%rdx
  cmp    $0x3,%r10
  jbe    L1
  lea    -0x4(%r10),%rdx
  shr    $0x2,%rdx
  lea    0x4(%rsi,%rdx,4),%rdx
L5:
  movzbl (%rsi),%ecx
  add    $0x4,%rsi
  add    %ecx,%eax
  movzbl -0x3(%rsi),%ecx
  add    %ecx,%r8d
  movzbl -0x2(%rsi),%ecx
  add    %ecx,%edi
  movzbl -0x1(%rsi),%ecx
  add    %ecx,%r9d
  cmp    %rdx,%rsi
  jne    L5
  and    $0x3,%r10d
L1:
  shl    $0x8,%edi
  shl    $0x10,%r8d
  add    %r8d,%edi
  shl    $0x18,%eax
  add    %r9d,%edi
  add    %edi,%eax
  cmp    $0x2,%r10
  je     L69
  cmp    $0x3,%r10
  je     L4
  cmp    $0x1,%r10
  je     L70
  ret
  nopl   0x0(%rax,%rax,1)
L4:
  movzbl 0x2(%rdx),%ecx
  shl    $0x8,%ecx
  add    %ecx,%eax
L69:
  movzbl 0x1(%rdx),%ecx
  shl    $0x10,%ecx
  add    %ecx,%eax
L70:
  movzbl (%rdx),%edx
  shl    $0x18,%edx
  add    %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %rdi,%rsi
  xor    %r9d,%r9d
  xor    %edi,%edi
  xor    %r8d,%r8d
  xor    %eax,%eax
  jmp    L51
```

assistant: ```c
unsigned int checksum(const char *zIn, size_t N) {
  const unsigned char *z = (const unsigned char *)zIn;
  unsigned sum0 = 0;
  unsigned sum1 = 0;
  unsigned sum2 = 0;
  unsigned sum3 = 0;
  while (N >= 16) {
    sum0 += ((unsigned)z[0] + z[4] + z[8] + z[12]);
    sum1 += ((unsigned)z[1] + z[5] + z[9] + z[13]);
    sum2 += ((unsigned)z[2] + z[6] + z[10] + z[14]);
    sum3 += ((unsigned)z[3] + z[7] + z[11] + z[15]);
    z += 16;
    N -= 16;
  }
  while (N >= 4) {
    sum0 += z[0];
    sum1 += z[1];
    sum2 += z[2];
    sum3 += z[3];
    z += 4;
    N -= 4;
  }
  sum3 += (sum2 << 8) + (sum1 << 16) + (sum0 << 24);
  switch (N) {
  case 3:
    sum3 += (z[2] << 8);
  case 2:
    sum3 += (z[1] << 16);
  case 1:
    sum3 += (z[0] << 24);
  default:;
  }
  return sum3;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movl   $0x1,-0x4(%rbp)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void main() {
  int32_t a;
  a = 1 && 2;
}
```
user: Converte this assembly snippet into its equivalent C code:

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

assistant: ```c
void main() {
  int32_t a;
  a = 1 && 2;
}
```
user: What would the C source look like for the assembly code provided here?

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

assistant: ```c
void main() {
  int32_t a;
  a = 1 && 2;
}
```
user: 请从给定的汇编指令推导出C源代码:

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

assistant: ```c
void main() {
  int32_t a;
  a = 1 && 2;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<zerrendaPantailaratu>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L0
L1:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L0:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L1
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D3
datatype: string

tool: D3: %d 

assistant: ```c
void zerrendaPantailaratu(int b[], int dim) {
  int i;
  for (i = 0; i < dim; i++)
    printf("%d ", b[i]);
  printf("\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<zerrendaPantailaratu>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %esi,%esi
  jle    L15
  mov    %rdi,%rbx
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%r12
  lea    D2(%rip),%rbp
L2:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%rbx
  cmp    %r12,%rbx
  jne    L2
L15:
  mov    $0xa,%edi
  call   <putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: parse_data
arguments: label: D2
type_infered: string

tool: D2: %d 

assistant: ```c
void zerrendaPantailaratu(int b[], int dim) {
  int i;
  for (i = 0; i < dim; i++)
    printf("%d ", b[i]);
  printf("\n");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<zerrendaPantailaratu>:
  endbr64
  test   %esi,%esi
  jle    L10
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D5(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L3:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L3
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L10:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

tool_calls: name: parse_data
arguments: tag: D5
type_guessed: string

tool: D5: %d 

assistant: ```c
void zerrendaPantailaratu(int b[], int dim) {
  int i;
  for (i = 0; i < dim; i++)
    printf("%d ", b[i]);
  printf("\n");
}
```
user: What is the c source code of the assembly code below:

```asm
<zerrendaPantailaratu>:
  endbr64
  test   %esi,%esi
  jle    L1
  lea    -0x1(%rsi),%eax
  push   %r12
  lea    0x4(%rdi,%rax,4),%r12
  push   %rbp
  lea    D7(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  xchg   %ax,%ax
L14:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L14
  pop    %rbx
  mov    $0xa,%edi
  pop    %rbp
  pop    %r12
  jmp    1060 <putchar@plt>
  nopl   0x0(%rax,%rax,1)
L1:
  mov    $0xa,%edi
  jmp    1060 <putchar@plt>
```

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

tool: D7: %d 

assistant: ```c
void zerrendaPantailaratu(int b[], int dim) {
  int i;
  for (i = 0; i < dim; i++)
    printf("%d ", b[i]);
  printf("\n");
}
```
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
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  jmp    L32
L29:
  cmpl   $0x1,-0x4(%rbp)
  jle    L18
  lea    D27(%rip),%rax
  jmp    L17
L18:
  lea    D20(%rip),%rax
L17:
  addq   $0x8,-0x10(%rbp)
  mov    -0x10(%rbp),%rdx
  mov    (%rdx),%rcx
  mov    %rax,%rdx
  mov    %rcx,%rsi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L32:
  subl   $0x1,-0x4(%rbp)
  cmpl   $0x0,-0x4(%rbp)
  jg     L29
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D27
type: string
name: extract_symbolic_data
arguments: tag: D20
type: string
name: extract_symbolic_data
arguments: tag: D13
type: string

tool: D27:  

tool: 

tool: D13: %s %s

assistant: ```c
int main(int argc, char *argv[]) {
  while (--argc > 0)
    printf("%s %s", *++argv, (argc > 1) ? " " : "");
  printf("\n");
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  lea    -0x1(%rdi),%ebx
  test   %ebx,%ebx
  jle    L18
  mov    %rsi,%rbp
  lea    D12(%rip),%r13
  lea    D13(%rip),%r12
  jmp    L0
L33:
  add    $0x8,%rbp
  mov    %r13,%rdx
  mov    0x0(%rbp),%rsi
  mov    %r12,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x1,%ebx
  je     L18
L0:
  cmp    $0x1,%ebx
  jne    L33
  mov    0x8(%rbp),%rsi
  lea    D24(%rip),%rdx
  lea    D13(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L18:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D12
type_guessed: string
name: extract_symbolic_data
arguments: tag: D13
type_guessed: string
name: extract_symbolic_data
arguments: tag: D24
type_guessed: string

tool: D12:  

tool: D13: %s %s

tool: 

assistant: ```c
int main(int argc, char *argv[]) {
  while (--argc > 0)
    printf("%s %s", *++argv, (argc > 1) ? " " : "");
  printf("\n");
  return 0;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%rdi),%ebx
  sub    $0x8,%rsp
  test   %ebx,%ebx
  jle    L24
  mov    %rsi,%rbp
  lea    D27(%rip),%r12
  lea    D12(%rip),%r13
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L28:
  mov    %r13,%rdx
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x8,%rbp
  call   <printf@plt>
  sub    $0x1,%ebx
L9:
  mov    0x8(%rbp),%rsi
  cmp    $0x1,%ebx
  jne    L28
  lea    D0(%rip),%rdx
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L24:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: tag: D27
type_guessed: string
name: unpack_data
arguments: tag: D12
type_guessed: string
name: unpack_data
arguments: tag: D0
type_guessed: string

tool: D27: %s %s

tool: D12:  

tool: 

assistant: ```c
int main(int argc, char *argv[]) {
  while (--argc > 0)
    printf("%s %s", *++argv, (argc > 1) ? " " : "");
  printf("\n");
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x1(%rdi),%ebx
  sub    $0x8,%rsp
  test   %ebx,%ebx
  jle    L29
  mov    %rsi,%rbp
  lea    D9(%rip),%r12
  lea    D10(%rip),%r13
  jmp    L15
  nopl   0x0(%rax,%rax,1)
L6:
  mov    %r13,%rdx
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x8,%rbp
  call   <printf@plt>
  sub    $0x1,%ebx
L15:
  mov    0x8(%rbp),%rsi
  cmp    $0x1,%ebx
  jne    L6
  lea    D13(%rip),%rdx
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
L29:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: read_data
arguments: label: D9
data_type: string
name: read_data
arguments: label: D10
data_type: string
name: read_data
arguments: label: D13
data_type: string

tool: D9: %s %s

tool: D10:  

tool: 

assistant: ```c
int main(int argc, char *argv[]) {
  while (--argc > 0)
    printf("%s %s", *++argv, (argc > 1) ? " " : "");
  printf("\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<glutStrokeFloat>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %rdi,-0x58(%rbp)
  movss  %xmm0,-0x5c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x5c(%rbp),%edx
  lea    -0x50(%rbp),%rax
  movd   %edx,%xmm0
  lea    D3(%rip),%rdx
  mov    $0x40,%esi
  mov    %rax,%rdi
  call   <snprintf@plt>
  lea    -0x50(%rbp),%rdx
  mov    -0x58(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <glutStrokeString@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D3
type_infered: string

