ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Translate the below assembly code into C programming language:

```asm
<submatrix>:
  endbr64
  push   %r15
  sub    %rsi,%rdx
  mov    %r9,%r15
  push   %r14
  push   %r13
  mov    %r8,%r13
  push   %r12
  mov    %rdi,%r12
  lea    0x10(,%rdx,8),%rdi
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  mov    %r8,%rbx
  sub    $0x8,%rsp
  sub    0x40(%rsp),%rbx
  call   <malloc@plt>
  mov    %rax,%r14
  test   %rax,%rax
  je     L23
L52:
  mov    $0x1,%eax
  sub    %r15,%rax
  lea    (%r14,%rax,8),%r8
  cmp    %r13,%rbp
  jg     L53
  lea    0x1(%rbp),%rcx
  mov    %r13,%rsi
  lea    0x8(%r14),%rax
  lea    0x0(,%rcx,8),%r9
  lea    0x0(,%rbx,4),%rdx
  sub    %rbp,%rsi
  lea    (%r12,%r9,1),%rdi
  cmp    %rdi,%rax
  je     L7
  cmp    $0x2,%rsi
  jbe    L7
  lea    0x1(%rsi),%rcx
  movq   %rdx,%xmm1
  mov    %rcx,%rsi
  punpcklqdq %xmm1,%xmm1
  shr    %rsi
  shl    $0x4,%rsi
  lea    0x8(%r14,%rsi,1),%rdi
  mov    %r12,%rsi
  sub    %r14,%rsi
  lea    -0x8(%rsi,%r9,1),%rsi
  nopl   0x0(%rax,%rax,1)
L3:
  movdqu -0x8(%rax,%rsi,1),%xmm0
  add    $0x10,%rax
  paddq  %xmm1,%xmm0
  movups %xmm0,-0x10(%rax)
  cmp    %rax,%rdi
  jne    L3
  mov    %rcx,%rax
  and    $0xfffffffffffffffe,%rax
  add    %rax,%rbp
  lea    (%r15,%rax,1),%r9
  cmp    %rax,%rcx
  je     L53
  mov    (%r12,%rbp,8),%rax
  add    %rdx,%rax
  mov    %rax,(%r8,%r9,8)
L53:
  add    $0x8,%rsp
  mov    %r8,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nop
L23:
  lea    D6(%rip),%rdi
  call   <nrerror@plt>
  jmp    L52
  nopl   0x0(%rax)
L7:
  sub    %rbp,%r15
  add    $0x1,%r13
  lea    (%r8,%r15,8),%rsi
  jmp    L13
  nopl   (%rax)
L95:
  add    $0x1,%rcx
L13:
  mov    (%r12,%rbp,8),%rax
  add    %rdx,%rax
  mov    %rax,(%rsi,%rbp,8)
  mov    %rcx,%rbp
  cmp    %r13,%rcx
  jne    L95
  jmp    L53
```

tool_calls: name: unpack_data
arguments: label: D6
type_guessed: string

tool: D6: allocation failure in submatrix()

assistant: ```c
float **submatrix(float **a, long oldrl, long oldrh, long oldcl, long oldch, long newrl, long newcl) {
  long i, j, nrow = oldrh - oldrl + 1, ncol = oldch - newcl;
  float **m;
  m = (float **)malloc((size_t)((nrow + 1) * sizeof(float *)));
  if (!m)
    nrerror("allocation failure in submatrix()");
  m += 1;
  m -= newrl;
  for (i = oldrl, j = newrl; i <= oldch; i++, j++)
    m[j] = a[i] + ncol;
  return m;
}
```
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
  movl   $0x0,-0x8(%rbp)
  jmp    L29
L14:
  addl   $0x1,-0x8(%rbp)
L29:
  cmpl   $0x3ff,-0x8(%rbp)
  jle    L14
  movl   $0x0,-0x4(%rbp)
  jmp    L38
L0:
  addl   $0x1,-0x4(%rbp)
L38:
  cmpl   $0x3ff,-0x4(%rbp)
  jle    L0
  mov    $0x0,%eax
  pop    %rbp
  ret
```

