ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 这段汇编代码用C语言表示是什么样的：

```asm
<coinChange>:
  endbr64
  push   %r13
  lea    0x1(%rdx),%r13d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x0(,%r13,4),%edi
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  movl   $0x0,(%rax)
  mov    %rax,%r10
  test   %ebx,%ebx
  jle    L61
  lea    0x4(%rax),%rdi
  lea    -0x1(%r12),%eax
  mov    $0x1,%esi
  lea    0x4(%rbp,%rax,4),%r8
  nopw   0x0(%rax,%rax,1)
L6:
  mov    %r13d,(%rdi)
  test   %r12d,%r12d
  jle    L57
  mov    %rbp,%rax
  nopl   0x0(%rax,%rax,1)
L1:
  mov    (%rax),%edx
  cmp    %esi,%edx
  jg     L46
  mov    %esi,%ecx
  sub    %edx,%ecx
  movslq %ecx,%rdx
  mov    (%r10,%rdx,4),%ecx
  mov    (%rdi),%edx
  lea    0x1(%rcx),%r9d
  cmp    %edx,%ecx
  cmovl  %r9d,%edx
  mov    %edx,(%rdi)
L46:
  add    $0x4,%rax
  cmp    %r8,%rax
  jne    L1
L57:
  lea    0x1(%rsi),%eax
  add    $0x4,%rdi
  cmp    %esi,%ebx
  je     L61
  mov    %eax,%esi
  jmp    L6
  nopw   0x0(%rax,%rax,1)
L61:
  movslq %ebx,%rax
  mov    (%r10,%rax,4),%r12d
  cmp    %ebx,%r12d
  jg     L39
L48:
  mov    %r10,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L39:
  mov    $0xffffffff,%r12d
  jmp    L48
```

assistant: ```c
int coinChange(int *coins, int coinsSize, int amount) {
  int *dp, i, j, ret;
  dp = (int *)malloc((amount + 1) * sizeof(int));
  dp[0] = 0;
  for (i = 1; i <= amount; ++i) {
    dp[i] = amount + 1;
    for (j = 0; j < coinsSize; ++j) {
      if (coins[j] <= i)
        dp[i] = ((dp[i]) < (dp[i - coins[j]] + 1) ? (dp[i]) : (dp[i - coins[j]] + 1));
    }
  }
  ret = dp[amount] <= amount ? dp[amount] : -1;
  free(dp);
  return ret;
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<coinChange>:
  endbr64
  push   %r13
  lea    0x1(%rdx),%r13d
  push   %r12
  mov    %esi,%r12d
  push   %rbp
  mov    %rdi,%rbp
  lea    0x0(,%r13,4),%edi
  push   %rbx
  mov    %edx,%ebx
  sub    $0x8,%rsp
  call   <malloc@plt>
  movl   $0x0,(%rax)
  mov    %rax,%r10
  test   %ebx,%ebx
  jle    L48
  lea    0x4(%rax),%rdi
  lea    -0x1(%r12),%eax
  mov    $0x1,%esi
  lea    0x4(%rbp,%rax,4),%r8
  nopw   0x0(%rax,%rax,1)
L32:
  mov    %r13d,(%rdi)
  test   %r12d,%r12d
  jle    L79
  mov    %rbp,%rax
  nopl   0x0(%rax,%rax,1)
L62:
  mov    (%rax),%edx
  cmp    %esi,%edx
  jg     L1
  mov    %esi,%ecx
  sub    %edx,%ecx
  movslq %ecx,%rdx
  mov    (%r10,%rdx,4),%ecx
  mov    (%rdi),%edx
  lea    0x1(%rcx),%r9d
  cmp    %edx,%ecx
  cmovl  %r9d,%edx
  mov    %edx,(%rdi)
L1:
  add    $0x4,%rax
  cmp    %rax,%r8
  jne    L62
L79:
  lea    0x1(%rsi),%eax
  add    $0x4,%rdi
  cmp    %esi,%ebx
  je     L48
  mov    %eax,%esi
  jmp    L32
  nopw   0x0(%rax,%rax,1)
L48:
  movslq %ebx,%rax
  mov    (%r10,%rax,4),%r12d
  cmp    %ebx,%r12d
  jg     L15
L49:
  mov    %r10,%rdi
  call   <free@plt>
  add    $0x8,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L15:
  mov    $0xffffffff,%r12d
  jmp    L49
```

assistant: ```c
int coinChange(int *coins, int coinsSize, int amount) {
  int *dp, i, j, ret;
  dp = (int *)malloc((amount + 1) * sizeof(int));
  dp[0] = 0;
  for (i = 1; i <= amount; ++i) {
    dp[i] = amount + 1;
    for (j = 0; j < coinsSize; ++j) {
      if (coins[j] <= i)
        dp[i] = ((dp[i]) < (dp[i - coins[j]] + 1) ? (dp[i]) : (dp[i - coins[j]] + 1));
    }
  }
  ret = dp[amount] <= amount ? dp[amount] : -1;
  free(dp);
  return ret;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<integerReplacement>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  cmpl   $0x7fffffff,-0x14(%rbp)
  jne    L17
  mov    $0x20,%eax
  jmp    L52
L17:
  cmpl   $0x1,-0x14(%rbp)
  jne    L45
  mov    $0x0,%eax
  jmp    L52
L45:
  mov    -0x14(%rbp),%eax
  and    $0x1,%eax
  test   %eax,%eax
  jne    L53
  mov    -0x14(%rbp),%eax
  sar    %eax
  mov    %eax,%edi
  call   <integerReplacement>
  add    $0x1,%eax
  jmp    L52
L53:
  mov    -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%edi
  call   <integerReplacement>
  add    $0x1,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edi
  call   <integerReplacement>
  add    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x8(%rbp),%eax
  cmp    %eax,%edx
  cmovle %edx,%eax
L52:
  leave
  ret
```

assistant: ```c
int integerReplacement(int n) {
  if (n == 2147483647)
    return 32;
  if (n == 1)
    return 0;
  if (n % 2 == 0)
    return 1 + integerReplacement(n >> 1);
  int plusOne = 1 + integerReplacement(n + 1);
  int minusOne = 1 + integerReplacement(n - 1);
  return plusOne < minusOne ? plusOne : minusOne;
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<integerReplacement>:
  endbr64
  cmp    $0x7fffffff,%edi
  je     L0
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%ebx
  mov    $0x0,%eax
  cmp    $0x1,%edi
  je     L5
  test   $0x1,%dil
  je     L23
  lea    0x1(%rdi),%edi
  call   <integerReplacement>
  mov    %eax,%ebp
  lea    -0x1(%rbx),%edi
  call   <integerReplacement>
  mov    %eax,%edx
  lea    0x1(%rbp),%eax
  add    $0x1,%edx
  cmp    %edx,%eax
  cmovg  %edx,%eax
L5:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L23:
  sar    %ebx
  mov    %ebx,%edi
  call   <integerReplacement>
  add    $0x1,%eax
  jmp    L5
L0:
  mov    $0x20,%eax
  ret
```

assistant: ```c
int integerReplacement(int n) {
  if (n == 2147483647)
    return 32;
  if (n == 1)
    return 0;
  if (n % 2 == 0)
    return 1 + integerReplacement(n >> 1);
  int plusOne = 1 + integerReplacement(n + 1);
  int minusOne = 1 + integerReplacement(n - 1);
  return plusOne < minusOne ? plusOne : minusOne;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<integerReplacement>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  cmp    $0x7fffffff,%edi
  je     L20
  mov    %edi,%ebx
  cmp    $0x1,%edi
  je     L27
  mov    $0x1,%eax
  xor    %r12d,%r12d
  test   $0x1,%dil
  je     L22
  jmp    L31
  nopw   0x0(%rax,%rax,1)
L39:
  add    $0x1,%eax
  test   $0x1,%bl
  jne    L31
L22:
  sar    %ebx
  mov    %eax,%r12d
  cmp    $0x1,%ebx
  jne    L39
L57:
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L31:
  lea    0x1(%rbx),%edi
  call   <integerReplacement>
  lea    -0x1(%rbx),%edi
  mov    %eax,%ebp
  call   <integerReplacement>
  lea    0x1(%rbp),%edx
  pop    %rbx
  pop    %rbp
  add    $0x1,%eax
  cmp    %eax,%edx
  cmovg  %eax,%edx
  add    %edx,%r12d
  mov    %r12d,%eax
  pop    %r12
  ret
L20:
  mov    $0x20,%r12d
  jmp    L57
L27:
  xor    %r12d,%r12d
  jmp    L57
```

