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

```asm
<cpuusage>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x34(%rbp)
  mov    $0x1,%esi
  mov    $0x186a0,%edi
  call   <get_times@plt>
  mov    %rax,-0x20(%rbp)
  mov    -0x20(%rbp),%rax
  imul   $0x2710,%rax,%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <get_times@plt>
  mov    %rax,-0x18(%rbp)
  movl   $0xa,-0x30(%rbp)
  movl   $0xa,-0x2c(%rbp)
  movl   $0x3e8,-0x28(%rbp)
  mov    -0x34(%rbp),%eax
  mov    %eax,%esi
  lea    D6(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x34(%rbp),%eax
  cltq
  imul   -0x18(%rbp),%rax
  mov    %rax,%rcx
  movabs $0xa3d70a3d70a3d70b,%rdx
  mov    %rcx,%rax
  imul   %rdx
  lea    (%rdx,%rcx,1),%rax
  sar    $0x6,%rax
  sar    $0x3f,%rcx
  mov    %rcx,%rdx
  sub    %rdx,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  sub    -0x10(%rbp),%rax
  imul   $0xf4240,%rax,%rax
  cqto
  idivq  -0x18(%rbp)
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L4:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <dosum@plt>
  mov    -0x24(%rbp),%eax
  mov    %eax,%edi
  call   <usleep@plt>
  mov    -0x2c(%rbp),%eax
  add    %eax,-0x30(%rbp)
  mov    $0x2e,%edi
  call   <putchar@plt>
  jmp    L4
```

tool_calls: name: extract_symbolic_data
arguments: tag: D6
type: string
name: extract_symbolic_data
arguments: tag: D24
type: string
name: extract_symbolic_data
arguments: tag: D5
type: string

tool: D6: cpu load usage: %d


tool: D24: cacl times %ld every second


tool: D5: sleep time %d us every second


assistant: ```c
int cpuusage(int usage) {
  long int times1 = get_times(100000, 1);
  long int times2 = get_times(times1 * 10000, 1000);
  int n = 10, step = 10;
  int precision = 1000;
  printf("cpu load usage: %d\n", usage);
  long int cacl_times = usage * times2 / 100;
  printf("cacl times %ld every second\n", cacl_times);
  long long int temp = 1000000 * (times2 - cacl_times) / times2;
  int sleep_time = temp;
  printf("sleep time %d us every second\n", sleep_time);
  while (1) {
    dosum(cacl_times);
    usleep(sleep_time);
    n += step;
    printf(".");
  }
  return 0;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<cpuusage>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x1,%esi
  mov    $0x186a0,%edi
  call   <get_times@plt>
  imul   $0x2710,%rax,%rdi
  mov    $0x3e8,%esi
  call   <get_times@plt>
  mov    %rax,%rbp
  mov    %ebx,%esi
  lea    D1(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movslq %ebx,%rcx
  imul   %rbp,%rcx
  movabs $0xa3d70a3d70a3d70b,%rdx
  mov    %rcx,%rax
  imul   %rdx
  lea    (%rdx,%rcx,1),%rbx
  sar    $0x6,%rbx
  sar    $0x3f,%rcx
  sub    %rcx,%rbx
  mov    %rbx,%rsi
  lea    D7(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %rbp,%rax
  sub    %rbx,%rax
  imul   $0xf4240,%rax,%rax
  cqto
  idiv   %rbp
  mov    %rax,%rbp
  mov    %eax,%esi
  lea    D17(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L5:
  mov    %rbx,%rdi
  call   <dosum@plt>
  mov    %ebp,%edi
  call   <usleep@plt>
  mov    $0x2e,%edi
  call   <putchar@plt>
  jmp    L5
```

tool_calls: name: extract_symbolic_data
arguments: label: D1
type_guessed: string
name: extract_symbolic_data
arguments: label: D7
type_guessed: string
name: extract_symbolic_data
arguments: label: D17
type_guessed: string

tool: D1: cpu load usage: %d


tool: D7: cacl times %ld every second


tool: D17: sleep time %d us every second


assistant: ```c
int cpuusage(int usage) {
  long int times1 = get_times(100000, 1);
  long int times2 = get_times(times1 * 10000, 1000);
  int n = 10, step = 10;
  int precision = 1000;
  printf("cpu load usage: %d\n", usage);
  long int cacl_times = usage * times2 / 100;
  printf("cacl times %ld every second\n", cacl_times);
  long long int temp = 1000000 * (times2 - cacl_times) / times2;
  int sleep_time = temp;
  printf("sleep time %d us every second\n", sleep_time);
  while (1) {
    dosum(cacl_times);
    usleep(sleep_time);
    n += step;
    printf(".");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<cpuusage>:
  endbr64
  push   %r12
  mov    $0x1,%esi
  push   %rbx
  mov    %edi,%ebx
  mov    $0x186a0,%edi
  sub    $0x8,%rsp
  call   <get_times@plt>
  mov    $0x3e8,%esi
  imul   $0x2710,%rax,%rdi
  call   <get_times@plt>
  mov    %ebx,%esi
  lea    D28(%rip),%rdi
  mov    %rax,%r12
  xor    %eax,%eax
  call   <printf@plt>
  movslq %ebx,%rcx
  lea    D16(%rip),%rdi
  movabs $0xa3d70a3d70a3d70b,%rdx
  imul   %r12,%rcx
  mov    %rcx,%rax
  imul   %rdx
  xor    %eax,%eax
  lea    (%rdx,%rcx,1),%rbx
  sar    $0x3f,%rcx
  sar    $0x6,%rbx
  sub    %rcx,%rbx
  mov    %rbx,%rsi
  call   <printf@plt>
  mov    %r12,%rax
  lea    D20(%rip),%rdi
  sub    %rbx,%rax
  imul   $0xf4240,%rax,%rax
  cqto
  idiv   %r12
  mov    %rax,%r12
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  nopl   0x0(%rax)
L4:
  mov    %rbx,%rdi
  call   <dosum@plt>
  mov    %r12d,%edi
  call   <usleep@plt>
  mov    $0x2e,%edi
  call   <putchar@plt>
  jmp    L4
```

tool_calls: name: unpack_data
arguments: tag: D28
datatype: string
name: unpack_data
arguments: tag: D16
datatype: string
name: unpack_data
arguments: tag: D20
datatype: string

tool: D28: cpu load usage: %d


tool: D16: cacl times %ld every second


tool: D20: sleep time %d us every second


assistant: ```c
int cpuusage(int usage) {
  long int times1 = get_times(100000, 1);
  long int times2 = get_times(times1 * 10000, 1000);
  int n = 10, step = 10;
  int precision = 1000;
  printf("cpu load usage: %d\n", usage);
  long int cacl_times = usage * times2 / 100;
  printf("cacl times %ld every second\n", cacl_times);
  long long int temp = 1000000 * (times2 - cacl_times) / times2;
  int sleep_time = temp;
  printf("sleep time %d us every second\n", sleep_time);
  while (1) {
    dosum(cacl_times);
    usleep(sleep_time);
    n += step;
    printf(".");
  }
  return 0;
}
```
user: Translate the below assembly code into C programming language:

```asm
<cpuusage>:
  endbr64
  push   %r12
  mov    $0x1,%esi
  push   %rbx
  mov    %edi,%ebx
  mov    $0x186a0,%edi
  sub    $0x8,%rsp
  call   <get_times@plt>
  mov    $0x3e8,%esi
  imul   $0x2710,%rax,%rdi
  call   <get_times@plt>
  mov    %ebx,%esi
  lea    D3(%rip),%rdi
  mov    %rax,%r12
  xor    %eax,%eax
  call   <printf@plt>
  movslq %ebx,%rcx
  lea    D19(%rip),%rdi
  movabs $0xa3d70a3d70a3d70b,%rdx
  imul   %r12,%rcx
  mov    %rcx,%rax
  imul   %rdx
  xor    %eax,%eax
  lea    (%rdx,%rcx,1),%rbx
  sar    $0x3f,%rcx
  sar    $0x6,%rbx
  sub    %rcx,%rbx
  mov    %rbx,%rsi
  call   <printf@plt>
  mov    %r12,%rax
  lea    D21(%rip),%rdi
  sub    %rbx,%rax
  imul   $0xf4240,%rax,%rax
  cqto
  idiv   %r12
  mov    %rax,%r12
  mov    %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  nopl   0x0(%rax)
L1:
  mov    %rbx,%rdi
  call   <dosum@plt>
  mov    %r12d,%edi
  call   <usleep@plt>
  mov    $0x2e,%edi
  call   <putchar@plt>
  jmp    L1
```

tool_calls: name: disassemble_data
arguments: data_tag: D3
type_guessed: string
name: disassemble_data
arguments: data_tag: D19
type_guessed: string
name: disassemble_data
arguments: data_tag: D21
type_guessed: string

tool: D3: cpu load usage: %d


tool: D19: cacl times %ld every second


tool: D21: sleep time %d us every second


assistant: ```c
int cpuusage(int usage) {
  long int times1 = get_times(100000, 1);
  long int times2 = get_times(times1 * 10000, 1000);
  int n = 10, step = 10;
  int precision = 1000;
  printf("cpu load usage: %d\n", usage);
  long int cacl_times = usage * times2 / 100;
  printf("cacl times %ld every second\n", cacl_times);
  long long int temp = 1000000 * (times2 - cacl_times) / times2;
  int sleep_time = temp;
  printf("sleep time %d us every second\n", sleep_time);
  while (1) {
    dosum(cacl_times);
    usleep(sleep_time);
    n += step;
    printf(".");
  }
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<remainderf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  movss  %xmm0,-0x14(%rbp)
  movss  %xmm1,-0x18(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0xc(%rbp),%rdx
  movss  -0x18(%rbp),%xmm0
  mov    -0x14(%rbp),%eax
  mov    %rdx,%rdi
  movaps %xmm0,%xmm1
  movd   %eax,%xmm0
  call   <remquof@plt>
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L3
  call   <__stack_chk_fail@plt>
L3:
  leave
  ret
```

assistant: ```c
float remainderf(float x, float y) {
  int q;
  return remquof(x, y, &q);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<remainderf>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <remquof@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x18,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
float remainderf(float x, float y) {
  int q;
  return remquof(x, y, &q);
}
```
user: What is the c source code of the assembly code below?

```asm
<remainderf>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <remquof@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
float remainderf(float x, float y) {
  int q;
  return remquof(x, y, &q);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<remainderf>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  call   <remquof@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x18,%rsp
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
float remainderf(float x, float y) {
  int q;
  return remquof(x, y, &q);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<pendfunc1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x4,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D0
type: string

tool: D0: in pendfunc1, x is %d


assistant: ```c
void pendfunc1(int x) {
  int y = x + 4;
  printf("in pendfunc1, x is %d\n", x);
}
```
user: Translate the below assembly code into C programming language:

```asm
<pendfunc1>:
  endbr64
  sub    $0x8,%rsp
  mov    %edi,%esi
  lea    D8(%rip),%rdi
  call   <printf@plt>
  add    $0x8,%rsp
  ret
```

tool_calls: name: extract_symbolic_data
arguments: tag: D8
datatype: string

tool: D8: in pendfunc1, x is %d


assistant: ```c
void pendfunc1(int x) {
  int y = x + 4;
  printf("in pendfunc1, x is %d\n", x);
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<pendfunc1>:
  endbr64
  mov    %edi,%esi
  lea    D8(%rip),%rdi
  jmp    1050 <printf@plt>
```

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

tool: D8: in pendfunc1, x is %d


assistant: ```c
void pendfunc1(int x) {
  int y = x + 4;
  printf("in pendfunc1, x is %d\n", x);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<pendfunc1>:
  endbr64
  mov    %edi,%esi
  lea    D4(%rip),%rdi
  jmp    1050 <printf@plt>
```

tool_calls: name: unpack_data
arguments: tag: D4
type_infered: string

tool: D4: in pendfunc1, x is %d


assistant: ```c
void pendfunc1(int x) {
  int y = x + 4;
  printf("in pendfunc1, x is %d\n", x);
}
```
user: What is the c source code of the assembly code below:

```asm
<fastcgi_cleanup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0xc(%rbp)
  lea    -0xc(%rbp),%rax
  mov    %rax,%rdi
  call   <wait@plt>
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
void fastcgi_cleanup() {
  int status = 0;
  wait(&status);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<fastcgi_cleanup>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x0,0x4(%rsp)
  lea    0x4(%rsp),%rdi
  call   <wait@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x18,%rsp
  ret
L9:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fastcgi_cleanup() {
  int status = 0;
  wait(&status);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<fastcgi_cleanup>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0x0,0x4(%rsp)
  call   <wait@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L6
  add    $0x18,%rsp
  ret
L6:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fastcgi_cleanup() {
  int status = 0;
  wait(&status);
}
```
user: What is the c source code of the assembly code below?

```asm
<fastcgi_cleanup>:
  endbr64
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdi
  movl   $0x0,0x4(%rsp)
  call   <wait@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L8
  add    $0x18,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void fastcgi_cleanup() {
  int status = 0;
  wait(&status);
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<numa_allocate_cpumask>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    $0x0,%eax
  call   <numa_num_possible_cpus@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <numa_bitmask_alloc@plt>
  leave
  ret
```

assistant: ```c
struct bitmask *numa_allocate_cpumask() {
  int ncpus = numa_num_possible_cpus();
  return numa_bitmask_alloc(ncpus);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<numa_allocate_cpumask>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <numa_num_possible_cpus@plt>
  mov    %eax,%edi
  call   <numa_bitmask_alloc@plt>
  add    $0x8,%rsp
  ret
```

assistant: ```c
struct bitmask *numa_allocate_cpumask() {
  int ncpus = numa_num_possible_cpus();
  return numa_bitmask_alloc(ncpus);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<numa_allocate_cpumask>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <numa_num_possible_cpus@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    1070 <numa_bitmask_alloc@plt>
```

assistant: ```c
struct bitmask *numa_allocate_cpumask() {
  int ncpus = numa_num_possible_cpus();
  return numa_bitmask_alloc(ncpus);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<numa_allocate_cpumask>:
  endbr64
  sub    $0x8,%rsp
  xor    %eax,%eax
  call   <numa_num_possible_cpus@plt>
  add    $0x8,%rsp
  mov    %eax,%edi
  jmp    1070 <numa_bitmask_alloc@plt>
```

assistant: ```c
struct bitmask *numa_allocate_cpumask() {
  int ncpus = numa_num_possible_cpus();
  return numa_bitmask_alloc(ncpus);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<rta_put_address>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  mov    (%rax),%eax
  cmp    $0x80,%eax
  je     L49
  cmp    $0x81,%eax
  jne    L42
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0x4,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <rta_put@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  jmp    L7
L49:
  mov    -0x1c(%rbp),%ecx
  mov    -0x18(%rbp),%rax
  mov    $0x10,%edx
  mov    %ecx,%esi
  mov    %rax,%rdi
  call   <rta_put@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    -0x28(%rbp),%rdx
  mov    (%rdx),%edx
  mov    %edx,(%rax)
  jmp    L7
L42:
  mov    $0xffffffff,%eax
  jmp    L33
L7:
  mov    $0x0,%eax
L33:
  leave
  ret
```

assistant: ```c
__attribute__((visibility("hidden"))) int rta_put_address(struct nlmsghdr *msg, int type, struct sockaddr *adr) {
  switch (adr->sa_family) {
  case AF_INET: {
    struct in_addr *i = rta_put(msg, type, 4);
    *i = ((struct sockaddr_in *)adr)->sin_addr;
    break;
  }
  case AF_INET6: {
    struct in6_addr *i6 = rta_put(msg, type, 16);
    *i6 = ((struct sockaddr_in6 *)adr)->sin6_addr;
    break;
  }
  default:
    return -1;
  }
  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
<rta_put_address>:
  endbr64
  push   %rbx
  mov    %rdx,%rbx
  mov    (%rdx),%eax
  cmp    $0x80,%eax
  je     L19
  mov    $0xffffffff,%edx
  cmp    $0x81,%eax
  jne    L29
  mov    $0x4,%edx
  call   <rta_put@plt>
  mov    (%rbx),%edx
  mov    %edx,(%rax)
  mov    $0x0,%edx
L29:
  mov    %edx,%eax
  pop    %rbx
  ret
L19:
  mov    $0x10,%edx
  call   <rta_put@plt>
  mov    (%rbx),%edx
  mov    %edx,(%rax)
  mov    $0x0,%edx
  jmp    L29
```

assistant: ```c
__attribute__((visibility("hidden"))) int rta_put_address(struct nlmsghdr *msg, int type, struct sockaddr *adr) {
  switch (adr->sa_family) {
  case AF_INET: {
    struct in_addr *i = rta_put(msg, type, 4);
    *i = ((struct sockaddr_in *)adr)->sin_addr;
    break;
  }
  case AF_INET6: {
    struct in6_addr *i6 = rta_put(msg, type, 16);
    *i6 = ((struct sockaddr_in6 *)adr)->sin6_addr;
    break;
  }
  default:
    return -1;
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<rta_put_address>:
  endbr64
  push   %rbx
  mov    (%rdx),%eax
  mov    %rdx,%rbx
  cmp    $0x80,%eax
  je     L19
  mov    $0xffffffff,%r8d
  cmp    $0x81,%eax
  jne    L15
  mov    $0x4,%edx
  call   <rta_put@plt>
  mov    (%rbx),%edx
  xor    %r8d,%r8d
  mov    %edx,(%rax)
L15:
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L19:
  mov    $0x10,%edx
  call   <rta_put@plt>
  mov    (%rbx),%edx
  xor    %r8d,%r8d
  mov    %edx,(%rax)
  mov    %r8d,%eax
  pop    %rbx
  ret
```