assistant: ```c
int main(void) {
#pragma acc parallel vector_length(128)
  {
#pragma acc loop vector
    for (int i = 0; i < 1024; i++)
      ;
#pragma acc loop vector
    for (int i = 0; i < 1024; i++)
      ;
  }
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  mov    $0x400,%eax
L14:
  sub    $0x1,%eax
  jne    L14
  mov    $0x400,%eax
L10:
  sub    $0x1,%eax
  jne    L10
  mov    $0x0,%eax
  ret
```

assistant: ```c
int main(void) {
#pragma acc parallel vector_length(128)
  {
#pragma acc loop vector
    for (int i = 0; i < 1024; i++)
      ;
#pragma acc loop vector
    for (int i = 0; i < 1024; i++)
      ;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
#pragma acc parallel vector_length(128)
  {
#pragma acc loop vector
    for (int i = 0; i < 1024; i++)
      ;
#pragma acc loop vector
    for (int i = 0; i < 1024; i++)
      ;
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  xor    %eax,%eax
  ret
```

assistant: ```c
int main(void) {
#pragma acc parallel vector_length(128)
  {
#pragma acc loop vector
    for (int i = 0; i < 1024; i++)
      ;
#pragma acc loop vector
    for (int i = 0; i < 1024; i++)
      ;
  }
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<iterate_number>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  movss  D12(%rip),%xmm0
  movss  %xmm0,-0x10(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  D12(%rip),%xmm0
  movss  %xmm0,-0x8(%rbp)
  movss  D12(%rip),%xmm0
  movss  %xmm0,-0xc(%rbp)
  jmp    L10
L4:
  movss  -0x10(%rbp),%xmm0
  addss  -0xc(%rbp),%xmm0
  movss  -0x8(%rbp),%xmm2
  movss  D14(%rip),%xmm1
  addss  %xmm2,%xmm1
  divss  %xmm1,%xmm0
  movss  -0x4(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x4(%rbp)
  movss  -0x10(%rbp),%xmm0
  addss  -0xc(%rbp),%xmm0
  movss  %xmm0,-0x10(%rbp)
  movss  -0xc(%rbp),%xmm1
  movss  D30(%rip),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0xc(%rbp)
  movss  -0x8(%rbp),%xmm1
  movss  D12(%rip),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x8(%rbp)
L10:
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x14(%rbp),%xmm0
  comiss -0x8(%rbp),%xmm0
  jae    L4
  movss  -0x4(%rbp),%xmm0
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D12
type: f32
name: parse_data
arguments: data_tag: D14
type: byte[4]
name: parse_data
arguments: data_tag: D30
type: byte[4]

