ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 下面的汇编指令对应的C源代码是什么：

```asm
<bigint_print>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    $0x28,%edi
  call   <putchar@plt>
  mov    -0x20(%rbp),%rax
  sub    $0x1,%rax
  mov    %rax,-0x8(%rbp)
  jmp    L5
L16:
  mov    -0x8(%rbp),%rdx
  mov    -0x20(%rbp),%rax
  sub    %rdx,%rax
  lea    -0x1(%rax),%rdx
  mov    -0x8(%rbp),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  subq   $0x1,-0x8(%rbp)
L5:
  cmpq   $0x0,-0x8(%rbp)
  jg     L16
  mov    -0x20(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%eax
  mov    %eax,%esi
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D7
data_type: string
name: read_data
arguments: label: D17
data_type: string

tool: D7: %u*2^(%u) + 

tool: D17: %u*2^(%u))

assistant: ```c
void bigint_print(const unsigned char *x, unsigned long xlen) {
  long i;
  printf("(");
  for (i = xlen - 1; i > 0; i--)
    printf("%u*2^(%u) + ", x[i], xlen - 1 - i);
  printf("%u*2^(%u))", x[0], xlen - 1);
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bigint_print>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    $0x28,%edi
  call   <putchar@plt>
  lea    -0x1(%rbp),%r14
  test   %r14,%r14
  jle    L5
  mov    %rbp,%rbx
  lea    D17(%rip),%r13
L4:
  mov    %rbp,%rdx
  sub    %rbx,%rdx
  movzbl -0x1(%r12,%rbx,1),%esi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  sub    $0x1,%rbx
  cmp    $0x1,%rbx
  jne    L4
L5:
  movzbl (%r12),%esi
  mov    %r14,%rdx
  lea    D11(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D17
type: string
name: disassemble_data
arguments: tag: D11
type: string

tool: D17: %u*2^(%u) + 

tool: D11: %u*2^(%u))

assistant: ```c
void bigint_print(const unsigned char *x, unsigned long xlen) {
  long i;
  printf("(");
  for (i = xlen - 1; i > 0; i--)
    printf("%u*2^(%u) + ", x[i], xlen - 1 - i);
  printf("%u*2^(%u))", x[0], xlen - 1);
}
```
user: What is the c source code of the assembly code below?

```asm
<bigint_print>:
  endbr64
  push   %r13
  lea    D6(%rip),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  sub    $0x1,%r12
  mov    %rdi,%rbp
  mov    $0x28,%edi
  push   %rbx
  mov    %r12,%rbx
  sub    $0x8,%rsp
  call   <putchar@plt>
  test   %r12,%r12
  jle    L13
  nopl   0x0(%rax)
L12:
  movzbl 0x0(%rbp,%rbx,1),%esi
  mov    %r12,%rdx
  xor    %eax,%eax
  mov    %r13,%rdi
  sub    %rbx,%rdx
  call   <printf@plt>
  sub    $0x1,%rbx
  jne    L12
L13:
  movzbl 0x0(%rbp),%esi
  add    $0x8,%rsp
  mov    %r12,%rdx
  xor    %eax,%eax
  pop    %rbx
  lea    D2(%rip),%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <printf@plt>
```

tool_calls: name: unpack_data
arguments: label: D6
type_guessed: string
name: unpack_data
arguments: label: D2
type_guessed: string

tool: D6: %u*2^(%u) + 

tool: D2: %u*2^(%u))

assistant: ```c
void bigint_print(const unsigned char *x, unsigned long xlen) {
  long i;
  printf("(");
  for (i = xlen - 1; i > 0; i--)
    printf("%u*2^(%u) + ", x[i], xlen - 1 - i);
  printf("%u*2^(%u))", x[0], xlen - 1);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<bigint_print>:
  endbr64
  push   %r13
  lea    D0(%rip),%r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  sub    $0x1,%r12
  mov    %rdi,%rbp
  mov    $0x28,%edi
  push   %rbx
  mov    %r12,%rbx
  sub    $0x8,%rsp
  call   <putchar@plt>
  test   %r12,%r12
  jle    L11
  nopl   0x0(%rax)
L4:
  movzbl 0x0(%rbp,%rbx,1),%esi
  mov    %r12,%rdx
  xor    %eax,%eax
  mov    %r13,%rdi
  sub    %rbx,%rdx
  call   <printf@plt>
  sub    $0x1,%rbx
  jne    L4
L11:
  movzbl 0x0(%rbp),%esi
  add    $0x8,%rsp
  mov    %r12,%rdx
  xor    %eax,%eax
  pop    %rbx
  lea    D12(%rip),%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  jmp    1070 <printf@plt>
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type_guessed: string
name: unpack_data
arguments: data_tag: D12
type_guessed: string

