ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<hIndex0>:
  endbr64
  mov    %esi,%r8d
  test   %esi,%esi
  jle    L36
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rsi
  nopl   0x0(%rax,%rax,1)
L17:
  mov    %rdi,%rax
  xor    %ecx,%ecx
  xor    %edx,%edx
  jmp    L11
  nopl   0x0(%rax)
L19:
  add    $0x4,%rax
  add    $0x1,%edx
  cmp    %rsi,%rax
  je     L74
L11:
  cmp    %r8d,(%rax)
  jg     L19
  sete   %r9b
  add    $0x4,%rax
  movzbl %r9b,%r9d
  add    %r9d,%ecx
  cmp    %rsi,%rax
  jne    L11
L74:
  cmp    %r8d,%edx
  jg     L50
  add    %edx,%ecx
  cmp    %r8d,%ecx
  jge    L36
L50:
  sub    $0x1,%r8d
  jne    L17
L36:
  mov    %r8d,%eax
  ret
```

assistant: ```c
int hIndex0(int *s, int size) {
  int i = size;
  for (; i > 0; i--) {
    int count = 0;
    int extra = 0;
    for (int j = 0; j < size; j++) {
      if (s[j] > i)
        count++;
      if (s[j] == i)
        extra++;
    }
    if (count <= i && count + extra >= i)
      break;
  }
  return i;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<hIndex0>:
  endbr64
  mov    %esi,%r9d
  test   %esi,%esi
  jle    L20
  mov    %esi,%eax
  push   %rbp
  mov    %esi,%r8d
  lea    -0x1(%rsi),%ebp
  shr    $0x2,%eax
  push   %rbx
  mov    %esi,%ebx
  pxor   %xmm6,%xmm6
  sub    $0x1,%eax
  and    $0xfffffffc,%ebx
  shl    $0x4,%rax
  lea    0x10(%rdi,%rax,1),%rcx
  nopl   0x0(%rax)
L153:
  cmp    $0x2,%ebp
  jbe    L65
  movd   %r8d,%xmm7
  pxor   %xmm2,%xmm2
  mov    %rdi,%rax
  pshufd $0x0,%xmm7,%xmm5
  movdqa %xmm2,%xmm3
  cs nopw 0x0(%rax,%rax,1)
L4:
  movdqu (%rax),%xmm0
  add    $0x10,%rax
  movdqa %xmm0,%xmm4
  pcmpeqd %xmm5,%xmm0
  pcmpgtd %xmm5,%xmm4
  movdqa %xmm4,%xmm1
  psubd  %xmm4,%xmm3
  pcmpeqd %xmm6,%xmm1
  pand   %xmm1,%xmm0
  psubd  %xmm0,%xmm2
  cmp    %rax,%rcx
  jne    L4
  movdqa %xmm3,%xmm0
  mov    %ebx,%esi
  psrldq $0x8,%xmm0
  paddd  %xmm0,%xmm3
  movdqa %xmm3,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm3
  movdqa %xmm2,%xmm0
  psrldq $0x8,%xmm0
  movd   %xmm3,%edx
  paddd  %xmm0,%xmm2
  movdqa %xmm2,%xmm0
  psrldq $0x4,%xmm0
  paddd  %xmm0,%xmm2
  movd   %xmm2,%eax
  cmp    %ebx,%r9d
  je     L8
L38:
  movslq %esi,%r10
  lea    0x0(,%r10,4),%r11
  cmp    %r8d,(%rdi,%r10,4)
  jle    L138
  add    $0x1,%edx
L71:
  lea    0x1(%rsi),%r10d
  cmp    %r9d,%r10d
  jge    L8
  cmp    0x4(%rdi,%r11,1),%r8d
  jge    L83
  add    $0x1,%edx
L150:
  add    $0x2,%esi
  cmp    %esi,%r9d
  jle    L8
  cmp    0x8(%rdi,%r11,1),%r8d
  jl     L48
  sete   %sil
  movzbl %sil,%esi
  add    %esi,%eax
L8:
  cmp    %edx,%r8d
  jl     L125
  add    %edx,%eax
  cmp    %r8d,%eax
  jge    L85
L125:
  sub    $0x1,%r8d
  jne    L153
L85:
  mov    %r8d,%eax
  pop    %rbx
  pop    %rbp
  ret
  nopl   0x0(%rax)
L138:
  sete   %r10b
  movzbl %r10b,%r10d
  add    %r10d,%eax
  jmp    L71
  nopl   (%rax)
L83:
  sete   %r10b
  movzbl %r10b,%r10d
  add    %r10d,%eax
  jmp    L150
  nopl   (%rax)
L48:
  add    $0x1,%edx
  jmp    L8
  nopl   (%rax)
L65:
  xor    %esi,%esi
  xor    %eax,%eax
  xor    %edx,%edx
  jmp    L38
L20:
  mov    %esi,%eax
  ret
```

assistant: ```c
int hIndex0(int *s, int size) {
  int i = size;
  for (; i > 0; i--) {
    int count = 0;
    int extra = 0;
    for (int j = 0; j < size; j++) {
      if (s[j] > i)
        count++;
      if (s[j] == i)
        extra++;
    }
    if (count <= i && count + extra >= i)
      break;
  }
  return i;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<eepromWriteMulti>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x4(%rbp)
  mov    %esi,-0x8(%rbp)
  mov    %rdx,-0x10(%rbp)
  mov    %ecx,-0x14(%rbp)
  jmp    L19
L13:
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%eax
  movzbl %al,%edx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%ecx
  mov    %ecx,-0x8(%rbp)
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <eepromWrite@plt>
L19:
  mov    -0x14(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x14(%rbp)
  test   %eax,%eax
  jne    L13
  nop
  nop
  leave
  ret
```

assistant: ```c
void eepromWriteMulti(int deviceAddress, unsigned int memAddress, unsigned char *values, unsigned int size) {
  while (size--)
    eepromWrite(0, memAddress++, *values);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<eepromWriteMulti>:
  endbr64
  test   %ecx,%ecx
  je     L18
  push   %r12
  push   %rbp
  push   %rbx
  mov    %esi,%ebx
  mov    %rdx,%r12
  lea    (%rsi,%rcx,1),%ebp
L16:
  movzbl (%r12),%edx
  mov    %ebx,%esi
  mov    $0x0,%edi
  call   <eepromWrite@plt>
  add    $0x1,%ebx
  cmp    %ebx,%ebp
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L18:
  ret
```

assistant: ```c
void eepromWriteMulti(int deviceAddress, unsigned int memAddress, unsigned char *values, unsigned int size) {
  while (size--)
    eepromWrite(0, memAddress++, *values);
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<eepromWriteMulti>:
  endbr64
  test   %ecx,%ecx
  je     L19
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  lea    (%rsi,%rcx,1),%ebp
  push   %rbx
  mov    %esi,%ebx
  nopl   0x0(%rax)
L6:
  movzbl (%r12),%edx
  mov    %ebx,%esi
  xor    %edi,%edi
  add    $0x1,%ebx
  call   <eepromWrite@plt>
  cmp    %ebx,%ebp
  jne    L6
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L19:
  ret
```

assistant: ```c
void eepromWriteMulti(int deviceAddress, unsigned int memAddress, unsigned char *values, unsigned int size) {
  while (size--)
    eepromWrite(0, memAddress++, *values);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<eepromWriteMulti>:
  endbr64
  test   %ecx,%ecx
  je     L9
  push   %r12
  mov    %rdx,%r12
  push   %rbp
  lea    (%rsi,%rcx,1),%ebp
  push   %rbx
  mov    %esi,%ebx
  nopl   0x0(%rax)
L19:
  movzbl (%r12),%edx
  mov    %ebx,%esi
  xor    %edi,%edi
  add    $0x1,%ebx
  call   <eepromWrite@plt>
  cmp    %ebx,%ebp
  jne    L19
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L9:
  ret
```

assistant: ```c
void eepromWriteMulti(int deviceAddress, unsigned int memAddress, unsigned char *values, unsigned int size) {
  while (size--)
    eepromWrite(0, memAddress++, *values);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<RandomFunc>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x18(%rbp),%rax
  movzwl (%rax),%eax
  xor    $0x1cdb,%ax
  mov    %ax,-0xa(%rbp)
  movw   $0x0,-0xc(%rbp)
  nop
  movzwl -0xa(%rbp),%eax
  shr    %ax
  xor    $0x6964,%ax
  mov    %eax,%edx
  mov    -0x20(%rbp),%rax
  mov    %dx,(%rax)
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L5
  call   <__stack_chk_fail@plt>
L5:
  leave
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local2;
  unsigned short local1;
  char copy11;
  {
    state[0UL] = input[0UL] ^ (unsigned short)7387;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      local2 = 0UL;
      while (local2 < (unsigned short)0) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy11;
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] >> (unsigned short)1) ^ (unsigned short)26980;
  }
}
```
user: Translate the below assembly code into C programming language:

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  shr    %ax
  xor    $0x6709,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local2;
  unsigned short local1;
  char copy11;
  {
    state[0UL] = input[0UL] ^ (unsigned short)7387;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      local2 = 0UL;
      while (local2 < (unsigned short)0) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy11;
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] >> (unsigned short)1) ^ (unsigned short)26980;
  }
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  shr    %ax
  xor    $0x6709,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local2;
  unsigned short local1;
  char copy11;
  {
    state[0UL] = input[0UL] ^ (unsigned short)7387;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      local2 = 0UL;
      while (local2 < (unsigned short)0) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy11;
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] >> (unsigned short)1) ^ (unsigned short)26980;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<RandomFunc>:
  endbr64
  movzwl (%rdi),%eax
  shr    %ax
  xor    $0x6709,%ax
  mov    %ax,(%rsi)
  ret
```

assistant: ```c
void RandomFunc(unsigned short input[1], unsigned short output[1]) {
  unsigned short state[1];
  unsigned short local2;
  unsigned short local1;
  char copy11;
  {
    state[0UL] = input[0UL] ^ (unsigned short)7387;
    local1 = 0UL;
    while (local1 < (unsigned short)0) {
      local2 = 0UL;
      while (local2 < (unsigned short)0) {
        copy11 = *((char *)(&state[0UL]) + 1);
        *((char *)(&state[0UL]) + 1) = *((char *)(&state[0UL]) + 0);
        *((char *)(&state[0UL]) + 0) = copy11;
        local2 += 2UL;
      }
      local1 += 2UL;
    }
    output[0UL] = (state[0UL] >> (unsigned short)1) ^ (unsigned short)26980;
  }
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<chomp>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L4
  mov    -0x8(%rbp),%rax
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L4:
  nop
  leave
  ret
```

assistant: ```c
void chomp(char *str) {
  size_t len = strlen(str);
  if (str[len - 1] == '\n')
    str[len - 1] = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<chomp>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    -0x1(%rbx,%rax,1),%rax
  cmpb   $0xa,(%rax)
  je     L4
L5:
  pop    %rbx
  ret
L4:
  movb   $0x0,(%rax)
  jmp    L5
```

assistant: ```c
void chomp(char *str) {
  size_t len = strlen(str);
  if (str[len - 1] == '\n')
    str[len - 1] = '\0';
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<chomp>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    -0x1(%rbx,%rax,1),%rax
  cmpb   $0xa,(%rax)
  je     L4
  pop    %rbx
  ret
  nopl   0x0(%rax)
L4:
  movb   $0x0,(%rax)
  pop    %rbx
  ret
```

assistant: ```c
void chomp(char *str) {
  size_t len = strlen(str);
  if (str[len - 1] == '\n')
    str[len - 1] = '\0';
}
```
user: What is the c source code of the assembly code below?