tool: D12: 1

tool: {"D14": [0x00,0x00,0x40,0x40]}

tool: {"D30": [0x00,0x00,0x00,0x40]}

assistant: ```c
float iterate_number(int num) {
  float aux = 1, impar, i;
  float acumulador = 0;
  for (i = 1, impar = 1; i <= num; i++) {
    acumulador = acumulador + ((aux + impar) / (i + 3));
    aux = aux + impar;
    impar = impar + 2;
  }
  return acumulador;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<iterate_number>:
  endbr64
  pxor   %xmm6,%xmm6
  cvtsi2ss %edi,%xmm6
  comiss D23(%rip),%xmm6
  jb     L3
  pxor   %xmm3,%xmm3
  movss  D23(%rip),%xmm0
  movaps %xmm0,%xmm2
  movaps %xmm0,%xmm1
  movss  D5(%rip),%xmm9
  movss  D7(%rip),%xmm8
  movaps %xmm0,%xmm7
L27:
  addss  %xmm2,%xmm1
  movaps %xmm0,%xmm5
  addss  %xmm9,%xmm5
  movaps %xmm1,%xmm4
  divss  %xmm5,%xmm4
  addss  %xmm4,%xmm3
  addss  %xmm8,%xmm2
  addss  %xmm7,%xmm0
  comiss %xmm0,%xmm6
  jae    L27
L16:
  movaps %xmm3,%xmm0
  ret
L3:
  pxor   %xmm3,%xmm3
  jmp    L16
```

tool_calls: name: parse_data
arguments: tag: D23
data_type: f32
name: parse_data
arguments: tag: D5
data_type: byte[4]
name: parse_data
arguments: tag: D7
data_type: byte[4]

tool: D23: 1

tool: {"D5": [0x00,0x00,0x40,0x40]}

tool: {"D7": [0x00,0x00,0x00,0x40]}

assistant: ```c
float iterate_number(int num) {
  float aux = 1, impar, i;
  float acumulador = 0;
  for (i = 1, impar = 1; i <= num; i++) {
    acumulador = acumulador + ((aux + impar) / (i + 3));
    aux = aux + impar;
    impar = impar + 2;
  }
  return acumulador;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<iterate_number>:
  endbr64
  pxor   %xmm7,%xmm7
  movss  D3(%rip),%xmm6
  pxor   %xmm3,%xmm3
  cvtsi2ss %edi,%xmm7
  comiss %xmm6,%xmm7
  jb     L5
  movss  D19(%rip),%xmm9
  movaps %xmm6,%xmm0
  movaps %xmm6,%xmm2
  movaps %xmm6,%xmm1
  movss  D15(%rip),%xmm8
  nopl   0x0(%rax,%rax,1)
L8:
  addss  %xmm2,%xmm1
  movaps %xmm0,%xmm5
  addss  %xmm6,%xmm0
  addss  %xmm9,%xmm5
  addss  %xmm8,%xmm2
  movaps %xmm1,%xmm4
  comiss %xmm0,%xmm7
  divss  %xmm5,%xmm4
  addss  %xmm4,%xmm3
  jae    L8
L5:
  movaps %xmm3,%xmm0
  ret
```

tool_calls: name: read_data
arguments: tag: D3
type_guessed: f32
name: read_data
arguments: tag: D19
type_guessed: byte[4]
name: read_data
arguments: tag: D15
type_guessed: byte[4]

tool: D3: 1

tool: {"D19": [0x00,0x00,0x40,0x40]}

tool: {"D15": [0x00,0x00,0x00,0x40]}

assistant: ```c
float iterate_number(int num) {
  float aux = 1, impar, i;
  float acumulador = 0;
  for (i = 1, impar = 1; i <= num; i++) {
    acumulador = acumulador + ((aux + impar) / (i + 3));
    aux = aux + impar;
    impar = impar + 2;
  }
  return acumulador;
}
```
user: Translate the below assembly code into C programming language:

```asm
<iterate_number>:
  endbr64
  pxor   %xmm7,%xmm7
  movss  D20(%rip),%xmm6
  pxor   %xmm3,%xmm3
  cvtsi2ss %edi,%xmm7
  comiss %xmm6,%xmm7
  jb     L11
  movss  D22(%rip),%xmm9
  movaps %xmm6,%xmm0
  movaps %xmm6,%xmm2
  movaps %xmm6,%xmm1
  movss  D9(%rip),%xmm8
  nopl   0x0(%rax,%rax,1)
L6:
  addss  %xmm2,%xmm1
  movaps %xmm0,%xmm5
  addss  %xmm6,%xmm0
  addss  %xmm9,%xmm5
  addss  %xmm8,%xmm2
  movaps %xmm1,%xmm4
  comiss %xmm0,%xmm7
  divss  %xmm5,%xmm4
  addss  %xmm4,%xmm3
  jae    L6
L11:
  movaps %xmm3,%xmm0
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D20
datatype: f32
name: retrieve_data
arguments: data_tag: D22
datatype: byte[4]
name: retrieve_data
arguments: data_tag: D9
datatype: byte[4]

tool: D20: 1

tool: {"D22": [0x00,0x00,0x40,0x40]}

tool: {"D9": [0x00,0x00,0x00,0x40]}

assistant: ```c
float iterate_number(int num) {
  float aux = 1, impar, i;
  float acumulador = 0;
  for (i = 1, impar = 1; i <= num; i++) {
    acumulador = acumulador + ((aux + impar) / (i + 3));
    aux = aux + impar;
    impar = impar + 2;
  }
  return acumulador;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<print_bits>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,%eax
  mov    %al,-0x14(%rbp)
  movl   $0x80,-0x8(%rbp)
  movzbl -0x14(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L26
L5:
  mov    -0x4(%rbp),%eax
  and    -0x8(%rbp),%eax
  test   %eax,%eax
  je     L39
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L25
L39:
  lea    D22(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L25:
  sarl   -0x8(%rbp)
L26:
  cmpl   $0x0,-0x8(%rbp)
  jg     L5
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D12
datatype: string
name: read_data
arguments: data_label: D22
datatype: string
name: read_data
arguments: data_label: D26
datatype: string

