Question

我有一个奇怪的记忆错误。

代码很简单：

void *to = calloc(2, sizeof(uint64_t));
... 
int add_symbol_to_symbol(void *from, void *to) {
  uintptr_t *fromSymbol;
  uintptr_t *toSymbol;
  uint64_t    i;

  fromSymbol = (uintptr_t*) from;
  toSymbol = (uintptr_t*) to;

  for (i = 0; i < 2; i++)
  {
    *toSymbol ^= *fromSymbol;
    toSymbol++;
    fromSymbol++;
  }

  return 0;
}

当我调试代码并打印出内存的地址和内存的值时，我可以看到toSymbol在xor出现之前在其内存中有非零值...但这取决于我如何打印输出数据。

GDB：

print toSymbol
$21 = (uintptr_t *) 0x650400
(gdb) print (toSymbol+1)
$23 = (uintptr_t *) 0x650408
(gdb) print *(toSymbol)
$25 = 0
(gdb) print *(toSymbol+1)
$24 = 4575657221408423936
(gdb) print *(unsigned long long)(toSymbol+1)
$26 = 0
(gdb) print *(unsigned long long*)(toSymbol+1)
$27 = 4575657221408423936
print *(uintptr_t*)(0x650408)
$30 = 4575657221408423936
(gdb) print *(uintptr_t)(0x650408)
$31 = 0

所以我的问题：

为什么print （uintptr_t ）（0x650408）显示垃圾数据？
为什么print *（uintptr）（0x650408）显示为零？（即，为什么将64位十六进制地址转换为64位值而不是64位地址会改变任何东西！？！）

无符号长long和无符号long long之间的区别*

0 个答案: