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

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D26(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    D23(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x10(%rsp),%r8d
  mov    0x14(%rsp),%r9d
  cmp    %r8d,%esi
  jg     L5
  cmp    %r9d,%esi
  jg     L8
  cmp    %r9d,%r8d
  jge    L1
L36:
  mov    %r8d,%esi
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L15
L5:
  cmp    %r9d,%r8d
  jg     L36
  cmp    %r9d,%esi
  jl     L8
L1:
  mov    %r9d,%esi
L8:
  lea    D27(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L15:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L35
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L35:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D26
data_type: string
name: unpack_data
arguments: label: D23
data_type: string
name: unpack_data
arguments: label: D27
data_type: string

tool: D26: Please enter 3 numbers separated by spaces > 

tool: D23: %d%d%d

tool: D27: %d is the median


assistant: ```c
int main() {
  int n1, n2, n3, small;
  printf("Please enter 3 numbers separated by spaces > ");
  scanf("%d%d%d", &n1, &n2, &n3);
  if (n1 <= n2) {
    small = n1;
    if (small > n3)
      printf("%d is the median\n", n1);
    else if (n3 > n2)
      printf("%d is the median\n", n2);
    else
      printf("%d is the median\n", n3);
  } else {
    small = n2;
    if (small > n3)
      printf("%d is the median\n", n2);
    else if (n3 > n1)
      printf("%d is the median\n", n1);
    else
      printf("%d is the median\n", n3);
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  sub    $0x28,%rsp
  lea    D3(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  lea    0x14(%rsp),%rcx
  lea    0x10(%rsp),%rdx
  lea    D9(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%esi
  mov    0x10(%rsp),%r8d
  mov    0x14(%rsp),%r9d
  cmp    %r8d,%esi
  jg     L29
  cmp    %r9d,%esi
  jg     L60
  cmp    %r9d,%r8d
  jge    L43
L59:
  mov    %r8d,%esi
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L30
L29:
  cmp    %r9d,%r8d
  jg     L59
  cmp    %r9d,%esi
  jl     L60
L43:
  mov    %r9d,%esi
L60:
  lea    D2(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L30:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  xor    %eax,%eax
  add    $0x28,%rsp
  ret
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D3
data_type: string
name: unpack_data
arguments: data_tag: D9
data_type: string
name: unpack_data
arguments: data_tag: D2
data_type: string

tool: D3: Please enter 3 numbers separated by spaces > 

tool: D9: %d%d%d

tool: D2: %d is the median


assistant: ```c
int main() {
  int n1, n2, n3, small;
  printf("Please enter 3 numbers separated by spaces > ");
  scanf("%d%d%d", &n1, &n2, &n3);
  if (n1 <= n2) {
    small = n1;
    if (small > n3)
      printf("%d is the median\n", n1);
    else if (n3 > n2)
      printf("%d is the median\n", n2);
    else
      printf("%d is the median\n", n3);
  } else {
    small = n2;
    if (small > n3)
      printf("%d is the median\n", n2);
    else if (n3 > n1)
      printf("%d is the median\n", n1);
    else
      printf("%d is the median\n", n3);
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<isLessOrEqual>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  shr    $0x1f,%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  shr    $0x1f,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x10(%rbp),%eax
  not    %eax
  shr    $0x1f,%eax
  movzbl %al,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  not    %eax
  and    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0xc(%rbp),%eax
  xor    -0x8(%rbp),%eax
  xor    $0x1,%eax
  and    -0x4(%rbp),%eax
  or     %edx,%eax
  pop    %rbp
  ret
```

assistant: ```c
int isLessOrEqual(int x, int y) {
  int sub = y + (~x + 1);
  int signx = (x >> 31) & 1;
  int signy = (y >> 31) & 1;
  int subsign = !(sub >> 31);
  return ((signx ^ signy) & signx) | ((signx ^ signy ^ 1) & subsign);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<isLessOrEqual>:
  endbr64
  mov    %edi,%ecx
  shr    $0x1f,%ecx
  mov    %esi,%edx
  shr    $0x1f,%edx
  mov    %ecx,%eax
  xor    %edx,%eax
  xor    $0x1,%eax
  cmp    %edi,%esi
  setge  %sil
  movzbl %sil,%esi
  and    %esi,%eax
  not    %edx
  and    %ecx,%edx
  or     %edx,%eax
  ret
```

assistant: ```c
int isLessOrEqual(int x, int y) {
  int sub = y + (~x + 1);
  int signx = (x >> 31) & 1;
  int signy = (y >> 31) & 1;
  int subsign = !(sub >> 31);
  return ((signx ^ signy) & signx) | ((signx ^ signy ^ 1) & subsign);
}
```
user: Translate the below assembly code into C programming language:

```asm
<isLessOrEqual>:
  endbr64
  mov    %edi,%ecx
  mov    %esi,%edx
  shr    $0x1f,%ecx
  shr    $0x1f,%edx
  mov    %ecx,%eax
  xor    %edx,%eax
  not    %edx
  xor    $0x1,%eax
  cmp    %edi,%esi
  setge  %sil
  and    %ecx,%edx
  movzbl %sil,%esi
  and    %esi,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
int isLessOrEqual(int x, int y) {
  int sub = y + (~x + 1);
  int signx = (x >> 31) & 1;
  int signy = (y >> 31) & 1;
  int subsign = !(sub >> 31);
  return ((signx ^ signy) & signx) | ((signx ^ signy ^ 1) & subsign);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<isLessOrEqual>:
  endbr64
  mov    %edi,%ecx
  mov    %esi,%edx
  shr    $0x1f,%ecx
  shr    $0x1f,%edx
  mov    %ecx,%eax
  xor    %edx,%eax
  not    %edx
  xor    $0x1,%eax
  cmp    %edi,%esi
  setge  %sil
  and    %ecx,%edx
  movzbl %sil,%esi
  and    %esi,%eax
  or     %edx,%eax
  ret
```

assistant: ```c
int isLessOrEqual(int x, int y) {
  int sub = y + (~x + 1);
  int signx = (x >> 31) & 1;
  int signy = (y >> 31) & 1;
  int subsign = !(sub >> 31);
  return ((signx ^ signy) & signx) | ((signx ^ signy ^ 1) & subsign);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0xa,-0x8(%rbp)
  jmp    L18
L37:
  movl   $0x0,-0x4(%rbp)
  jmp    L32
L17:
  lea    D16(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L32:
  mov    -0x4(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jl     L17
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  subl   $0x1,-0x8(%rbp)
L18:
  cmpl   $0x0,-0x8(%rbp)
  jg     L37
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D16
data_type: string
name: unpack_data
arguments: data_tag: D5
data_type: string

tool: D16: *

tool: D5: 


assistant: ```c
int main(void) {
  int i, j;
  for (i = 10; i > 0; i--) {
    for (j = 0; j < i; j++) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0xa,%ebp
  lea    D5(%rip),%r12
  lea    D8(%rip),%r13
  jmp    L34
L44:
  mov    %r12,%rdi
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebp,%ebx
  jne    L44
L41:
  mov    %r13,%rdi
  call   <printf@plt>
  sub    $0x1,%ebp
  je     L0
L34:
  mov    $0x0,%ebx
  test   %ebp,%ebp
  jg     L44
  jmp    L41
L0:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D5
datatype: string
name: extract_symbolic_data
arguments: data_label: D8
datatype: string

tool: D5: *

tool: D8: 


assistant: ```c
int main(void) {
  int i, j;
  for (i = 10; i > 0; i--) {
    for (j = 0; j < i; j++) {
      printf("*");
    }
    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
  lea    D10(%rip),%r13
  push   %r12
  lea    D5(%rip),%r12
  push   %rbp
  mov    $0xa,%ebp
  push   %rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L10:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L12:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jg     L12
  mov    %r13,%rdi
  call   <printf@plt>
  sub    $0x1,%ebp
  jne    L10
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D10
data_type: string
name: unpack_data
arguments: data_label: D5
data_type: string

tool: D10: 


tool: D5: *

assistant: ```c
int main(void) {
  int i, j;
  for (i = 10; i > 0; i--) {
    for (j = 0; j < i; j++) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r13
  lea    D17(%rip),%r13
  push   %r12
  lea    D8(%rip),%r12
  push   %rbp
  mov    $0xa,%ebp
  push   %rbx
  sub    $0x8,%rsp
  nopl   0x0(%rax)
L11:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L13:
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  cmp    %ebx,%ebp
  jg     L13
  mov    %r13,%rdi
  call   <printf@plt>
  sub    $0x1,%ebp
  jne    L11
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D17
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D8
type_guessed: string

tool: D17: 


tool: D8: *

assistant: ```c
int main(void) {
  int i, j;
  for (i = 10; i > 0; i--) {
    for (j = 0; j < i; j++) {
      printf("*");
    }
    printf("\n");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bitmap_zero>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,%edi
  call   <__builtin_constant_p@plt>
  test   %rax,%rax
  je     L13
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  jg     L13
  mov    -0x18(%rbp),%rax
  movq   $0x0,(%rax)
  jmp    L23
L13:
  mov    -0x1c(%rbp),%eax
  cltq
  add    $0x3f,%rax
  shr    $0x6,%rax
  shl    $0x3,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  nop
L23:
  nop
  leave
  ret
```

