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
<addbuffer>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L15
L0:
  mov    -0x4(%rbp),%eax
  cltq
  lea    (%rax,%rax,1),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  mov    %eax,%ecx
  mov    -0x4(%rbp),%eax
  cltq
  lea    (%rax,%rax,1),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzwl (%rax),%eax
  lea    (%rcx,%rax,1),%edx
  mov    -0x4(%rbp),%eax
  cltq
  lea    (%rax,%rax,1),%rcx
  mov    -0x18(%rbp),%rax
  add    %rcx,%rax
  mov    %dx,(%rax)
  addl   $0x1,-0x4(%rbp)
L15:
  mov    -0x4(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L0
  nop
  nop
  pop    %rbp
  ret
```

assistant: ```c
void addbuffer(short int *b1, short int *b2, int l) {
  int i;
  for (i = 0; i < l; i++) {
    b1[i] += b2[i];
  }
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<addbuffer>:
  endbr64
  test   %edx,%edx
  jle    L9
  mov    %edx,%edx
  mov    $0x0,%eax
L17:
  movzwl (%rsi,%rax,2),%ecx
  add    %cx,(%rdi,%rax,2)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L17
L9:
  ret
```

assistant: ```c
void addbuffer(short int *b1, short int *b2, int l) {
  int i;
  for (i = 0; i < l; i++) {
    b1[i] += b2[i];
  }
}
```
user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<addbuffer>:
  endbr64
  test   %edx,%edx
  jle    L1
  movslq %edx,%rdx
  xor    %eax,%eax
  nopl   (%rax)
L2:
  movzwl (%rsi,%rax,2),%ecx
  add    %cx,(%rdi,%rax,2)
  add    $0x1,%rax
  cmp    %rax,%rdx
  jne    L2
L1:
  ret
```

assistant: ```c
void addbuffer(short int *b1, short int *b2, int l) {
  int i;
  for (i = 0; i < l; i++) {
    b1[i] += b2[i];
  }
}
```
user: 这段汇编代码用C语言表示是什么样的？

```asm
<addbuffer>:
  endbr64
  mov    %rdi,%rcx
  test   %edx,%edx
  jle    L74
  lea    0x2(%rsi),%r9
  mov    %edx,%r8d
  lea    -0x1(%rdx),%eax
  sub    %r9,%rdi
  cmp    $0xc,%rdi
  jbe    L101
  cmp    $0x2,%eax
  jbe    L101
  cmp    $0x6,%eax
  jbe    L31
  mov    %edx,%edi
  xor    %eax,%eax
  shr    $0x3,%edi
  shl    $0x4,%rdi
  nopl   0x0(%rax,%rax,1)
L1:
  movdqu (%rcx,%rax,1),%xmm0
  movdqu (%rsi,%rax,1),%xmm2
  paddw  %xmm2,%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %rdi,%rax
  jne    L1
  mov    %edx,%eax
  and    $0xfffffff8,%eax
  mov    %eax,%edi
  cmp    %eax,%edx
  je     L74
  mov    %edx,%r8d
  sub    %eax,%r8d
  lea    -0x1(%r8),%r9d
  cmp    $0x2,%r9d
  jbe    L64
L42:
  lea    (%rcx,%rax,2),%r9
  movq   (%rsi,%rax,2),%xmm0
  mov    %r8d,%eax
  movq   (%r9),%xmm1
  and    $0xfffffffc,%eax
  add    %eax,%edi
  paddw  %xmm1,%xmm0
  movq   %xmm0,(%r9)
  cmp    %eax,%r8d
  je     L74
L64:
  movslq %edi,%r8
  lea    (%r8,%r8,1),%rax
  movzwl (%rsi,%r8,2),%r8d
  add    %r8w,(%rcx,%rax,1)
  lea    0x1(%rdi),%r8d
  cmp    %r8d,%edx
  jle    L74
  add    $0x2,%edi
  movzwl 0x2(%rsi,%rax,1),%r8d
  add    %r8w,0x2(%rcx,%rax,1)
  cmp    %edi,%edx
  jle    L74
  movzwl 0x4(%rsi,%rax,1),%edx
  add    %dx,0x4(%rcx,%rax,1)
  ret
  xchg   %ax,%ax
L101:
  mov    %edx,%edx
  xor    %eax,%eax
  nopl   0x0(%rax)
L67:
  movzwl (%rsi,%rax,2),%edi
  add    %di,(%rcx,%rax,2)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L67
L74:
  ret
L31:
  xor    %eax,%eax
  xor    %edi,%edi
  jmp    L42
```

assistant: ```c
void addbuffer(short int *b1, short int *b2, int l) {
  int i;
  for (i = 0; i < l; i++) {
    b1[i] += b2[i];
  }
}
```
user: Converte this assembly snippet into its equivalent C code:

```asm
<multiply78>:
  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    L1
L8:
  movl   $0x0,-0x3c(%rbp)
  jmp    L33
L19:
  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)
L33:
  mov    -0x3c(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L19
  addl   $0x1,-0x40(%rbp)
L1:
  mov    -0x40(%rbp),%esi
  cmp    -0x44(%rbp),%esi
  jl     L8
  nop
  nop
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
```

