ylfeng/ReF-Decompile-dataset huggingface dataset:
user: What would the C source look like for the assembly code provided here:

```asm
<bitmap_and>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbp
  mov    %rsi,%r12
  mov    %rdx,%r13
  mov    %ecx,%ebx
  mov    %ecx,%edi
  call   <__builtin_constant_p@plt>
  cmp    $0x40,%ebx
  jg     L27
  test   %rax,%rax
  jne    L28
L27:
  mov    %ebx,%ecx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <__bitmap_and@plt>
L7:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L28:
  mov    (%r12),%rax
  and    0x0(%r13),%rax
  mov    %rax,0x0(%rbp)
  setne  %al
  movzbl %al,%eax
  jmp    L7
```

assistant: ```c
__attribute__((always_)) int bitmap_and(unsigned long *dst, const unsigned long *src1, const unsigned long *src2, int nbits) {
  if ((__builtin_constant_p(nbits) && (nbits) <= 64))
    return (*dst = *src1 & *src2) != 0;
  return __bitmap_and(dst, src1, src2, nbits);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bitmap_and>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  mov    %ecx,%edi
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <__builtin_constant_p@plt>
  cmp    $0x40,%r12d
  jg     L18
  test   %rax,%rax
  jne    L6
L18:
  add    $0x8,%rsp
  mov    %r12d,%ecx
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <__bitmap_and@plt>
  nopl   0x0(%rax)
L6:
  mov    0x0(%rbp),%rax
  and    0x0(%r13),%rax
  mov    %rax,(%r14)
  setne  %al
  add    $0x8,%rsp
  pop    %rbp
  movzbl %al,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
__attribute__((always_)) int bitmap_and(unsigned long *dst, const unsigned long *src1, const unsigned long *src2, int nbits) {
  if ((__builtin_constant_p(nbits) && (nbits) <= 64))
    return (*dst = *src1 & *src2) != 0;
  return __bitmap_and(dst, src1, src2, nbits);
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<bitmap_and>:
  endbr64
  push   %r14
  mov    %rdi,%r14
  mov    %ecx,%edi
  push   %r13
  mov    %rdx,%r13
  push   %r12
  mov    %ecx,%r12d
  push   %rbp
  mov    %rsi,%rbp
  sub    $0x8,%rsp
  call   <__builtin_constant_p@plt>
  cmp    $0x40,%r12d
  jg     L5
  test   %rax,%rax
  jne    L15
L5:
  add    $0x8,%rsp
  mov    %r12d,%ecx
  mov    %r13,%rdx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1060 <__bitmap_and@plt>
  nopl   0x0(%rax)
L15:
  mov    0x0(%rbp),%rax
  and    0x0(%r13),%rax
  mov    %rax,(%r14)
  setne  %al
  add    $0x8,%rsp
  pop    %rbp
  movzbl %al,%eax
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
__attribute__((always_)) int bitmap_and(unsigned long *dst, const unsigned long *src1, const unsigned long *src2, int nbits) {
  if ((__builtin_constant_p(nbits) && (nbits) <= 64))
    return (*dst = *src1 & *src2) != 0;
  return __bitmap_and(dst, src1, src2, nbits);
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<bitcount>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x55555555,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shr    %eax
  and    $0x55555555,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  and    $0x33333333,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  shr    $0x2,%eax
  and    $0x33333333,%eax
  add    %edx,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x4,%eax
  mov    %eax,%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  and    $0xf0f0f0f,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x8,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  shr    $0x10,%eax
  add    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  movzbl %al,%eax
  pop    %rbp
  ret
```

assistant: ```c
int bitcount(unsigned int a) {
  a = (a & 0x55555555) + ((a >> 1) & 0x55555555);
  a = (a & 0x33333333) + ((a >> 2) & 0x33333333);
  a = (a + (a >> 4)) & 0x0f0f0f0f;
  a = (a + (a >> 8));
  a = (a + (a >> 16));
  return a & 0xff;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bitcount>:
  endbr64
  mov    %edi,%eax
  and    $0x55555555,%eax
  shr    %edi
  and    $0x55555555,%edi
  add    %edi,%eax
  mov    %eax,%edx
  and    $0x33333333,%edx
  shr    $0x2,%eax
  and    $0x33333333,%eax
  add    %eax,%edx
  mov    %edx,%eax
  shr    $0x4,%eax
  add    %edx,%eax
  and    $0xf0f0f0f,%eax
  mov    %eax,%edx
  shr    $0x8,%edx
  add    %eax,%edx
  mov    %edx,%eax
  shr    $0x10,%eax
  add    %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int bitcount(unsigned int a) {
  a = (a & 0x55555555) + ((a >> 1) & 0x55555555);
  a = (a & 0x33333333) + ((a >> 2) & 0x33333333);
  a = (a + (a >> 4)) & 0x0f0f0f0f;
  a = (a + (a >> 8));
  a = (a + (a >> 16));
  return a & 0xff;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<bitcount>:
  endbr64
  mov    %edi,%eax
  shr    %edi
  and    $0x55555555,%eax
  and    $0x55555555,%edi
  add    %edi,%eax
  mov    %eax,%edx
  shr    $0x2,%eax
  and    $0x33333333,%eax
  and    $0x33333333,%edx
  add    %eax,%edx
  mov    %edx,%eax
  shr    $0x4,%eax
  add    %edx,%eax
  and    $0xf0f0f0f,%eax
  mov    %eax,%edx
  shr    $0x8,%edx
  add    %eax,%edx
  mov    %edx,%eax
  shr    $0x10,%eax
  add    %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int bitcount(unsigned int a) {
  a = (a & 0x55555555) + ((a >> 1) & 0x55555555);
  a = (a & 0x33333333) + ((a >> 2) & 0x33333333);
  a = (a + (a >> 4)) & 0x0f0f0f0f;
  a = (a + (a >> 8));
  a = (a + (a >> 16));
  return a & 0xff;
}
```
user: Translate the below assembly code into C programming language:

```asm
<bitcount>:
  endbr64
  mov    %edi,%eax
  shr    %edi
  and    $0x55555555,%eax
  and    $0x55555555,%edi
  add    %edi,%eax
  mov    %eax,%edx
  shr    $0x2,%eax
  and    $0x33333333,%eax
  and    $0x33333333,%edx
  add    %eax,%edx
  mov    %edx,%eax
  shr    $0x4,%eax
  add    %edx,%eax
  and    $0xf0f0f0f,%eax
  mov    %eax,%edx
  shr    $0x8,%edx
  add    %eax,%edx
  mov    %edx,%eax
  shr    $0x10,%eax
  add    %edx,%eax
  movzbl %al,%eax
  ret
```

assistant: ```c
int bitcount(unsigned int a) {
  a = (a & 0x55555555) + ((a >> 1) & 0x55555555);
  a = (a & 0x33333333) + ((a >> 2) & 0x33333333);
  a = (a + (a >> 4)) & 0x0f0f0f0f;
  a = (a + (a >> 8));
  a = (a + (a >> 16));
  return a & 0xff;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<maxProfit>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    -0x18(%rbp),%rax
  mov    (%rax),%eax
  mov    %eax,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x1,-0x4(%rbp)
  jmp    L12
L4:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0xc(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <MIN@plt>
  mov    %eax,-0xc(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  sub    -0xc(%rbp),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <MAX@plt>
  mov    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L12:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L4
  mov    -0x8(%rbp),%eax
  leave
  ret
```

assistant: ```c
int maxProfit(int *a, int n) {
  int min = a[0];
  int max = 0;
  for (int i = 1; i < n; ++i) {
    min = MIN(min, a[i]);
    max = MAX(max, a[i] - min);