assistant: ```c
void bitmap_zero(unsigned long *dst, int nbits) {
  if ((__builtin_constant_p(nbits) && (nbits) <= __WORDSIZE))
    *dst = 0UL;
  else {
    int len = (((nbits) + (8 * sizeof(long)) - 1) / (8 * sizeof(long))) * sizeof(unsigned long);
    memset(dst, 0, len);
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bitmap_zero>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %esi,%ebx
  mov    %esi,%edi
  call   <__builtin_constant_p@plt>
  test   %rax,%rax
  je     L5
  mov    D8(%rip),%rax
  cmp    %ebx,(%rax)
  jl     L5
  movq   $0x0,0x0(%rbp)
  jmp    L19
L5:
  movslq %ebx,%rdx
  add    $0x3f,%rdx
  shr    $0x6,%rdx
  shl    $0x3,%edx
  mov    $0x0,%esi
  mov    %rbp,%rdi
  call   <memset@plt>
L19:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void bitmap_zero(unsigned long *dst, int nbits) {
  if ((__builtin_constant_p(nbits) && (nbits) <= __WORDSIZE))
    *dst = 0UL;
  else {
    int len = (((nbits) + (8 * sizeof(long)) - 1) / (8 * sizeof(long))) * sizeof(unsigned long);
    memset(dst, 0, len);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bitmap_zero>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <__builtin_constant_p@plt>
  test   %rax,%rax
  je     L11
  mov    D8(%rip),%rax
  cmp    %ebx,(%rax)
  jl     L11
  movq   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L11:
  movslq %ebx,%rdx
  add    $0x8,%rsp
  mov    %rbp,%rdi
  xor    %esi,%esi
  add    $0x3f,%rdx
  pop    %rbx
  pop    %rbp
  shr    $0x6,%rdx
  shl    $0x3,%edx
  jmp    1060 <memset@plt>
```

assistant: ```c
void bitmap_zero(unsigned long *dst, int nbits) {
  if ((__builtin_constant_p(nbits) && (nbits) <= __WORDSIZE))
    *dst = 0UL;
  else {
    int len = (((nbits) + (8 * sizeof(long)) - 1) / (8 * sizeof(long))) * sizeof(unsigned long);
    memset(dst, 0, len);
  }
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<bitmap_zero>:
  endbr64
  push   %rbp
  mov    %rdi,%rbp
  mov    %esi,%edi
  push   %rbx
  mov    %esi,%ebx
  sub    $0x8,%rsp
  call   <__builtin_constant_p@plt>
  test   %rax,%rax
  je     L4
  mov    D1(%rip),%rax
  cmp    %ebx,(%rax)
  jl     L4
  movq   $0x0,0x0(%rbp)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
  nopl   (%rax)
L4:
  movslq %ebx,%rdx
  add    $0x8,%rsp
  mov    %rbp,%rdi
  xor    %esi,%esi
  add    $0x3f,%rdx
  pop    %rbx
  pop    %rbp
  shr    $0x6,%rdx
  shl    $0x3,%edx
  jmp    1060 <memset@plt>
```

assistant: ```c
void bitmap_zero(unsigned long *dst, int nbits) {
  if ((__builtin_constant_p(nbits) && (nbits) <= __WORDSIZE))
    *dst = 0UL;
  else {
    int len = (((nbits) + (8 * sizeof(long)) - 1) / (8 * sizeof(long))) * sizeof(unsigned long);
    memset(dst, 0, len);
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<readLength>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  mov    %rax,%rdi
  call   <atoi@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%rax
  lea    0x6(%rax),%rdx
  mov    -0x18(%rbp),%rax
  mov    %rdx,(%rax)
  mov    -0x4(%rbp),%eax
  leave
  ret
```

assistant: ```c
int readLength(char const **readPtr) {
  int length = atoi(*readPtr);
  *readPtr += 6;
  return length;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<readLength>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%rdi
  call   <atoi@plt>
  addq   $0x6,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int readLength(char const **readPtr) {
  int length = atoi(*readPtr);
  *readPtr += 6;
  return length;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<readLength>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%rdi
  call   <atoi@plt>
  addq   $0x6,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int readLength(char const **readPtr) {
  int length = atoi(*readPtr);
  *readPtr += 6;
  return length;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<readLength>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  mov    (%rdi),%rdi
  call   <atoi@plt>
  addq   $0x6,(%rbx)
  pop    %rbx
  ret
```

assistant: ```c
int readLength(char const **readPtr) {
  int length = atoi(*readPtr);
  *readPtr += 6;
  return length;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<ascii_isalpha>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x60,-0x4(%rbp)
  jle    L34
  cmpb   $0x7a,-0x4(%rbp)
  jg     L34
  mov    $0x1,%eax
  jmp    L36
L34:
  cmpb   $0x40,-0x4(%rbp)
  jle    L51
  cmpb   $0x5a,-0x4(%rbp)
  jg     L51
  mov    $0x1,%eax
  jmp    L36
L51:
  mov    $0x0,%eax
L36:
  pop    %rbp
  ret
```

assistant: ```c
unsigned int ascii_isalpha(char c) {
  if (c >= 'a' && c <= 'z')
    return 1;
  if (c >= 'A' && c <= 'Z')
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<ascii_isalpha>:
  endbr64
  and    $0xffffffdf,%edi
  sub    $0x41,%edi
  cmp    $0x19,%dil
  setbe  %al
  movzbl %al,%eax
  ret
```

assistant: ```c
unsigned int ascii_isalpha(char c) {
  if (c >= 'a' && c <= 'z')
    return 1;
  if (c >= 'A' && c <= 'Z')
    return 1;
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<ascii_isalpha>:
  endbr64
  and    $0xffffffdf,%edi
  xor    %eax,%eax
  sub    $0x41,%edi
  cmp    $0x19,%dil
  setbe  %al
  ret
```

assistant: ```c
unsigned int ascii_isalpha(char c) {
  if (c >= 'a' && c <= 'z')
    return 1;
  if (c >= 'A' && c <= 'Z')
    return 1;
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<ascii_isalpha>:
  endbr64
  and    $0xffffffdf,%edi
  xor    %eax,%eax
  sub    $0x41,%edi
  cmp    $0x19,%dil
  setbe  %al
  ret
```

