ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What is the c source code of the assembly code below:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x61000(%rsp),%r11
L44:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L44
  sub    $0xbd0,%rsp
  mov    $0x186a0,%esi
  lea    D3(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x61bc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x61a90(%rsp),%rbx
  mov    %rbp,%rdi
  call   <preparePrimeSieve@plt>
  jmp    L16
  nopl   0x0(%rax)
L20:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <findMaxPrimeSubstr@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
L16:
  mov    %rbx,%rdi
  call   <gets@plt>
  cmpb   $0x30,0x61a90(%rsp)
  jne    L20
  cmpb   $0x0,0x61a91(%rsp)
  jne    L20
  mov    0x61bc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x61bd0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D3
data_type: string

tool: D3: %d


assistant: ```c
int main(void) {
  int primeSieve[100000 + 1];
  char strNum[300 + 1];
  preparePrimeSieve(primeSieve, 100000);
  while (1) {
    gets(strNum);
    if (strNum[0] == '0' && strNum[1] == '\0')
      break;
    printf("%d\n", findMaxPrimeSubstr(strNum, primeSieve));
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x61000(%rsp),%r11
L35:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L35
  sub    $0xbd0,%rsp
  mov    $0x186a0,%esi
  lea    D2(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x61bc8(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  lea    0x61a90(%rsp),%rbx
  mov    %rbp,%rdi
  call   <preparePrimeSieve@plt>
  jmp    L11
  nopl   0x0(%rax)
L42:
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <findMaxPrimeSubstr@plt>
  mov    %r12,%rdi
  mov    %eax,%esi
  call   <printf@plt>
L11:
  mov    %rbx,%rdi
  call   <gets@plt>
  cmpb   $0x30,0x61a90(%rsp)
  jne    L42
  cmpb   $0x0,0x61a91(%rsp)
  jne    L42
  mov    0x61bc8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x61bd0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L15:
  call   <__stack_chk_fail@plt>
```

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

tool: D2: %d


assistant: ```c
int main(void) {
  int primeSieve[100000 + 1];
  char strNum[300 + 1];
  preparePrimeSieve(primeSieve, 100000);
  while (1) {
    gets(strNum);
    if (strNum[0] == '0' && strNum[1] == '\0')
      break;
    printf("%d\n", findMaxPrimeSubstr(strNum, primeSieve));
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x3,-0x14(%rbp)
L85:
  lea    D20(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x18(%rbp),%eax
  test   %eax,%eax
  jle    L41
  mov    -0x18(%rbp),%eax
  mov    %eax,%esi
  lea    D36(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L41:
  movl   $0x2,-0x10(%rbp)
  jmp    L71
L70:
  movl   $0x2,-0xc(%rbp)
  jmp    L78
L86:
  mov    -0x14(%rbp),%eax
  cltd
  idivl  -0xc(%rbp)
  mov    %edx,%eax
  test   %eax,%eax
  je     L77
  addl   $0x1,-0xc(%rbp)
L78:
  mov    -0x14(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jg     L86
  jmp    L55
L77:
  nop
L55:
  mov    -0xc(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jne    L18
  mov    -0x14(%rbp),%eax
  mov    %eax,%esi
  lea    D28(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x10(%rbp)
L18:
  addl   $0x1,-0x14(%rbp)
L71:
  mov    -0x18(%rbp),%eax
  cmp    %eax,-0x10(%rbp)
  jle    L70
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L85
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D20
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D39
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D36
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D14
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D28
type_guessed: string

tool: D20: How many prime numbers?

tool: D39: %d

tool: D36: First %d prime numbers:


tool: D14: 2 

tool: D28: %d 

assistant: ```c
int main() {
  int n, i = 3, count, c;
  while (1) {
    printf("How many prime numbers?\n");
    scanf("%d", &n);
    if (n >= 1) {
      printf("First %d prime numbers:\n", n);
      printf("2 ");
    }
    for (count = 2; count <= n;) {
      for (c = 2; c <= i - 1; c++) {
        if (i % c == 0)
          break;
      }
      if (c == i) {
        printf("%d ", i);
        count++;
      }
      i++;
    }
    printf("\n");
  }
  return 0;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  mov    $0x3,%ebx
  lea    D19(%rip),%r13
  lea    D18(%rip),%r12
  lea    D29(%rip),%r14
  jmp    L52
L114:
  lea    D40(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D20(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L82
L106:
  je     L74
L112:
  add    $0x1,%ebx
  cmp    %ebp,0x4(%rsp)
  jl     L25
L77:
  cmp    $0x2,%ebx
  jle    L106
  test   $0x1,%bl
  je     L112
  mov    $0x2,%ecx
L28:
  add    $0x1,%ecx
  cmp    %ebx,%ecx
  je     L74
  mov    %ebx,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  jne    L28
  jmp    L112
L74:
  mov    %ebx,%esi
  mov    %r14,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebp
  jmp    L112
L25:
  mov    $0xa,%edi
  call   <putchar@plt>
L52:
  mov    %r13,%rdi
  call   <puts@plt>
  lea    0x4(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  mov    0x4(%rsp),%esi
  test   %esi,%esi
  jg     L114
L82:
  mov    $0x2,%ebp
  cmpl   $0x1,0x4(%rsp)
  jg     L77
  jmp    L25
```

tool_calls: name: extract_symbolic_data
arguments: tag: D19
data_type: string
name: extract_symbolic_data
arguments: tag: D18
data_type: string
name: extract_symbolic_data
arguments: tag: D29
data_type: string
name: extract_symbolic_data
arguments: tag: D40
data_type: string
name: extract_symbolic_data
arguments: tag: D20
data_type: string

tool: D19: How many prime numbers?

tool: D18: %d

tool: D29: %d 

tool: D40: First %d prime numbers:


tool: D20: 2 

assistant: ```c
int main() {
  int n, i = 3, count, c;
  while (1) {
    printf("How many prime numbers?\n");
    scanf("%d", &n);
    if (n >= 1) {
      printf("First %d prime numbers:\n", n);
      printf("2 ");
    }
    for (count = 2; count <= n;) {
      for (c = 2; c <= i - 1; c++) {
        if (i % c == 0)
          break;
      }
      if (c == i) {
        printf("%d ", i);
        count++;
      }
      i++;
    }
    printf("\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x3,%r15d
  push   %r14
  lea    D44(%rip),%r14
  push   %r13
  lea    D14(%rip),%r13
  push   %r12
  lea    D9(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D15(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rbp
  jmp    L18
  nopl   0x0(%rax,%rax,1)
L35:
  mov    $0xa,%edi
  call   <putchar@plt>
L18:
  mov    %r12,%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%esi
  test   %esi,%esi
  jle    L35
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r13,%rdi
  call   <printf@plt>
  cmpl   $0x1,0x14(%rsp)
  jle    L35
  mov    $0x2,%r9d
L28:
  mov    $0x2,%ecx
  nopl   0x0(%rax,%rax,1)
L63:
  mov    %r15d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L0
  add    $0x1,%ecx
  cmp    %r15d,%ecx
  jne    L63
  mov    %r15d,%esi
  lea    D39(%rip),%rdi
  xor    %eax,%eax
  mov    %r9d,0xc(%rsp)
  call   <printf@plt>
  mov    0xc(%rsp),%r9d
  add    $0x1,%r9d
L0:
  add    $0x1,%r15d
  cmp    %r9d,0x14(%rsp)
  jge    L28
  jmp    L35
```

tool_calls: name: unpack_data
arguments: data_tag: D44
type_infered: string
name: unpack_data
arguments: data_tag: D14
type_infered: string
name: unpack_data
arguments: data_tag: D9
type_infered: string
name: unpack_data
arguments: data_tag: D15
type_infered: string
name: unpack_data
arguments: data_tag: D39
type_infered: string

tool: D44: First %d prime numbers:


tool: D14: 2 

tool: D9: How many prime numbers?

tool: D15: %d

tool: D39: %d 

assistant: ```c
int main() {
  int n, i = 3, count, c;
  while (1) {
    printf("How many prime numbers?\n");
    scanf("%d", &n);
    if (n >= 1) {
      printf("First %d prime numbers:\n", n);
      printf("2 ");
    }
    for (count = 2; count <= n;) {
      for (c = 2; c <= i - 1; c++) {
        if (i % c == 0)
          break;
      }
      if (c == i) {
        printf("%d ", i);
        count++;
      }
      i++;
    }
    printf("\n");
  }
  return 0;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x3,%r15d
  push   %r14
  lea    D43(%rip),%r14
  push   %r13
  lea    D44(%rip),%r13
  push   %r12
  lea    D0(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D40(%rip),%rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0x14(%rsp),%rbp
  jmp    L31
  nopl   0x0(%rax,%rax,1)
L20:
  mov    $0xa,%edi
  call   <putchar@plt>
L31:
  mov    %r12,%rdi
  call   <puts@plt>
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%esi
  test   %esi,%esi
  jle    L20
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x14(%rsp),%esi
  cmp    $0x1,%esi
  jle    L20
  mov    $0x2,%r9d
L51:
  mov    $0x2,%ecx
  nopl   (%rax)
L28:
  mov    %r15d,%eax
  cltd
  idiv   %ecx
  test   %edx,%edx
  je     L11
  add    $0x1,%ecx
  cmp    %r15d,%ecx
  jne    L28
  mov    %r15d,%esi
  lea    D13(%rip),%rdi
  xor    %eax,%eax
  mov    %r9d,0xc(%rsp)
  call   <printf@plt>
  mov    0xc(%rsp),%r9d
  mov    0x14(%rsp),%esi
  add    $0x1,%r9d
L11:
  add    $0x1,%r15d
  cmp    %esi,%r9d
  jle    L51
  jmp    L20
```

tool_calls: name: disassemble_data
arguments: label: D43
type_infered: string
name: disassemble_data
arguments: label: D44
type_infered: string
name: disassemble_data
arguments: label: D0
type_infered: string
name: disassemble_data
arguments: label: D40
type_infered: string
name: disassemble_data
arguments: label: D13
type_infered: string

tool: D43: First %d prime numbers:


tool: D44: 2 

tool: D0: How many prime numbers?

tool: D40: %d

tool: D13: %d 

assistant: ```c
int main() {
  int n, i = 3, count, c;
  while (1) {
    printf("How many prime numbers?\n");
    scanf("%d", &n);
    if (n >= 1) {
      printf("First %d prime numbers:\n", n);
      printf("2 ");
    }
    for (count = 2; count <= n;) {
      for (c = 2; c <= i - 1; c++) {
        if (i % c == 0)
          break;
      }
      if (c == i) {
        printf("%d ", i);
        count++;
      }
      i++;
    }
    printf("\n");
  }
  return 0;
}
```
user: What is the c source code of the assembly code below:

```asm
<srl_int64_to_bytes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    $0x38,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  mov    %dl,(%rax)
  mov    $0x30,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  add    $0x1,%rax
  mov    %dl,(%rax)
  mov    $0x28,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  add    $0x2,%rax
  mov    %dl,(%rax)
  mov    $0x20,%edx
  mov    -0x4(%rbp),%eax
  mov    %edx,%ecx
  sar    %cl,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  add    $0x3,%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  shr    $0x18,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  add    $0x4,%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  sar    $0x10,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  add    $0x5,%rax
  mov    %dl,(%rax)
  mov    -0x4(%rbp),%eax
  sar    $0x8,%eax
  mov    %eax,%edx
  mov    -0x10(%rbp),%rax
  add    $0x6,%rax
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rax
  add    $0x7,%rax
  mov    -0x4(%rbp),%edx
  mov    %dl,(%rax)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void srl_int64_to_bytes(int64_t i, char *buf) {
  buf[0] = (i >> 56) & 0xFF;
  buf[1] = (i >> 48) & 0xFF;
  buf[2] = (i >> 40) & 0xFF;
  buf[3] = (i >> 32) & 0xFF;
  buf[4] = (i >> 24) & 0xFF;
  buf[5] = (i >> 16) & 0xFF;
  buf[6] = (i >> 8) & 0xFF;
  buf[7] = i & 0xFF;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<srl_int64_to_bytes>:
  endbr64
  mov    %edi,%eax
  mov    %edi,%edx
  mov    $0x38,%ecx
  sar    %cl,%edx
  mov    %dl,(%rsi)
  mov    %edi,%edx
  mov    $0x30,%ecx
  sar    %cl,%edx
  mov    %dl,0x1(%rsi)
  mov    %edi,%edx
  mov    $0x28,%ecx
  sar    %cl,%edx
  mov    %dl,0x2(%rsi)
  mov    %edi,%edx
  mov    $0x20,%ecx
  sar    %cl,%edx
  mov    %dl,0x3(%rsi)
  mov    %edi,%edx
  shr    $0x18,%edx
  mov    %dl,0x4(%rsi)
  mov    %edi,%edx
  sar    $0x10,%edx
  mov    %dl,0x5(%rsi)
  movzbl %ah,%ecx
  mov    %cl,0x6(%rsi)
  mov    %dil,0x7(%rsi)
  ret
```

assistant: ```c
void srl_int64_to_bytes(int64_t i, char *buf) {
  buf[0] = (i >> 56) & 0xFF;
  buf[1] = (i >> 48) & 0xFF;
  buf[2] = (i >> 40) & 0xFF;
  buf[3] = (i >> 32) & 0xFF;
  buf[4] = (i >> 24) & 0xFF;
  buf[5] = (i >> 16) & 0xFF;
  buf[6] = (i >> 8) & 0xFF;
  buf[7] = i & 0xFF;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<srl_int64_to_bytes>:
  endbr64
  mov    %edi,%eax
  mov    $0x38,%ecx
  sar    %cl,%eax
  mov    $0x30,%ecx
  mov    %al,(%rsi)
  mov    %edi,%eax
  sar    %cl,%eax
  mov    $0x28,%ecx
  mov    %al,0x1(%rsi)
  mov    %edi,%eax
  sar    %cl,%eax
  mov    $0x20,%ecx
  mov    %al,0x2(%rsi)
  mov    %edi,%eax
  bswap  %edi
  sar    %cl,%eax
  mov    %edi,0x4(%rsi)
  mov    %al,0x3(%rsi)
  ret
```

