ylfeng/ReF-Decompile-dataset huggingface dataset:
user: Converte this assembly snippet into its equivalent C code:

```asm
<CG_GetBlockCvarForTimer>:
  endbr64
  lea    D38(%rip),%rax
  cmp    $0x128,%edi
  je     L5
  jg     L23
  lea    D10(%rip),%rax
  cmp    $0x112,%edi
  je     L5
  cmp    $0x127,%edi
  lea    D34(%rip),%rax
  lea    D8(%rip),%rdx
  cmove  %rdx,%rax
  ret
L23:
  cmp    $0x129,%edi
  lea    D34(%rip),%rax
  lea    D23(%rip),%rdx
  cmove  %rdx,%rax
L5:
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D38
type_infered: string
name: disassemble_data
arguments: tag: D10
type_infered: string
name: disassemble_data
arguments: tag: D34
type_infered: string
name: disassemble_data
arguments: tag: D8
type_infered: string
name: disassemble_data
arguments: tag: D23
type_infered: string

tool: D38: ui_assaultblocked3

tool: D10: ui_assaultblocked1

tool: D34: wrong

tool: D8: ui_assaultblocked2

tool: D23: ui_assaultblocked4

assistant: ```c
const char *CG_GetBlockCvarForTimer(int timer) {
  switch (timer) {
  case 274:
    return "ui_assaultblocked1";
  case 295:
    return "ui_assaultblocked2";
  case 296:
    return "ui_assaultblocked3";
  case 297:
    return "ui_assaultblocked4";
  default:
    return "wrong";
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<CG_GetBlockCvarForTimer>:
  endbr64
  lea    D46(%rip),%rax
  cmp    $0x128,%edi
  je     L1
  jg     L2
  lea    D17(%rip),%rax
  cmp    $0x112,%edi
  je     L1
  cmp    $0x127,%edi
  lea    D8(%rip),%rdx
  lea    D15(%rip),%rax
  cmove  %rdx,%rax
  ret
  nopl   (%rax)
L2:
  cmp    $0x129,%edi
  lea    D15(%rip),%rax
  lea    D27(%rip),%rdx
  cmove  %rdx,%rax
L1:
  ret
```

tool_calls: name: extract_symbolic_data
arguments: label: D46
datatype: string
name: extract_symbolic_data
arguments: label: D17
datatype: string
name: extract_symbolic_data
arguments: label: D8
datatype: string
name: extract_symbolic_data
arguments: label: D15
datatype: string
name: extract_symbolic_data
arguments: label: D27
datatype: string

tool: D46: ui_assaultblocked3

tool: D17: ui_assaultblocked1

tool: D8: ui_assaultblocked2

tool: D15: wrong

tool: D27: ui_assaultblocked4

assistant: ```c
const char *CG_GetBlockCvarForTimer(int timer) {
  switch (timer) {
  case 274:
    return "ui_assaultblocked1";
  case 295:
    return "ui_assaultblocked2";
  case 296:
    return "ui_assaultblocked3";
  case 297:
    return "ui_assaultblocked4";
  default:
    return "wrong";
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<CG_GetBlockCvarForTimer>:
  endbr64
  lea    D6(%rip),%rax
  cmp    $0x128,%edi
  je     L15
  jg     L14
  lea    D57(%rip),%rax
  cmp    $0x112,%edi
  je     L15
  cmp    $0x127,%edi
  lea    D4(%rip),%rdx
  lea    D15(%rip),%rax
  cmove  %rdx,%rax
  ret
  nopl   (%rax)
L14:
  cmp    $0x129,%edi
  lea    D15(%rip),%rax
  lea    D47(%rip),%rdx
  cmove  %rdx,%rax
L15:
  ret
```

tool_calls: name: read_data
arguments: label: D6
data_type: string
name: read_data
arguments: label: D57
data_type: string
name: read_data
arguments: label: D4
data_type: string
name: read_data
arguments: label: D15
data_type: string
name: read_data
arguments: label: D47
data_type: string

tool: D6: ui_assaultblocked3

tool: D57: ui_assaultblocked1

tool: D4: ui_assaultblocked2

tool: D15: wrong

tool: D47: ui_assaultblocked4

assistant: ```c
const char *CG_GetBlockCvarForTimer(int timer) {
  switch (timer) {
  case 274:
    return "ui_assaultblocked1";
  case 295:
    return "ui_assaultblocked2";
  case 296:
    return "ui_assaultblocked3";
  case 297:
    return "ui_assaultblocked4";
  default:
    return "wrong";
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bcache_write>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <device_block_size@plt>
  mov    %eax,-0x4(%rbp)
  movl   $0x0,-0x10(%rbp)
  jmp    L35
L11:
  mov    -0x28(%rbp),%edx
  mov    -0x10(%rbp),%eax
  add    %eax,%edx
  mov    -0x10(%rbp),%eax
  imul   -0x4(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x20(%rbp),%rax
  add    %rax,%rcx
  mov    -0x18(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <bcache_write_block@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jle    L0
  addl   $0x1,-0x8(%rbp)
  addl   $0x1,-0x10(%rbp)
L35:
  mov    -0x10(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L11
  jmp    L52
L0:
  nop
L52:
  cmpl   $0x0,-0x8(%rbp)
  jle    L43
  mov    -0x8(%rbp),%eax
  jmp    L18
L43:
  mov    -0xc(%rbp),%eax
L18:
  leave
  ret
```

assistant: ```c
int bcache_write(struct device *device, const char *data, int blocks, int offset) {
  int i, r;
  int count = 0;
  int bs = device_block_size(device);
  for (i = 0; i < blocks; i++) {
    r = bcache_write_block(device, &data[i * bs], offset + i);
    if (r < 1)
      break;
    count++;
  }
  if (count > 0) {
    return count;
  } else {
    return r;
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<bcache_write>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %edx,%r14d
  mov    %ecx,%r15d
  call   <device_block_size@plt>
  test   %r14d,%r14d
  jle    L37
  movslq %eax,%r12
  mov    $0x0,%ebx
L1:
  lea    (%r15,%rbx,1),%edx
  mov    %rbp,%rsi
  mov    %r13,%rdi
  call   <bcache_write_block@plt>
  test   %eax,%eax
  jle    L30
  add    $0x1,%ebx
  add    %r12,%rbp
  cmp    %ebx,%r14d
  jne    L1
  mov    %r14d,%ebx
L30:
  test   %ebx,%ebx
  cmovle %eax,%ebx
L15:
  mov    %ebx,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  mov    $0x0,%ebx
  jmp    L15
```

assistant: ```c
int bcache_write(struct device *device, const char *data, int blocks, int offset) {
  int i, r;
  int count = 0;
  int bs = device_block_size(device);
  for (i = 0; i < blocks; i++) {
    r = bcache_write_block(device, &data[i * bs], offset + i);
    if (r < 1)
      break;
    count++;
  }
  if (count > 0) {
    return count;
  } else {
    return r;
  }
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<bcache_write>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <device_block_size@plt>
  test   %r12d,%r12d
  jle    L5
  movslq %eax,%rbx
  xor    %r14d,%r14d
  jmp    L49
L33:
  add    $0x1,%r14d
  add    %rbx,%r15
  cmp    %r14d,%r12d
  je     L1
L49:
  lea    0x0(%r13,%r14,1),%edx
  mov    %r15,%rsi
  mov    %rbp,%rdi
  call   <bcache_write_block@plt>
  test   %eax,%eax
  jg     L33
  test   %r14d,%r14d
  cmovne %r14d,%eax
L36:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L1:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L5:
  xor    %eax,%eax
  jmp    L36
```

assistant: ```c
int bcache_write(struct device *device, const char *data, int blocks, int offset) {
  int i, r;
  int count = 0;
  int bs = device_block_size(device);
  for (i = 0; i < blocks; i++) {
    r = bcache_write_block(device, &data[i * bs], offset + i);
    if (r < 1)
      break;
    count++;
  }
  if (count > 0) {
    return count;
  } else {
    return r;
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<bcache_write>:
  endbr64
  push   %r15
  mov    %rsi,%r15
  push   %r14
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %edx,%r12d
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <device_block_size@plt>
  test   %r12d,%r12d
  jle    L27
  movslq %eax,%rbx
  xor    %r14d,%r14d
  jmp    L32
L2:
  add    $0x1,%r14d
  add    %rbx,%r15
  cmp    %r14d,%r12d
  je     L26
L32:
  lea    0x0(%r13,%r14,1),%edx
  mov    %r15,%rsi
  mov    %rbp,%rdi
  call   <bcache_write_block@plt>
  test   %eax,%eax
  jg     L2
  test   %r14d,%r14d
  cmovne %r14d,%eax
L10:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L26:
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L27:
  xor    %eax,%eax
  jmp    L10
```

assistant: ```c
int bcache_write(struct device *device, const char *data, int blocks, int offset) {
  int i, r;
  int count = 0;
  int bs = device_block_size(device);
  for (i = 0; i < blocks; i++) {
    r = bcache_write_block(device, &data[i * bs], offset + i);
    if (r < 1)
      break;
    count++;
  }
  if (count > 0) {
    return count;
  } else {
    return r;
  }
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<glmInterpolateVector2f>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  movss  %xmm0,-0x2c(%rbp)
  movss  D2(%rip),%xmm0
  subss  -0x2c(%rbp),%xmm0
  movss  %xmm0,-0x4(%rbp)
  mov    -0x20(%rbp),%rax
  movss  (%rax),%xmm0
  movaps %xmm0,%xmm1
  mulss  -0x4(%rbp),%xmm1
  mov    -0x28(%rbp),%rax
  movss  (%rax),%xmm0
  mulss  -0x2c(%rbp),%xmm0
  addss  %xmm1,%xmm0
  mov    -0x18(%rbp),%rax
  movss  %xmm0,(%rax)
  mov    -0x20(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  movaps %xmm0,%xmm1
  mulss  -0x4(%rbp),%xmm1
  mov    -0x28(%rbp),%rax
  add    $0x4,%rax
  movss  (%rax),%xmm0
  mulss  -0x2c(%rbp),%xmm0
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  addss  %xmm1,%xmm0
  movss  %xmm0,(%rax)
  nop
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D2
type_guessed: f32

tool: D2: 1

assistant: ```c
void glmInterpolateVector2f(float result[2], const float v1[2], const float v2[2], float a) {
  float b = 1.0 - a;
  result[0] = b * v1[0] + a * v2[0];
  result[1] = b * v1[1] + a * v2[1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<glmInterpolateVector2f>:
  endbr64
  movss  D6(%rip),%xmm1
  subss  %xmm0,%xmm1
  movaps %xmm1,%xmm2
  mulss  (%rsi),%xmm2
  movaps %xmm0,%xmm3
  mulss  (%rdx),%xmm3
  addss  %xmm3,%xmm2
  movss  %xmm2,(%rdi)
  mulss  0x4(%rsi),%xmm1
  mulss  0x4(%rdx),%xmm0
  addss  %xmm0,%xmm1
  movss  %xmm1,0x4(%rdi)
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D6
data_type: f32

tool: D6: 1

assistant: ```c
void glmInterpolateVector2f(float result[2], const float v1[2], const float v2[2], float a) {
  float b = 1.0 - a;
  result[0] = b * v1[0] + a * v2[0];
  result[1] = b * v1[1] + a * v2[1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<glmInterpolateVector2f>:
  endbr64
  movss  D1(%rip),%xmm1
  movss  (%rsi),%xmm2
  movss  (%rdx),%xmm3
  subss  %xmm0,%xmm1
  mulss  %xmm0,%xmm3
  mulss  %xmm1,%xmm2
  addss  %xmm3,%xmm2
  movss  %xmm2,(%rdi)
  mulss  0x4(%rsi),%xmm1
  mulss  0x4(%rdx),%xmm0
  addss  %xmm0,%xmm1
  movss  %xmm1,0x4(%rdi)
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D1
type: f32

tool: D1: 1

assistant: ```c
void glmInterpolateVector2f(float result[2], const float v1[2], const float v2[2], float a) {
  float b = 1.0 - a;
  result[0] = b * v1[0] + a * v2[0];
  result[1] = b * v1[1] + a * v2[1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<glmInterpolateVector2f>:
  endbr64
  movss  D2(%rip),%xmm1
  movss  (%rsi),%xmm2
  movss  (%rdx),%xmm3
  subss  %xmm0,%xmm1
  mulss  %xmm0,%xmm3
  mulss  %xmm1,%xmm2
  addss  %xmm3,%xmm2
  movss  %xmm2,(%rdi)
  mulss  0x4(%rsi),%xmm1
  mulss  0x4(%rdx),%xmm0
  addss  %xmm0,%xmm1
  movss  %xmm1,0x4(%rdi)
  ret
```

tool_calls: name: read_data
arguments: tag: D2
type: f32

tool: D2: 1

assistant: ```c
void glmInterpolateVector2f(float result[2], const float v1[2], const float v2[2], float a) {
  float b = 1.0 - a;
  result[0] = b * v1[0] + a * v2[0];
  result[1] = b * v1[1] + a * v2[1];
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <getid@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <f@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <printid@plt>
  nop
  leave
  ret
```

assistant: ```c
void main() {
  int n;
  int fakt;
  getid(n);
  fakt = f(n);
  printid(fakt);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    $0x0,%ebx
  mov    %ebx,%edi
  call   <getid@plt>
  mov    %ebx,%edi
  call   <f@plt>
  mov    %eax,%edi
  call   <printid@plt>
  pop    %rbx
  ret
```

