为什么传递给函数的指针值没有变化? [英] Why does not the value of the pointer passed to the function change?

问题描述

我将指针作为参数传递给函数.在函数内部，我对其进行了更改->它已更改.但是之后的输出与函数调用之前的输出相同.指针没有改变.怎么了?

I pass the pointer as a parameter to the function. Inside the function I change it -> it has changed. But after the output is the same as before the function call. The pointer has not changed. What's wrong?

void GetFrom(PVOID lpBuffer){
lpBuffer = malloc(12);
memset(lpBuffer, 0, 12);
printf("%p",pointer); // 0000028D46DECE50
}

PVOID pointer = 0x0;
printf("%p",pointer); // 000000C2628FEFE4
GetFromMap(pointer);
printf("%p",pointer); // 000000C2628FEFE4

在调试中，我看到指针的值在函数内部发生了变化.

In debugging, I saw that the value of the pointer changed inside the function.

推荐答案

请注意，lpBuffer = malloc(12)只是更改了局部变量lpBuffer的值，但这对已传递给GetFrom的变量没有影响. .要更改调用方变量中指针的值，您需要将指针传递给该指针.

Note that lpBuffer = malloc(12) just changes the value of local variable lpBuffer, but this has no effect to the variable which's value has been passed to GetFrom. To change the value of the pointer in the caller`s variable, you need to pass a pointer to that pointer.

void GetFrom(PVOID *lpBuffer){
  *lpBuffer = malloc(12);
  memset(*lpBuffer, 0, 12);
  printf("%p",*lpBuffer); // !=0x0
}

PVOID pointer = 0x0;
GetFromMap(&pointer);
printf("%p",pointer); // should be != 0x0

BTW:您可以使用calloc代替malloc和后续的memset to 0，它使用0-

BTW: instead of malloc and a subsequent memset to 0, you could use calloc, which implicitly fills the allocated memory with 0-

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