tool: D0: %u*2^(%u) + 

tool: D12: %u*2^(%u))

assistant: ```c
void bigint_print(const unsigned char *x, unsigned long xlen) {
  long i;
  printf("(");
  for (i = xlen - 1; i > 0; i--)
    printf("%u*2^(%u) + ", x[i], xlen - 1 - i);
  printf("%u*2^(%u))", x[0], xlen - 1);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x11223344,%eax
  mov    $0x1a2b3c4d,%edx
  clc
  mov    $0xaabbccdd,%ecx
  cmovb  %eax,%edx
  mov    %edx,-0x10(%rbp)
  mov    $0x11223344,%eax
  mov    $0x1a2b3c4d,%edx
  stc
  mov    $0xaabbccdd,%ecx
  cmovb  %eax,%edx
  mov    %edx,-0xc(%rbp)
  mov    -0x10(%rbp),%eax
  cmp    $0x1a2b3c4d,%eax
  je     L14
  mov    $0x1,%eax
  (bad)
L14:
  mov    -0xc(%rbp),%eax
  cmp    $0x11223344,%eax
  je     L19
  mov    $0x1,%eax
  (bad)
L19:
  mov    $0x0,%eax
  (bad)
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L18
  call   <__stack_chk_fail@plt>
L18:
  leave
  ret
```

assistant: ```c
int main() {
  int x, y;
  __asm__ __volatile__("mov $0x11223344, %%eax\n\t"
                       "mov $0x1a2b3c4d, %%edx\n\t"
                       "clc\n\t"
                       "mov $0xaabbccdd, %%ecx\n\t"
                       "cmovc %%eax, %%edx\n\t"
                       "mov %%edx, %0\n\t"
                       "mov $0x11223344, %%eax\n\t"
                       "mov $0x1a2b3c4d, %%edx\n\t"
                       "stc\n\t"
                       "mov $0xaabbccdd, %%ecx\n\t"
                       "cmovc %%eax, %%edx\n\t"
                       "mov %%edx, %1\n\t"
                       : "=m"(x), "=m"(y)
                       :
                       : "eax", "ecx", "edx");
  do {
    if (!(x == 0x1a2b3c4d))
      asm volatile(".byte 0xd6" : : "a"(1));
  } while (0);
  do {
    if (!(y == 0x11223344))
      asm volatile(".byte 0xd6" : : "a"(1));
  } while (0);
  asm volatile(".byte 0xd6" : : "a"(0));
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x11223344,%eax
  mov    $0x1a2b3c4d,%edx
  clc
  mov    $0xaabbccdd,%ecx
  cmovb  %eax,%edx
  mov    %edx,(%rsp)
  mov    $0x11223344,%eax
  mov    $0x1a2b3c4d,%edx
  stc
  mov    $0xaabbccdd,%ecx
  cmovb  %eax,%edx
  mov    %edx,0x4(%rsp)
  cmpl   $0x1a2b3c4d,(%rsp)
  je     L7
  mov    $0x1,%eax
  (bad)
L7:
  cmpl   $0x11223344,0x4(%rsp)
  je     L12
  mov    $0x1,%eax
  (bad)
L12:
  mov    $0x0,%eax
  (bad)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L23
  mov    $0x0,%eax
  add    $0x18,%rsp
  ret
L23:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int x, y;
  __asm__ __volatile__("mov $0x11223344, %%eax\n\t"
                       "mov $0x1a2b3c4d, %%edx\n\t"
                       "clc\n\t"
                       "mov $0xaabbccdd, %%ecx\n\t"
                       "cmovc %%eax, %%edx\n\t"
                       "mov %%edx, %0\n\t"
                       "mov $0x11223344, %%eax\n\t"
                       "mov $0x1a2b3c4d, %%edx\n\t"
                       "stc\n\t"
                       "mov $0xaabbccdd, %%ecx\n\t"
                       "cmovc %%eax, %%edx\n\t"
                       "mov %%edx, %1\n\t"
                       : "=m"(x), "=m"(y)
                       :
                       : "eax", "ecx", "edx");
  do {
    if (!(x == 0x1a2b3c4d))
      asm volatile(".byte 0xd6" : : "a"(1));
  } while (0);
  do {
    if (!(y == 0x11223344))
      asm volatile(".byte 0xd6" : : "a"(1));
  } while (0);
  asm volatile(".byte 0xd6" : : "a"(0));
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x11223344,%eax
  mov    $0x1a2b3c4d,%edx
  clc
  mov    $0xaabbccdd,%ecx
  cmovb  %eax,%edx