assistant: ```c
void main() {
  int n;
  int fakt;
  getid(n);
  fakt = f(n);
  printid(fakt);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  call   <getid@plt>
  mov    %ebp,%edi
  call   <f@plt>
  pop    %rbp
  mov    %eax,%edi
  jmp    1080 <printid@plt>
```

assistant: ```c
void main() {
  int n;
  int fakt;
  getid(n);
  fakt = f(n);
  printid(fakt);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %rbp
  xor    %ebp,%ebp
  mov    %ebp,%edi
  call   <getid@plt>
  mov    %ebp,%edi
  call   <f@plt>
  pop    %rbp
  mov    %eax,%edi
  jmp    1080 <printid@plt>
```

assistant: ```c
void main() {
  int n;
  int fakt;
  getid(n);
  fakt = f(n);
  printid(fakt);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<calloc_2d>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  mov    -0x28(%rbp),%rax
  shl    $0x3,%rax
  mov    %rax,%rdi
  call   <malloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L39
  mov    -0x28(%rbp),%rax
  imul   -0x30(%rbp),%rax
  mov    -0x38(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <calloc@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  je     L33
  movq   $0x0,-0x10(%rbp)
  jmp    L32
L6:
  mov    -0x10(%rbp),%rax
  imul   -0x30(%rbp),%rax
  imul   -0x38(%rbp),%rax
  mov    %rax,%rcx
  mov    -0x10(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0x8(%rbp),%rdx
  add    %rcx,%rdx
  mov    %rdx,(%rax)
  addq   $0x1,-0x10(%rbp)
L32:
  mov    -0x10(%rbp),%rax
  cmp    -0x28(%rbp),%rax
  jb     L6
  jmp    L39
L33:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  movq   $0x0,-0x18(%rbp)
L39:
  mov    -0x18(%rbp),%rax
  leave
  ret
```

assistant: ```c
void **calloc_2d(size_t n, size_t m, size_t size) {
  char **ret = (char **)malloc(n * sizeof(char *));
  if (ret) {
    char *area = (char *)calloc(n * m, size);
    if (area) {
      for (size_t i = 0; i < n; ++i)
        ret[i] = (char *)&area[i * m * size];
    } else {
      free(ret);
      ret = NULL;
    }
  }
  return (void **)ret;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<calloc_2d>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r14
  mov    %rsi,%rbx
  mov    %rdx,%r15
  lea    0x0(,%rdi,8),%r12
  mov    %r12,%rdi
  call   <malloc@plt>
  mov    %rax,%r13
  test   %rax,%rax
  jne    L32
L50:
  mov    %r13,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L32:
  mov    %r14,%rdi
  imul   %rbx,%rdi
  mov    %r15,%rsi
  call   <calloc@plt>
  test   %rax,%rax
  je     L53
  test   %r14,%r14
  je     L50
  imul   %r15,%rbx
  mov    %r13,%rdx
  add    %r13,%r12
L31:
  mov    %rax,(%rdx)
  add    %rbx,%rax
  add    $0x8,%rdx
  cmp    %r12,%rdx
  jne    L31
  jmp    L50
L53:
  mov    %r13,%rdi
  call   <free@plt>
  mov    $0x0,%r13d
  jmp    L50
```