  }
  return max;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<maxProfit>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%ebx
  cmp    $0x1,%esi
  jle    L7
  lea    0x4(%rdi),%rbp
  lea    -0x2(%rsi),%eax
  lea    0x8(%rdi,%rax,4),%r13
  mov    $0x0,%r12d
L4:
  mov    0x0(%rbp),%esi
  mov    %ebx,%edi
  call   <MIN@plt>
  mov    %eax,%ebx
  mov    0x0(%rbp),%esi
  sub    %eax,%esi
  mov    %r12d,%edi
  call   <MAX@plt>
  mov    %eax,%r12d
  add    $0x4,%rbp
  cmp    %r13,%rbp
  jne    L4
L22:
  mov    %r12d,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L7:
  mov    $0x0,%r12d
  jmp    L22
```

assistant: ```c
int maxProfit(int *a, int n) {
  int min = a[0];
  int max = 0;
  for (int i = 1; i < n; ++i) {
    min = MIN(min, a[i]);
    max = MAX(max, a[i] - min);
  }
  return max;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<maxProfit>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%ebp
  cmp    $0x1,%esi
  jle    L12
  lea    -0x2(%rsi),%eax
  lea    0x4(%rdi),%rbx
  xor    %r12d,%r12d
  lea    0x8(%rdi,%rax,4),%r13
  nopl   0x0(%rax)
L6:
  mov    (%rbx),%esi
  mov    %ebp,%edi
  add    $0x4,%rbx
  call   <MIN@plt>
  mov    -0x4(%rbx),%esi
  mov    %r12d,%edi
  mov    %eax,%ebp
  sub    %eax,%esi
  call   <MAX@plt>
  mov    %eax,%r12d
  cmp    %r13,%rbx
  jne    L6
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L12:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int maxProfit(int *a, int n) {
  int min = a[0];
  int max = 0;
  for (int i = 1; i < n; ++i) {
    min = MIN(min, a[i]);
    max = MAX(max, a[i] - min);
  }
  return max;
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<maxProfit>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%ebp
  cmp    $0x1,%esi
  jle    L14
  lea    -0x2(%rsi),%eax
  lea    0x4(%rdi),%rbx
  xor    %r12d,%r12d
  lea    0x8(%rdi,%rax,4),%r13
  nopl   0x0(%rax)
L9:
  mov    (%rbx),%esi
  mov    %ebp,%edi
  add    $0x4,%rbx
  call   <MIN@plt>
  mov    -0x4(%rbx),%esi
  mov    %r12d,%edi
  mov    %eax,%ebp
  sub    %eax,%esi
  call   <MAX@plt>
  mov    %eax,%r12d
  cmp    %r13,%rbx
  jne    L9
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopw   0x0(%rax,%rax,1)
L14:
  add    $0x8,%rsp
  xor    %r12d,%r12d
  pop    %rbx
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int maxProfit(int *a, int n) {
  int min = a[0];
  int max = 0;
  for (int i = 1; i < n; ++i) {
    min = MIN(min, a[i]);
    max = MAX(max, a[i] - min);
  }
  return max;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<bbp_parse_hex>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cltq
  mov    %rax,-0x8(%rbp)
  movq   $0x0,-0x10(%rbp)
  jmp    L19
L4:
  mov    -0x10(%rbp),%rax
  lea    (%rax,%rax,1),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <bbp_hex2byte@plt>
  mov    %al,-0x12(%rbp)
  mov    -0x10(%rbp),%rax
  add    %rax,%rax
  lea    0x1(%rax),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%edi
  call   <bbp_hex2byte@plt>
  mov    %al,-0x11(%rbp)
  movzbl -0x12(%rbp),%eax
  shl    $0x4,%eax
  mov    %eax,%edx
  movzbl -0x11(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0x28(%rbp),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addq   $0x1,-0x10(%rbp)
L19:
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jb     L4
  nop
  nop
  leave
  ret
```

assistant: ```c
void bbp_parse_hex(uint8_t *v, const char *str) {
  const size_t count = strlen(str) / 2;
  size_t i;
  for (i = 0; i < count; ++i) {
