ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<commands>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  cmp    $0xf,%rdi
  jbe    L60
  sub    $0x10,%rbx
  mov    $0x1,%ebp
L60:
  mov    $0x20,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   $0xffffffffffffffef,%rbx
  je     L35
  cmp    $0x7,%rbx
  ja     L11
  mov    $0x10,%r11d
  mov    $0x3,%edi
  xor    %ecx,%ecx
  xor    %eax,%eax
  mov    $0x8,%r9d
  mov    $0x2,%r8d
  mov    $0x1,%edx
  cmp    $0x3,%rbx
  jbe    L12
L31:
  lea    D24(%rip),%rsi
  mov    %edx,%r10d
  sub    $0x4,%rbx
  mov    %rsi,(%r12,%rcx,1)
  mov    %eax,%esi
  mov    %r9,%rcx
  movslq %edx,%rax
  mov    %r11,%r9
  mov    %r8d,%edx
  mov    %edi,%r8d
  mov    $0x1,%edi
L16:
  cmp    $0x1,%rbx
  jbe    L9
  lea    D15(%rip),%rax
  mov    %edi,%esi
  sub    $0x2,%rbx
  mov    %r10d,%edi
  mov    %rax,(%r12,%rcx,1)
  movslq %edx,%rax
  mov    %r9,%rcx
  mov    %r8d,%edx
L9:
  test   %rbx,%rbx
  je     L33
  lea    D38(%rip),%rax
  mov    %edi,%esi
  mov    %rax,(%r12,%rcx,1)
  movslq %edx,%rax
L33:
  test   %ebp,%ebp
  jne    L45
  test   %esi,%esi
  je     L45
  lea    -0x1(%rsi),%edx
  lea    (%r12,%rax,4),%rax
  mov    %r12,%rbx
  lea    0x8(%r12,%rdx,8),%rbp
  lea    -0x8(%rax,%rax,1),%r13
L109:
  mov    %r13,%rsi
  mov    %rbx,%rdi
  sub    %rbx,%rsi
  add    $0x8,%rbx
  call   <swap@plt>
  cmp    %rbx,%rbp
  jne    L109
L45:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L11:
  lea    D1(%rip),%rax
  sub    $0x8,%rbx
  mov    $0x4,%edi
  mov    $0x18,%r11d
  mov    %rax,(%r12)
  mov    $0x10,%r9d
  mov    $0x3,%r8d
  mov    $0x2,%edx
  mov    $0x8,%ecx
  mov    $0x1,%eax
  cmp    $0x3,%rbx
  ja     L31
L12:
  mov    %eax,%edi
  xor    %esi,%esi
  mov    $0x1,%r10d
  jmp    L16
  cs nopw 0x0(%rax,%rax,1)
L35:
  movq   $0x0,(%rax)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: parse_data
arguments: tag: D24
datatype: string
name: parse_data
arguments: tag: D15
datatype: string
name: parse_data
arguments: tag: D38
datatype: string
name: parse_data
arguments: tag: D1
datatype: string

tool: D24: close your eyes

tool: D15: double blink

tool: D38: wink

tool: D1: jump

assistant: ```c
const char **commands(size_t number) {
  bool reverse = false;
  if (number >= 16) {
    reverse = true;
    number -= 16;
  }
  char **result = malloc(sizeof(char *) * 4);
  if (number == 0 || number == 16) {
    result[0] = NULL;
    return (const char **)result;
  }
  int idx = 0;
  if (number >= 8) {
    result[idx++] = "jump";
    number -= 8;
  }
  if (number >= 4) {
    result[idx++] = "close your eyes";
    number -= 4;
  }
  if (number >= 2) {
    result[idx++] = "double blink";
    number -= 2;
  }
  if (number >= 1) {
    result[idx++] = "wink";
    number -= 1;
  }
  if (!reverse) {
    for (int i = 0; i < idx / 2; ++i)
      swap(&result[i], &result[idx - i - 1]);
  }
  return (const char **)result;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<htmlstrip>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  jmp    L2
L10:
  mov    -0x8(%rbp),%rax
  lea    D0(%rip),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strspn@plt>
  cltq
  add    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L2
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x3f,(%rax)
L2:
  cmpq   $0x0,-0x8(%rbp)
  je     L23
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L10
L23:
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D0
type: string

tool: D0: 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ,. 

assistant: ```c
void htmlstrip(char *s) {
  while (s && *s) {
    s += strspn(s, "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ,. ");
    if (*s)
      *s++ = '?';
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<htmlstrip>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  lea    D8(%rip),%rbp
L1:
  test   %rbx,%rbx
  je     L38
  cmpb   $0x0,(%rbx)
  je     L38
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <strspn@plt>
  cltq
  add    %rax,%rbx
  cmpb   $0x0,(%rbx)
  je     L1
  movb   $0x3f,(%rbx)
  lea    0x1(%rbx),%rbx
  jmp    L1
L38:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: read_data
arguments: data_label: D8
datatype: string

tool: D8: 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ,. 

assistant: ```c
void htmlstrip(char *s) {
  while (s && *s) {
    s += strspn(s, "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ,. ");
    if (*s)
      *s++ = '?';
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<htmlstrip>:
  endbr64
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L43
  cs nopw 0x0(%rax,%rax,1)
L8:
  cmpb   $0x0,(%rbx)
  je     L1
  mov    %rbx,%rdi
  mov    %rbp,%rsi
  call   <strspn@plt>
  cltq
  add    %rax,%rbx
  cmpb   $0x0,(%rbx)
  jne    L48
L43:
  test   %rbx,%rbx
  jne    L8
L1:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L48:
  movb   $0x3f,(%rbx)
  add    $0x1,%rbx
  jmp    L43
```

tool_calls: name: retrieve_data
arguments: data_tag: D4
data_type: string

tool: D4: 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ,. 

assistant: ```c
void htmlstrip(char *s) {
  while (s && *s) {
    s += strspn(s, "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ,. ");
    if (*s)
      *s++ = '?';
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<htmlstrip>:
  endbr64
  push   %rbp
  lea    D8(%rip),%rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  jmp    L13
  cs nopw 0x0(%rax,%rax,1)
L45:
  cmpb   $0x0,(%rbx)
  je     L22
  mov    %rbx,%rdi
  mov    %rbp,%rsi
  call   <strspn@plt>
  cltq
  add    %rax,%rbx
  cmpb   $0x0,(%rbx)
  jne    L12
L13:
  test   %rbx,%rbx
  jne    L45
L22:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L12:
  movb   $0x3f,(%rbx)
  add    $0x1,%rbx
  jmp    L13
```

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

tool: D8: 0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ,. 