assistant: ```c
unsigned int ascii_isalpha(char c) {
  if (c >= 'a' && c <= 'z')
    return 1;
  if (c >= 'A' && c <= 'Z')
    return 1;
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ft_itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %edi,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  shr    $0x1f,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  mov    %rax,-0x10(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <numlength@plt>
  mov    %eax,%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L44
  mov    $0x0,%eax
  jmp    L63
L44:
  cmpq   $0x0,-0x10(%rbp)
  jne    L50
  mov    $0x30,%edx
  jmp    L54
L50:
  mov    $0x0,%edx
L54:
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  cmpq   $0x0,-0x10(%rbp)
  jns    L43
  mov    -0x8(%rbp),%rax
  movb   $0x2d,(%rax)
  negq   -0x10(%rbp)
L43:
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  subl   $0x1,-0x14(%rbp)
  jmp    L9
L30:
  mov    -0x10(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  mov    %rcx,%rsi
  sar    $0x3f,%rsi
  sub    %rsi,%rax
  mov    %rax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %edx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rcx
  movabs $0x6666666666666667,%rdx
  mov    %rcx,%rax
  imul   %rdx
  mov    %rdx,%rax
  sar    $0x2,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  subl   $0x1,-0x14(%rbp)
L9:
  cmpq   $0x0,-0x10(%rbp)
  jg     L30
  mov    -0x8(%rbp),%rax
L63:
  leave
  ret
```

assistant: ```c
char *ft_itoa(int n) {
  int i;
  char *str;
  long long nb;
  i = (n < 0) ? 1 : 0;
  nb = n;
  i += numlength(n);
  str = (char *)(malloc(i + 1));
  if (!str)
    return (NULL);
  str[0] = (nb == 0) ? '0' : 0;
  if (nb < 0) {
    str[0] = '-';
    nb *= -1;
  }
  str[i] = '\0';
  i--;
  while (nb > 0) {
    str[i] = ((nb % 10) + '0');
    nb /= 10;
    i--;
  }
  return (str);
}
```
user: Translate the below assembly code into C programming language:

```asm
<ft_itoa>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
  movslq %edi,%rbx
  call   <numlength@plt>
  shr    $0x1f,%ebp
  add    %eax,%ebp
  lea    0x1(%rbp),%edi
  call   <malloc@plt>
  test   %rax,%rax
  je     L49
  mov    %rax,%r8
  test   %rbx,%rbx
  je     L26
  jns    L54
  movb   $0x2d,(%rax)
  neg    %rbx
  movslq %ebp,%rdx
  movb   $0x0,(%rax,%rdx,1)
  sub    $0x1,%ebp
  jmp    L30
L54:
  movb   $0x0,(%rax)
  movslq %ebp,%rdx
  movb   $0x0,(%rax,%rdx,1)
  sub    $0x1,%ebp
  test   %rbx,%rbx
  jle    L60
L30:
  movslq %ebp,%rbp
  lea    0x0(%rbp,%rax,1),%rcx
  movabs $0x6666666666666667,%rdi
L52:
  mov    %rbx,%rax
  imul   %rdi
  sar    $0x2,%rdx
  mov    %rbx,%rax
  sar    $0x3f,%rax
  sub    %rax,%rdx
  lea    (%rdx,%rdx,4),%rsi
  add    %rsi,%rsi
  mov    %rbx,%rax
  sub    %rsi,%rax
  add    $0x30,%eax
  mov    %al,(%rcx)
  mov    %rbx,%rax
  mov    %rdx,%rbx
  sub    $0x1,%rcx
  cmp    $0x9,%rax
  jg     L52
L60:
  mov    %r8,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L49:
  mov    $0x0,%r8d
  jmp    L60
L26:
  movb   $0x30,(%rax)
  movslq %ebp,%rbp
  movb   $0x0,(%rax,%rbp,1)
  jmp    L60
```