  mov    %edx,(%rsp)
  mov    $0x11223344,%eax
  mov    $0x1a2b3c4d,%edx
  stc
  mov    $0xaabbccdd,%ecx
  cmovb  %eax,%edx
  mov    %edx,0x4(%rsp)
  cmpl   $0x1a2b3c4d,(%rsp)
  je     L12
  mov    $0x1,%eax
  (bad)
L12:
  cmpl   $0x11223344,0x4(%rsp)
  je     L21
  mov    $0x1,%eax
  (bad)
L21:
  xor    %eax,%eax
  (bad)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L0:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int x, y;
  __asm__ __volatile__("mov $0x11223344, %%eax\n\t"
                       "mov $0x1a2b3c4d, %%edx\n\t"
                       "clc\n\t"
                       "mov $0xaabbccdd, %%ecx\n\t"
                       "cmovc %%eax, %%edx\n\t"
                       "mov %%edx, %0\n\t"
                       "mov $0x11223344, %%eax\n\t"
                       "mov $0x1a2b3c4d, %%edx\n\t"
                       "stc\n\t"
                       "mov $0xaabbccdd, %%ecx\n\t"
                       "cmovc %%eax, %%edx\n\t"
                       "mov %%edx, %1\n\t"
                       : "=m"(x), "=m"(y)
                       :
                       : "eax", "ecx", "edx");
  do {
    if (!(x == 0x1a2b3c4d))
      asm volatile(".byte 0xd6" : : "a"(1));
  } while (0);
  do {
    if (!(y == 0x11223344))
      asm volatile(".byte 0xd6" : : "a"(1));
  } while (0);
  asm volatile(".byte 0xd6" : : "a"(0));
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x11223344,%eax
  mov    $0x1a2b3c4d,%edx
  clc
  mov    $0xaabbccdd,%ecx
  cmovb  %eax,%edx
  mov    %edx,(%rsp)
  mov    $0x11223344,%eax
  mov    $0x1a2b3c4d,%edx
  stc
  mov    $0xaabbccdd,%ecx
  cmovb  %eax,%edx
  mov    %edx,0x4(%rsp)
  cmpl   $0x1a2b3c4d,(%rsp)
  je     L9
  mov    $0x1,%eax
  (bad)
L9:
  cmpl   $0x11223344,0x4(%rsp)
  je     L20
  mov    $0x1,%eax
  (bad)
L20:
  xor    %eax,%eax
  (bad)
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L27
  xor    %eax,%eax
  add    $0x18,%rsp
  ret
L27:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int main() {
  int x, y;
  __asm__ __volatile__("mov $0x11223344, %%eax\n\t"
                       "mov $0x1a2b3c4d, %%edx\n\t"
                       "clc\n\t"
                       "mov $0xaabbccdd, %%ecx\n\t"
                       "cmovc %%eax, %%edx\n\t"
                       "mov %%edx, %0\n\t"
                       "mov $0x11223344, %%eax\n\t"
                       "mov $0x1a2b3c4d, %%edx\n\t"
                       "stc\n\t"
                       "mov $0xaabbccdd, %%ecx\n\t"
                       "cmovc %%eax, %%edx\n\t"
                       "mov %%edx, %1\n\t"
                       : "=m"(x), "=m"(y)
                       :
                       : "eax", "ecx", "edx");
  do {
    if (!(x == 0x1a2b3c4d))
      asm volatile(".byte 0xd6" : : "a"(1));
  } while (0);
  do {
    if (!(y == 0x11223344))
      asm volatile(".byte 0xd6" : : "a"(1));
  } while (0);
  asm volatile(".byte 0xd6" : : "a"(0));
  return 0;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<ckupper>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x4(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L4
  mov    $0x0,%eax
  jmp    L23
L16:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <islower@plt>
  test   %eax,%eax
  je     L20
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <toupper@plt>
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %dl,(%rax)
L20:
  addq   $0x1,-0x18(%rbp)
  addl   $0x1,-0x4(%rbp)
L4:
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L16
  mov    -0x4(%rbp),%eax
L23:
  leave
  ret
```

assistant: ```c
int ckupper(s)
char *s;
{
  int n = 0;
  if (!s)
    return (0);
  while (*s) {
    if (islower(*s))
      *s = (char)toupper(*s);
    s++, n++;
  }
  return (n);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<ckupper>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L41
  mov    %rdi,%rbp
  movzbl (%rdi),%eax
  test   %al,%al
  je     L10
  mov    $0x0,%r12d
  jmp    L60
L8:
  add    $0x1,%rbp
  add    $0x1,%r12d
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L21
L60:
  movsbl %al,%ebx
  mov    %ebx,%edi
  call   <islower@plt>
  test   %eax,%eax
  je     L8
  mov    %ebx,%edi
  call   <toupper@plt>
  mov    %al,0x0(%rbp)
  jmp    L8
L41:
  mov    $0x0,%r12d
L21:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L10:
  mov    $0x0,%r12d
  jmp    L21
```

assistant: ```c
int ckupper(s)
char *s;
{
  int n = 0;
  if (!s)
    return (0);
  while (*s) {
    if (islower(*s))
      *s = (char)toupper(*s);
    s++, n++;
  }
  return (n);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<ckupper>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L6
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  xor    %r12d,%r12d
  test   %dil,%dil
  je     L29
  nopl   0x0(%rax,%rax,1)
L27:
  movsbl %dil,%ebp
  mov    %ebp,%edi
  call   <islower@plt>
  test   %eax,%eax
  je     L2
  mov    %ebp,%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
L2:
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  add    $0x1,%r12d
  test   %dil,%dil
  jne    L27
L29:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L6:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int ckupper(s)
char *s;
{
  int n = 0;
  if (!s)
    return (0);
  while (*s) {
    if (islower(*s))
      *s = (char)toupper(*s);
    s++, n++;
  }
  return (n);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<ckupper>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  test   %rdi,%rdi
  je     L39
  mov    %rdi,%rbx
  movzbl (%rdi),%edi
  xor    %r12d,%r12d
  test   %dil,%dil
  je     L4
  nopl   0x0(%rax,%rax,1)
L22:
  movsbl %dil,%ebp
  mov    %ebp,%edi
  call   <islower@plt>
  test   %eax,%eax
  je     L10
  mov    %ebp,%edi
  call   <toupper@plt>
  mov    %al,(%rbx)
L10:
  movzbl 0x1(%rbx),%edi
  add    $0x1,%rbx
  add    $0x1,%r12d
  test   %dil,%dil
  jne    L22
L4:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax)
L39:
  xor    %r12d,%r12d
  pop    %rbx
  pop    %rbp
  mov    %r12d,%eax
  pop    %r12
  ret
```

assistant: ```c
int ckupper(s)
char *s;
{
  int n = 0;
  if (!s)
    return (0);
  while (*s) {
    if (islower(*s))
      *s = (char)toupper(*s);
    s++, n++;
  }
  return (n);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<check>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    %edx,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  mov    -0x10(%rbp),%rcx
  mov    -0x8(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <memcmp@plt>
  test   %rax,%rax
  je     L2
  mov    $0x0,%eax
  call   <abort@plt>
L2:
  nop
  leave
  ret
```

assistant: ```c
void check(char *test, char *match, int n) {
  if (memcmp(test, match, n))
    abort();
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<check>:
  endbr64
  sub    $0x8,%rsp
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L12
L4:
  add    $0x8,%rsp
  ret
L12:
  mov    $0x0,%eax
  call   <abort@plt>
  jmp    L4
```

assistant: ```c
void check(char *test, char *match, int n) {
  if (memcmp(test, match, n))
    abort();
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<check>:
  endbr64
  sub    $0x8,%rsp
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L7
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <abort@plt>
```

assistant: ```c
void check(char *test, char *match, int n) {
  if (memcmp(test, match, n))
    abort();
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<check>:
  endbr64
  sub    $0x8,%rsp
  call   <memcmp@plt>
  test   %rax,%rax
  jne    L6
  add    $0x8,%rsp
  ret
  nopw   0x0(%rax,%rax,1)
L6:
  xor    %eax,%eax
  add    $0x8,%rsp
  jmp    1060 <abort@plt>
```

assistant: ```c
void check(char *test, char *match, int n) {
  if (memcmp(test, match, n))
    abort();
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<list_length>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L49
  mov    $0x0,%eax
  jmp    L17
L49:
  movl   $0x0,-0x4(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L45
L56:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    $0xffffffff,%eax
  jne    L27
  mov    -0x4(%rbp),%eax
  jmp    L17
L27:
  addl   $0x1,-0x4(%rbp)
L45:
  cmpl   $0x14,-0x4(%rbp)
  jle    L56
  mov    $0x0,%eax
L17:
  pop    %rbp
  ret
```

assistant: ```c
int list_length(int *list) {
  if (!list)
    return 0;
  int i = 0;
  for (i = 0; i < 20 + 1; i++) {
    if (list[i] == -1)
      return i;
  }
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<list_length>:
  endbr64
  test   %rdi,%rdi
  je     L6
  mov    $0x0,%eax
L32:
  mov    %eax,%edx
  cmpl   $0xffffffff,(%rdi,%rax,4)
  je     L1
  add    $0x1,%rax
  cmp    $0x15,%rax
  jne    L32
  mov    $0x0,%edx
L1:
  mov    %edx,%eax
  ret
L6:
  mov    $0x0,%edx
  jmp    L1
```

assistant: ```c
int list_length(int *list) {
  if (!list)
    return 0;
  int i = 0;
  for (i = 0; i < 20 + 1; i++) {
    if (list[i] == -1)
      return i;
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<list_length>:
  endbr64
  test   %rdi,%rdi
  je     L25
  xor    %eax,%eax
  jmp    L10
  nopl   (%rax)
L21:
  add    $0x1,%rax
  cmp    $0x15,%rax
  je     L25
L10:
  cmpl   $0xffffffff,(%rdi,%rax,4)
  mov    %eax,%r8d
  jne    L21
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L25:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int list_length(int *list) {
  if (!list)
    return 0;
  int i = 0;
  for (i = 0; i < 20 + 1; i++) {
    if (list[i] == -1)
      return i;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<list_length>:
  endbr64
  test   %rdi,%rdi
  je     L38
  xor    %eax,%eax
  jmp    L16
  nopl   (%rax)
L12:
  add    $0x1,%rax
  cmp    $0x15,%rax
  je     L38
L16:
  cmpl   $0xffffffff,(%rdi,%rax,4)
  mov    %eax,%r8d
  jne    L12
  mov    %r8d,%eax
  ret
  nopw   0x0(%rax,%rax,1)
L38:
  xor    %r8d,%r8d
  mov    %r8d,%eax
  ret
```

assistant: ```c
int list_length(int *list) {
  if (!list)
    return 0;
  int i = 0;
  for (i = 0; i < 20 + 1; i++) {
    if (list[i] == -1)
      return i;
  }
  return 0;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<solve>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movq   $0x0,-0x8(%rbp)
  movl   $0x2,-0x10(%rbp)
  jmp    L61
L48:
  movl   $0x2,-0xc(%rbp)
  jmp    L54
L58:
  mov    -0x10(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  jne    L45
  movl   $0x0,-0x14(%rbp)
  jmp    L40
L45:
  movl   $0x1,-0x14(%rbp)
  addl   $0x1,-0xc(%rbp)
L54:
  mov    -0x10(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt@plt>
  cmp    %eax,-0xc(%rbp)
  jle    L58
L40:
  cmpl   $0x0,-0x14(%rbp)
  je     L8
  mov    -0x10(%rbp),%eax
  cltq
  add    %rax,-0x8(%rbp)
L8:
  addl   $0x1,-0x10(%rbp)
L61:
  cmpl   $0x1e8480,-0x10(%rbp)
  jle    L48
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
long solve() {
  long sum = 0;
  int is_prime;
  for (int i = 2; i <= 2e6; i++) {
    for (int j = 2; j <= sqrt(i); j++) {
      if (i % j == 0) {
        is_prime = 0;
        break;
      } else {
        is_prime = 1;
      }
    }
    if (is_prime) {
      sum += i;
    }
  }
  return sum;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<solve>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x2,%r14d
  mov    $0x0,%r15d
  mov    $0x1,%r13d
  jmp    L0
L19:
  lea    (%r15,%r14,1),%rax
  test   %r12d,%r12d
  cmovne %rax,%r15
L12:
  add    $0x1,%r14
  cmp    $0x1e8481,%r14
  je     L11
L0:
  mov    %r14d,%ebp
  mov    $0x2,%ebx
L14:
  mov    %ebp,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jl     L19
  mov    %ebp,%eax
  cltd
  idiv   %ebx
  mov    %edx,%r12d
  test   %edx,%edx
  je     L12
  add    $0x1,%ebx
  mov    %r13d,%r12d
  jmp    L14
L11:
  mov    %r15,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
long solve() {
  long sum = 0;
  int is_prime;
  for (int i = 2; i <= 2e6; i++) {
    for (int j = 2; j <= sqrt(i); j++) {
      if (i % j == 0) {
        is_prime = 0;
        break;
      } else {
        is_prime = 1;
      }
    }
    if (is_prime) {
      sum += i;
    }
  }
  return sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<solve>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    $0x2,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  nopl   (%rax)
L9:
  mov    %r13d,%r12d
  mov    $0x2,%ebx
  jmp    L3
  nopw   0x0(%rax,%rax,1)
L23:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  mov    %edx,%ebp
  test   %edx,%edx
  je     L26
  add    $0x1,%ebx
  mov    $0x1,%ebp
L3:
  mov    %r12d,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jge    L23
  lea    (%r14,%r13,1),%rax
  test   %ebp,%ebp
  cmovne %rax,%r14
L26:
  add    $0x1,%r13
  cmp    $0x1e8481,%r13
  jne    L9
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
long solve() {
  long sum = 0;
  int is_prime;
  for (int i = 2; i <= 2e6; i++) {
    for (int j = 2; j <= sqrt(i); j++) {
      if (i % j == 0) {
        is_prime = 0;
        break;
      } else {
        is_prime = 1;
      }
    }
    if (is_prime) {
      sum += i;
    }
  }
  return sum;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<solve>:
  endbr64
  push   %r14
  xor    %r14d,%r14d
  push   %r13
  mov    $0x2,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  nopl   (%rax)
L10:
  mov    %r13d,%r12d
  mov    $0x2,%ebx
  jmp    L34
  nopw   0x0(%rax,%rax,1)
L20:
  mov    %r12d,%eax
  cltd
  idiv   %ebx
  mov    %edx,%ebp
  test   %edx,%edx
  je     L18
  add    $0x1,%ebx
  mov    $0x1,%ebp
L34:
  mov    %r12d,%edi
  call   <sqrt@plt>
  cmp    %ebx,%eax
  jge    L20
  lea    (%r14,%r13,1),%rax
  test   %ebp,%ebp
  cmovne %rax,%r14
L18:
  add    $0x1,%r13
  cmp    $0x1e8481,%r13
  jne    L10
  pop    %rbx
  mov    %r14,%rax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
long solve() {
  long sum = 0;
  int is_prime;
  for (int i = 2; i <= 2e6; i++) {
    for (int j = 2; j <= sqrt(i); j++) {
      if (i % j == 0) {
        is_prime = 0;
        break;
      } else {
        is_prime = 1;
      }
    }
    if (is_prime) {
      sum += i;
    }
  }
  return sum;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<HSMMCSDResponseGet>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L7
L10:
  mov    -0x4(%rbp),%eax
  add    $0x84,%eax
  lea    0x0(,%rax,4),%edx
  mov    -0x14(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%eax
  mov    %rax,%rcx
  mov    -0x4(%rbp),%eax
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  mov    (%rcx),%eax
  mov    %eax,(%rdx)
  addl   $0x1,-0x4(%rbp)
L7:
  cmpl   $0x3,-0x4(%rbp)
  jbe    L10
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void HSMMCSDResponseGet(unsigned int baseAddr, unsigned int *rsp) {
  unsigned int i;
  for (i = 0; i <= 3; i++) {
    rsp[i] = (*((volatile unsigned int *)(baseAddr + (0x210 + (i * 4)))));
  }
}
```