assistant: ```c
void srl_int64_to_bytes(int64_t i, char *buf) {
  buf[0] = (i >> 56) & 0xFF;
  buf[1] = (i >> 48) & 0xFF;
  buf[2] = (i >> 40) & 0xFF;
  buf[3] = (i >> 32) & 0xFF;
  buf[4] = (i >> 24) & 0xFF;
  buf[5] = (i >> 16) & 0xFF;
  buf[6] = (i >> 8) & 0xFF;
  buf[7] = i & 0xFF;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<srl_int64_to_bytes>:
  endbr64
  mov    %edi,%edx
  mov    $0x38,%ecx
  mov    %edi,%eax
  push   %r14
  sar    %cl,%edx
  mov    $0x30,%ecx
  push   %rbx
  mov    %edi,%ebx
  sar    %cl,%eax
  sar    $0x8,%ebx
  mov    %edi,%r8d
  mov    %edi,%r9d
  mov    %eax,%r14d
  movzbl %dil,%eax
  mov    %edi,%r10d
  mov    %edi,%r11d
  shl    $0x8,%rax
  movzbl %bl,%edi
  sar    $0x10,%r11d
  mov    $0x28,%ecx
  or     %rdi,%rax
  movzbl %r11b,%edi
  shr    $0x18,%r10d
  movzbl %dl,%edx
  shl    $0x8,%rax
  sar    %cl,%r8d
  mov    $0x20,%ecx
  pop    %rbx
  or     %rdi,%rax
  movzbl %r10b,%edi
  sar    %cl,%r9d
  movzbl %r14b,%ecx
  shl    $0x8,%rax
  pop    %r14
  or     %rdi,%rax
  movzbl %r9b,%edi
  shl    $0x8,%rax
  or     %rdi,%rax
  movzbl %r8b,%edi
  shl    $0x8,%rax
  or     %rdi,%rax
  shl    $0x8,%rax
  or     %rcx,%rax
  shl    $0x8,%rax
  or     %rdx,%rax
  mov    %rax,(%rsi)
  ret
```

assistant: ```c
void srl_int64_to_bytes(int64_t i, char *buf) {
  buf[0] = (i >> 56) & 0xFF;
  buf[1] = (i >> 48) & 0xFF;
  buf[2] = (i >> 40) & 0xFF;
  buf[3] = (i >> 32) & 0xFF;
  buf[4] = (i >> 24) & 0xFF;
  buf[5] = (i >> 16) & 0xFF;
  buf[6] = (i >> 8) & 0xFF;
  buf[7] = i & 0xFF;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<skipToken>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %rdi,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %rax,%rdi
  call   <skipSpace@plt>
  mov    %rax,-0x8(%rbp)
  jmp    L26
L11:
  addq   $0x1,-0x8(%rbp)
L26:
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L0
  mov    -0x8(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <isspace@plt>
  test   %eax,%eax
  je     L11
L0:
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *skipToken(char *ptr) {
  ptr = skipSpace(ptr);
  while ((((*ptr) != '\0') && (!isspace(*ptr)))) {
    ptr++;
  }
  return ptr;
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<skipToken>:
  endbr64
  push   %rbx
  call   <skipSpace@plt>
  mov    %rax,%rbx
  movzbl (%rax),%edi
  test   %dil,%dil
  je     L10
L26:
  movsbl %dil,%edi
  call   <isspace@plt>
  test   %eax,%eax
  jne    L10
  add    $0x1,%rbx
  movzbl (%rbx),%edi
  test   %dil,%dil
  jne    L26
L10:
  mov    %rbx,%rax
  pop    %rbx
  ret
```

assistant: ```c
char *skipToken(char *ptr) {
  ptr = skipSpace(ptr);
  while ((((*ptr) != '\0') && (!isspace(*ptr)))) {
    ptr++;
  }
  return ptr;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<skipToken>:
  endbr64
  push   %r12
  call   <skipSpace@plt>
  movsbl (%rax),%edi
  mov    %rax,%r12
  test   %dil,%dil
  jne    L23
  jmp    L2
  nopl   0x0(%rax,%rax,1)
L28:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  test   %dil,%dil
  je     L2
L23:
  call   <isspace@plt>
  test   %eax,%eax
  je     L28
L2:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *skipToken(char *ptr) {
  ptr = skipSpace(ptr);
  while ((((*ptr) != '\0') && (!isspace(*ptr)))) {
    ptr++;
  }
  return ptr;
}
```
user: Translate the below assembly code into C programming language:

