Question

这在x86中编译很好，但是当我在x64配置中使用它时，x和y变量在我尝试访问时没有地址？是否需要某种填充以对齐更大的地址？使用MSVC ..

#define ARR_SIZE 25

typedef struct {
    unsigned int x;
    unsigned int y;
}Stuff;

void allocateArray(Stuff *stuffArr) {

    Stuff *stuff = malloc(sizeof (Stuff) * ARR_SIZE);

    for (int i = 0; i < ARR_SIZE; i++) {
        (*(stuff + i)) = (Stuff) { i, i + i };
    }

    for (int i = 0; i < ARR_SIZE; i++) {
        printf("%d : %d\n", (stuff + i)->x, (stuff + i)->y);
    }

    stuffArr = stuff;
}

void deallocateArray(Stuff *stuffArr) {
    free(stuffArr);
}

int main(){
    Stuff * stuff = NULL;

    allocateArray(stuff);
    deallocateArray(stuff);

    return 0;
}

Answer 1

正如user3386109所说，代码不正确。您可能期望allocateArray()函数返回已分配的指针，而您正在按值传递指针，以便stuff内的变量main()不会更新。

你可以：

将allocateArray()签名更改为void allocateArray(Stuff **stuffArr)
将allocateArray()签名更改为Stuff *allocateArray()

（第二个将更加惯用和明确）。

我把它写成：

Stuff *allocateArray(size_t count) {
    Stuff *stuff = (Stuff *) malloc(sizeof (Stuff) * count);

    if (! stuff)
        return NULL;

    for (int i = 0; i < count; i++) {
        stuff[i].x = i;
        stuff[i].y = 2 * i;

        printf("%d : %d\n", stuff[i].x, stuff[i].y);
    }

    return stuff;
}

void deallocateArray(Stuff *stuffArr) {
    if (stuffArr)
        free(stuffArr);
}

int main(){
    Stuff * stuff = allocateArray(ARR_SIZE);
    deallocateArray(stuff);

    return 0;
}

Answer 2

当您在allocateArray中传递内容时，您创建了一个局部变量，当您在函数末尾重新设置它时，main中的变量不会更新

这应该有效，你刚刚在allocateArray函数中丢失了指针

#define ARR_SIZE 25

typedef struct {
    unsigned int x;
    unsigned int y;
}Stuff;

Stuff *allocateArray() {

   Stuff *stuff = malloc(sizeof (Stuff) * ARR_SIZE);

   for (int i = 0; i < ARR_SIZE; i++) {
       (*(stuff + i)) = (Stuff) { i, i + i };
   }

   for (int i = 0; i < ARR_SIZE; i++) {
       printf("%d : %d\n", (stuff + i)->x, (stuff + i)->y);
   }

   return stuff;
}

void deallocateArray(Stuff *stuffArr) {
  free(stuffArr);
}

int main(){
  Stuff * stuff = NULL;

  stuff = allocateArray();
  deallocateArray(stuff);

  return 0;
}

Answer 3

我将您的代码复制并粘贴到Visual Studio 2015中，并在x86和x64中进行编译，并且两次都得到完全相同的输出，但是就像user3386109所说的那样，您实际上并没有更改stuff中的变量main 1}}。

你也可以使用数组索引而不是像这样的指针添加整数。

for (int i = 0; i < ARR_SIZE; i++) {
    stuff[i] = (Stuff) { i, i + i };
}

for (int i = 0; i < ARR_SIZE; i++) {
    printf("%d : %d\n", stuff[i].x, stuff[i].y);
}

Answer 4

好问题是它在x86中是如何工作的。（这是未定义的行为）。试试这个：

#define ARR_SIZE 25
#include  "stdlib.h"
#include  "stdio.h"
typedef struct {
    unsigned int x;
    unsigned int y;
}Stuff;

void * allocateArray(Stuff *stuffArr) { //this row

    Stuff *stuff = (Stuff*)malloc(sizeof (Stuff) * ARR_SIZE);

    for (int i = 0; i < ARR_SIZE; i++) {
        (*(stuff + i)) = (Stuff) { i, i + i };
    }

    for (int i = 0; i < ARR_SIZE; i++) {
        printf("%d : %d\n", (stuff + i)->x, (stuff + i)->y);
    }

    stuffArr = stuff;
    return stuff; //this row
}

void deallocateArray(Stuff *stuffArr) {
    free(stuffArr);
}

int main(){
    Stuff * stuff = NULL;
    printf("%d\n",stuff);
    stuff=(Stuff*)allocateArray(stuff); ///this row
    printf("%p\n",stuff);
    deallocateArray(stuff);
    printf("%p\n",stuff);
    return 0;
}

如何编译x64

4 个答案: