移动一种类型和另一种类型的值是否违反了严格的别名?

是否违反了严格的别名规则,使用uint32移动任何类型的项目,然后读取它们?如果是这样,它是否也违反了从uint32数组到任何类型数组的memcpy的严格别名规则,然后将元素读回?

下面的代码示例演示了这两种情况:

#include <assert.h>
#include <stdio.h>
#include <stdint.h>
#include <string.h>

int main(void) {
    const char *strings[5] = {
        "zero", "one", "two", "three", "four"
    };
    uint32_t buffer[5];
    int i;

    assert(sizeof(const char*) == sizeof(uint32_t));

    memcpy(buffer, strings, sizeof(buffer));

    //twiddle with the buffer a bit
    buffer[0] = buffer[3];
    buffer[2] = buffer[4];
    buffer[3] = buffer[1];

    //Does this violate strict aliasing?
    const char **buffer_cc = (const char**)buffer;
    printf("Test 1:\n");
    for (i=0; i<5; i++)
        printf("\t%s ", buffer_cc[i]);
    printf("\n");

    //How about this?
    memcpy(strings, buffer, sizeof(strings));
    printf("Test 2:\n");
    for (i=0; i<5; i++)
        printf("\t%s ", strings[i]);
    printf("\n");

    return 0;
}

请忽略我对32位平台的假设。另外,如果元素的大小与uint32-t不同,我知道要填充它们并复制uint32-t的正确数字。我的问题是这样做是否违反了严格的别名。