assistant: ```c
int integerReplacement(int n) {
  if (n == 2147483647)
    return 32;
  if (n == 1)
    return 0;
  if (n % 2 == 0)
    return 1 + integerReplacement(n >> 1);
  int plusOne = 1 + integerReplacement(n + 1);
  int minusOne = 1 + integerReplacement(n - 1);
  return plusOne < minusOne ? plusOne : minusOne;
}
```
user: What is the c source code of the assembly code below:

```asm
<printf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x4e0,%rsp
  mov    %rdi,-0x4d8(%rbp)
  mov    %rsi,-0xa8(%rbp)
  mov    %rdx,-0xa0(%rbp)
  mov    %rcx,-0x98(%rbp)
  mov    %r8,-0x90(%rbp)
  mov    %r9,-0x88(%rbp)
  test   %al,%al
  je     L17
  movaps %xmm0,-0x80(%rbp)
  movaps %xmm1,-0x70(%rbp)
  movaps %xmm2,-0x60(%rbp)
  movaps %xmm3,-0x50(%rbp)
  movaps %xmm4,-0x40(%rbp)
  movaps %xmm5,-0x30(%rbp)
  movaps %xmm6,-0x20(%rbp)
  movaps %xmm7,-0x10(%rbp)
L17:
  mov    %fs:0x28,%rax
  mov    %rax,-0xb8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x4c8(%rbp)
  mov    -0x4d8(%rbp),%rdx
  mov    -0x4c4(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <va_start@plt>
  mov    -0x4c4(%rbp),%edx
  mov    -0x4d8(%rbp),%rcx
  lea    -0x4c0(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <vsprintf@plt>
  mov    %eax,-0x4c8(%rbp)
  mov    -0x4c8(%rbp),%edx
  lea    -0x4c0(%rbp),%rax
  mov    %rax,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  mov    -0x4c4(%rbp),%eax
  mov    %eax,%edi
  call   <va_end@plt>
  mov    -0x4c8(%rbp),%eax
  mov    -0xb8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int printf(const char *format, ...) {
  va_list val;
  int printed = 0;
  char printf_buf[1024];
  va_start(val, format);
  printed = vsprintf(printf_buf, format, val);
  write(1, printf_buf, printed);
  va_end(val);
  return printed;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<printf>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x410,%rsp
  mov    %rdi,%rbx
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rdi,%rsi
  mov    $0x0,%ebp
  mov    %ebp,%edi
  call   <va_start@plt>
  mov    %rsp,%r12
  mov    %ebp,%edx
  mov    %rbx,%rsi
  mov    %r12,%rdi
  call   <vsprintf@plt>
  mov    %eax,%ebx
  mov    %eax,%edx
  mov    %r12,%rsi
  mov    $0x1,%edi
  call   <write@plt>
  mov    %ebp,%edi
  call   <va_end@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  mov    %ebx,%eax
  add    $0x410,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int printf(const char *format, ...) {
  va_list val;
  int printed = 0;
  char printf_buf[1024];
  va_start(val, format);
  printed = vsprintf(printf_buf, format, val);
  write(1, printf_buf, printed);
  va_end(val);
  return printed;
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<printf>:
  endbr64
  push   %r14
  mov    %rdi,%rsi
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <vsprintf@plt>
  mov    $0x1,%edi
  mov    %r14,%rsi
  mov    %eax,%edx
  mov    %eax,%r12d
  call   <write@plt>
  mov    %r13d,%edi
  call   <va_end@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L2
  add    $0x418,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L2:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int printf(const char *format, ...) {
  va_list val;