assistant: ```c
void multiply78(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: Converte this assembly snippet into its equivalent C code:

```asm
<multiply78>:
  endbr64
  test   %edi,%edi
  jle    L2
  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
L10:
  mov    %rbp,%r9
  mov    %rcx,%rdi
  mov    %r8,%rsi
  mov    $0x0,%eax
L19:
  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    L19
  add    $0x1,%r13d
  add    %r12,%rbp
  add    %r12,%rcx
  add    %r12,%r8
  cmp    %r13d,%ebx
  jne    L10
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L2:
  ret
```

assistant: ```c
void multiply78(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: What would the C source look like for the assembly code provided here:

```asm
<multiply78>:
  endbr64
  test   %edi,%edi
  jle    L5
  movslq %edi,%r10
  push   %rbx
  mov    %esi,%r9d
  xor    %r11d,%r11d
  lea    0x0(,%r10,4),%rbx
  nopw   0x0(%rax,%rax,1)
L28:
  xor    %eax,%eax
  nopw   0x0(%rax,%rax,1)
L17:
  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    L17
  add    $0x1,%r11d
  add    %rbx,%rdx
  add    %rbx,%rcx
  add    %rbx,%r8
  cmp    %r11d,%edi
  jne    L28
  pop    %rbx
  ret
L5:
  ret
```