assistant: ```c
void htmlstrip(char *s) {
  while (s && *s) {
    s += strspn(s, "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ,. ");
    if (*s)
      *s++ = '?';
  }
}
```
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   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0xc(%rbp)
  lea    D2(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x14(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x10(%rbp)
  jmp    L19
L14:
  mov    -0x10(%rbp),%edx
  mov    -0x10(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <pow@plt>
  pxor   %xmm2,%xmm2
  cvtsi2ss %rax,%xmm2
  movss  %xmm2,-0x24(%rbp)
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <fact@plt>
  movss  -0x24(%rbp),%xmm1
  divss  %xmm0,%xmm1
  movss  -0xc(%rbp),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  addl   $0x1,-0x10(%rbp)
L19:
  mov    -0x14(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L14
  pxor   %xmm3,%xmm3
  cvtss2sd -0xc(%rbp),%xmm3
  movq   %xmm3,%rax
  movq   %rax,%xmm0
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D2
datatype: string
name: unpack_data
arguments: tag: D23
datatype: string
name: unpack_data
arguments: tag: D0
datatype: string

tool: D2: Enter the length of the series: 

tool: D23: %d

tool: D0: Sum of the series: %.2f


assistant: ```c
int main() {
  int n, i;
  float sum = 0;
  printf("Enter the length of the series: ");
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    sum += ((float)pow(i, i)) / fact(i);
  printf("Sum of the series: %.2f\n", sum);
}
```
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    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D22(%rip),%rdi
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  cmpl   $0x0,0x14(%rsp)
  jle    L28
  pxor   %xmm3,%xmm3
  movss  %xmm3,0xc(%rsp)
  mov    $0x1,%ebx
L32:
  mov    %ebx,%esi
  mov    %ebx,%edi
  call   <pow@plt>
  mov    %rax,%rbp
  mov    %ebx,%edi
  call   <fact@plt>
  movaps %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2ss %rbp,%xmm0
  divss  %xmm1,%xmm0
  addss  0xc(%rsp),%xmm0
  movss  %xmm0,0xc(%rsp)
  add    $0x1,%ebx
  cmp    %ebx,0x14(%rsp)
  jge    L32
L17:
  pxor   %xmm0,%xmm0
  cvtss2sd 0xc(%rsp),%xmm0
  lea    D8(%rip),%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  mov    $0x0,%eax
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  ret
L28:
  pxor   %xmm4,%xmm4
  movss  %xmm4,0xc(%rsp)
  jmp    L17
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D22
type: string
name: disassemble_data
arguments: data_label: D6
type: string
name: disassemble_data
arguments: data_label: D8
type: string

tool: D22: Enter the length of the series: 

tool: D6: %d

tool: D8: Sum of the series: %.2f


assistant: ```c
int main() {
  int n, i;
  float sum = 0;
  printf("Enter the length of the series: ");
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    sum += ((float)pow(i, i)) / fact(i);
  printf("Sum of the series: %.2f\n", sum);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rdi
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D6(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  pxor   %xmm1,%xmm1
  test   %eax,%eax
  jle    L7
  mov    $0x1,%ebx
  nopl   0x0(%rax,%rax,1)
L20:
  mov    %ebx,%esi
  mov    %ebx,%edi
  movss  %xmm1,0xc(%rsp)
  call   <pow@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  mov    %rax,%rbp
  call   <fact@plt>
  movss  0xc(%rsp),%xmm1
  movaps %xmm0,%xmm2
  pxor   %xmm0,%xmm0
  cvtsi2ss %rbp,%xmm0
  divss  %xmm2,%xmm0
  addss  %xmm0,%xmm1
  cmp    %ebx,0x14(%rsp)
  jge    L20
L7:
  pxor   %xmm0,%xmm0
  lea    D17(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm1,%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D2
type_infered: string
name: parse_data
arguments: data_tag: D6
type_infered: string
name: parse_data
arguments: data_tag: D17
type_infered: string

tool: D2: Enter the length of the series: 

tool: D6: %d

tool: D17: Sum of the series: %.2f


assistant: ```c
int main() {
  int n, i;
  float sum = 0;
  printf("Enter the length of the series: ");
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    sum += ((float)pow(i, i)) / fact(i);
  printf("Sum of the series: %.2f\n", sum);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  lea    D9(%rip),%rdi
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D3(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  pxor   %xmm1,%xmm1
  test   %eax,%eax
  jle    L22
  mov    $0x1,%ebx
  nopl   0x0(%rax,%rax,1)
L20:
  mov    %ebx,%esi
  mov    %ebx,%edi
  movss  %xmm1,0xc(%rsp)
  call   <pow@plt>
  mov    %ebx,%edi
  add    $0x1,%ebx
  mov    %rax,%rbp
  call   <fact@plt>
  movss  0xc(%rsp),%xmm1
  movaps %xmm0,%xmm2
  pxor   %xmm0,%xmm0
  cvtsi2ss %rbp,%xmm0
  divss  %xmm2,%xmm0
  addss  %xmm0,%xmm1
  cmp    %ebx,0x14(%rsp)
  jge    L20
L22:
  pxor   %xmm0,%xmm0
  lea    D19(%rip),%rdi
  mov    $0x1,%eax
  cvtss2sd %xmm1,%xmm0
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L5
  add    $0x28,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_tag: D9
data_type: string
name: retrieve_data
arguments: data_tag: D3
data_type: string
name: retrieve_data
arguments: data_tag: D19
data_type: string

tool: D9: Enter the length of the series: 

tool: D3: %d

tool: D19: Sum of the series: %.2f


assistant: ```c
int main() {
  int n, i;
  float sum = 0;
  printf("Enter the length of the series: ");
  scanf("%d", &n);
  for (i = 1; i <= n; i++)
    sum += ((float)pow(i, i)) / fact(i);
  printf("Sum of the series: %.2f\n", sum);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pretvorba>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x68,%rsp
  mov    %edi,-0x64(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x54(%rbp)
  movq   $0x0,-0x40(%rbp)
  movq   $0x0,-0x38(%rbp)
  movq   $0x0,-0x30(%rbp)
  movq   $0x0,-0x28(%rbp)
  movq   $0x0,-0x20(%rbp)
  movl   $0x0,-0x4c(%rbp)
  movl   $0x0,-0x48(%rbp)
  mov    -0x64(%rbp),%eax
  mov    %eax,-0x44(%rbp)
L55:
  mov    -0x64(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x64(%rbp)
  addl   $0x1,-0x48(%rbp)
  addl   $0x1,-0x54(%rbp)
  cmpl   $0x0,-0x64(%rbp)
  jg     L55
  mov    -0x44(%rbp),%eax
  mov    %eax,-0x64(%rbp)
  movl   $0x0,-0x54(%rbp)
  jmp    L5
L9:
  mov    -0x48(%rbp),%eax
  sub    $0x1,%eax
  sub    -0x54(%rbp),%eax
  mov    %eax,%edx
  mov    -0x64(%rbp),%eax
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  shr    $0x1f,%ecx
  add    %ecx,%eax
  and    $0x1,%eax
  sub    %ecx,%eax
  mov    %eax,%ecx
  movslq %edx,%rax
  mov    %ecx,-0x40(%rbp,%rax,4)
  mov    -0x64(%rbp),%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x64(%rbp)
  addl   $0x1,-0x54(%rbp)
L5:
  mov    -0x54(%rbp),%eax
  cmp    -0x48(%rbp),%eax
  jl     L9
  movl   $0x0,-0x54(%rbp)
  mov    -0x48(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x50(%rbp)
  jmp    L33
L35:
  mov    -0x54(%rbp),%eax
  cltq
  mov    -0x40(%rbp,%rax,4),%ebx
  mov    -0x50(%rbp),%eax
  mov    %eax,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%edi
  call   <round@plt>
  imul   %ebx,%eax
  add    %eax,-0x4c(%rbp)
  addl   $0x1,-0x54(%rbp)
  subl   $0x1,-0x50(%rbp)
L33:
  mov    -0x54(%rbp),%eax
  cmp    -0x48(%rbp),%eax
  jl     L35
  mov    -0x4c(%rbp),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L12
  call   <__stack_chk_fail@plt>
L12:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
int pretvorba(int broj) {
  int i = 0, j, binarni[10] = {}, rezultat = 0, br_znam = 0, temp;
  temp = broj;
  do {
    broj = broj / 2;
    br_znam++;
    i++;
  } while (broj > 0);
  broj = temp;
  for (i = 0; i < br_znam; i++) {
    binarni[(br_znam - 1) - i] = broj % 2;
    broj = broj / 2;
  }
  for (i = 0, j = br_znam - 1; i < br_znam; i++, j--) {
    rezultat += binarni[i] * round(pow(10, j));
  }
  return rezultat;
}
```
user: Translate the below assembly code into C programming language:

```asm
<pretvorba>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %edi,%edx
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movq   $0x0,(%rsp)
  movq   $0x0,0x8(%rsp)
  movq   $0x0,0x10(%rsp)
  movq   $0x0,0x18(%rsp)
  movq   $0x0,0x20(%rsp)
  mov    %edi,%eax
  mov    $0x0,%ecx
L42:
  mov    %eax,%esi
  mov    %eax,%edi
  shr    $0x1f,%edi
  add    %edi,%eax
  sar    %eax
  mov    %ecx,%ebx
  add    $0x1,%ecx
  cmp    $0x1,%esi
  jg     L42
  test   %ecx,%ecx
  jle    L32
  movslq %ebx,%rax
  lea    (%rsp,%rax,4),%rsi
  mov    $0x0,%ecx
L13:
  mov    %edx,%edi
  shr    $0x1f,%edi
  lea    (%rdx,%rdi,1),%eax
  and    $0x1,%eax
  sub    %edi,%eax
  mov    %eax,(%rsi)
  mov    %edi,%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  mov    %ecx,%eax
  add    $0x1,%ecx
  sub    $0x4,%rsi
  cmp    %eax,%ebx
  jg     L13
  mov    $0x0,%ebp
  mov    $0x0,%r13d
L36:
  mov    (%rsp,%rbp,4),%r12d
  mov    %ebx,%esi
  sub    %ebp,%esi
  mov    $0xa,%edi
  call   <pow@plt>
  mov    %eax,%edi
  call   <round@plt>
  imul   %eax,%r12d
  add    %r12d,%r13d
  add    $0x1,%rbp
  lea    -0x1(%rbp),%eax
  cmp    %ebx,%eax
  jl     L36
L58:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  mov    %r13d,%eax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L32:
  mov    $0x0,%r13d
  jmp    L58
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int pretvorba(int broj) {
  int i = 0, j, binarni[10] = {}, rezultat = 0, br_znam = 0, temp;
  temp = broj;
  do {
    broj = broj / 2;
    br_znam++;
    i++;
  } while (broj > 0);
  broj = temp;
  for (i = 0; i < br_znam; i++) {
    binarni[(br_znam - 1) - i] = broj % 2;
    broj = broj / 2;
  }
  for (i = 0, j = br_znam - 1; i < br_znam; i++, j--) {
    rezultat += binarni[i] * round(pow(10, j));
  }
  return rezultat;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<pretvorba>:
  endbr64
  push   %r13
  pxor   %xmm0,%xmm0
  mov    %edi,%edx
  xor    %esi,%esi
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  movaps %xmm0,(%rsp)
  mov    %edi,%eax
  movq   $0x0,0x20(%rsp)
  movaps %xmm0,0x10(%rsp)
  nopw   0x0(%rax,%rax,1)
L14:
  mov    %eax,%ecx
  shr    $0x1f,%eax
  mov    %esi,%ebx
  lea    0x1(%rsi),%esi
  add    %ecx,%eax
  sar    %eax
  cmp    $0x1,%ecx
  jg     L14
  movslq %ebx,%rax
  mov    %rsp,%r13
  lea    -0x4(%rsp),%rdi
  lea    0x0(%r13,%rax,4),%rcx
  nopl   0x0(%rax,%rax,1)
L23:
  mov    %edx,%esi
  sub    $0x4,%rcx
  shr    $0x1f,%esi
  lea    (%rdx,%rsi,1),%eax
  and    $0x1,%eax
  sub    %esi,%eax
  mov    %eax,0x4(%rcx)
  mov    %esi,%eax
  add    %edx,%eax
  sar    %eax
  mov    %eax,%edx
  cmp    %rdi,%rcx
  jne    L23
  xor    %r12d,%r12d
  nopl   0x0(%rax)
L15:
  mov    %ebx,%esi
  mov    $0xa,%edi
  mov    0x0(%r13),%ebp
  sub    $0x1,%ebx
  call   <pow@plt>
  add    $0x4,%r13
  mov    %eax,%edi
  call   <round@plt>
  imul   %eax,%ebp
  add    %ebp,%r12d
  cmp    $0xffffffff,%ebx
  jne    L15
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  add    $0x38,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L35:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int pretvorba(int broj) {
  int i = 0, j, binarni[10] = {}, rezultat = 0, br_znam = 0, temp;
  temp = broj;
  do {
    broj = broj / 2;
    br_znam++;
    i++;
  } while (broj > 0);
  broj = temp;
  for (i = 0; i < br_znam; i++) {
    binarni[(br_znam - 1) - i] = broj % 2;
    broj = broj / 2;
  }
  for (i = 0, j = br_znam - 1; i < br_znam; i++, j--) {
    rezultat += binarni[i] * round(pow(10, j));
  }
  return rezultat;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<calloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *calloc(size_t size) {
  void *mem = malloc(size);
  memset(mem, 0, size);
  return mem;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<calloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  call   <malloc@plt>
  mov    %rax,%rbx
  mov    %rbp,%rdx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *calloc(size_t size) {
  void *mem = malloc(size);
  memset(mem, 0, size);
  return mem;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<calloc>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13,%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memset@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *calloc(size_t size) {
  void *mem = malloc(size);
  memset(mem, 0, size);
  return mem;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<calloc>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  push   %r12
  sub    $0x8,%rsp
  call   <malloc@plt>
  mov    %r13,%rdx
  xor    %esi,%esi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <memset@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void *calloc(size_t size) {
  void *mem = malloc(size);
  memset(mem, 0, size);
  return mem;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<lexer_char_upper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x60,-0x4(%rbp)
  jle    L0
  cmpb   $0x7a,-0x4(%rbp)
  jg     L0
  movzbl -0x4(%rbp),%eax
  sub    $0x20,%eax
  jmp    L14
L0:
  movzbl -0x4(%rbp),%eax
L14:
  pop    %rbp
  ret
```

assistant: ```c
char lexer_char_upper(char c) {
  if (c >= 'a' && c <= 'z')
    return (char)('A' + (c - 'a'));
  return c;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<lexer_char_upper>:
  endbr64
  lea    -0x61(%rdi),%edx
  lea    -0x20(%rdi),%eax
  cmp    $0x1a,%dl
  cmovae %edi,%eax
  ret
```

assistant: ```c
char lexer_char_upper(char c) {
  if (c >= 'a' && c <= 'z')
    return (char)('A' + (c - 'a'));
  return c;
}
```
user: Translate the below assembly code into C programming language:

```asm
<lexer_char_upper>:
  endbr64
  lea    -0x61(%rdi),%edx
  lea    -0x20(%rdi),%eax
  cmp    $0x1a,%dl
  cmovae %edi,%eax
  ret
```

assistant: ```c
char lexer_char_upper(char c) {
  if (c >= 'a' && c <= 'z')
    return (char)('A' + (c - 'a'));
  return c;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<lexer_char_upper>:
  endbr64
  lea    -0x61(%rdi),%edx
  lea    -0x20(%rdi),%eax
  cmp    $0x1a,%dl
  cmovae %edi,%eax
  ret
```

assistant: ```c
char lexer_char_upper(char c) {
  if (c >= 'a' && c <= 'z')
    return (char)('A' + (c - 'a'));
  return c;
}
```
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    $0x50,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x64,-0x40(%rbp)
  movl   $0xc8,-0x30(%rbp)
  movl   $0x12c,-0x20(%rbp)
  lea    -0x30(%rbp),%rax
  mov    %rax,-0x38(%rbp)
  lea    -0x20(%rbp),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x38(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0x44(%rbp)
  mov    -0x44(%rbp),%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,%esi
  lea    D13(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D12
data_type: string
name: disassemble_data
arguments: data_tag: D13
data_type: string

tool: D12: %i 

tool: D13: %i


assistant: ```c
int main(void) {
  struct entry {
    int value;
    struct entry *next;
  };
  struct entry n1, n2, n3;
  int i;
  n1.value = 100;
  n2.value = 200;
  n3.value = 300;
  n1.next = &n2;
  n2.next = &n3;
  i = n1.next->value;
  printf("%i ", i);
  printf("%i\n", n2.next->value);
  return 0;
}
```