  int printed = 0;
  char printf_buf[1024];
  va_start(val, format);
  printed = vsprintf(printf_buf, format, val);
  write(1, printf_buf, printed);
  va_end(val);
  return printed;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<printf>:
  endbr64
  push   %r14
  mov    %rdi,%rsi
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  mov    %r13d,%edi
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%r14
  call   <va_start@plt>
  mov    %r13d,%edx
  mov    %rbp,%rsi
  mov    %r14,%rdi
  call   <vsprintf@plt>
  mov    $0x1,%edi
  mov    %r14,%rsi
  mov    %eax,%edx
  mov    %eax,%r12d
  call   <write@plt>
  mov    %r13d,%edi
  call   <va_end@plt>
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L4
  add    $0x418,%rsp
  mov    %r12d,%eax
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L4:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int printf(const char *format, ...) {
  va_list val;
  int printed = 0;
  char printf_buf[1024];
  va_start(val, format);
  printed = vsprintf(printf_buf, format, val);
  write(1, printf_buf, printed);
  va_end(val);
  return printed;
}
```
user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    %rsi,-0x20(%rbp)
  jmp    L7
L6:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <putchar@plt>
L7:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,-0x4(%rbp)
  mov    D6(%rip),%rax
  mov    (%rax),%eax
  cmp    %eax,-0x4(%rbp)
  jne    L6
  mov    $0x0,%eax
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c;
  while ((c = getchar()) != EOF)
    putchar(c);
  return 0;
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  sub    $0x8,%rsp
  jmp    L17
L14:
  call   <putchar@plt>
L17:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  mov    D4(%rip),%rax
  cmp    %edi,(%rax)
  jne    L14
  mov    $0x0,%eax
  add    $0x8,%rsp
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c;
  while ((c = getchar()) != EOF)
    putchar(c);
  return 0;
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %rbx
  mov    D2(%rip),%rbx
  jmp    L17
  xchg   %ax,%ax
L13:
  call   <putchar@plt>
L17:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L13
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c;
  while ((c = getchar()) != EOF)
    putchar(c);
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<main>:
  endbr64
  push   %rbx
  mov    D6(%rip),%rbx
  jmp    L6
  xchg   %ax,%ax
L14:
  call   <putchar@plt>
L6:
  xor    %eax,%eax
  call   <getchar@plt>
  mov    %eax,%edi
  cmp    %eax,(%rbx)
  jne    L14
  xor    %eax,%eax
  pop    %rbx
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  int c;
  while ((c = getchar()) != EOF)
    putchar(c);
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<prefix_sum>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  mov    %rdx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L3
L8:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rdx)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  add    %eax,-0x8(%rbp)
  addl   $0x1,-0x4(%rbp)
L3:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L8
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  mov    -0x8(%rbp),%eax
  mov    %eax,(%rdx)
  nop
  pop    %rbp
  ret
```

