ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r14
  lea    D18(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  movabs $0x3f00000000,%rax
  lea    0xc(%rsp),%r12
  lea    0x18(%rsp),%r13
  mov    %rax,0x10(%rsp)
  mov    $0x80,%eax
L13:
  shl    $0x8,%eax
  mov    $0x8,%ebp
  xor    %eax,%ebx
  nopl   (%rax)
L8:
  mov    %ebx,%edx
  add    %ebx,%ebx
  mov    %r14,%rdi
  mov    %ebx,%eax
  and    $0x8000,%edx
  xor    $0x1021,%eax
  test   %edx,%edx
  cmovne %eax,%ebx
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  sub    $0x1,%ebp
  jne    L8
  add    $0x4,%r12
  cmp    %r13,%r12
  je     L14
  mov    (%r12),%eax
  jmp    L13
  cs nopw 0x0(%rax,%rax,1)
L14:
  mov    %ebx,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  movzbl %bl,%edx
  movzbl %bh,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D18
datatype: string
name: parse_data
arguments: data_tag: D8
datatype: string

tool: D18: %u


tool: D8: 0x%02X 0x%02X


assistant: ```c
int main(int argc, char **argv) {
  uint8_t bytes[3];
  bytes[0] = 0x80;
  bytes[1] = 0;
  bytes[2] = 63;
  uint32_t crc;
  for (int i = 0; i < 3; i++) {
    crc = crc ^ (bytes[i] << 8);
    for (int j = 0; j < 8; j++) {
      if (crc & 0x8000) {
        crc = (crc << 1) ^ 0x1021;
      } else {
        crc = crc << 1;
      }
      printf("%u\n", crc);
    }
  }
  printf("%u\n", crc);
  printf("0x%02X 0x%02X\n", (crc >> 8) & 0xFF, crc & 0xFF);
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<main>:
  endbr64
  push   %r14
  lea    D23(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    D24(%rip),%rax
  movl   $0x3f,0x10(%rsp)
  lea    0x8(%rsp),%r12
  lea    0x14(%rsp),%r13
  mov    %rax,0x8(%rsp)
L20:
  mov    (%r12),%eax
  mov    $0x8,%ebp
  shl    $0x8,%eax
  xor    %eax,%ebx
  nopl   0x0(%rax,%rax,1)
L10:
  mov    %ebx,%edx
  add    %ebx,%ebx
  mov    %r14,%rdi
  mov    %ebx,%eax
  and    $0x8000,%edx
  xor    $0x1021,%eax
  test   %edx,%edx
  cmovne %eax,%ebx
  xor    %eax,%eax
  mov    %ebx,%esi
  call   <printf@plt>
  sub    $0x1,%ebp
  jne    L10
  add    $0x4,%r12
  cmp    %r12,%r13
  jne    L20
  mov    %ebx,%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  movzbl %bl,%edx
  movzbl %bh,%esi
  lea    D27(%rip),%rdi
  call   <printf@plt>
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L25
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L25:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: label: D23
type: string
name: read_data
arguments: label: D24
type: byte[8]
name: read_data
arguments: label: D27
type: string

tool: D23: %u


tool: {"D24": [0x80,0x00,0x00,0x00,0x00,0x00,0x00,0x00]}

tool: D27: 0x%02X 0x%02X


assistant: ```c
int main(int argc, char **argv) {
  uint8_t bytes[3];
  bytes[0] = 0x80;
  bytes[1] = 0;
  bytes[2] = 63;
  uint32_t crc;
  for (int i = 0; i < 3; i++) {
    crc = crc ^ (bytes[i] << 8);
    for (int j = 0; j < 8; j++) {
      if (crc & 0x8000) {
        crc = (crc << 1) ^ 0x1021;
      } else {
        crc = crc << 1;
      }
      printf("%u\n", crc);
    }
  }
  printf("%u\n", crc);
  printf("0x%02X 0x%02X\n", (crc >> 8) & 0xFF, crc & 0xFF);
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<gint>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D9(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    $0xa,%eax
  leave
  ret
```

tool_calls: name: parse_data
arguments: tag: D9
data_type: string

tool: D9: int is %d


assistant: ```c
int gint(int v) {
  printf("int is %d\n", v);
  return 10;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<gint>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D4(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D4
data_type: string

tool: D4: int is %d


assistant: ```c
int gint(int v) {
  printf("int is %d\n", v);
  return 10;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<gint>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%eax
  add    $0x8,%rsp
  ret
```

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

tool: D8: int is %d


assistant: ```c
int gint(int v) {
  printf("int is %d\n", v);
  return 10;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<gint>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    $0xa,%eax
  add    $0x8,%rsp
  ret
```

tool_calls: name: parse_data
arguments: tag: D1
type_guessed: string

tool: D1: int is %d


assistant: ```c
int gint(int v) {
  printf("int is %d\n", v);
  return 10;
}
```
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    $0x20,%rsp
  movsd  D57(%rip),%xmm0
  movsd  %xmm0,-0x8(%rbp)
  movl   $0x1,-0x1c(%rbp)
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D86(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movsd  D87(%rip),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  divsd  -0x8(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rdx
  mov    -0x1c(%rbp),%eax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  jmp    L8
L4:
  addl   $0x1,-0x1c(%rbp)
  movsd  -0x18(%rbp),%xmm0
  addsd  %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x10(%rbp),%xmm0
  addsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm0
  movapd %xmm0,%xmm1
  divsd  -0x8(%rbp),%xmm1
  movsd  -0x10(%rbp),%xmm0
  mov    -0x18(%rbp),%rdx
  mov    -0x1c(%rbp),%eax
  movapd %xmm1,%xmm2
  movapd %xmm0,%xmm1
  movq   %rdx,%xmm0
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
L8:
  cmpl   $0x3f,-0x1c(%rbp)
  jle    L4
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D57
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D4
data_type: string
name: extract_symbolic_data
arguments: data_tag: D17
data_type: string
name: extract_symbolic_data
arguments: data_tag: D42
data_type: string
name: extract_symbolic_data
arguments: data_tag: D86
data_type: string
name: extract_symbolic_data
arguments: data_tag: D87
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D28
data_type: string
name: extract_symbolic_data
arguments: data_tag: D68
data_type: string

tool: D57: 2e+16

tool: D4: square      grains        total      

tool: D17: fraction of 

tool: D42:             added         grains     

tool: D86: world total

tool: D87: 1

tool: D28: %4d %13.2e %12.2e %12.2e


tool: D68: That's all.

assistant: ```c
int main(void) {
  const double CROP = 2E16;
  double current, total;
  int count = 1;
  printf("square      grains        total      ");
  printf("fraction of \n");
  printf("            added         grains     ");
  printf("world total\n");
  total = current = 1.0;
  printf("%4d %13.2e %12.2e %12.2e\n", count, current, total, total / CROP);
  while (count < 64) {
    count = count + 1;
    current = 2.0 * current;
    total = total + current;
    printf("%4d %13.2e %12.2e %12.2e\n", count, current, total, total / CROP);
  }
  printf("That's all.\n");
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  lea    D95(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D58(%rip),%rdi
  call   <puts@plt>
  lea    D98(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D52(%rip),%rdi
  call   <puts@plt>
  movsd  D46(%rip),%xmm2
  mov    D100(%rip),%rax
  movq   %rax,%xmm1
  movq   %rax,%xmm0
  mov    $0x1,%esi
  lea    D44(%rip),%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  mov    $0x1,%ebp
  movabs $0x3ff0000000000000,%rbx
  mov    D100(%rip),%rax
  mov    %rax,0x8(%rsp)
  lea    D44(%rip),%r12
L0:
  add    $0x1,%ebp
  movsd  0x8(%rsp),%xmm4
  movapd %xmm4,%xmm3
  addsd  %xmm4,%xmm3
  movapd %xmm3,%xmm0
  movsd  %xmm3,0x8(%rsp)
  movq   %rbx,%xmm5
  addsd  %xmm3,%xmm5
  movq   %xmm5,%rbx
  movapd %xmm5,%xmm2
  divsd  D22(%rip),%xmm2
  movapd %xmm5,%xmm1
  mov    %ebp,%esi
  mov    %r12,%rdi
  mov    $0x3,%eax
  call   <printf@plt>
  cmp    $0x40,%ebp
  jne    L0
  lea    D11(%rip),%rdi
  call   <puts@plt>
  mov    $0x0,%eax
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
```

tool_calls: name: read_data
arguments: label: D95
type_infered: string
name: read_data
arguments: label: D58
type_infered: string
name: read_data
arguments: label: D98
type_infered: string
name: read_data
arguments: label: D52
type_infered: string
name: read_data
arguments: label: D46
type_infered: byte[8]
name: read_data
arguments: label: D100
type_infered: f64
name: read_data
arguments: label: D44
type_infered: string
name: read_data
arguments: label: D22
type_infered: f64
name: read_data
arguments: label: D11
type_infered: string

tool: D95: square      grains        total      

tool: D58: fraction of 

tool: D98:             added         grains     

tool: D52: world total

tool: {"D46": [0xbc,0x89,0xd8,0x97,0xb2,0xd2,0x8c,0x3c]}

tool: D100: 1

tool: D44: %4d %13.2e %12.2e %12.2e


tool: D22: 2e+16

tool: D11: That's all.

assistant: ```c
int main(void) {
  const double CROP = 2E16;
  double current, total;
  int count = 1;
  printf("square      grains        total      ");
  printf("fraction of \n");
  printf("            added         grains     ");
  printf("world total\n");
  total = current = 1.0;
  printf("%4d %13.2e %12.2e %12.2e\n", count, current, total, total / CROP);
  while (count < 64) {
    count = count + 1;
    current = 2.0 * current;
    total = total + current;
    printf("%4d %13.2e %12.2e %12.2e\n", count, current, total, total / CROP);
  }
  printf("That's all.\n");
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D97(%rip),%rdi
  xor    %eax,%eax
  lea    D25(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x18,%rsp
  call   <printf@plt>
  lea    D38(%rip),%rdi
  call   <puts@plt>
  lea    D78(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D4(%rip),%rdi
  call   <puts@plt>
  movsd  D52(%rip),%xmm1
  mov    $0x1,%esi
  mov    %rbp,%rdi
  movsd  D96(%rip),%xmm2
  mov    $0x3,%eax
  movapd %xmm1,%xmm0
  call   <printf@plt>
  mov    D52(%rip),%rax
  movq   %rax,%xmm1
  movq   %rax,%xmm0
  nopl   0x0(%rax)
L5:
  addsd  %xmm0,%xmm0
  add    $0x1,%ebx
  mov    %rbp,%rdi
  mov    $0x3,%eax
  mov    %ebx,%esi
  addsd  %xmm0,%xmm1
  movsd  %xmm0,(%rsp)
  movapd %xmm1,%xmm2
  movsd  %xmm1,0x8(%rsp)
  divsd  D32(%rip),%xmm2
  call   <printf@plt>
  cmp    $0x40,%ebx
  movsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  jne    L5
  lea    D21(%rip),%rdi
  call   <puts@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: parse_data
arguments: tag: D97
type: string
name: parse_data
arguments: tag: D25
type: string
name: parse_data
arguments: tag: D38
type: string
name: parse_data
arguments: tag: D78
type: string
name: parse_data
arguments: tag: D4
type: string
name: parse_data
arguments: tag: D52
type: f64
name: parse_data
arguments: tag: D96
type: byte[8]
name: parse_data
arguments: tag: D32
type: f64
name: parse_data
arguments: tag: D21
type: string

tool: D97: square      grains        total      

tool: D25: %4d %13.2e %12.2e %12.2e


tool: D38: fraction of 

tool: D78:             added         grains     

tool: D4: world total

tool: D52: 1

tool: {"D96": [0xbc,0x89,0xd8,0x97,0xb2,0xd2,0x8c,0x3c]}

tool: D32: 2e+16

tool: D21: That's all.

assistant: ```c
int main(void) {
  const double CROP = 2E16;
  double current, total;
  int count = 1;
  printf("square      grains        total      ");
  printf("fraction of \n");
  printf("            added         grains     ");
  printf("world total\n");
  total = current = 1.0;
  printf("%4d %13.2e %12.2e %12.2e\n", count, current, total, total / CROP);
  while (count < 64) {
    count = count + 1;
    current = 2.0 * current;
    total = total + current;
    printf("%4d %13.2e %12.2e %12.2e\n", count, current, total, total / CROP);
  }
  printf("That's all.\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   %rbp
  lea    D30(%rip),%rdi
  xor    %eax,%eax
  lea    D13(%rip),%rbp
  push   %rbx
  mov    $0x1,%ebx
  sub    $0x18,%rsp
  call   <printf@plt>
  lea    D88(%rip),%rdi
  call   <puts@plt>
  lea    D6(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D33(%rip),%rdi
  call   <puts@plt>
  movsd  D50(%rip),%xmm1
  mov    $0x1,%esi
  mov    %rbp,%rdi
  movsd  D81(%rip),%xmm2
  mov    $0x3,%eax
  movapd %xmm1,%xmm0
  call   <printf@plt>
  mov    D50(%rip),%rax
  movq   %rax,%xmm1
  movq   %rax,%xmm0
  nopl   0x0(%rax)
L5:
  addsd  %xmm0,%xmm0
  add    $0x1,%ebx
  mov    %rbp,%rdi
  mov    $0x3,%eax
  mov    %ebx,%esi
  addsd  %xmm0,%xmm1
  movsd  %xmm0,(%rsp)
  movapd %xmm1,%xmm2
  movsd  %xmm1,0x8(%rsp)
  divsd  D66(%rip),%xmm2
  call   <printf@plt>
  cmp    $0x40,%ebx
  movsd  (%rsp),%xmm0
  movsd  0x8(%rsp),%xmm1
  jne    L5
  lea    D8(%rip),%rdi
  call   <puts@plt>
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D30
type: string
name: extract_symbolic_data
arguments: data_label: D13
type: string
name: extract_symbolic_data
arguments: data_label: D88
type: string
name: extract_symbolic_data
arguments: data_label: D6
type: string
name: extract_symbolic_data
arguments: data_label: D33
type: string
name: extract_symbolic_data
arguments: data_label: D50
type: f64
name: extract_symbolic_data
arguments: data_label: D81
type: byte[8]
name: extract_symbolic_data
arguments: data_label: D66
type: f64
name: extract_symbolic_data
arguments: data_label: D8
type: string

tool: D30: square      grains        total      

tool: D13: %4d %13.2e %12.2e %12.2e


tool: D88: fraction of 

tool: D6:             added         grains     

tool: D33: world total

tool: D50: 1

tool: {"D81": [0xbc,0x89,0xd8,0x97,0xb2,0xd2,0x8c,0x3c]}

tool: D66: 2e+16

tool: D8: That's all.

assistant: ```c
int main(void) {
  const double CROP = 2E16;
  double current, total;
  int count = 1;
  printf("square      grains        total      ");
  printf("fraction of \n");
  printf("            added         grains     ");
  printf("world total\n");
  total = current = 1.0;
  printf("%4d %13.2e %12.2e %12.2e\n", count, current, total, total / CROP);
  while (count < 64) {
    count = count + 1;
    current = 2.0 * current;
    total = total + current;
    printf("%4d %13.2e %12.2e %12.2e\n", count, current, total, total / CROP);
  }
  printf("That's all.\n");
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<asc2nibble>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,%eax
  mov    %al,-0x4(%rbp)
  cmpb   $0x2f,-0x4(%rbp)
  jle    L71
  cmpb   $0x39,-0x4(%rbp)
  jg     L71
  movsbl -0x4(%rbp),%eax
  sub    $0x30,%eax
  jmp    L39
L71:
  cmpb   $0x40,-0x4(%rbp)
  jle    L72
  cmpb   $0x46,-0x4(%rbp)
  jg     L72
  movsbl -0x4(%rbp),%eax
  sub    $0x37,%eax
  jmp    L39
L72:
  cmpb   $0x60,-0x4(%rbp)
  jle    L80
  cmpb   $0x66,-0x4(%rbp)
  jg     L80
  movsbl -0x4(%rbp),%eax
  sub    $0x57,%eax
  jmp    L39
L80:
  mov    $0x10,%eax
L39:
  pop    %rbp
  ret
```

assistant: ```c
int asc2nibble(char c) {
  if ((c >= '0') && (c <= '9'))
    return c - '0';
  if ((c >= 'A') && (c <= 'F'))
    return c - 'A' + 10;
  if ((c >= 'a') && (c <= 'f'))
    return c - 'a' + 10;
  return 16;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<asc2nibble>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L0
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L7
  lea    -0x61(%rdi),%edx
  movsbl %dil,%eax
  sub    $0x57,%eax
  cmp    $0x6,%dl
  mov    $0x10,%edx
  cmovae %edx,%eax
  ret
L0:
  movsbl %dil,%edi
  lea    -0x30(%rdi),%eax
  ret
L7:
  movsbl %dil,%edi
  lea    -0x37(%rdi),%eax
  ret
```

assistant: ```c
int asc2nibble(char c) {
  if ((c >= '0') && (c <= '9'))
    return c - '0';
  if ((c >= 'A') && (c <= 'F'))
    return c - 'A' + 10;
  if ((c >= 'a') && (c <= 'f'))
    return c - 'a' + 10;
  return 16;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<asc2nibble>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L19
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L9
  lea    -0x61(%rdi),%edx
  lea    -0x57(%rdi),%eax
  cmp    $0x6,%dl
  movsbl %al,%eax
  mov    $0x10,%edx
  cmovae %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  sub    $0x37,%edi
  movsbl %dil,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  movsbl %al,%eax
  ret
```