tool: D3: %.2f

assistant: ```c
void glutStrokeFloat(void *fontface, float i) {
  char buf[64];
  snprintf(buf, 64, "%.2f", i);
  glutStrokeString(fontface, buf);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<glutStrokeFloat>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    D5(%rip),%rdx
  mov    $0x40,%esi
  mov    %rbp,%rdi
  call   <snprintf@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <glutStrokeString@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x58,%rsp
  pop    %rbx
  pop    %rbp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D5
type_infered: string

tool: D5: %.2f

assistant: ```c
void glutStrokeFloat(void *fontface, float i) {
  char buf[64];
  snprintf(buf, 64, "%.2f", i);
  glutStrokeString(fontface, buf);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<glutStrokeFloat>:
  endbr64
  push   %r12
  mov    $0x40,%esi
  lea    D8(%rip),%rdx
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <snprintf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <glutStrokeString@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x58,%rsp
  pop    %rbp
  pop    %r12
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D8
data_type: string

tool: D8: %.2f

assistant: ```c
void glutStrokeFloat(void *fontface, float i) {
  char buf[64];
  snprintf(buf, 64, "%.2f", i);
  glutStrokeString(fontface, buf);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<glutStrokeFloat>:
  endbr64
  push   %r12
  mov    $0x40,%esi
  lea    D3(%rip),%rdx
  push   %rbp
  mov    %rdi,%rbp
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x48(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r12
  mov    %r12,%rdi
  call   <snprintf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <glutStrokeString@plt>
  mov    0x48(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x58,%rsp
  pop    %rbp
  pop    %r12
  ret
L6:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D3
type_infered: string

tool: D3: %.2f

assistant: ```c
void glutStrokeFloat(void *fontface, float i) {
  char buf[64];
  snprintf(buf, 64, "%.2f", i);
  glutStrokeString(fontface, buf);
}
```
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
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D37(%rip),%rax
  mov    %rax,%rsi
  lea    D33(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D54(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movss  -0x18(%rbp),%xmm1
  movss  -0x18(%rbp),%xmm0
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x14(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0x14(%rbp),%xmm0
  movss  -0x18(%rbp),%xmm1
  pxor   %xmm2,%xmm2
  cvtss2sd %xmm1,%xmm2
  movq   %xmm2,%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D69(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  movss  -0x14(%rbp),%xmm0
  addss  %xmm0,%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0x10(%rbp),%xmm0
  movss  -0x18(%rbp),%xmm1
  pxor   %xmm3,%xmm3
  cvtss2sd %xmm1,%xmm3
  movq   %xmm3,%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  movss  -0x18(%rbp),%xmm0
  mulss  -0x10(%rbp),%xmm0
  movss  D44(%rip),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  pxor   %xmm0,%xmm0
  cvtss2sd -0xc(%rbp),%xmm0
  movss  -0x18(%rbp),%xmm1
  pxor   %xmm4,%xmm4
  cvtss2sd %xmm1,%xmm4
  movq   %xmm4,%rax
  movapd %xmm0,%xmm1
  movq   %rax,%xmm0
  lea    D32(%rip),%rax
  mov    %rax,%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D37
data_type: string
name: extract_symbolic_data
arguments: tag: D33
data_type: string
name: extract_symbolic_data
arguments: tag: D54
data_type: string
name: extract_symbolic_data
arguments: tag: D69
data_type: string
name: extract_symbolic_data
arguments: tag: D49
data_type: string
name: extract_symbolic_data
arguments: tag: D44
data_type: byte[4]
name: extract_symbolic_data
arguments: tag: D32
data_type: string

tool: D37: Inserire il valore del raggio: 

tool: D33: %s

tool: D54: %f

tool: D69: 


Il diametro del cerchio avente raggio %.2f e' pari a: %.2f

tool: D49: 

La circonferenza del cerchio avente raggio %.2f e' pari a: %.2f

tool: {"D44": [0x00,0x00,0x00,0x40]}

tool: D32: 

L'area del cerchio avente raggio %.2f e' pari a: %.2f


assistant: ```c
int main(void) {
  float raggio;
  float diametro;
  float circonferenza;
  float area;
  printf("%s", "Inserire il valore del raggio: ");
  scanf("%f", &raggio);
  printf("\n\n\nIl diametro del cerchio avente raggio %.2f e' pari a: %.2f", raggio, diametro = (raggio * raggio));
  printf("\n\nLa circonferenza del cerchio avente raggio %.2f e' pari a: %.2f", raggio, circonferenza = (diametro * (float)2));
  printf("\n\nL'area del cerchio avente raggio %.2f e' pari a: %.2f\n", raggio, area = ((circonferenza * raggio) / 2));
  return 0;
}
```