assistant: ```c
__attribute__((visibility("hidden"))) int rta_put_address(struct nlmsghdr *msg, int type, struct sockaddr *adr) {
  switch (adr->sa_family) {
  case AF_INET: {
    struct in_addr *i = rta_put(msg, type, 4);
    *i = ((struct sockaddr_in *)adr)->sin_addr;
    break;
  }
  case AF_INET6: {
    struct in6_addr *i6 = rta_put(msg, type, 16);
    *i6 = ((struct sockaddr_in6 *)adr)->sin6_addr;
    break;
  }
  default:
    return -1;
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<rta_put_address>:
  endbr64
  push   %rbx
  mov    (%rdx),%eax
  mov    %rdx,%rbx
  cmp    $0x80,%eax
  je     L1
  mov    $0xffffffff,%r8d
  cmp    $0x81,%eax
  jne    L4
  mov    $0x4,%edx
  call   <rta_put@plt>
  mov    (%rbx),%edx
  xor    %r8d,%r8d
  mov    %edx,(%rax)
L4:
  mov    %r8d,%eax
  pop    %rbx
  ret
  nopl   0x0(%rax)
L1:
  mov    $0x10,%edx
  call   <rta_put@plt>
  mov    (%rbx),%edx
  xor    %r8d,%r8d
  mov    %edx,(%rax)
  mov    %r8d,%eax
  pop    %rbx
  ret
```

assistant: ```c
__attribute__((visibility("hidden"))) int rta_put_address(struct nlmsghdr *msg, int type, struct sockaddr *adr) {
  switch (adr->sa_family) {
  case AF_INET: {
    struct in_addr *i = rta_put(msg, type, 4);
    *i = ((struct sockaddr_in *)adr)->sin_addr;
    break;
  }
  case AF_INET6: {
    struct in6_addr *i6 = rta_put(msg, type, 16);
    *i6 = ((struct sockaddr_in6 *)adr)->sin6_addr;
    break;
  }
  default:
    return -1;
  }
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<block_strdup>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L15
  mov    $0x0,%eax
  jmp    L7
L15:
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  add    $0x1,%rax
  mov    %rax,%rdi
  call   <block_malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x18(%rbp),%rdx
  mov    -0x8(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
  mov    -0x8(%rbp),%rax
L7:
  leave
  ret
```

assistant: ```c
char *block_strdup(char *from) {
  char *result;
  if (!from)
    return NULL;
  result = block_malloc(strlen(from) + 1);
  strcpy(result, from);
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<block_strdup>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  test   %rdi,%rdi
  je     L7
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <block_malloc@plt>
  mov    %rax,%rbp
  mov    %rbx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L13:
  mov    %rbp,%rax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L7:
  mov    %rdi,%rbp
  jmp    L13
```

assistant: ```c
char *block_strdup(char *from) {
  char *result;
  if (!from)
    return NULL;
  result = block_malloc(strlen(from) + 1);
  strcpy(result, from);
  return result;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<block_strdup>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L8
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <block_malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L8:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *block_strdup(char *from) {
  char *result;
  if (!from)
    return NULL;
  result = block_malloc(strlen(from) + 1);
  strcpy(result, from);
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<block_strdup>:
  endbr64
  push   %r12
  push   %rbp
  sub    $0x8,%rsp
  test   %rdi,%rdi
  je     L7
  mov    %rdi,%rbp
  call   <strlen@plt>
  lea    0x1(%rax),%rdi
  call   <block_malloc@plt>
  mov    %rbp,%rsi
  mov    %rax,%rdi
  mov    %rax,%r12
  call   <strcpy@plt>
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L7:
  xor    %r12d,%r12d
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbp
  pop    %r12
  ret
```

assistant: ```c
char *block_strdup(char *from) {
  char *result;
  if (!from)
    return NULL;
  result = block_malloc(strlen(from) + 1);
  strcpy(result, from);
  return result;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x0,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <dummy01@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <dummy02@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <dummy03@plt>
  mov    %eax,-0x4(%rbp)
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  i = dummy01(i);
  i = dummy02(i);
  i = dummy03(i);
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  mov    $0x0,%edi
  call   <dummy01@plt>
  mov    %eax,%edi
  call   <dummy02@plt>
  mov    %eax,%edi
  call   <dummy03@plt>
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  i = dummy01(i);
  i = dummy02(i);
  i = dummy03(i);
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <dummy01@plt>
  mov    %eax,%edi
  call   <dummy02@plt>
  mov    %eax,%edi
  call   <dummy03@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  i = dummy01(i);
  i = dummy02(i);
  i = dummy03(i);
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  xor    %edi,%edi
  call   <dummy01@plt>
  mov    %eax,%edi
  call   <dummy02@plt>
  mov    %eax,%edi
  call   <dummy03@plt>
  xor    %eax,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(void) {
  int i = 0;
  i = dummy01(i);
  i = dummy02(i);
  i = dummy03(i);
  return 0;
}
```