assistant: ```c
void prefix_sum(int *array, int n, int *out) {
  int sum = 0;
  int i;
  for (i = 0; i < n; ++i) {
    out[i] = sum;
    sum += array[i];
  }
  out[n] = sum;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<prefix_sum>:
  endbr64
  test   %esi,%esi
  jle    L17
  mov    %esi,%r8d
  mov    $0x0,%eax
  mov    $0x0,%ecx
L15:
  mov    %ecx,(%rdx,%rax,4)
  add    (%rdi,%rax,4),%ecx
  add    $0x1,%rax
  cmp    %r8,%rax
  jne    L15
L4:
  movslq %esi,%rsi
  mov    %ecx,(%rdx,%rsi,4)
  ret
L17:
  mov    $0x0,%ecx
  jmp    L4
```

assistant: ```c
void prefix_sum(int *array, int n, int *out) {
  int sum = 0;
  int i;
  for (i = 0; i < n; ++i) {
    out[i] = sum;
    sum += array[i];
  }
  out[n] = sum;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<prefix_sum>:
  endbr64
  test   %esi,%esi
  jle    L3
  movslq %esi,%rsi
  xor    %eax,%eax
  xor    %ecx,%ecx
  nop
L18:
  mov    %ecx,(%rdx,%rax,4)
  add    (%rdi,%rax,4),%ecx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L18
  mov    %ecx,(%rdx,%rsi,4)
  ret
  nopl   0x0(%rax,%rax,1)
L3:
  xor    %ecx,%ecx
  movslq %esi,%rsi
  mov    %ecx,(%rdx,%rsi,4)
  ret
```

assistant: ```c
void prefix_sum(int *array, int n, int *out) {
  int sum = 0;
  int i;
  for (i = 0; i < n; ++i) {
    out[i] = sum;
    sum += array[i];
  }
  out[n] = sum;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<prefix_sum>:
  endbr64
  test   %esi,%esi
  jle    L9
  movslq %esi,%rsi
  xor    %eax,%eax
  xor    %ecx,%ecx
  nop
L2:
  mov    %ecx,(%rdx,%rax,4)
  add    (%rdi,%rax,4),%ecx
  add    $0x1,%rax
  cmp    %rsi,%rax
  jne    L2
  mov    %ecx,(%rdx,%rsi,4)
  ret
  nopl   0x0(%rax,%rax,1)
L9:
  xor    %ecx,%ecx
  movslq %esi,%rsi
  mov    %ecx,(%rdx,%rsi,4)
  ret
```

assistant: ```c
void prefix_sum(int *array, int n, int *out) {
  int sum = 0;
  int i;
  for (i = 0; i < n; ++i) {
    out[i] = sum;
    sum += array[i];
  }
  out[n] = sum;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<arm_float>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  movl   $0x2,-0x8(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt_i@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <print_num@plt>
  mov    -0x8(%rbp),%eax
  mov    %eax,%edi
  call   <sqrt_d@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D14(%rip),%rax
  mov    %rax,%rdi
  call   <print_num@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D14
data_type: string

tool: D14: Square root is:

assistant: ```c
void arm_float(void) {
  int val;
  int num = 2;
  val = sqrt_i(num);
  print_num("Square root is:", val);
  val = sqrt_d(num);
  print_num("Square root is:", val);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<arm_float>:
  endbr64
  push   %rbx
  mov    $0x2,%edi
  call   <sqrt_i@plt>
  mov    %eax,%esi
  lea    D3(%rip),%rbx
  mov    %rbx,%rdi
  call   <print_num@plt>
  mov    $0x2,%edi
  call   <sqrt_d@plt>
  mov    %eax,%esi
  mov    %rbx,%rdi
  call   <print_num@plt>
  pop    %rbx
  ret
```

tool_calls: name: parse_data
arguments: data_tag: D3
datatype: string

tool: D3: Square root is:

