c - 为什么我在通过 int 转换后得到 EXC_BAD_ACCESS？

Question

在下面的代码中，我得到一个 EXC_BAD_ACCESS当我调用打印顶点时，没有明显原因的错误。指针指向完全​​相同的位置，但是当我传入 vp2 时它以某种方式崩溃了。

#include <stdio.h>

typedef struct {
    float x;
    float y;
    float z;
} Vertex;

void printVertex(Vertex *v);

int main (int argc, const char * argv[])
{
    Vertex v = {1,0,2};
    int memL = (int)&v;
    Vertex *vp = &v;

    printf("Memory Location: %i\n", memL);
    printf("Memory Pointed to by Pointer: %i\n", (int)vp);

    Vertex *vp2 = (Vertex *)memL;
    printf("Memory Pointed to by Pointer from memory location: %i\n", (int)vp2);

    printVertex(vp);  // Executes normally
    printVertex(vp2); // EXC_BAD_ACCESS

    return 0;
}

void printVertex(Vertex *v)
{
    printf("Vertex[%f,%f,%f]\n", v->x, v->y, v->z); // EXC_BAD_ACCESS when vp2 passed in
}

输出:

Memory Location: 1606416816
Memory Pointed to by Pointer: 1606416816
Memory Pointed to by Pointer from memory location: 1606416816
Vertex[1.000000,0.000000,2.000000]
EXC_BAD_ACCESS Error

Answer 1

int memL = (int)&v;

可能会截断 v 的地址.你可以试试:

intptr_t memL = (intptr_t)&v;

看看是不是这种情况。如果截断是一个问题，这不应该崩溃。