```asm
<chomp>:
  endbr64
  push   %rbx
  mov    %rdi,%rbx
  call   <strlen@plt>
  lea    -0x1(%rbx,%rax,1),%rax
  cmpb   $0xa,(%rax)
  je     L2
  pop    %rbx
  ret
  nopl   0x0(%rax)
L2:
  movb   $0x0,(%rax)
  pop    %rbx
  ret
```

assistant: ```c
void chomp(char *str) {
  size_t len = strlen(str);
  if (str[len - 1] == '\n')
    str[len - 1] = '\0';
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<getinput_>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x90,%rsp
  mov    %rdi,-0x78(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %edx,-0x84(%rbp)
  mov    %ecx,-0x88(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x6c(%rbp)
  jmp    L89
L71:
  mov    -0x6c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x78(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x6c(%rbp),%eax
  cltq
  mov    %dl,-0x60(%rbp,%rax,1)
  addl   $0x1,-0x6c(%rbp)
L89:
  cmpl   $0x4f,-0x6c(%rbp)
  jg     L42
  mov    -0x6c(%rbp),%eax
  cmp    -0x84(%rbp),%eax
  jl     L71
L42:
  mov    -0x6c(%rbp),%eax
  cltq
  movb   $0x0,-0x60(%rbp,%rax,1)
  lea    -0x60(%rbp),%rax
  mov    %rax,%rdi
  call   <readline@plt>
  mov    %rax,-0x68(%rbp)
  cmpq   $0x0,-0x68(%rbp)
  je     L72
  mov    -0x68(%rbp),%rax
  mov    %rax,%rdi
  call   <add_history@plt>
L72:
  cmpq   $0x0,-0x68(%rbp)
  je     L59
  movl   $0x0,-0x6c(%rbp)
  jmp    L58
L18:
  mov    -0x6c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  mov    -0x6c(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x80(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x6c(%rbp)
L58:
  mov    -0x6c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x68(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L73
  mov    -0x6c(%rbp),%eax
  cmp    -0x88(%rbp),%eax
  jl     L18
  jmp    L73
L59:
  movl   $0x0,-0x6c(%rbp)
  jmp    L73
L116:
  mov    -0x6c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x6c(%rbp)
  movslq %eax,%rdx
  mov    -0x80(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
L73:
  mov    -0x6c(%rbp),%eax
  cmp    -0x88(%rbp),%eax
  jl     L116
  cmpq   $0x0,-0x68(%rbp)
  je     L63
  mov    -0x68(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
L63:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L69
  call   <__stack_chk_fail@plt>
L69:
  leave
  ret
```

assistant: ```c
void getinput_(char *promptstr, char *line, int promptlen, int linelen) {
  char *line_read;
  char str[80 + 1];
  int i;
  for (i = 0; i < 80 && i < promptlen; i++)
    str[i] = promptstr[i];
  str[i] = '\0';
  line_read = readline(str);
  if (line_read)
    add_history(line_read);
  if (line_read)
    for (i = 0; line_read[i] && i < linelen; i++)
      line[i] = line_read[i];
  else
    i = 0;
  while (i < linelen)
    line[i++] = ' ';
  if (line_read)
    free(line_read);
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<getinput_>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x60,%rsp
  mov    %rsi,%rbp
  mov    %ecx,%ebx
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  test   %edx,%edx
  jle    L5
  mov    $0x1,%eax
  lea    -0x1(%rsp),%r8
L149:
  movzbl -0x1(%rdi,%rax,1),%ecx
  mov    %cl,(%r8,%rax,1)
  mov    %eax,%ecx
  add    $0x1,%rax
  cmp    $0x4f,%ecx
  jg     L11
  cmp    %ecx,%edx
  jg     L149
L11:
  movslq %ecx,%rcx
  movb   $0x0,(%rsp,%rcx,1)
  mov    %rsp,%rdi
  call   <readline@plt>
  mov    %rax,%r12
  test   %rax,%rax
  je     L120
  mov    %rax,%rdi
  call   <add_history@plt>
  movzbl (%r12),%ecx
  test   %ebx,%ebx
  jle    L124
  test   %cl,%cl
  je     L124
  mov    $0x1,%edx
L119:
  mov    %cl,-0x1(%rbp,%rdx,1)
  mov    %edx,%eax
  movzbl (%r12,%rdx,1),%ecx
  cmp    %edx,%ebx
  setg   %dil
  test   %cl,%cl
  setne  %sil
  add    $0x1,%rdx
  test   %sil,%dil
  jne    L119
L72:
  cmp    %eax,%ebx
  jle    L118
L108:
  cltq
L142:
  movb   $0x20,0x0(%rbp,%rax,1)
  add    $0x1,%rax
  cmp    %eax,%ebx
  jg     L142
  test   %r12,%r12
  je     L105