assistant: ```c
int asc2nibble(char c) {
  if ((c >= '0') && (c <= '9'))
    return c - '0';
  if ((c >= 'A') && (c <= 'F'))
    return c - 'A' + 10;
  if ((c >= 'a') && (c <= 'f'))
    return c - 'a' + 10;
  return 16;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<asc2nibble>:
  endbr64
  lea    -0x30(%rdi),%eax
  cmp    $0x9,%al
  jbe    L13
  lea    -0x41(%rdi),%eax
  cmp    $0x5,%al
  jbe    L1
  lea    -0x61(%rdi),%edx
  lea    -0x57(%rdi),%eax
  cmp    $0x6,%dl
  movsbl %al,%eax
  mov    $0x10,%edx
  cmovae %edx,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L1:
  sub    $0x37,%edi
  movsbl %dil,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L13:
  movsbl %al,%eax
  ret
```

assistant: ```c
int asc2nibble(char c) {
  if ((c >= '0') && (c <= '9'))
    return c - '0';
  if ((c >= 'A') && (c <= 'F'))
    return c - 'A' + 10;
  if ((c >= 'a') && (c <= 'f'))
    return c - 'a' + 10;
  return 16;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<streq>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L20
L17:
  mov    -0x18(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x18(%rbp)
  movzbl (%rax),%ecx
  mov    -0x20(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x20(%rbp)
  movzbl (%rax),%eax
  cmp    %al,%cl
  je     L7
  mov    $0x0,%eax
  jmp    L10
L7:
  addl   $0x1,-0x4(%rbp)
L20:
  mov    -0x20(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L17
  mov    -0x4(%rbp),%eax
L10:
  pop    %rbp
  ret
```

assistant: ```c
int streq(char str1[], char str2[]) {
  int k;
  k = 0;
  while (*str2) {
    if ((*str1++) != (*str2++))
      return 0;
    ++k;
  }
  return k;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<streq>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L24
  mov    $0x0,%eax
L35:
  cmp    %dl,(%rdi,%rax,1)
  jne    L23
  lea    0x1(%rax),%ecx
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  jne    L35
  jmp    L45
L24:
  mov    $0x0,%ecx
  jmp    L45
L23:
  mov    $0x0,%ecx
L45:
  mov    %ecx,%eax
  ret
```

assistant: ```c
int streq(char str1[], char str2[]) {
  int k;
  k = 0;
  while (*str2) {
    if ((*str1++) != (*str2++))
      return 0;
    ++k;
  }
  return k;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<streq>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L29
  xor    %eax,%eax
  jmp    L37
  nop
L24:
  lea    0x1(%rax),%r8d
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  je     L28
L37:
  cmp    %dl,(%rdi,%rax,1)
  je     L24
L29:
  xor    %r8d,%r8d
L28:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int streq(char str1[], char str2[]) {
  int k;
  k = 0;
  while (*str2) {
    if ((*str1++) != (*str2++))
      return 0;
    ++k;
  }
  return k;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<streq>:
  endbr64
  movzbl (%rsi),%edx
  test   %dl,%dl
  je     L34
  xor    %eax,%eax
  jmp    L18
  nop
L15:
  lea    0x1(%rax),%r8d
  add    $0x1,%rax
  movzbl (%rsi,%rax,1),%edx
  test   %dl,%dl
  je     L17
L18:
  cmp    %dl,(%rdi,%rax,1)
  je     L15
L34:
  xor    %r8d,%r8d
L17:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int streq(char str1[], char str2[]) {
  int k;
  k = 0;
  while (*str2) {
    if ((*str1++) != (*str2++))
      return 0;
    ++k;
  }
  return k;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<le_to_float>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <le_to_u32@plt>
  mov    %eax,-0x4(%rbp)
  movss  -0x4(%rbp),%xmm0
  leave
  ret
```

assistant: ```c
float le_to_float(const uint8_t *buf) {
  union {
    uint32_t u;
    float f;
  } convert;
  convert.u = le_to_u32(buf);
  return convert.f;
}
```
user: What is the c source code of the assembly code below?

```asm
<le_to_float>:
  endbr64
  sub    $0x8,%rsp
  call   <le_to_u32@plt>
  movd   %eax,%xmm0
  add    $0x8,%rsp
  ret
```

assistant: ```c
float le_to_float(const uint8_t *buf) {
  union {
    uint32_t u;
    float f;
  } convert;
  convert.u = le_to_u32(buf);
  return convert.f;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<le_to_float>:
  endbr64
  sub    $0x8,%rsp
  call   <le_to_u32@plt>
  add    $0x8,%rsp
  movd   %eax,%xmm0
  ret
```

assistant: ```c
float le_to_float(const uint8_t *buf) {
  union {
    uint32_t u;
    float f;
  } convert;
  convert.u = le_to_u32(buf);
  return convert.f;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<le_to_float>:
  endbr64
  sub    $0x8,%rsp
  call   <le_to_u32@plt>
  add    $0x8,%rsp
  movd   %eax,%xmm0
  ret
```

assistant: ```c
float le_to_float(const uint8_t *buf) {
  union {
    uint32_t u;
    float f;
  } convert;
  convert.u = le_to_u32(buf);
  return convert.f;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<fn4>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %edx,-0x1c(%rbp)
  mov    -0x1c(%rbp),%eax
  not    %eax
  and    -0x14(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x18(%rbp),%eax
  and    -0x1c(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x8(%rbp),%eax
  or     -0x4(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int fn4(int a, int b, int m) {
  int tem1 = ~m & a;
  int tem2 = b & m;
  return tem1 | tem2;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<fn4>:
  endbr64
  xor    %edi,%esi
  and    %edx,%esi
  mov    %esi,%eax
  xor    %edi,%eax
  ret
```

assistant: ```c
int fn4(int a, int b, int m) {
  int tem1 = ~m & a;
  int tem2 = b & m;
  return tem1 | tem2;
}
```
user: What is the c source code of the assembly code below:

```asm
<fn4>:
  endbr64
  xor    %edi,%esi
  and    %edx,%esi
  mov    %esi,%eax
  xor    %edi,%eax
  ret
```

assistant: ```c
int fn4(int a, int b, int m) {
  int tem1 = ~m & a;
  int tem2 = b & m;
  return tem1 | tem2;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<fn4>:
  endbr64
  xor    %edi,%esi
  and    %edx,%esi
  mov    %esi,%eax
  xor    %edi,%eax
  ret
```

assistant: ```c
int fn4(int a, int b, int m) {
  int tem1 = ~m & a;
  int tem2 = b & m;
  return tem1 | tem2;
}
```