assistant: ```c
char *ft_itoa(int n) {
  int i;
  char *str;
  long long nb;
  i = (n < 0) ? 1 : 0;
  nb = n;
  i += numlength(n);
  str = (char *)(malloc(i + 1));
  if (!str)
    return (NULL);
  str[0] = (nb == 0) ? '0' : 0;
  if (nb < 0) {
    str[0] = '-';
    nb *= -1;
  }
  str[i] = '\0';
  i--;
  while (nb > 0) {
    str[i] = ((nb % 10) + '0');
    nb /= 10;
    i--;
  }
  return (str);
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<ft_itoa>:
  endbr64
  push   %rbp
  push   %rbx
  movslq %edi,%rbx
  mov    %ebx,%edi
  mov    %rbx,%rbp
  shr    $0x1f,%ebp
  sub    $0x8,%rsp
  call   <numlength@plt>
  add    %eax,%ebp
  lea    0x1(%rbp),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L11
  movslq %ebp,%rax
  add    %r8,%rax
  test   %rbx,%rbx
  je     L19
  jns    L18
  movb   $0x2d,(%r8)
  neg    %rbx
L38:
  movabs $0xcccccccccccccccd,%rsi
  lea    -0x1(%rbp),%ecx
  movb   $0x0,(%rax)
  movslq %ecx,%rcx
  add    %r8,%rcx
  cs nopw 0x0(%rax,%rax,1)
L44:
  mov    %rbx,%rax
  sub    $0x1,%rcx
  mul    %rsi
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rbx
  add    $0x30,%ebx
  mov    %bl,0x1(%rcx)
  mov    %rdx,%rbx
  test   %rdx,%rdx
  jne    L44
L11:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
L19:
  movb   $0x30,(%r8)
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
L18:
  movb   $0x0,(%r8)
  jmp    L38
```

assistant: ```c
char *ft_itoa(int n) {
  int i;
  char *str;
  long long nb;
  i = (n < 0) ? 1 : 0;
  nb = n;
  i += numlength(n);
  str = (char *)(malloc(i + 1));
  if (!str)
    return (NULL);
  str[0] = (nb == 0) ? '0' : 0;
  if (nb < 0) {
    str[0] = '-';
    nb *= -1;
  }
  str[i] = '\0';
  i--;
  while (nb > 0) {
    str[i] = ((nb % 10) + '0');
    nb /= 10;
    i--;
  }
  return (str);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ft_itoa>:
  endbr64
  push   %rbp
  push   %rbx
  movslq %edi,%rbx
  mov    %ebx,%edi
  mov    %rbx,%rbp
  shr    $0x1f,%ebp
  sub    $0x8,%rsp
  call   <numlength@plt>
  add    %eax,%ebp
  lea    0x1(%rbp),%edi
  call   <malloc@plt>
  mov    %rax,%r8
  test   %rax,%rax
  je     L13
  movslq %ebp,%rax
  add    %r8,%rax
  test   %rbx,%rbx
  je     L42
  jns    L11
  movb   $0x2d,(%r8)
  neg    %rbx
L44:
  movabs $0xcccccccccccccccd,%rsi
  lea    -0x1(%rbp),%ecx
  movb   $0x0,(%rax)
  movslq %ecx,%rcx
  add    %r8,%rcx
  cs nopw 0x0(%rax,%rax,1)
L32:
  mov    %rbx,%rax
  sub    $0x1,%rcx
  mul    %rsi
  shr    $0x3,%rdx
  lea    (%rdx,%rdx,4),%rax
  add    %rax,%rax
  sub    %rax,%rbx
  add    $0x30,%ebx
  mov    %bl,0x1(%rcx)
  mov    %rdx,%rbx
  test   %rdx,%rdx
  jne    L32
L13:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
L42:
  movb   $0x30,(%r8)
  movb   $0x0,(%rax)
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  ret
L11:
  movb   $0x0,(%r8)
  jmp    L44
```

assistant: ```c
char *ft_itoa(int n) {
  int i;
  char *str;
  long long nb;
  i = (n < 0) ? 1 : 0;
  nb = n;
  i += numlength(n);
  str = (char *)(malloc(i + 1));
  if (!str)
    return (NULL);
  str[0] = (nb == 0) ? '0' : 0;
  if (nb < 0) {
    str[0] = '-';
    nb *= -1;
  }
  str[i] = '\0';
  i--;
  while (nb > 0) {
    str[i] = ((nb % 10) + '0');
    nb /= 10;
    i--;
  }
  return (str);
}
```