assistant: ```c
void arm_float(void) {
  int val;
  int num = 2;
  val = sqrt_i(num);
  print_num("Square root is:", val);
  val = sqrt_d(num);
  print_num("Square root is:", val);
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<arm_float>:
  endbr64
  push   %rbp
  mov    $0x2,%edi
  lea    D2(%rip),%rbp
  call   <sqrt_i@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <print_num@plt>
  mov    $0x2,%edi
  call   <sqrt_d@plt>
  mov    %rbp,%rdi
  pop    %rbp
  mov    %eax,%esi
  jmp    1090 <print_num@plt>
```

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

tool: D2: Square root is:

assistant: ```c
void arm_float(void) {
  int val;
  int num = 2;
  val = sqrt_i(num);
  print_num("Square root is:", val);
  val = sqrt_d(num);
  print_num("Square root is:", val);
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<arm_float>:
  endbr64
  push   %rbp
  mov    $0x2,%edi
  lea    D4(%rip),%rbp
  call   <sqrt_i@plt>
  mov    %rbp,%rdi
  mov    %eax,%esi
  call   <print_num@plt>
  mov    $0x2,%edi
  call   <sqrt_d@plt>
  mov    %rbp,%rdi
  pop    %rbp
  mov    %eax,%esi
  jmp    1090 <print_num@plt>
```

tool_calls: name: parse_data
arguments: data_tag: D4
type_infered: string

tool: D4: Square root is:

assistant: ```c
void arm_float(void) {
  int val;
  int num = 2;
  val = sqrt_i(num);
  print_num("Square root is:", val);
  val = sqrt_d(num);
  print_num("Square root is:", val);
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<BuscaBinaria>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %edi,-0x14(%rbp)
  mov    %esi,-0x18(%rbp)
  mov    %rdx,-0x20(%rbp)
  movl   $0xffffffff,-0xc(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L3
L24:
  mov    -0xc(%rbp),%edx
  mov    -0x8(%rbp),%eax
  add    %edx,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  cmp    %eax,-0x14(%rbp)
  jle    L22
  mov    -0x4(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  jmp    L3
L22:
  mov    -0x4(%rbp),%eax
  mov    %eax,-0x8(%rbp)
L3:
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  cmp    %eax,-0xc(%rbp)
  jl     L24
  mov    -0x8(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int BuscaBinaria(int x, int n, int v[]) {
  int e, m, d;
  e = -1;
  d = n;
  while (e < d - 1) {
    m = (e + d) / 2;
    if (v[m] < x)
      e = m;
    else
      d = m;
  }
  return d;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<BuscaBinaria>:
  endbr64
  mov    %esi,%eax
  mov    $0xffffffff,%r8d
L3:
  lea    -0x1(%rax),%esi
L15:
  cmp    %r8d,%esi
  jle    L9
  lea    (%r8,%rax,1),%r9d
  mov    %r9d,%ecx
  shr    $0x1f,%ecx
  add    %r9d,%ecx
  sar    %ecx
  movslq %ecx,%r9
  cmp    %edi,(%rdx,%r9,4)
  jge    L23
  mov    %ecx,%r8d
  jmp    L15
L23:
  mov    %ecx,%eax
  jmp    L3
L9:
  ret
```

assistant: ```c
int BuscaBinaria(int x, int n, int v[]) {
  int e, m, d;
  e = -1;
  d = n;
  while (e < d - 1) {
    m = (e + d) / 2;
    if (v[m] < x)
      e = m;
    else
      d = m;
  }
  return d;
}
```
user: Demonstrate the C code that generates the assembly output below:

```asm
<BuscaBinaria>:
  endbr64
  mov    %esi,%eax
  mov    $0xffffffff,%r8d
L33:
  lea    -0x1(%rax),%esi
  jmp    L15
  nopl   0x0(%rax)
L11:
  lea    (%r8,%rax,1),%r9d
  mov    %r9d,%ecx
  shr    $0x1f,%ecx
  add    %r9d,%ecx
  sar    %ecx
  movslq %ecx,%r9
  cmp    %edi,(%rdx,%r9,4)
  jge    L4
  mov    %ecx,%r8d
L15:
  cmp    %esi,%r8d
  jl     L11
  ret
  nopl   0x0(%rax)
L4:
  mov    %ecx,%eax
  jmp    L33
```