```asm
<skipToken>:
  endbr64
  push   %r12
  call   <skipSpace@plt>
  movsbl (%rax),%edi
  mov    %rax,%r12
  test   %dil,%dil
  jne    L12
  jmp    L39
  nopl   0x0(%rax,%rax,1)
L20:
  movsbl 0x1(%r12),%edi
  add    $0x1,%r12
  test   %dil,%dil
  je     L39
L12:
  call   <isspace@plt>
  test   %eax,%eax
  je     L20
L39:
  mov    %r12,%rax
  pop    %r12
  ret
```

assistant: ```c
char *skipToken(char *ptr) {
  ptr = skipSpace(ptr);
  while ((((*ptr) != '\0') && (!isspace(*ptr)))) {
    ptr++;
  }
  return ptr;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<multiply226>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %edi,-0x44(%rbp)
  mov    %esi,-0x48(%rbp)
  mov    %rdx,-0x50(%rbp)
  mov    %rcx,-0x58(%rbp)
  mov    %r8,-0x60(%rbp)
  mov    -0x44(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r14
  mov    $0x0,%r15d
  mov    -0x44(%rbp),%edx
  movslq %edx,%rcx
  sub    $0x1,%rcx
  mov    %rcx,-0x30(%rbp)
  movslq %edx,%rcx
  mov    %rcx,%r12
  mov    $0x0,%r13d
  mov    -0x44(%rbp),%ecx
  movslq %ecx,%rsi
  sub    $0x1,%rsi
  mov    %rsi,-0x28(%rbp)
  movslq %ecx,%rsi
  mov    %rsi,%r10
  mov    $0x0,%r11d
  movl   $0x0,-0x40(%rbp)
  jmp    L12
L11:
  movl   $0x0,-0x3c(%rbp)
  jmp    L23
L38:
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %eax,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x50(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%esi
  imul   -0x48(%rbp),%esi
  mov    %esi,%r9d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %edx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x58(%rbp),%rsi
  add    %rsi,%rdi
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    (%rdi,%rsi,4),%r8d
  mov    -0x40(%rbp),%esi
  movslq %esi,%rdi
  movslq %ecx,%rsi
  imul   %rdi,%rsi
  lea    0x0(,%rsi,4),%rdi
  mov    -0x60(%rbp),%rsi
  add    %rsi,%rdi
  add    %r9d,%r8d
  mov    -0x3c(%rbp),%esi
  movslq %esi,%rsi
  mov    %r8d,(%rdi,%rsi,4)
  addl   $0x1,-0x3c(%rbp)
L23:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L38
  addl   $0x1,-0x40(%rbp)
L12:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L11
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply226(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<multiply226>:
  endbr64
  test   %edi,%edi
  jle    L4
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebx
  mov    %esi,%r10d
  movslq %edi,%r12
  shl    $0x2,%r12
  mov    %rdx,%rbp
  mov    $0x0,%r13d
  mov    %edi,%r11d
L19:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L5:
  mov    %r10d,%edx
  imul   (%r9,%rax,4),%edx
  add    (%rdi,%rax,4),%edx
  mov    %edx,(%rsi,%rax,4)
  add    $0x1,%rax
  cmp    %rax,%r11
  jne    L5
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L19
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L4:
  ret
```

assistant: ```c
void multiply226(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<multiply226>:
  endbr64
  test   %edi,%edi
  jle    L17
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L20:
  mov    (%rdx,%rax,4),%esi
  imul   %r9d,%esi
  add    (%rcx,%rax,4),%esi
  mov    %esi,(%r8,%rax,4)
  add    $0x1,%rax
  cmp    %r10,%rax
  jne    L20
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L4
  pop    %rbx
  ret
L17:
  ret
```

assistant: ```c
void multiply226(int n, int a, int tab1[n][n], int tab2[n][n], int tab3[n][n]) {
  int i, j, k;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      tab3[i][j] = a * tab1[i][j] + tab2[i][j];
    }
  }
}
```