将一种类型的值与另一种类型的值移动是否违反严格的别名? [英] Does moving values of one type with another type violate strict aliasing?

问题描述

使用uint32_t移动任何类型的项目然后将其读回是否违反严格的别名规则?如果是这样，是否也违反了严格的别名规则，即从uint32_ts数组到任何类型的数组进行memcpy，然后再读回元素?

Does it violate strict aliasing rules to move items of any type around using uint32_t, then read them back? If so, does it also violate strict aliasing rules to memcpy from an array of uint32_ts to an array of any type, then read the elements back?

以下代码示例演示了两种情况:

The following code sample demonstrates both cases:

#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正确数量的uint32_t.我的问题集中在这样做是否违反严格的别名.

Please disregard my assumption of a 32-bit platform. Also, if the elements aren't the same size as uint32_t, I know to pad them and copy the correct number of uint32_t's. My question focuses on whether or not doing so violates strict aliasing.

推荐答案

第一个循环确实违反了严格的别名-它通过类型为char *的左值访问uint32_t对象.但是，在这种特定情况下，很难看到任何优化器都会给您带来问题.如果您对其进行了一些更改，则您将执行以下操作:

The first loop does technically violate strict aliasing - it accesses uint32_t objects through an lvalue of type char *. It's hard to see how any optimiser would cause you a problem in this specific case, though. If you altered it a little so you were doing something like:

printf("\t%s ", buffer_cc[0]);
buffer[0] = buffer[3];
printf("\t%s ", buffer_cc[0]);

您可能看到相同的字符串打印了两次-因为优化器将在其权限内仅将buffer_cc[0]加载到寄存器一次，因为第二行仅修改了uint32_t.

You might see the same string printed twice - since the optimiser would be within its rights to only load buffer_cc[0] into a register once, because the second line is only modifying an object of type uint32_t.

第二个循环，可以将它们重新返回.

The second loop, that memcpys them back, is fine.

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