assistant: ```c
int BuscaBinaria(int x, int n, int v[]) {
  int e, m, d;
  e = -1;
  d = n;
  while (e < d - 1) {
    m = (e + d) / 2;
    if (v[m] < x)
      e = m;
    else
      d = m;
  }
  return d;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<BuscaBinaria>:
  endbr64
  mov    %esi,%eax
  mov    $0xffffffff,%r8d
L4:
  lea    -0x1(%rax),%esi
  jmp    L18
  nopl   0x0(%rax)
L7:
  lea    (%r8,%rax,1),%r9d
  mov    %r9d,%ecx
  shr    $0x1f,%ecx
  add    %r9d,%ecx
  sar    %ecx
  movslq %ecx,%r9
  cmp    %edi,(%rdx,%r9,4)
  jge    L16
  mov    %ecx,%r8d
L18:
  cmp    %esi,%r8d
  jl     L7
  ret
  nopl   0x0(%rax)
L16:
  mov    %ecx,%eax
  jmp    L4
```

assistant: ```c
int BuscaBinaria(int x, int n, int v[]) {
  int e, m, d;
  e = -1;
  d = n;
  while (e < d - 1) {
    m = (e + d) / 2;
    if (v[m] < x)
      e = m;
    else
      d = m;
  }
  return d;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<div>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %ecx,%eax
  mov    %edi,%ecx
  mov    %cl,-0x14(%rbp)
  mov    %esi,%ecx
  mov    %cl,-0x18(%rbp)
  mov    %dl,-0x1c(%rbp)
  mov    %al,-0x20(%rbp)
  cmpb   $0x0,-0x14(%rbp)
  js     L20
  cmpb   $0x0,-0x1c(%rbp)
  js     L20
  movzbl -0x18(%rbp),%edx
  movzbl -0x20(%rbp),%eax
  add    %edx,%eax
  mov    %al,-0x1(%rbp)
  movsbl -0x1(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  movzbl -0x1(%rbp),%eax
  cmp    -0x18(%rbp),%al
  jge    L14
  movzbl -0x14(%rbp),%eax
  add    $0x1,%eax
  mov    %al,-0x14(%rbp)
L14:
  movzbl -0x1(%rbp),%eax
  mov    %al,-0x18(%rbp)
  movzbl -0x14(%rbp),%edx
  movzbl -0x1c(%rbp),%eax
  add    %edx,%eax
  mov    %al,-0x1(%rbp)
L20:
  movsbl -0x18(%rbp),%edx
  movsbl -0x14(%rbp),%eax
  mov    %edx,%esi
  mov    %eax,%edi
  call   <makeShort@plt>
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D0
datatype: string

tool: D0: t %d


assistant: ```c
short div(char a, char b, char c, char d) {
  char t;
  if (a >= 0 && c >= 0) {
    t = b + d;
    printf("t %d\n", t);
    if (t < b) {
      a = a + 1;
    }
    b = t;
    t = a + c;
    if (t < a && t < c) {
    }
  }
  return makeShort(a, b);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<div>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r13d
  or     %dil,%dl
  jns    L13
L17:
  movsbl %sil,%esi
  movsbl %r13b,%edi
  call   <makeShort@plt>
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L13:
  mov    %edi,%ebx
  mov    %esi,%r12d
  lea    (%rcx,%rsi,1),%ebp
  movsbl %bpl,%esi
  lea    D1(%rip),%rdi
  call   <printf@plt>
  mov    %ebp,%esi
  cmp    %bpl,%r12b
  jle    L17
  lea    0x1(%rbx),%r13d
  jmp    L17
```

tool_calls: name: unpack_data
arguments: data_label: D1
type_guessed: string

tool: D1: t %d


assistant: ```c
short div(char a, char b, char c, char d) {
  char t;
  if (a >= 0 && c >= 0) {
    t = b + d;
    printf("t %d\n", t);
    if (t < b) {
      a = a + 1;
    }
    b = t;
    t = a + c;
    if (t < a && t < c) {
    }
  }
  return makeShort(a, b);
}
```