assistant: ```c
void multiply78(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: 下面的汇编指令对应的C源代码是什么:

```asm
<multiply78>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  movslq %edi,%rbp
  push   %rbx
  test   %ebp,%ebp
  jle    L57
  mov    %rbp,%r11
  mov    %esi,%ebx
  lea    0x4(%rdx),%r10
  mov    %r8,%rax
  mov    %r11d,%edx
  mov    %r11d,%esi
  lea    0x4(%rcx),%r9
  shl    $0x2,%rbp
  and    $0xfffffffc,%esi
  shr    $0x2,%edx
  movd   %ebx,%xmm4
  xor    %r12d,%r12d
  lea    0x1(%rsi),%ecx
  lea    0x2(%rsi),%edi
  mov    %esi,-0x24(%rsp)
  movslq %esi,%r13
  lea    -0x1(%rdx),%r8d
  mov    %r11d,%esi
  lea    -0x1(%r11),%edx
  mov    %ecx,-0x20(%rsp)
  pshufd $0x0,%xmm4,%xmm2
  mov    %edi,-0x1c(%rsp)
  add    $0x1,%r8
  movslq %ecx,%r14
  mov    %edx,-0x28(%rsp)
  movdqa %xmm2,%xmm3
  shl    $0x4,%r8
  movslq %edi,%r15
  mov    %rsi,-0x10(%rsp)
  psrlq  $0x20,%xmm3
  nopl   (%rax)
L69:
  mov    %rax,%rdx
  lea    -0x4(%r10),%rcx
  lea    -0x4(%r9),%rsi
  sub    %r10,%rdx
  cmp    $0x8,%rdx
  mov    %rax,%rdx
  seta   %dil
  sub    %r9,%rdx
  cmp    $0x8,%rdx
  seta   %dl
  test   %dl,%dil
  je     L7
  xor    %edx,%edx
  cmpl   $0x2,-0x28(%rsp)
  jbe    L7
  nopl   0x0(%rax)
L38:
  movdqu (%rcx,%rdx,1),%xmm0
  movdqu (%rcx,%rdx,1),%xmm1
  psrlq  $0x20,%xmm0
  pmuludq %xmm2,%xmm1
  pmuludq %xmm3,%xmm0
  pshufd $0x8,%xmm1,%xmm1
  pshufd $0x8,%xmm0,%xmm0
  punpckldq %xmm0,%xmm1
  movdqu (%rsi,%rdx,1),%xmm0
  paddd  %xmm1,%xmm0
  movups %xmm0,(%rax,%rdx,1)
  add    $0x10,%rdx
  cmp    %r8,%rdx
  jne    L38
  cmp    -0x24(%rsp),%r11d
  je     L79
  mov    (%rcx,%r13,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r13,4),%edx
  mov    %edx,(%rax,%r13,4)
  cmp    -0x20(%rsp),%r11d
  jle    L79
  mov    (%rcx,%r14,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r14,4),%edx
  mov    %edx,(%rax,%r14,4)
  cmp    -0x1c(%rsp),%r11d
  jle    L79
  mov    (%rcx,%r15,4),%edx
  imul   %ebx,%edx
  add    (%rsi,%r15,4),%edx
  mov    %edx,(%rax,%r15,4)
L79:
  add    $0x1,%r12d
  add    %rbp,%rax
  add    %rbp,%r10
  add    %rbp,%r9
  cmp    %r12d,%r11d
  jne    L69
L57:
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  cs nopw 0x0(%rax,%rax,1)
L7:
  mov    %r8,-0x18(%rsp)
  mov    -0x10(%rsp),%r8
  xor    %edx,%edx
  nopl   0x0(%rax)
L87:
  mov    (%rcx,%rdx,4),%edi
  imul   %ebx,%edi
  add    (%rsi,%rdx,4),%edi
  mov    %edi,(%rax,%rdx,4)
  add    $0x1,%rdx
  cmp    %rdx,%r8
  jne    L87
  mov    -0x18(%rsp),%r8
  jmp    L79
```

assistant: ```c
void multiply78(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
<findMaxConsecutiveOnes>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x18(%rbp)
  mov    %esi,-0x1c(%rbp)
  movl   $0x0,-0xc(%rbp)
  movl   $0x0,-0x8(%rbp)
  movl   $0x0,-0x4(%rbp)
  jmp    L30
L17:
  mov    -0x4(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  test   %eax,%eax
  je     L24
  addl   $0x1,-0x8(%rbp)
  jmp    L32
L24:
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L53
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L53:
  movl   $0x0,-0x8(%rbp)
L32:
  addl   $0x1,-0x4(%rbp)
L30:
  mov    -0x4(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jl     L17
  mov    -0x8(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jle    L52
  mov    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
L52:
  mov    -0xc(%rbp),%eax
  pop    %rbp
  ret
```

assistant: ```c
int findMaxConsecutiveOnes(int *nums, int numsSize) {
  int maxLen = 0;
  int tmpLen = 0;
  int i;
  for (i = 0; i < numsSize; i++) {
    if (nums[i]) {
      tmpLen++;
    } else {
      if (tmpLen > maxLen) {
        maxLen = tmpLen;
      }
      tmpLen = 0;
    }
  }
  if (tmpLen > maxLen) {
    maxLen = tmpLen;
  }
  return maxLen;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<findMaxConsecutiveOnes>:
  endbr64
  test   %esi,%esi
  jle    L9
  mov    %rdi,%rdx
  lea    -0x1(%rsi),%eax
  lea    0x4(%rdi,%rax,4),%rdi
  mov    $0x0,%ecx
  mov    $0x0,%eax
  jmp    L29
L38:
  cmp    %ecx,%eax
  cmovl  %ecx,%eax
  mov    %esi,%ecx
L18:
  add    $0x4,%rdx
  cmp    %rdi,%rdx
  je     L12
L29:
  mov    (%rdx),%esi
  test   %esi,%esi
  je     L38
  add    $0x1,%ecx
  jmp    L18
L9:
  mov    $0x0,%ecx
  mov    $0x0,%eax
L12:
  cmp    %ecx,%eax
  cmovl  %ecx,%eax
  ret
```

