【发布时间】:2021-02-25 12:00:13
【问题描述】:
嗯,事情很简单,我明白了
warning: ‘void* memset(void*, int, size_t)’ clearing an object of non-trivial type ‘struct FormatHashBuffers(CBlock*, char*, char*, char*)::<unnamed>’; use assignment or value-initialization instead [-Wclass-memaccess] memset(&tmp, 0, sizeof(tmp)); 关于这个函数和 idk 为什么,当我用 g++ 5 构建时没有警告,但是当我用 7.1 或 8.5 构建时我得到警告,知道为什么或如何解决它吗?提前致谢。
void FormatHashBuffers(CBlock* pblock, char* pmidstate, char* pdata,
char* phash1) {
//
// Pre-build hash buffers
//
struct
{
struct unnamed2
{
int nVersion;
uint256 hashPrevBlock;
uint256 hashMerkleRoot;
unsigned int nTime;
unsigned int nBits;
unsigned int nNonce;
}
block;
unsigned char pchPadding0[64];
uint256 hash1;
unsigned char pchPadding1[64];
}
tmp;
memset(&tmp, 0, sizeof(tmp));
tmp.block.nVersion = pblock->nVersion;
tmp.block.hashPrevBlock = pblock->hashPrevBlock;
tmp.block.hashMerkleRoot = pblock->hashMerkleRoot;
tmp.block.nTime = pblock->nTime;
tmp.block.nBits = pblock->nBits;
tmp.block.nNonce = pblock->nNonce;
FormatHashBlocks(&tmp.block, sizeof(tmp.block));
FormatHashBlocks(&tmp.hash1, sizeof(tmp.hash1));
// Byte swap all the input buffer
for (unsigned int i = 0; i < sizeof(tmp) / 4; i++)
((unsigned int*)&tmp)[i] = ByteReverse(((unsigned int*)&tmp)[i]);
// Precalc the first half of the first hash, which stays constant
SHA256Transform(pmidstate, &tmp.block, pSHA256InitState);
memcpy(pdata, &tmp.block, 128);
memcpy(phash1, &tmp.hash1, 64);
}
【问题讨论】:
-
为什么不在你的类中初始化成员变量呢?
-
警告是在版本 5 和版本 7.1 之间的某个时间添加的。 (开发人员在特殊情况下添加警告的时间有限。)您可以通过不以未定义行为方式使用
memset来解决问题。