assistant: ```c
void **calloc_2d(size_t n, size_t m, size_t size) {
  char **ret = (char **)malloc(n * sizeof(char *));
  if (ret) {
    char *area = (char *)calloc(n * m, size);
    if (area) {
      for (size_t i = 0; i < n; ++i)
        ret[i] = (char *)&area[i * m * size];
    } else {
      free(ret);
      ret = NULL;
    }
  }
  return (void **)ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<calloc_2d>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x0(,%rdi,8),%rbp
  push   %rbx
  mov    %rbp,%rdi
  mov    %rsi,%rbx
  call   <malloc@plt>
  test   %rax,%rax
  jne    L14
L23:
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L14:
  mov    %r12,%rdi
  mov    %r13,%rsi
  mov    %rax,%r14
  imul   %rbx,%rdi
  call   <calloc@plt>
  test   %rax,%rax
  je     L39
  test   %r12,%r12
  je     L23
  imul   %r13,%rbx
  mov    %r14,%rdx
  lea    0x0(%rbp,%r14,1),%rcx
  nopl   0x0(%rax,%rax,1)
L36:
  mov    %rax,(%rdx)
  add    $0x8,%rdx
  add    %rbx,%rax
  cmp    %rcx,%rdx
  jne    L36
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L39:
  mov    %r14,%rdi
  xor    %r14d,%r14d
  call   <free@plt>
  jmp    L23
```

assistant: ```c
void **calloc_2d(size_t n, size_t m, size_t size) {
  char **ret = (char **)malloc(n * sizeof(char *));
  if (ret) {
    char *area = (char *)calloc(n * m, size);
    if (area) {
      for (size_t i = 0; i < n; ++i)
        ret[i] = (char *)&area[i * m * size];
    } else {
      free(ret);
      ret = NULL;
    }
  }
  return (void **)ret;
}
```
user: Translate the below assembly code into C programming language:

```asm
<calloc_2d>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %rdi,%r12
  push   %rbp
  lea    0x0(,%rdi,8),%rbp
  push   %rbx
  mov    %rbp,%rdi
  mov    %rsi,%rbx
  call   <malloc@plt>
  test   %rax,%rax
  jne    L13
L26:
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L13:
  mov    %r12,%rdi
  mov    %r13,%rsi
  mov    %rax,%r14
  imul   %rbx,%rdi
  call   <calloc@plt>
  test   %rax,%rax
  je     L47
  test   %r12,%r12
  je     L26
  imul   %r13,%rbx
  mov    %r14,%rdx
  lea    0x0(%rbp,%r14,1),%rcx
  nopl   0x0(%rax,%rax,1)
L32:
  mov    %rax,(%rdx)
  add    $0x8,%rdx
  add    %rbx,%rax
  cmp    %rcx,%rdx
  jne    L32
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax,%rax,1)
L47:
  mov    %r14,%rdi
  xor    %r14d,%r14d
  call   <free@plt>
  jmp    L26
```

assistant: ```c
void **calloc_2d(size_t n, size_t m, size_t size) {
  char **ret = (char **)malloc(n * sizeof(char *));
  if (ret) {
    char *area = (char *)calloc(n * m, size);
    if (area) {
      for (size_t i = 0; i < n; ++i)
        ret[i] = (char *)&area[i * m * size];
    } else {
      free(ret);
      ret = NULL;
    }
  }
  return (void **)ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<test1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %esi,-0xc(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <__builtin_assume@plt>
  addl   $0x1,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  setne  %al
  movzbl %al,%eax
  mov    %eax,%edi
  call   <__builtin_assume@plt>
  mov    -0x8(%rbp),%rax
  mov    (%rax),%eax
  leave
  ret
```

assistant: ```c
int test1(int *a, int i) {
  __builtin_assume(a != 0);
  __builtin_assume(++i != 0);
  return a[0];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<test1>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  mov    %esi,%ebp
  test   %rdi,%rdi
  setne  %dil
  movzbl %dil,%edi
  call   <__builtin_assume@plt>
  cmp    $0xffffffff,%ebp
  setne  %dil
  movzbl %dil,%edi
  call   <__builtin_assume@plt>
  mov    (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int test1(int *a, int i) {
  __builtin_assume(a != 0);
  __builtin_assume(++i != 0);
  return a[0];
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<test1>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  sub    $0x8,%rsp
  test   %rbx,%rbx
  setne  %dil
  call   <__builtin_assume@plt>
  xor    %edi,%edi
  cmp    $0xffffffff,%ebp
  setne  %dil
  call   <__builtin_assume@plt>
  mov    (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int test1(int *a, int i) {
  __builtin_assume(a != 0);
  __builtin_assume(++i != 0);
  return a[0];
}
```
user: What is the c source code of the assembly code below:

```asm
<test1>:
  endbr64
  push   %rbp
  mov    %esi,%ebp
  push   %rbx
  mov    %rdi,%rbx
  xor    %edi,%edi
  sub    $0x8,%rsp
  test   %rbx,%rbx
  setne  %dil
  call   <__builtin_assume@plt>
  xor    %edi,%edi
  cmp    $0xffffffff,%ebp
  setne  %dil
  call   <__builtin_assume@plt>
  mov    (%rbx),%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
int test1(int *a, int i) {
  __builtin_assume(a != 0);
  __builtin_assume(++i != 0);
  return a[0];
}
```