assistant: ```c
int findMaxConsecutiveOnes(int *nums, int numsSize) {
  int maxLen = 0;
  int tmpLen = 0;
  int i;
  for (i = 0; i < numsSize; i++) {
    if (nums[i]) {
      tmpLen++;
    } else {
      if (tmpLen > maxLen) {
        maxLen = tmpLen;
      }
      tmpLen = 0;
    }
  }
  if (tmpLen > maxLen) {
    maxLen = tmpLen;
  }
  return maxLen;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<findMaxConsecutiveOnes>:
  endbr64
  test   %esi,%esi
  jle    L32
  lea    -0x1(%rsi),%eax
  xor    %edx,%edx
  lea    0x4(%rdi,%rax,4),%rcx
  xor    %eax,%eax
  jmp    L20
  cs nopw 0x0(%rax,%rax,1)
L30:
  add    $0x4,%rdi
  add    $0x1,%edx
  cmp    %rcx,%rdi
  je     L31
L20:
  mov    (%rdi),%esi
  test   %esi,%esi
  jne    L30
  cmp    %edx,%eax
  cmovl  %edx,%eax
  add    $0x4,%rdi
  xor    %edx,%edx
  cmp    %rcx,%rdi
  jne    L20
L31:
  cmp    %edx,%eax
  cmovl  %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L32:
  xor    %eax,%eax
  ret
```

assistant: ```c
int findMaxConsecutiveOnes(int *nums, int numsSize) {
  int maxLen = 0;
  int tmpLen = 0;
  int i;
  for (i = 0; i < numsSize; i++) {
    if (nums[i]) {
      tmpLen++;
    } else {
      if (tmpLen > maxLen) {
        maxLen = tmpLen;
      }
      tmpLen = 0;
    }
  }
  if (tmpLen > maxLen) {
    maxLen = tmpLen;
  }
  return maxLen;
}
```
user: What would the C source look like for the assembly code provided here:

```asm
<findMaxConsecutiveOnes>:
  endbr64
  test   %esi,%esi
  jle    L12
  lea    -0x1(%rsi),%eax
  xor    %edx,%edx
  lea    0x4(%rdi,%rax,4),%rcx
  xor    %eax,%eax
  jmp    L18
  cs nopw 0x0(%rax,%rax,1)
L5:
  add    $0x4,%rdi
  add    $0x1,%edx
  cmp    %rdi,%rcx
  je     L47
L18:
  mov    (%rdi),%esi
  test   %esi,%esi
  jne    L5
  cmp    %edx,%eax
  cmovl  %edx,%eax
  add    $0x4,%rdi
  xor    %edx,%edx
  cmp    %rdi,%rcx
  jne    L18
L47:
  cmp    %edx,%eax
  cmovl  %edx,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L12:
  xor    %eax,%eax
  ret
```

assistant: ```c
int findMaxConsecutiveOnes(int *nums, int numsSize) {
  int maxLen = 0;
  int tmpLen = 0;
  int i;
  for (i = 0; i < numsSize; i++) {
    if (nums[i]) {
      tmpLen++;
    } else {
      if (tmpLen > maxLen) {
        maxLen = tmpLen;
      }
      tmpLen = 0;
    }
  }
  if (tmpLen > maxLen) {
    maxLen = tmpLen;
  }
  return maxLen;
}
```
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    $0x10,%rsp
  movl   $0x64,-0x4(%rbp)
  jmp    L15
L10:
  mov    -0x4(%rbp),%eax
  mov    %eax,%esi
  lea    D1(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  subl   $0x4,-0x4(%rbp)
L15:
  cmpl   $0x0,-0x4(%rbp)
  jne    L10
  mov    $0x0,%eax
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D1
data_type: string

tool: D1: value of a: %d


assistant: ```c
int main() {
  for (unsigned int a = 100; a > 0; a -= 4) {
    printf("value of a: %d\n", a);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x64,%ebx
  lea    D8(%rip),%rbp
L2:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x4,%ebx
  jne    L2
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

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

tool: D8: value of a: %d


assistant: ```c
int main() {
  for (unsigned int a = 100; a > 0; a -= 4) {
    printf("value of a: %d\n", a);
  }
  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
  push   %rbp
  lea    D4(%rip),%rbp
  push   %rbx
  mov    $0x64,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x4,%ebx
  jne    L3
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

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

tool: D4: value of a: %d


assistant: ```c
int main() {
  for (unsigned int a = 100; a > 0; a -= 4) {
    printf("value of a: %d\n", a);
  }
  return 0;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<main>:
  endbr64
  push   %rbp
  lea    D2(%rip),%rbp
  push   %rbx
  mov    $0x64,%ebx
  sub    $0x8,%rsp
  cs nopw 0x0(%rax,%rax,1)
L3:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  call   <printf@plt>
  sub    $0x4,%ebx
  jne    L3
  add    $0x8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  ret
```

