对结构在内存中的位置感到困惑

Question

我正在探索指针，目前正在搞乱结构指针。目前，我有这段代码:

#include <stdlib.h>
#include <stdio.h>

struct point{
double x;
double y;
};

int main(){
    char *pc;
    int *pi;
    struct point *pp;

    //char pointer
    printf("%p\n\n", (void*)pc);
    pc++;
    printf("%p\n\n", (void*)pc);

    //int pointer
    printf("%p\n\n", (void*)pi);
    pi += 2;
    printf("%p\n\n", (void*)pi);

    //struct pointer
    printf("%p\n\n", (void*)pp);
    pp -= 3;
    printf("%p\n\n", (void*)pp);
}

这段代码的输出是这样的:

0x104978036

0x104978037

0x7fff5ec4bbc8

0x7fff5ec4bbd0

0x0

0xffffffffffffffd0

我理解前 4 个输出，使用 char 和 int 指针算法，但是我很困惑为什么它返回 0x0 struct指针的内存地址？另外，如果我想要内存中的地址，double y，我将如何打印它？

Answer 1

您的代码调用了未定义的行为，因为您使用的所有指针都未初始化!

在启用警告的情况下编译(例如 GCC 的 -Wall)，将得到:

prog.c: In function 'main':
prog.c:15:5: warning: 'pc' is used uninitialized in this function [-Wuninitialized]
     printf("%p\n\n", (void*)pc);
     ^~~~~~~~~~~~~~~~~~~~~~~~~~~
prog.c:20:5: warning: 'pi' is used uninitialized in this function [-Wuninitialized]
     printf("%p\n\n", (void*)pi);
     ^~~~~~~~~~~~~~~~~~~~~~~~~~~
prog.c:25:5: warning: 'pp' is used uninitialized in this function [-Wuninitialized]
     printf("%p\n\n", (void*)pp);
     ^~~~~~~~~~~~~~~~~~~~~~~~~~~

---

像这样初始化你的指针，例如:

char c = 'a';
char *pc = &c;
int i = 5;
int *pi = &i;
struct point p;
struct point *pp = &p;

并且可以获得:

0x7ffe629c90d7 // address of the charatcer

0x7ffe629c90d8 // sizeof(char) is 1

0x7ffe629c90d0 // address of the integer

0x7ffe629c90d8 // sizeof(int) is 8 in this system

0x7ffe629c90c0 // address of the struct

0x7ffe629c9090 // sizeof(struct point) is 16 in this system
// 0x7ffe629c90c0 - 0x7ffe629c9090 = 0x30 -> Decimal: 48 = 3 * 16