tool: D12: 1

tool: D22: 0

tool: D26:  

assistant: ```c
void print_bits(unsigned char octet) {
  int z = 128, oct = octet;
  while (z > 0) {
    if (oct & z)
      printf("1");
    else
      printf("0");
    z >>= 1;
  }
  printf(" ");
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<print_bits>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  movzbl %dil,%r12d
  mov    $0x8,%ebp
  mov    $0x80,%ebx
  lea    D16(%rip),%r14
  lea    D2(%rip),%r13
  jmp    L33
L38:
  mov    %r14,%rdi
  call   <printf@plt>
L1:
  sar    %ebx
  sub    $0x1,%ebp
  je     L35
L33:
  test   %ebx,%r12d
  je     L38
  mov    %r13,%rdi
  call   <printf@plt>
  jmp    L1
L35:
  lea    D1(%rip),%rdi
  call   <printf@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: read_data
arguments: data_tag: D16
type: string
name: read_data
arguments: data_tag: D2
type: string
name: read_data
arguments: data_tag: D1
type: string

tool: D16: 0

tool: D2: 1

tool: D1:  

assistant: ```c
void print_bits(unsigned char octet) {
  int z = 128, oct = octet;
  while (z > 0) {
    if (oct & z)
      printf("1");
    else
      printf("0");
    z >>= 1;
  }
  printf(" ");
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<print_bits>:
  endbr64
  push   %r14
  lea    D2(%rip),%r14
  push   %r13
  lea    D15(%rip),%r13
  push   %r12
  movzbl %dil,%r12d
  push   %rbp
  mov    $0x8,%ebp
  push   %rbx
  mov    $0x80,%ebx
  jmp    L3
  nopw   0x0(%rax,%rax,1)
L27:
  mov    %r13,%rdi
  sar    %ebx
  call   <printf@plt>
  sub    $0x1,%ebp
  je     L23
L3:
  test   %ebx,%r12d
  jne    L27
  mov    %r14,%rdi
  sar    %ebx
  call   <printf@plt>
  sub    $0x1,%ebp
  jne    L3
L23:
  pop    %rbx
  lea    D0(%rip),%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D2
type: string
name: extract_symbolic_data
arguments: data_label: D15
type: string
name: extract_symbolic_data
arguments: data_label: D0
type: string

tool: D2: 0

tool: D15: 1

tool: D0:  

assistant: ```c
void print_bits(unsigned char octet) {
  int z = 128, oct = octet;
  while (z > 0) {
    if (oct & z)
      printf("1");
    else
      printf("0");
    z >>= 1;
  }
  printf(" ");
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<print_bits>:
  endbr64
  push   %r14
  lea    D20(%rip),%r14
  push   %r13
  lea    D16(%rip),%r13
  push   %r12
  movzbl %dil,%r12d
  push   %rbp
  mov    $0x8,%ebp
  push   %rbx
  mov    $0x80,%ebx
  jmp    L16
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %r13,%rdi
  sar    %ebx
  call   <printf@plt>
  sub    $0x1,%ebp
  je     L27
L16:
  test   %ebx,%r12d
  jne    L20
  mov    %r14,%rdi
  sar    %ebx
  call   <printf@plt>
  sub    $0x1,%ebp
  jne    L16
L27:
  pop    %rbx
  lea    D13(%rip),%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1050 <printf@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D20
data_type: string
name: extract_symbolic_data
arguments: data_tag: D16
data_type: string
name: extract_symbolic_data
arguments: data_tag: D13
data_type: string

tool: D20: 0

tool: D16: 1

tool: D13:  