    const char hi = bbp_hex2byte(str[i * 2]);
    const char lo = bbp_hex2byte(str[i * 2 + 1]);
    v[i] = hi * 16 + lo;
  }
}
```
user: 下面的汇编指令对应的C源代码是什么:

```asm
<bbp_parse_hex>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%r13
  mov    %rsi,%rbp
  mov    %rsi,%rdi
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rax,1),%eax
  sar    %eax
  cltq
  test   %rax,%rax
  je     L6
  mov    %r13,%r12
  add    $0x1,%rbp
  add    %rax,%r13
L5:
  movsbl -0x1(%rbp),%edi
  call   <bbp_hex2byte@plt>
  mov    %eax,%ebx
  movsbl 0x0(%rbp),%edi
  call   <bbp_hex2byte@plt>
  shl    $0x4,%ebx
  add    %eax,%ebx
  mov    %bl,(%r12)
  add    $0x1,%r12
  add    $0x2,%rbp
  cmp    %r13,%r12
  jne    L5
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void bbp_parse_hex(uint8_t *v, const char *str) {
  const size_t count = strlen(str) / 2;
  size_t i;
  for (i = 0; i < count; ++i) {
    const char hi = bbp_hex2byte(str[i * 2]);
    const char lo = bbp_hex2byte(str[i * 2 + 1]);
    v[i] = hi * 16 + lo;
  }
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<bbp_parse_hex>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rax,1),%r12d
  sar    %r12d
  movslq %r12d,%r12
  test   %r12,%r12
  je     L10
  add    $0x1,%rbp
  add    %r13,%r12
  nopw   0x0(%rax,%rax,1)
L1:
  movsbl -0x1(%rbp),%edi
  add    $0x1,%r13
  add    $0x2,%rbp
  call   <bbp_hex2byte@plt>
  movsbl -0x2(%rbp),%edi
  mov    %eax,%ebx
  call   <bbp_hex2byte@plt>
  shl    $0x4,%ebx
  add    %eax,%ebx
  mov    %bl,-0x1(%r13)
  cmp    %r12,%r13
  jne    L1
L10:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void bbp_parse_hex(uint8_t *v, const char *str) {
  const size_t count = strlen(str) / 2;
  size_t i;
  for (i = 0; i < count; ++i) {
    const char hi = bbp_hex2byte(str[i * 2]);
    const char lo = bbp_hex2byte(str[i * 2 + 1]);
    v[i] = hi * 16 + lo;
  }
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<bbp_parse_hex>:
  endbr64
  push   %r13
  mov    %rdi,%r13
  mov    %rsi,%rdi
  push   %r12
  push   %rbp
  mov    %rsi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  mov    %eax,%edx
  shr    $0x1f,%edx
  lea    (%rdx,%rax,1),%r12d
  sar    %r12d
  movslq %r12d,%r12
  test   %r12,%r12
  je     L6
  add    $0x1,%rbp
  add    %r13,%r12
  nopw   0x0(%rax,%rax,1)
L19:
  movsbl -0x1(%rbp),%edi
  add    $0x1,%r13
  add    $0x2,%rbp
  call   <bbp_hex2byte@plt>
  movsbl -0x2(%rbp),%edi
  mov    %eax,%ebx
  call   <bbp_hex2byte@plt>
  shl    $0x4,%ebx
  add    %eax,%ebx
  mov    %bl,-0x1(%r13)
  cmp    %r12,%r13
  jne    L19
L6:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void bbp_parse_hex(uint8_t *v, const char *str) {
  const size_t count = strlen(str) / 2;
  size_t i;
  for (i = 0; i < count; ++i) {
    const char hi = bbp_hex2byte(str[i * 2]);
    const char lo = bbp_hex2byte(str[i * 2 + 1]);
    v[i] = hi * 16 + lo;
  }
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<read_line>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  movq   $0x0,-0x10(%rbp)
  movq   $0xa,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%esi
  mov    $0x0,%edi
  call   <realloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L15
  mov    -0x18(%rbp),%rax
  jmp    L41
L0:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  mov    %dl,(%rax)
  mov    -0x10(%rbp),%rax
  cmp    -0x8(%rbp),%rax
  jne    L15
  addq   $0x10,-0x8(%rbp)
  mov    -0x8(%rbp),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  jne    L15
  mov    -0x18(%rbp),%rax
  jmp    L41
