c - 基于内存对齐将字节数组转换为 int 是否安全？

Question

我最近想通过字节移位和逻辑或将一个 4 字节数组的内容分配到一个 4 字节变量中。

我想检查直接将数组转换为 int* 并获取值是否有效。我写了一个小测试程序:

int main(int argc, char** argv) {
    int out = 0;
    char in[4] = {0xFF,0xFF,0xFF,0xFF};
    char *in2 = NULL;


    printf("out = %d\n",out);

    out = (int)in[0] << 24 | (int)in[1] << 16 | (int)in[2] << 8 | (int)in[3];
    printf("out = %d\n",out);
    out = 0;

    out = *((int*)in);
    printf("out = %d\n",out);

    in2 = (char*)malloc(4);
    for(int i = 0; i < 4; i++)
        in2[i] = 0xFF;


    out = 0;
    // Byte-Shifting & OR
    out = (int)in2[0] << 24 | (int)in2[1] << 16 | (int)in2[2] << 8 | (int)in2[3]; 
    printf("out = %d\n",out);
    out = 0;

    //Direct assignment by recasting
    out = *((int*)in2); 
    printf("out = %d\n",out);

    return 0;
}

输出如预期的那样等于 -1 :

out = 0 out = -1 out = -1 out = -1 out = -1

但我担心内存对齐和此方法的安全性:它安全吗？如果不是，是由于编译器规范还是某种未定义的行为？

Answer 1

But I'm worried about memory alignment and the safety of this method : is it safe ? If not, is it due to compiler specification or some kind of undefined behavior ?

不，这不安全:

out = *((int*)in);

是未定义的行为，它违反了 C 别名规则，可能会执行未对齐的访问。