tool_calls: name: retrieve_data
arguments: data_label: D2
type_infered: string

tool: D2: value of a: %d


assistant: ```c
int main() {
  for (unsigned int a = 100; a > 0; a -= 4) {
    printf("value of a: %d\n", a);
  }
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的：

```asm
<be64dec_vect>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %rdx,-0x38(%rbp)
  movq   $0x0,-0x18(%rbp)
  jmp    L4
L3:
  mov    -0x18(%rbp),%rax
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    -0x18(%rbp),%rdx
  lea    0x0(,%rdx,4),%rcx
  mov    -0x28(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %rax,%rdi
  call   <_sha512_be64dec@plt>
  mov    %eax,(%rbx)
  addq   $0x1,-0x18(%rbp)
L4:
  mov    -0x38(%rbp),%rax
  shr    $0x3,%rax
  cmp    %rax,-0x18(%rbp)
  jb     L3
  nop
  nop
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
void be64dec_vect(uint64_t *dst, const unsigned char *src, size_t len) {
  size_t i;
  for (i = 0; i < len / 8; i++) {
    dst[i] = _sha512_be64dec(src + i * 8);
  }
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<be64dec_vect>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdx,%r13
  shr    $0x3,%r13
  cmp    $0x7,%rdx
  jbe    L18
  mov    %rdi,%r12
  mov    %rsi,%rbp
  mov    $0x0,%ebx
L19:
  mov    %rbp,%rdi
  call   <_sha512_be64dec@plt>
  mov    %eax,(%r12,%rbx,4)
  add    $0x1,%rbx
  add    $0x8,%rbp
  cmp    %r13,%rbx
  jb     L19
L18:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
void be64dec_vect(uint64_t *dst, const unsigned char *src, size_t len) {
  size_t i;
  for (i = 0; i < len / 8; i++) {
    dst[i] = _sha512_be64dec(src + i * 8);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<be64dec_vect>:
  endbr64
  shr    $0x3,%rdx
  je     L16
  push   %r12
  lea    (%rsi,%rdx,8),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopl   0x0(%rax,%rax,1)
L18:
  mov    %rbx,%rdi
  add    $0x8,%rbx
  add    $0x4,%rbp
  call   <_sha512_be64dec@plt>
  mov    %eax,-0x4(%rbp)
  cmp    %r12,%rbx
  jne    L18
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L16:
  ret
```

assistant: ```c
void be64dec_vect(uint64_t *dst, const unsigned char *src, size_t len) {
  size_t i;
  for (i = 0; i < len / 8; i++) {
    dst[i] = _sha512_be64dec(src + i * 8);
  }
}
```
user: 请从给定的汇编指令推导出C源代码:

```asm
<be64dec_vect>:
  endbr64
  shr    $0x3,%rdx
  je     L10
  push   %r12
  lea    (%rsi,%rdx,8),%r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  mov    %rsi,%rbx
  nopl   0x0(%rax,%rax,1)
L16:
  mov    %rbx,%rdi
  add    $0x8,%rbx
  add    $0x4,%rbp
  call   <_sha512_be64dec@plt>
  mov    %eax,-0x4(%rbp)
  cmp    %r12,%rbx
  jne    L16
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   (%rax)
L10:
  ret
```

assistant: ```c
void be64dec_vect(uint64_t *dst, const unsigned char *src, size_t len) {
  size_t i;
  for (i = 0; i < len / 8; i++) {
    dst[i] = _sha512_be64dec(src + i * 8);
  }
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<priorrune>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  mov    %rdi,-0x8(%rbp)
  mov    %rsi,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movzbl (%rax),%edx
  mov    -0x8(%rbp),%rax
  mov    %dl,(%rax)
  mov    $0x1,%eax
  pop    %rbp
  ret
```

assistant: ```c
int priorrune(char *rp, char *s) {
  *rp = s[0];
  return 1;
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<priorrune>:
  endbr64
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int priorrune(char *rp, char *s) {
  *rp = s[0];
  return 1;
}
```
user: Translate the below assembly code into C programming language:

```asm
<priorrune>:
  endbr64
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int priorrune(char *rp, char *s) {
  *rp = s[0];
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<priorrune>:
  endbr64
  movzbl (%rsi),%eax
  mov    %al,(%rdi)
  mov    $0x1,%eax
  ret
```

assistant: ```c
int priorrune(char *rp, char *s) {
  *rp = s[0];
  return 1;
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<myvalgrad>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L5
L17:
  mov    -0x1c(%rbp),%eax
  add    $0x1,%eax
  pxor   %xmm0,%xmm0
  cvtsi2sd %eax,%xmm0
  movsd  %xmm0,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  movq   %rax,%xmm0
  call   <sqrt@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movq   %rax,%xmm0
  call   <exp@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  movapd %xmm0,%xmm1
  mulsd  -0x10(%rbp),%xmm1
  movsd  -0x8(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  -0x18(%rbp),%xmm1
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  mov    -0x1c(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movsd  -0x8(%rbp),%xmm0
  subsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L5:
  mov    -0x1c(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L17
  movsd  -0x18(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  leave
  ret
```

assistant: ```c
double myvalgrad(double *g, double *x, int n) {
  double ex, f, t;
  int i;
  f = (double)0;
  for (i = 0; i < n; i++) {
    t = i + 1;
    t = sqrt(t);
    ex = exp(x[i]);
    f += ex - t * x[i];
    g[i] = ex - t;
  }
  return (f);
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<myvalgrad>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x10,%rsp
  test   %edx,%edx
  jle    L38
  mov    %rdi,%rbp
  mov    %rsi,%r12
  lea    0x1(%rdx),%r13d
  mov    $0x1,%ebx
  pxor   %xmm6,%xmm6
  movsd  %xmm6,(%rsp)
L48:
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebx,%xmm0
  pxor   %xmm4,%xmm4
  ucomisd %xmm0,%xmm4
  ja     L31
  sqrtsd %xmm0,%xmm0
  movsd  %xmm0,0x8(%rsp)
L3:
  mov    -0x8(%r12,%rbx,8),%r14
  movq   %r14,%xmm0
  call   <exp@plt>
  movq   %r14,%xmm1
  movsd  0x8(%rsp),%xmm3
  mulsd  %xmm3,%xmm1
  movapd %xmm0,%xmm2
  subsd  %xmm1,%xmm2
  addsd  (%rsp),%xmm2
  movsd  %xmm2,(%rsp)
  subsd  %xmm3,%xmm0
  movsd  %xmm0,-0x8(%rbp,%rbx,8)
  add    $0x1,%rbx
  cmp    %r13,%rbx
  jne    L48
L25:
  movsd  (%rsp),%xmm0
  add    $0x10,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L31:
  call   <sqrt@plt>
  movsd  %xmm0,0x8(%rsp)
  jmp    L3
L38:
  pxor   %xmm7,%xmm7
  movsd  %xmm7,(%rsp)
  jmp    L25
```

assistant: ```c
double myvalgrad(double *g, double *x, int n) {
  double ex, f, t;
  int i;
  f = (double)0;
  for (i = 0; i < n; i++) {
    t = i + 1;
    t = sqrt(t);
    ex = exp(x[i]);
    f += ex - t * x[i];
    g[i] = ex - t;
  }
  return (f);
}
```