L118:
  mov    %r12,%rdi
  call   <free@plt>
L105:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L33
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  mov    $0x0,%ecx
  jmp    L11
L124:
  mov    $0x0,%eax
  jmp    L72
L120:
  mov    $0x0,%eax
  test   %ebx,%ebx
  jg     L108
  jmp    L105
L33:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void getinput_(char *promptstr, char *line, int promptlen, int linelen) {
  char *line_read;
  char str[80 + 1];
  int i;
  for (i = 0; i < 80 && i < promptlen; i++)
    str[i] = promptstr[i];
  str[i] = '\0';
  line_read = readline(str);
  if (line_read)
    add_history(line_read);
  if (line_read)
    for (i = 0; line_read[i] && i < linelen; i++)
      line[i] = line_read[i];
  else
    i = 0;
  while (i < linelen)
    line[i++] = ' ';
  if (line_read)
    free(line_read);
}
```
user: Translate the below assembly code into C programming language:

```asm
<getinput_>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  test   %edx,%edx
  jle    L128
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
  nopl   0x0(%rax,%rax,1)
L49:
  movzbl -0x1(%rdi,%rax,1),%ecx
  mov    %cl,(%rsi,%rax,1)
  movslq %eax,%rcx
  add    $0x1,%rax
  cmp    $0x4f,%ecx
  jg     L42
  cmp    %ecx,%edx
  jg     L49
L42:
  mov    %rsp,%rdi
  movb   $0x0,(%rsp,%rcx,1)
  call   <readline@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L93
  mov    %rax,%rdi
  call   <add_history@plt>
  movzbl 0x0(%rbp),%ecx
  test   %cl,%cl
  je     L15
  lea    0x1(%rbx),%edx
  mov    $0x1,%eax
  test   %ebx,%ebx
  jg     L148
  nopl   0x0(%rax,%rax,1)
L111:
  mov    %rbp,%rdi
  call   <free@plt>
L90:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L108
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L121:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L111
L148:
  mov    %cl,-0x1(%r12,%rax,1)
  movzbl 0x0(%rbp,%rax,1),%ecx
  movslq %eax,%rsi
  test   %cl,%cl
  jne    L121
L4:
  cmp    %ebx,%esi
  jge    L111
  lea    -0x1(%rbx),%edx
  lea    (%r12,%rsi,1),%rdi
  mov    %edx,%ebx
  sub    %esi,%ebx
  mov    $0x20,%esi
  lea    0x1(%rbx),%rdx
  call   <memset@plt>
  jmp    L111
  nopl   0x0(%rax)
L93:
  test   %ebx,%ebx
  jle    L90
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L108
  add    $0x60,%rsp
  mov    %ebx,%edx
  mov    %r12,%rdi
  mov    $0x20,%esi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10c0 <memset@plt>
  nopl   0x0(%rax,%rax,1)
L128:
  xor    %ecx,%ecx
  jmp    L42
L15:
  xor    %esi,%esi
  jmp    L4
L108:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void getinput_(char *promptstr, char *line, int promptlen, int linelen) {
  char *line_read;
  char str[80 + 1];
  int i;
  for (i = 0; i < 80 && i < promptlen; i++)
    str[i] = promptstr[i];
  str[i] = '\0';
  line_read = readline(str);
  if (line_read)
    add_history(line_read);
  if (line_read)
    for (i = 0; line_read[i] && i < linelen; i++)
      line[i] = line_read[i];
  else
    i = 0;
  while (i < linelen)
    line[i++] = ' ';
  if (line_read)
    free(line_read);
}
```
user: What is the c source code of the assembly code below?

```asm
<getinput_>:
  endbr64
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x60,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  test   %edx,%edx
  jle    L47
  mov    $0x1,%eax
  lea    -0x1(%rsp),%rsi
  nopl   0x0(%rax,%rax,1)
L28:
  movzbl -0x1(%rdi,%rax,1),%ecx
  mov    %cl,(%rsi,%rax,1)
  movslq %eax,%rcx
  add    $0x1,%rax
  cmp    $0x4f,%ecx
  jg     L130
  cmp    %edx,%ecx
  jl     L28
L130:
  mov    %rsp,%rdi
  movb   $0x0,(%rsp,%rcx,1)
  call   <readline@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L22
  mov    %rax,%rdi
  call   <add_history@plt>
  movzbl 0x0(%rbp),%ecx
  test   %cl,%cl
  je     L68
  lea    0x1(%rbx),%edx
  mov    $0x1,%eax
  test   %ebx,%ebx
  jg     L101
  nopl   0x0(%rax,%rax,1)
L141:
  mov    %rbp,%rdi
  call   <free@plt>
L95:
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x60,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L20:
  add    $0x1,%rax
  cmp    %rax,%rdx
  je     L141
L101:
  mov    %cl,-0x1(%r12,%rax,1)
  movzbl 0x0(%rbp,%rax,1),%ecx
  movslq %eax,%rsi
  test   %cl,%cl
  jne    L20
L61:
  cmp    %ebx,%esi
  jge    L141
  lea    -0x1(%rbx),%edx
  lea    (%r12,%rsi,1),%rdi
  mov    %edx,%ebx
  sub    %esi,%ebx
  mov    $0x20,%esi
  lea    0x1(%rbx),%rdx
  call   <memset@plt>
  jmp    L141
  nopl   0x0(%rax)
L22:
  test   %ebx,%ebx
  jle    L95
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x60,%rsp
  mov    %ebx,%edx
  mov    %r12,%rdi
  mov    $0x20,%esi
  pop    %rbx
  pop    %rbp
  pop    %r12
  jmp    10c0 <memset@plt>
  nopl   0x0(%rax,%rax,1)
L47:
  xor    %ecx,%ecx
  jmp    L130
L68:
  xor    %esi,%esi
  jmp    L61
L7:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void getinput_(char *promptstr, char *line, int promptlen, int linelen) {
  char *line_read;
  char str[80 + 1];
  int i;
  for (i = 0; i < 80 && i < promptlen; i++)
    str[i] = promptstr[i];
  str[i] = '\0';
  line_read = readline(str);
  if (line_read)
    add_history(line_read);
  if (line_read)
    for (i = 0; line_read[i] && i < linelen; i++)
      line[i] = line_read[i];
  else
    i = 0;
  while (i < linelen)
    line[i++] = ' ';
  if (line_read)
    free(line_read);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<printMatrixCustom>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %rdx,-0x28(%rbp)
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movl   $0x0,-0x8(%rbp)
  jmp    L3
L16:
  movl   $0x0,-0x4(%rbp)
  jmp    L35
L28:
  mov    -0x8(%rbp),%eax
  cltq
  imul   -0x28(%rbp),%rax
  mov    %rax,%rdx
  mov    -0x4(%rbp),%eax
  cltq
  add    %rdx,%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  addl   $0x1,-0x4(%rbp)
L35:
  mov    -0x4(%rbp),%eax
  cltq
  cmp    %rax,-0x28(%rbp)
  ja     L28
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,-0x8(%rbp)
L3:
  mov    -0x8(%rbp),%eax
  cltq
  cmp    %rax,-0x20(%rbp)
  ja     L16
  mov    $0xa,%edi
  call   <putchar@plt>
  nop
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: label: D17
type: string
name: retrieve_data
arguments: label: D14
type: string

tool: D17: 
 *************** MATRIX ****************


tool: D14:  %.4f 

assistant: ```c
void printMatrixCustom(double *matrix, size_t size_r, size_t size_c) {
  printf("\n *************** MATRIX ****************\n\n");
  int i, j;
  for (i = 0; i < size_r; i++) {
    for (j = 0; j < size_c; ++j) {
      printf(" %.4f ", matrix[i * size_c + j]);
    }
    printf("\n");
  }
  printf("\n");
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<printMatrixCustom>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r15
  mov    %rdx,%r12
  lea    D6(%rip),%rdi
  call   <puts@plt>
  test   %r15,%r15
  je     L12
  lea    0x0(,%r12,8),%rax
  mov    %rax,0x8(%rsp)
  mov    $0x0,%r14d
  lea    D13(%rip),%r13
  jmp    L45
L46:
  movsd  0x0(%rbp,%rbx,8),%xmm0
  mov    %r13,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %r12,%rbx
  jne    L46
L9:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x1,%r14
  add    0x8(%rsp),%rbp
  cmp    %r14,%r15
  je     L12
L45:
  mov    $0x0,%ebx
  test   %r12,%r12
  jne    L46
  jmp    L9
L12:
  mov    $0xa,%edi
  call   <putchar@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: retrieve_data
arguments: data_tag: D6
data_type: string
name: retrieve_data
arguments: data_tag: D13
data_type: string

tool: D6: 
 *************** MATRIX ****************


tool: D13:  %.4f 

assistant: ```c
void printMatrixCustom(double *matrix, size_t size_r, size_t size_c) {
  printf("\n *************** MATRIX ****************\n\n");
  int i, j;
  for (i = 0; i < size_r; i++) {
    for (j = 0; j < size_c; ++j) {
      printf(" %.4f ", matrix[i * size_c + j]);
    }
    printf("\n");
  }
  printf("\n");
}
```