assistant: ```c
void print_bits(unsigned char octet) {
  int z = 128, oct = octet;
  while (z > 0) {
    if (oct & z)
      printf("1");
    else
      printf("0");
    z >>= 1;
  }
  printf(" ");
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<addBinary>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  mov    %rsi,%r13
  call   <strlen@plt>
  mov    %eax,%r12d
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    %eax,%ebp
  lea    0x1(%r12),%eax
  lea    0x1(%rbp),%r14d
  cmp    %ebp,%r12d
  cmovg  %eax,%r14d
  lea    0x1(%r14),%edi
  call   <malloc@plt>
  mov    %rax,%rdi
  movslq %r14d,%r14
  movb   $0x0,(%rax,%r14,1)
  movb   $0x0,-0x1(%rax,%r14,1)
  lea    -0x1(%r12),%edx
  lea    -0x1(%rbp),%ecx
  mov    %r12d,%eax
  neg    %eax
  js     L47
  test   %ecx,%ecx
  js     L12
L47:
  movslq %edx,%rdx
  mov    $0x0,%esi
  mov    $0x0,%r10d
  mov    $0x0,%r8d
  movslq %ebp,%rbp
  movslq %r12d,%r12
  sub    %r12,%rbp
  lea    0x0(%r13,%rbp,1),%r11
  jmp    L65
L89:
  movsbl (%rbx,%rdx,1),%eax
  sub    $0x30,%eax
L17:
  mov    %r8d,%r12d
  test   %ecx,%ecx
  js     L5
  movsbl (%r11,%rdx,1),%r12d
  sub    $0x30,%r12d
L5:
  mov    %eax,%ebp
  xor    %r10d,%ebp
  xor    %r12d,%ebp
  add    $0x30,%ebp
  mov    %bpl,(%rdi,%rsi,1)
  add    %r10d,%eax
  add    %r12d,%eax
  cmp    $0x1,%eax
  setg   %r12b
  setg   %r10b
  movzbl %r10b,%r10d
  sub    $0x1,%ecx
  mov    %edx,%eax
  neg    %eax
  shr    $0x1f,%eax
  mov    %ecx,%ebp
  not    %ebp
  shr    $0x1f,%ebp
  or     %ebp,%eax
  add    $0x1,%rsi
  sub    $0x1,%rdx
  or     %r12b,%al
  je     L7
L65:
  mov    %esi,%r9d
  mov    %r8d,%eax
  test   %edx,%edx
  jns    L89
  jmp    L17
L7:
  test   %r9d,%r9d
  jle    L12
  movslq %r9d,%r9
  mov    $0x0,%eax
L64:
  movzbl (%rdi,%rax,1),%edx
  movzbl (%rdi,%r9,1),%ecx
  mov    %cl,(%rdi,%rax,1)
  mov    %dl,(%rdi,%r9,1)
  add    $0x1,%rax
  sub    $0x1,%r9
  cmp    %r9d,%eax
  jl     L64
L12:
  mov    %rdi,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *addBinary(char *a, char *b) {
  int len1 = strlen(a);
  int len2 = strlen(b);
  int len = len1 > len2 ? len1 + 1 : len2 + 1;
  char *result = malloc(len + 1);
  result[len] = '\0';
  result[len - 1] = '\0';
  int i, j, carry = 0;
  len = 0;
  for (i = len1 - 1, j = len2 - 1; carry || i >= 0 || j >= 0; i--, j--) {
    int na = i >= 0 ? a[i] - '0' : 0;
    int nb = j >= 0 ? b[j] - '0' : 0;
    result[len++] = (na ^ nb ^ carry) + '0';
    carry = carry + na + nb >= 2 ? 1 : 0;
  }
  for (i = 0, j = len - 1; i < j; i++, j--) {
    char c = result[i];
    result[i] = result[j];
    result[j] = c;
  }
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<addBinary>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  movslq %eax,%rbp
  call   <strlen@plt>
  movslq %eax,%r12
  lea    0x1(%rbp),%eax
  cmp    %r12d,%ebp
  lea    0x1(%r12),%r13d
  cmovg  %eax,%r13d
  lea    0x1(%r13),%edi
  movslq %r13d,%r13
  call   <malloc@plt>
  lea    -0x1(%rbp),%r9d
  lea    -0x1(%r12),%ecx
  xor    %esi,%esi
  sub    %r12,%rbp
  movb   $0x0,(%rax,%r13,1)
  mov    %rax,%r10
  movslq %ecx,%rcx
  movb   $0x0,-0x1(%rax,%r13,1)
  xor    %edi,%edi
  lea    (%r14,%rbp,1),%r11
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L2:
  test   %ecx,%ecx
  js     L48
  mov    %edi,%edx
  xor    %eax,%eax