L15:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <fgetc@plt>
  mov    %eax,-0x1c(%rbp)
  mov    D8(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x1c(%rbp)
  je     L49
  cmpl   $0xa,-0x1c(%rbp)
  jne    L0
L49:
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x10(%rbp)
  mov    -0x18(%rbp),%rdx
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %eax,%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <realloc@plt>
L41:
  leave
  ret
```

assistant: ```c
char *read_line(FILE *fp) {
  char *str;
  int ch;
  size_t len = 0;
  size_t size = 10;
  str = realloc(NULL, sizeof(char) * size);
  if (!str)
    return str;
  while (EOF != (ch = fgetc(fp)) && ch != '\n') {
    str[len++] = ch;
    if (len == size) {
      str = realloc(str, sizeof(char) * (size += 16));
      if (!str)
        return str;
    }
  }
  str[len++] = '\0';
  return realloc(str, sizeof(char) * len);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<read_line>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%r13
  mov    $0xa,%esi
  mov    $0x0,%edi
  call   <realloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  je     L34
  mov    $0xa,%r12d
  mov    $0x0,%ebx
  jmp    L1
L36:
  mov    %r14,%rbx
L1:
  mov    %r13,%rdi
  call   <fgetc@plt>
  mov    D9(%rip),%rdx
  cmp    %eax,(%rdx)
  je     L25
  cmp    $0xa,%eax
  je     L25
  lea    0x1(%rbx),%r14
  mov    %al,0x0(%rbp,%rbx,1)
  cmp    %r14,%r12
  jne    L36
  add    $0x10,%r12
  mov    %r12d,%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L36
  jmp    L34
L25:
  movb   $0x0,0x0(%rbp,%rbx,1)
  lea    0x1(%rbx),%esi
  mov    %rbp,%rdi
  call   <realloc@plt>
  mov    %rax,%rbp
L34:
  mov    %rbp,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
```

assistant: ```c
char *read_line(FILE *fp) {
  char *str;
  int ch;
  size_t len = 0;
  size_t size = 10;
  str = realloc(NULL, sizeof(char) * size);
  if (!str)
    return str;
  while (EOF != (ch = fgetc(fp)) && ch != '\n') {
    str[len++] = ch;
    if (len == size) {
      str = realloc(str, sizeof(char) * (size += 16));
      if (!str)
        return str;
    }
  }
  str[len++] = '\0';
  return realloc(str, sizeof(char) * len);
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<read_line>:
  endbr64
  push   %r14
  mov    $0xa,%esi
  push   %r13
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  push   %rbp
  push   %rbx
  call   <realloc@plt>
  test   %rax,%rax
  je     L4
  mov    D3(%rip),%r13
  mov    %rax,%rbp
  mov    $0xa,%ebx
  xor    %r14d,%r14d
  nopw   0x0(%rax,%rax,1)
L11:
  mov    %r12,%rdi
  call   <fgetc@plt>
  lea    0x0(%rbp,%r14,1),%rdx
  add    $0x1,%r14
  cmp    %eax,0x0(%r13)
  je     L21
  cmp    $0xa,%eax
  je     L21
  mov    %al,(%rdx)
  cmp    %r14,%rbx
  jne    L11
  add    $0x10,%rbx
  mov    %rbp,%rdi
  mov    %ebx,%esi
  call   <realloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L11
L4:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L21:
  movb   $0x0,(%rdx)
  mov    %r14d,%esi
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1070 <realloc@plt>
```

assistant: ```c
char *read_line(FILE *fp) {
  char *str;
  int ch;
  size_t len = 0;
  size_t size = 10;
  str = realloc(NULL, sizeof(char) * size);
  if (!str)
    return str;
  while (EOF != (ch = fgetc(fp)) && ch != '\n') {
    str[len++] = ch;
    if (len == size) {
      str = realloc(str, sizeof(char) * (size += 16));
      if (!str)
        return str;
    }
  }
  str[len++] = '\0';
  return realloc(str, sizeof(char) * len);
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<read_line>:
  endbr64
  push   %r14
  mov    $0xa,%esi
  push   %r13
  push   %r12
  mov    %rdi,%r12
  xor    %edi,%edi
  push   %rbp
  push   %rbx
  call   <realloc@plt>
  test   %rax,%rax
  je     L25
  mov    D8(%rip),%r13
  mov    %rax,%rbp
  mov    $0xa,%ebx
  xor    %r14d,%r14d
  nopw   0x0(%rax,%rax,1)
L39:
  mov    %r12,%rdi
  call   <fgetc@plt>
  lea    0x0(%rbp,%r14,1),%rdx
  add    $0x1,%r14
  cmp    %eax,0x0(%r13)
  je     L23
  cmp    $0xa,%eax
  je     L23
  mov    %al,(%rdx)
  cmp    %r14,%rbx
  jne    L39
  add    $0x10,%rbx
  mov    %rbp,%rdi
  mov    %ebx,%esi
  call   <realloc@plt>
  mov    %rax,%rbp
  test   %rax,%rax
  jne    L39
L25:
  pop    %rbx
  xor    %eax,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  nopl   0x0(%rax)
L23:
  movb   $0x0,(%rdx)
  mov    %r14d,%esi
  pop    %rbx
  mov    %rbp,%rdi
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  jmp    1070 <realloc@plt>
```

assistant: ```c
char *read_line(FILE *fp) {
  char *str;
  int ch;
  size_t len = 0;
  size_t size = 10;
  str = realloc(NULL, sizeof(char) * size);
  if (!str)
    return str;
  while (EOF != (ch = fgetc(fp)) && ch != '\n') {
    str[len++] = ch;
    if (len == size) {
      str = realloc(str, sizeof(char) * (size += 16));
      if (!str)
        return str;
    }
  }
  str[len++] = '\0';
  return realloc(str, sizeof(char) * len);
}
```
user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<syscall>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    -0x14(%rbp),%eax
  int    $0x50
  mov    %eax,-0xc(%rbp)
  mov    -0xc(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int syscall(int number) {
  int result;
  asm volatile("movl %0, %%eax" ::"m"(number));
  asm volatile("int $80");
  asm volatile("movl %%eax, %0" : "=m"(result));
  return result;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<syscall>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    0xc(%rsp),%eax
  int    $0x50
  mov    %eax,0x14(%rsp)
  mov    0x14(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L3
  add    $0x28,%rsp
  ret
L3:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int syscall(int number) {
  int result;
  asm volatile("movl %0, %%eax" ::"m"(number));
  asm volatile("int $80");
  asm volatile("movl %%eax, %0" : "=m"(result));
  return result;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<syscall>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    0xc(%rsp),%eax
  int    $0x50
  mov    %eax,0x14(%rsp)
  mov    0x14(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L5
  add    $0x28,%rsp
  ret
L5:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int syscall(int number) {
  int result;
  asm volatile("movl %0, %%eax" ::"m"(number));
  asm volatile("int $80");
  asm volatile("movl %%eax, %0" : "=m"(result));
  return result;
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<syscall>:
  endbr64
  sub    $0x28,%rsp
  mov    %edi,0xc(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  mov    0xc(%rsp),%eax
  int    $0x50
  mov    %eax,0x14(%rsp)
  mov    0x14(%rsp),%eax
  mov    0x18(%rsp),%rdx
  sub    %fs:0x28,%rdx
  jne    L8
  add    $0x28,%rsp
  ret
L8:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int syscall(int number) {
  int result;
  asm volatile("movl %0, %%eax" ::"m"(number));
  asm volatile("int $80");
  asm volatile("movl %%eax, %0" : "=m"(result));
  return result;
}
```