L33:
  movsbl (%rbx,%rcx,1),%r8d
  sub    $0x30,%r8d
  xor    %r8d,%eax
  add    %r8d,%edx
L16:
  xor    %edi,%eax
  xor    %edi,%edi
  add    $0x30,%eax
  cmp    $0x1,%edx
  setg   %dil
  mov    %al,(%r10,%rsi,1)
  sub    $0x1,%r9d
  add    $0x1,%rsi
  sub    $0x1,%rcx
L1:
  test   %r9d,%r9d
  jns    L82
  test   %edi,%edi
  je     L2
  test   %r9d,%r9d
  js     L56
L82:
  movsbl (%r11,%rcx,1),%edx
  sub    $0x30,%edx
L87:
  mov    %edx,%eax
  add    %edi,%edx
  test   %ecx,%ecx
  js     L16
  jmp    L33
  nopl   (%rax)
L48:
  lea    -0x1(%rsi),%ecx
  xor    %eax,%eax
  movslq %ecx,%rdx
  test   %ecx,%ecx
  jle    L64
  nopl   0x0(%rax)
L77:
  movzbl (%r10,%rax,1),%ecx
  movzbl (%r10,%rdx,1),%esi
  mov    %sil,(%r10,%rax,1)
  add    $0x1,%rax
  mov    %cl,(%r10,%rdx,1)
  sub    $0x1,%rdx
  cmp    %eax,%edx
  jg     L77
L64:
  pop    %rbx
  mov    %r10,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L56:
  xor    %edx,%edx
  jmp    L87
```

assistant: ```c
char *addBinary(char *a, char *b) {
  int len1 = strlen(a);
  int len2 = strlen(b);
  int len = len1 > len2 ? len1 + 1 : len2 + 1;
  char *result = malloc(len + 1);
  result[len] = '\0';
  result[len - 1] = '\0';
  int i, j, carry = 0;
  len = 0;
  for (i = len1 - 1, j = len2 - 1; carry || i >= 0 || j >= 0; i--, j--) {
    int na = i >= 0 ? a[i] - '0' : 0;
    int nb = j >= 0 ? b[j] - '0' : 0;
    result[len++] = (na ^ nb ^ carry) + '0';
    carry = carry + na + nb >= 2 ? 1 : 0;
  }
  for (i = 0, j = len - 1; i < j; i++, j--) {
    char c = result[i];
    result[i] = result[j];
    result[j] = c;
  }
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<addBinary>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rsi,%rbx
  call   <strlen@plt>
  mov    %rbx,%rdi
  movslq %eax,%rbp
  call   <strlen@plt>
  movslq %eax,%r12
  lea    0x1(%rbp),%eax
  cmp    %r12d,%ebp
  lea    0x1(%r12),%r13d
  cmovg  %eax,%r13d
  lea    0x1(%r13),%edi
  movslq %r13d,%r13
  call   <malloc@plt>
  lea    -0x1(%rbp),%r9d
  lea    -0x1(%r12),%ecx
  xor    %esi,%esi
  sub    %r12,%rbp
  movb   $0x0,(%rax,%r13,1)
  mov    %rax,%r10
  movslq %ecx,%rcx
  movb   $0x0,-0x1(%rax,%r13,1)
  xor    %edi,%edi
  lea    (%r14,%rbp,1),%r11
  jmp    L87
  nopw   0x0(%rax,%rax,1)
L98:
  test   %ecx,%ecx
  js     L26
  mov    %edi,%edx
  xor    %eax,%eax
L75:
  movsbl (%rbx,%rcx,1),%r8d
  sub    $0x30,%r8d
  xor    %r8d,%eax
  add    %r8d,%edx
L24:
  xor    %edi,%eax
  xor    %edi,%edi
  add    $0x30,%eax
  cmp    $0x1,%edx
  setg   %dil
  mov    %al,(%r10,%rsi,1)
  sub    $0x1,%r9d
  add    $0x1,%rsi
  sub    $0x1,%rcx
L87:
  test   %r9d,%r9d
  jns    L41
  test   %edi,%edi
  je     L98
  test   %r9d,%r9d
  js     L34
L41:
  movsbl (%r11,%rcx,1),%edx
  sub    $0x30,%edx
L30:
  mov    %edx,%eax
  add    %edi,%edx
  test   %ecx,%ecx
  js     L24
  jmp    L75
  nopl   (%rax)
L26:
  lea    -0x1(%rsi),%ecx
  xor    %eax,%eax
  movslq %ecx,%rdx
  test   %ecx,%ecx
  jle    L47
  nopl   0x0(%rax)
L13:
  movzbl (%r10,%rax,1),%ecx
  movzbl (%r10,%rdx,1),%esi
  mov    %sil,(%r10,%rax,1)
  add    $0x1,%rax
  mov    %cl,(%r10,%rdx,1)
  sub    $0x1,%rdx
  cmp    %edx,%eax
  jl     L13
L47:
  pop    %rbx
  mov    %r10,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L34:
  xor    %edx,%edx
  jmp    L30
```

assistant: ```c
char *addBinary(char *a, char *b) {
  int len1 = strlen(a);
  int len2 = strlen(b);
  int len = len1 > len2 ? len1 + 1 : len2 + 1;
  char *result = malloc(len + 1);
  result[len] = '\0';
  result[len - 1] = '\0';
  int i, j, carry = 0;
  len = 0;
  for (i = len1 - 1, j = len2 - 1; carry || i >= 0 || j >= 0; i--, j--) {
    int na = i >= 0 ? a[i] - '0' : 0;
    int nb = j >= 0 ? b[j] - '0' : 0;
    result[len++] = (na ^ nb ^ carry) + '0';
    carry = carry + na + nb >= 2 ? 1 : 0;
  }
  for (i = 0, j = len - 1; i < j; i++, j--) {
    char c = result[i];
    result[i] = result[j];
    result[j] = c;
  }
  return result;
}
```
user: Translate the below assembly code into C programming language:

```asm
<Malloc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  jmp    L26
L21:
  mov    $0x0,%eax
  call   <flushentry@plt>
  test   %rax,%rax
  je     L36
L26:
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L21
  jmp    L7
L36:
  nop
L7:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
void *Malloc(int size) {
  void *retval;
  while ((retval = malloc(size)) == NULL)
    if (flushentry() == 0)
      break;
  return (retval);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<Malloc>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebp
L14:
  mov    %ebp,%edi
  call   <malloc@plt>
  mov    %rax,%rbx
  test   %rax,%rax
  jne    L18
  mov    $0x0,%eax
  call   <flushentry@plt>
  test   %rax,%rax
  jne    L14
L18:
  mov    %rbx,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void *Malloc(int size) {
  void *retval;
  while ((retval = malloc(size)) == NULL)
    if (flushentry() == 0)
      break;
  return (retval);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<Malloc>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  jmp    L23
  nop
L4:
  xor    %eax,%eax
  call   <flushentry@plt>
  test   %rax,%rax
  je     L16
L23:
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L4
L16:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
void *Malloc(int size) {
  void *retval;
  while ((retval = malloc(size)) == NULL)
    if (flushentry() == 0)
      break;
  return (retval);
}
```
user: What is the c source code of the assembly code below:

```asm
<Malloc>:
  endbr64
  push   %r12
  push   %rbx
  mov    %edi,%ebx
  sub    $0x8,%rsp
  jmp    L23
  nop
L29:
  xor    %eax,%eax
  call   <flushentry@plt>
  test   %rax,%rax
  je     L17
L23:
  mov    %ebx,%edi
  call   <malloc@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L29
L17:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %r12
  ret
```

assistant: ```c
void *Malloc(int size) {
  void *retval;
  while ((retval = malloc(size)) == NULL)
    if (flushentry() == 0)
      break;
  return (retval);
}
```